text stringlengths 1 2.12k | source dict |
|---|---|
$\displaystyle \large \frac{255!}{232! \cdot 255^{24}} \cdot \frac{24!}{2! \cdot 22!}$
Notice that this is exactly 12 times the agreed upon value, given approximately as 0.0324, because:
$\displaystyle \large \frac{24!}{2! \cdot 22!} = 12 \cdot 23$
Please permit me to:
1) Show an inconsistency in the analysis of the agreed upon result.
2) Explain my concept of the solution.
3) Demonstrate the concept using what I think is the simplest example.
4) Provide a little background as to my interest in this.
So, for 1):
Quote:
How many ways can I draw 23 differently numbered cards from that same set with replacement? Answer: $\displaystyle \large \prod_{j=1}^{23}(256 - j).$ Do you agree? How many ways can I draw one of 23 specific numbers from that set? Answer: 23.
So we have, I think:
$\displaystyle 255 \cdot 254 \cdot 253 \cdot \ldots \cdot 234 \cdot 233 \cdot 23$
If that is the case, then in the simplified example given earlier in the thread:
Quote:
Two numbers the same (be careful: the different number can come first, second or third) = $\displaystyle 6∗5∗3=90.$
Why isn't that 3 a 2?
By the logic given in the first quote above, we can match 2 specific numbers here, not 3. But as is cautioned, it is not the number of numbers that is important, it is the number of places that is important. This appears to be inconsistent to me.
So for 2):
I would word it differently: It is the number of combinations of 3 numbers taken two at a time (the two types are numbers that match and the numbers that do not match):
$\displaystyle \large \frac{3!}{2! \cdot 1!}=3$
So to explain my solution, for the first number we have our choice of 255. Then let's match it 'right off the bat' with the second number so we have a choice of 1, then we can't match any more so we count down from there:
$\displaystyle \large 255 \cdot 1 \cdot 254 \cdot 253 \cdot \ldots \cdot 234 \cdot 233$ | {
"domain": "mymathforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109534209825,
"lm_q1q2_score": 0.8706115679179115,
"lm_q2_score": 0.8840392817460332,
"openwebmath_perplexity": 1050.170538587273,
"openwebmath_score": 0.807102620601654,
"tags": null,
"url": "http://mymathforum.com/probability-statistics/340911-what-probability-getting-same-number-twice-2.html"
} |
$\displaystyle \large 255 \cdot 1 \cdot 254 \cdot 253 \cdot \ldots \cdot 234 \cdot 233$
Obviously, we still have two types of numbers, the matching ones, and the non-matching ones, which gives the multiplication by:
$\displaystyle \large \frac{24!}{2! \cdot 22!} = 12 \cdot 23$
for the various combinations.
For 3) let's try to simplify things as much as we can. Let's have 3 symbols and take 4 of the symbols and calculate how many combinations exactly two of the symbols appear in.
I think that it is:
$\displaystyle \large 3 \cdot 1 \cdot 2 \cdot 1 \cdot \frac{4!}{2! \cdot 2!}=36$
If I am not mistaken, by the method of calculation as described in the thread the number is:
$\displaystyle \large 3 \cdot 2 \cdot 1 \cdot 3=18$
So let's just list them using symbols 0,1,2. Taking 2 as the repeated symbol, in ascending order we have:
0122
0212
0221
1022
1202
1220
2012
2021
2102
2120
2201
2210
So we have 12 combinations with the 2 repeated. We can have the 0 and 1 repeated as well so we have 3*12=36 (not 18 ).
For 4), I became interested in this type of thing because at my place of employment, in order to log in remotely, we have to enter a code that is chosen at random, 6 each of the digits 0-9.
I noticed that it was relatively rare to have no repeated digits so I tried to figure out the odds of getting exactly two repeating digits (among other things). I wrote programs in Ruby (which I will gladly share but I doubt anyone uses Ruby here) to run random numbers and looking at the percentage, and also to calculate exactly by running through all of the combinations and looking at the percentage.
I am pretty sure that I used reasoning similar to that given earlier in the thread but the calculated results did not match the programs' outputs until I used the methods given above. | {
"domain": "mymathforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109534209825,
"lm_q1q2_score": 0.8706115679179115,
"lm_q2_score": 0.8840392817460332,
"openwebmath_perplexity": 1050.170538587273,
"openwebmath_score": 0.807102620601654,
"tags": null,
"url": "http://mymathforum.com/probability-statistics/340911-what-probability-getting-same-number-twice-2.html"
} |
Caveat: My 'exactly two matching numbers' routine may be flawed but for base 3 with 4 symbols taken, it does give 36, in agreement with the result above. For base 10 and 6 numbers taken it gives 453600 which agrees with:
$\displaystyle \large 10 \cdot 1 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot \frac{6!}{2! \cdot 4!}$
My random number calculating program calculates ~0.39 for the stated problem, in agreement with the calculation (and Denis' initial result, which I think is correct).
Tags number, probability
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post JDSimpkins Probability and Statistics 9 April 27th, 2015 07:01 AM meghraj Probability and Statistics 9 November 8th, 2012 07:00 PM eliotjames Advanced Statistics 3 October 7th, 2012 04:02 PM ershi Number Theory 113 August 18th, 2012 05:45 AM hoyy1kolko Probability and Statistics 3 June 4th, 2011 05:38 PM
Contact - Home - Forums - Cryptocurrency Forum - Top | {
"domain": "mymathforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109534209825,
"lm_q1q2_score": 0.8706115679179115,
"lm_q2_score": 0.8840392817460332,
"openwebmath_perplexity": 1050.170538587273,
"openwebmath_score": 0.807102620601654,
"tags": null,
"url": "http://mymathforum.com/probability-statistics/340911-what-probability-getting-same-number-twice-2.html"
} |
Friday, February 3, 2023
HomeCSIRUGC NETThe probability that a ticketless traveler is caught during a trip is...
# The probability that a ticketless traveler is caught during a trip is 0.1. If the traveler makes 4 trips , the probability that he/she will be caught during at least one of the trips is:
## Question:
The probability that a ticketless traveler is caught during a trip is 0.1. If the traveler makes 4 trips , the probability that he/she will be caught during at least one of the trips is:
1. $1-(0.9)^4$
2. $(1-0.9)^4$
3. $1-(1-0.9)^4$
4. $(0.9)^4$
The probability that a ticketless traveler is caught during a trip is 0.1.
$therefore p=0.1$ and $q=1-p=1-0.1=0.9$
Since, traveler makes 4 trips, $n=4$.
Let X be the random variable that the traveler is caught during a trip.
Hence, $X leadsto B(n=4, p=0.1)$ and $X=0, 1, 2, 3, 4$.
$therefore P[X=x]=$ $^4C_x$ $(0.1)^x(0.9)^{4-x}$
$therefore$ the probability that he/she will be caught during at least one of the trips
=1-{probability that he/she will be caught during at most one of the trips}.
$=1- P[X=0]$
$=1-$ $^4C_0$ $(0.1)^0(0.9)^4$
$=1-(0.9)^4$
RELATED ARTICLES
### The product of 2 numbers is 1575 and their quotient is $frac{9}{7}$. Then the sum of the numbers is
1. This is rightly named as the age of traveler-centricity and with the evolution of the new era of personalized travel; it is leading to research and development of a host of new so-called intelligent services. The command-and-control perspectives of traveling have changed a lot from the past and the focus has shifted more on the traveler and the productivity of each trip. It has become essential to maintain that the travelers have the greatest return on investment on each trip. Travel 101
### Dimension theorem of a quotient space: If W be a subspace of a finite dimensional vector space V over ( mathbb{R} ) then (...
$${}$$ | {
"domain": "brainbalancemathematics.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.984810954312569,
"lm_q1q2_score": 0.8706115626886181,
"lm_q2_score": 0.8840392756357326,
"openwebmath_perplexity": 889.0210695918074,
"openwebmath_score": 0.6206298470497131,
"tags": null,
"url": "https://brainbalancemathematics.com/the-probability-that-a-ticketless-traveler-is-caught-during-a-trip-is-0-1-if-the-traveler-makes-4-trips-the-probability-that-he-she-will-be-caught-during-at-least-one-of-the-trips-is/"
} |
# Probabilities of archers A and B hitting the bull's eye
1. Jan 10, 2013
### jinhuit95
1. The problem statement, all variables and given/known data
Two archers A and B take turns to shoot, with archer A taking the first shot. The probabilities of archers A and B hitting the bull's eye is 1/6 and 1/5 respectively. Show that the probability of archer A hitting the bull's eye first is 1/2.
2. Relevant equations
3. The attempt at a solution
Well, i thought about drawing tree diagram but the problem is I have no idea when to stop because I don't know when he will hit.
2. Jan 10, 2013
### dx
Re: Probability
First, what are the possible ways for this to happen?
A could hit the bulls eye on his first attempt. Or, A misses, B misses and then A hits the bulls eye. Or, A misses, B misses, A misses, B misses and then A hits the bulls eye... and so on.
Find the probability for each of these ways (for A to hit the bull's eye first), and add them up.
You will need to use the following:
1 + a2 + a3 + ... = 1/(1-a) for |a| < 1
3. Jan 10, 2013
### jinhuit95
Re: Probability
the problem is, if he hits on the first attempt that's 1/6. so If i go on, it will be 1/6 + ( 5/6 * 4/5 * 5/6 * 4/5 * 5/6 * 4/5....) I have no idea when to stop. Moreover, the product of probability he misses will be come smaller and smaller and that number + 1/6 will not give me half i think.
4. Jan 10, 2013
### dx
Re: Probability
Let's take the case when A gets it on his second try: A misses, B misses, and then A hits. What is the probability for this?
5. Jan 10, 2013
### jinhuit95
Re: Probability
that will be 5/6 * 4/5 * 1/6 right?
6. Jan 10, 2013
### jinhuit95
Re: Probability
I'm sorry but what's this?
'1 + a2 + a3 + ... = 1/(1-a) for |a| < 1'
7. Jan 10, 2013
### dx
Re: Probability
Yes, correct.
Now what is the probability that A gets it on his 3rd try? (A misses, B misses, A misses, B misses, A hits bull's eye)
Ignore that for the moment. We'll get to it soon. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109503004294,
"lm_q1q2_score": 0.8706115621504746,
"lm_q2_score": 0.884039278690883,
"openwebmath_perplexity": 1557.7460466744242,
"openwebmath_score": 0.7214272618293762,
"tags": null,
"url": "https://www.physicsforums.com/threads/probabilities-of-archers-a-and-b-hitting-the-bulls-eye.663461/"
} |
Ignore that for the moment. We'll get to it soon.
8. Jan 10, 2013
### jinhuit95
Re: Probability
5/6* 4/5 * 5/6 * 4/5 * 1/6
9. Jan 10, 2013
### dx
Re: Probability
Correct.
So do you see the pattern?
First try: 1/6
Second try: (1/6)(4/6)
Third try: (1/6)(4/6)2
Fourth try: (1/6)(4/6)3
...
What is the probability that he gets in on the n'th try? Each time you are simply multiplying the previous one by (5/6)(4/5) = 4/6
10. Jan 10, 2013
### jinhuit95
Re: Probability
1/6 * 4/6 ^ n-1!
11. Jan 10, 2013
### dx
Re: Probability
Exactly.
Now you simply add them up. Let a = 4/6
(1/6) + (1/6)a + (1/6)a2 + (1/6)a3...
which is (1/6)(1 + a + a2 + a3 + a4...)
This is where you use the formula I mentioned before.
12. Jan 10, 2013
### jinhuit95
Re: Probability
Got it! Thanks alot!!
13. Jan 10, 2013
### dx
Re: Probability
No problem.
14. Jan 10, 2013
### Ray Vickson
Re: Probability
Here is another approach: let x = the probability that A will win, given that A is just about to shoot.
If A hits the target on his first shot, he wins; the probability of this is 1/6. If A misses his first shot (prob. = 5/6), then for A to win it must be the case that B misses his first shot (prob = 4/5); at that point, A is just about to shoot and we are back to the starting point (win prob. = x at that point). So, altogether we have
$$x = \frac{1}{6} + \frac{5}{6} \frac{4}{5}\, x = \frac{1}{6} + \frac{2}{3} x.$$
Solving, we get x = 1/2.
15. Jan 10, 2013
### CAF123
Re: Probability
@Ray
Is there a reason why doing: $$x = \frac{1}{6} + \frac{2}{3}x + \frac{5}{6} \frac{4}{5} \frac{5}{6} \frac{4}{5}x,$$ and solving is incorrect? (I am just considering the possibility of both A and B losing until A hits on his third go).
16. Jan 10, 2013
### Joffan
Re: Probability
My friend Kim and I had a similar competition, but we're much worse. I am slightly worse than Kim, so I went first. On any one shot, my chance of hitting the bull is 1/39, and Kim's chance is 1/38. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109503004294,
"lm_q1q2_score": 0.8706115621504746,
"lm_q2_score": 0.884039278690883,
"openwebmath_perplexity": 1557.7460466744242,
"openwebmath_score": 0.7214272618293762,
"tags": null,
"url": "https://www.physicsforums.com/threads/probabilities-of-archers-a-and-b-hitting-the-bulls-eye.663461/"
} |
Show that, once again, this is an even competition - my chance of getting the bull's eye first is 1/2.
17. Jan 10, 2013
### CAF123
Re: Probability
Are you stuck? If so, just use exactly the same method as you used to solve the first one. Alternatively, say you win on your $i$th shot. This means you (or Kim) cannot have hit the bull on the previous $i-1$ attempts.
What is the prob of you not hitting the bull on the $i-1$ attempts?
What is the prob of Kim not hitting the bull on her $i-1$ attempts?
What is the prob of you hitting on the $i$th trial given that you and Kim failed on the $i-1$ trials?
By the independence of trials you can then multiply these together and sum.
(This is really the same method by dx just reworded a bit differently)
18. Jan 10, 2013
### haruspex
Re: Probability
Just use Ray's method again. It's much the simplest way to solve these problems. Did you understand Ray's method?
19. Jan 10, 2013
### Joffan
Re: Probability
Sorry for the misunderstanding guys. I was giving an extension question to illustrate the reasoning; I'm not stuck.
20. Jan 10, 2013
### Ray Vickson
Re: Probability
We can see it is incorrect by solving it and getting the wrong answer (x = -3/2 instead of x = 1/2). To understand WHY it is wrong, go back and look at the definition of x and the argument I gave for the equation satisfied by x. The point is that x already includes all the possibilities of winning on the first, third, fifth, .... go.
However, it would be correct to write
$$x = \frac{1}{6} + \frac{2}{3}\frac{1}{6} + \frac{2}{3}\frac{2}{3}\,x$$
for similar reasons. This equation does have the correct solution x = 1/2.
Last edited: Jan 10, 2013 | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9848109503004294,
"lm_q1q2_score": 0.8706115621504746,
"lm_q2_score": 0.884039278690883,
"openwebmath_perplexity": 1557.7460466744242,
"openwebmath_score": 0.7214272618293762,
"tags": null,
"url": "https://www.physicsforums.com/threads/probabilities-of-archers-a-and-b-hitting-the-bulls-eye.663461/"
} |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 17 Oct 2018, 20:07
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# A farmer has an apple orchard consisting of Fuji and Gala
Author Message
TAGS:
### Hide Tags
Intern
Joined: 23 Apr 2010
Posts: 7
A farmer has an apple orchard consisting of Fuji and Gala [#permalink]
### Show Tags
Updated on: 12 Aug 2012, 07:01
3
12
00:00
Difficulty:
65% (hard)
Question Stats:
70% (02:56) correct 30% (03:10) wrong based on 696 sessions
### HideShow timer Statistics
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
A. 22
B. 33
C. 55
D. 77
E. 88
_________________
"Choose to chance the rapids and dance the tides"
Originally posted by iamseer on 26 Apr 2010, 15:13.
Last edited by Bunuel on 12 Aug 2012, 07:01, edited 1 time in total.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8386
Location: Pune, India
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
02 Apr 2011, 07:04
3
3
jay121 wrote:
Please tell me how should I calculate 187*0.85=220 within 30 seconds?
Is there any specific trick for that?
Even if i split it up and say: 0.1x +0.75x = 187 I have difficulties to solve it fast? Any tipps?
Thanks
10% are cross pollinated and 75% are pure Fuji so 85% are Fuji which gives you | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972768020107,
"lm_q2_score": 0.8705972768020107,
"openwebmath_perplexity": 9131.965340582265,
"openwebmath_score": 0.5062430500984192,
"tags": null,
"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html"
} |
Thanks
10% are cross pollinated and 75% are pure Fuji so 85% are Fuji which gives you
$$(\frac{85}{100})*x = 187$$
Now it is obvious that number of trees has to be an integer so 85 and 187 need to have come common factor.
187 isn't divisible by 2 (not even), by 3(1+8+7 = 16 so not divisible by 3), by 7 (since 7*2 = 14, you have 47 left which will not go by 7) but it is divisible by 11 (11*1 = 11 and 11*7 = 77). So you split 187 into 11*17.
Now 85 is 17*5.
Now the equation becomes:
$$(\frac{17*5}{100})*x = 11*17$$
17 gets canceled and 5 gets canceled with 100 leaving a 20. So you get x = 220
_________________
Karishma
Veritas Prep GMAT Instructor
GMAT self-study has never been more personalized or more fun. Try ORION Free!
##### General Discussion
Manager
Joined: 12 Jan 2010
Posts: 246
Schools: DukeTuck,Kelogg,Darden
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
26 Apr 2010, 15:17
3
1
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
Let the total trees be x
3/4 are pure Fuji = 3x/4
10% cross pollinated = x/10
now The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187
3x/4 + x/10 = 187
solve this x = 220
220-187 = 33 are the pure Gala trees.
_________________
Run towards the things that make you uncomfortable daily. The greatest risk is not taking risks
http://gmatclub.com/forum/from-690-to-730-q50-v38-97356.html
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2606
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972768020107,
"lm_q2_score": 0.8705972768020107,
"openwebmath_perplexity": 9131.965340582265,
"openwebmath_score": 0.5062430500984192,
"tags": null,
"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html"
} |
### Show Tags
Updated on: 26 Apr 2010, 15:21
1
2
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
Let f = pure fuji , g = pure gala and c - cross pollinated.
c = 10% of x where x is total trees.
c = .1x
also 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220
220 - 187 = pure gala = 33.
PS: While posting the questions please post options and source.
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
Originally posted by gurpreetsingh on 26 Apr 2010, 15:19.
Last edited by gurpreetsingh on 26 Apr 2010, 15:21, edited 1 time in total.
Manager
Joined: 12 Jan 2010
Posts: 246
Schools: DukeTuck,Kelogg,Darden
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
26 Apr 2010, 15:21
gurpreetsingh wrote:
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
Let f = pure fuji , g = pure gala and c - cross pollinated.
c = 10% of x where x is total trees.
c = .1x
also 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220
220 - 187 = pure gala = 33.
Hi shouldnt it be 33, since 187 includes the fuji and the ones that cross pollinated???
_________________ | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972768020107,
"lm_q2_score": 0.8705972768020107,
"openwebmath_perplexity": 9131.965340582265,
"openwebmath_score": 0.5062430500984192,
"tags": null,
"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html"
} |
Run towards the things that make you uncomfortable daily. The greatest risk is not taking risks
http://gmatclub.com/forum/from-690-to-730-q50-v38-97356.html
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2606
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
26 Apr 2010, 15:23
Silvers wrote:
gurpreetsingh wrote:
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
Let f = pure fuji , g = pure gala and c - cross pollinated.
c = 10% of x where x is total trees.
c = .1x
also 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220
220 - 187 = pure gala = 33.
Hi shouldnt it be 33, since 187 includes the fuji and the ones that cross pollinated???
Its 33 only check again what you have quoted.
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
GMAT Club Premium Membership - big benefits and savings
Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
Manager
Joined: 22 Oct 2009
Posts: 235
GMAT 1: 760 Q49 V44
GPA: 3.88
### Show Tags
04 Jul 2010, 11:16
1
First - I want to point out the fact that the Google ad on this particular thread is an advertisement that says, "Apple trees are on sale now!"
Classic.
I'm not sure how to do this without simply setting up the formulas and hacking your way through the algebra. I'm guessing from your post ("need to understand the easiest way to solve this one") that you don't want this method. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972768020107,
"lm_q2_score": 0.8705972768020107,
"openwebmath_perplexity": 9131.965340582265,
"openwebmath_score": 0.5062430500984192,
"tags": null,
"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html"
} |
Anyone else?
Manager
Joined: 16 Feb 2010
Posts: 181
### Show Tags
04 Jul 2010, 11:52
yup....no need for the 4 equations method
f+g= T
0.1T = c
0.75T = F
f+c= 187
you can figure out the rest.....i hope there is a simpler way of doing it....
Intern
Joined: 03 Jun 2010
Posts: 21
### Show Tags
05 Jul 2010, 13:10
2
Gala apples - G
Fuji apples - F
C - Cross Pollinated apples
Total apples - X
Given:
F = 3/4X
X = G + F + cross pollinated
Fuji + Cross pollinated (10 % of all apples) = 187
Solution:
3/4X + 1/10X = 187
Hence X = 220
X = G + F + cross pollinated
220 = G + 187
Hence G = 33.
Director
Joined: 21 Dec 2009
Posts: 534
Concentration: Entrepreneurship, Finance
### Show Tags
05 Jul 2010, 14:05
zisis wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
(A) 22
(B) 33
(C) 55
(D) 77
(E) 88
OA:
please explain method.....need to understand the easiest way to solve this one
I got confused with the highlighted part; how would the question be framed if the part in bold where to mean
pure fuji plus cross-pollinated fuji(not including cross-pollinated Gala)? Am I reading too much meaning to the question?
_________________
KUDOS me if you feel my contribution has helped you.
Intern
Joined: 18 Dec 2010
Posts: 4
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
01 Apr 2011, 21:04
Please tell me how should I calculate 187*0.85=220 within 30 seconds?
Is there any specific trick for that?
Even if i split it up and say: 0.1x +0.75x = 187 I have difficulties to solve it fast? Any tipps?
Thanks
Retired Moderator
Joined: 16 Nov 2010
Posts: 1437
Location: United States (IN)
Concentration: Strategy, Technology
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972768020107,
"lm_q2_score": 0.8705972768020107,
"openwebmath_perplexity": 9131.965340582265,
"openwebmath_score": 0.5062430500984192,
"tags": null,
"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html"
} |
### Show Tags
01 Apr 2011, 21:35
1
Let x be the number of apple trees
0.1x of apple trees are cross-pollinated
3x/4 = Fuji
3x/4 + 0.1x = 187
=> 0.85x = 187
=> 0.05x = 11
=> x = 1100/5 = 220
Now x/4 - 0.1x = Pure Gala
= 55 - 22 = 33
_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)
GMAT Club Premium Membership - big benefits and savings
Director
Joined: 01 Feb 2011
Posts: 668
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
02 Apr 2011, 13:24
Lets assume F stands for pure Fuji, G for pure Gala , CP for cross pollinated and T for total trees.
F + CP = 187
=> 3T/4 + 10T/100 = 187 = > T = 220
T = F+G+CP = > G = 33
Retired Moderator
Joined: 20 Dec 2010
Posts: 1835
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
02 Apr 2011, 13:48
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
10% of his trees cross pollinated i.e. Cross Pollinated = 0.1*Total
The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187 i.e. Pure Fuji+ Cross Pollinated=187
3/4 of all his trees are pure Fuji i.e. Pure Fuji=0.75*Total
Pure Fuji+ Cross Pollinated=0.75*Total+0.1*Total=0.85*Total
0.85*Total=187
Total=187/0.85
Pure Gala = Total - (Pure Fuji+ Cross Pollinated) = 187/0.85-187=187((1/0.85)-1)=187(0.15/0.85)=187(3/17)=11*3=33
Ans: 33 Pure Gala Apples
_________________
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1055
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
01 May 2011, 23:35
0.1x + 0.75x = 187
x = 220 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972768020107,
"lm_q2_score": 0.8705972768020107,
"openwebmath_perplexity": 9131.965340582265,
"openwebmath_score": 0.5062430500984192,
"tags": null,
"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html"
} |
### Show Tags
01 May 2011, 23:35
0.1x + 0.75x = 187
x = 220
number of G trees = 220-187 = 33
Hence B
Intern
Joined: 15 Nov 2011
Posts: 20
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
12 Aug 2012, 05:53
Hello,
I got this question today on my MGMAT CAT and tried first to solve it through a chart for overlapping sets, but then realized that this is not possible. I solved the question afterwards as in the posts here equation system). How can I identify that the chart won't work here?
Thanks!
Intern
Joined: 31 May 2012
Posts: 10
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
22 Sep 2012, 14:17
rocketscience wrote:
Hello,
I got this question today on my MGMAT CAT and tried first to solve it through a chart for overlapping sets, but then realized that this is not possible. I solved the question afterwards as in the posts here equation system). How can I identify that the chart won't work here?
Thanks!
2X2 works as well. You just need to set "neither" as 0 because apples have to be either G or F.
-------F -----nF
G----.1x-----y
nG---.75x---0
------187----?---x
From here solve for x and y will be the difference between x and 187.
Manager
Joined: 02 Nov 2009
Posts: 112
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
04 Oct 2012, 15:48
this is the explanatio i understood
can anyone explain y it should be 55-22
Thank
subhashghosh wrote:
Let x be the number of apple trees
0.1x of apple trees are cross-pollinated
3x/4 = Fuji
3x/4 + 0.1x = 187
=> 0.85x = 187
=> 0.05x = 11
=> x = 1100/5 = 220
Now x/4 - 0.1x = Pure Gala
= 55 - 22 = 33
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8386
Location: Pune, India
Re: A farmer has an apple orchard: MGMAT, PS [#permalink]
### Show Tags
04 Oct 2012, 21:16
venmic wrote:
this is the explanatio i understood
can anyone explain y it should be 55-22
Thank
subhashghosh wrote:
Let x be the number of apple trees | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972768020107,
"lm_q2_score": 0.8705972768020107,
"openwebmath_perplexity": 9131.965340582265,
"openwebmath_score": 0.5062430500984192,
"tags": null,
"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html"
} |
Thank
subhashghosh wrote:
Let x be the number of apple trees
0.1x of apple trees are cross-pollinated
3x/4 = Fuji
3x/4 + 0.1x = 187
=> 0.85x = 187
=> 0.05x = 11
=> x = 1100/5 = 220
Now x/4 - 0.1x = Pure Gala
= 55 - 22 = 33
3/4 (i.e. 75%) are pure Fuji and 10% are cross so 15% are pure Gala.
So once you get x, you can calculate pure Gala as 15% of x = 15/100 * 220 = 33
or you can say Pure Gala = 25% of x (Gala apples) - 10% of x (cross) = 55 - 22 = 33 (as done above)
_________________
Karishma
Veritas Prep GMAT Instructor
GMAT self-study has never been more personalized or more fun. Try ORION Free!
Intern
Joined: 12 Jun 2012
Posts: 36
Re: A farmer has an apple orchard consisting of Fuji and Gala [#permalink]
### Show Tags
05 Oct 2012, 00:45
2
1
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?
A. 22
B. 33
C. 55
D. 77
E. 88
THE QUICK METHOD...
Fuji + Cross = 187
10% are cross
75% are Fuji
so 85% = 187
We want to know what the 15% is
Divide our percent by 10, 8.5% = 18.7
Double it, 17% = 38
We need 15 percent and it is pretty obvious 33 fits the bill
_________________
Re: A farmer has an apple orchard consisting of Fuji and Gala &nbs [#permalink] 05 Oct 2012, 00:45
Go to page 1 2 Next [ 28 posts ]
Display posts from previous: Sort by | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972768020107,
"lm_q2_score": 0.8705972768020107,
"openwebmath_perplexity": 9131.965340582265,
"openwebmath_score": 0.5062430500984192,
"tags": null,
"url": "https://gmatclub.com/forum/a-farmer-has-an-apple-orchard-consisting-of-fuji-and-gala-93322.html"
} |
Find all School-related info fast with the new School-Specific MBA Forum
It is currently 31 May 2016, 06:03
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# The probability that a visitor at the mall buys a pack of
Author Message
TAGS:
### Hide Tags
Senior Manager
Joined: 05 Oct 2008
Posts: 272
Followers: 3
Kudos [?]: 278 [4] , given: 22
The probability that a visitor at the mall buys a pack of [#permalink]
### Show Tags
20 Oct 2009, 06:44
4
KUDOS
16
This post was
BOOKMARKED
00:00
Difficulty:
(N/A)
Question Stats:
68% (02:04) correct 32% (01:01) wrong based on 137 sessions
### HideShow timer Statistics
Three for comparison: How do we decide what formula to use? Can someone explain the logic please..
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
How do we decide whether the solution is
3/10* 3/10 * 7/10 = .063
OR
3/10* 3/10 * 7/10 *3 = 0.189
Likewise the following question:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior.If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15
How do we decide whether the solution is
60/1000 * 1/800 = 3/40,000
OR
60/1000 * 1/800 * 2 = 3/20,000
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* $$\frac{2}{3}$$
* $$\frac{1}{2}$$
* $$\frac{2}{5}$$
* $$\frac{1}{3}$$
This has yet another method to calculate!!
Manager
Joined: 25 Aug 2009
Posts: 175
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Followers: 11
Kudos [?]: 161 [8] , given: 3
### Show Tags
20 Oct 2009, 13:46
8
KUDOS
Answer to the first question -
0.3*0.3*0.7 is equivalent to saying the the first person picks candy and second person picks candy and third person doesn't pick candy.
however this is a different case compared to
0.3*0.7*0.3 is equivalent to saying the the first person picks candy and second person doesn't pick candy and third person picks candy.
which in turn is a fifferent case compared to
0.7*0.3*0.3 (I'll skip the verbose comments)
Answer to the second question -
Probability of picking a sibling from the junior class is 60/1000
and Probability of picking the corresponding sibling from the senior class is 1/800.
However in this case if we flip the case we still end up with the same sibling pair since the order is not important. Hence we don't multiply by 2
Answer to the third question -
We just list the various cases where the sum will be 8
6 2
2 6
5 3
3 5
4 4
Total = 5
5 occurs in 2 cases.
hence prob is 2/5.
_________________
Rock On
Math Expert
Joined: 02 Sep 2009
Posts: 33103
Followers: 5782
Kudos [?]: 70969 [19] , given: 9857
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
Kudos [?]: 70969 [19] , given: 9857
### Show Tags
20 Oct 2009, 21:06
19
KUDOS
Expert's post
8
This post was
BOOKMARKED
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
Solution: P(B=2)=3!/2!*0.3^2*0.7=0.189
The point here is not about to multiply or not by 3, the point is to find # of ways favorable scenario to occur: in our case we are asked to find the probability of 2 out of 3 visitors to buy the candy.
3 visitors, 2 out of them buy the candy, it can occur in 3 ways: BBN, BNB, NBB --> =3!/2!=3. We are dividing by 2! because B1 and B2 are identical for us, combinations between them aren’t important. Meaning that favorable scenario: B1, B2, N and B2, B1, N is the same: two first visitors bought the candy and the third didn’t.
NOTE: P(B=2) is the same probability as the P(N=1), as if exactly two bought, means that exactly one didn’t.
Let’s consider some similar examples:
1. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly one visitors will buy a pack of candy?
The same here favorable scenarios are: NNB, NBN, BNN – total of three. 3!/2! because again two visitors who didn’t bought the candy are identical for us: N1,N2,B is the same scenario as N2,N1,B – first two visitors didn’t buy the candy and the third one did.
So, the answer for this case would be: P(N=2)=3!/2!*0.7^2*0.3=0.441
NOTE: P(N=2) is the same probability as the P(B=1), as if exactly two didn’t buy, means that exactly one did.
2. The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that at least one visitors will buy a pack of candy?
At least ONE buys, means that buys exactly one OR exactly two OR exactly three: | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
At least ONE buys, means that buys exactly one OR exactly two OR exactly three:
P(B>=1)=P(B=1)+P(B=2)+P(B=3)=3!/2!*0.3*0.7^2+3!/2!*0.3^2*0.7+3!/3!*0.3^3=0.441+0.189+0.027=0.657
P(B=1) --> 0.3*0.7^2 (one bought, two didn’t) multiplied by combinations of BNN=3!/2!=3 (Two identical N’s)
P(B=2) --> 0.3^2*0.7 (two bought, one didn’t) multiplied by combinations of BBN=3!/2!=3 (Two identical B’s)
P(B=3) --> 0.3^3 (three bought) multiplied by combinations of BBB=3!/3!=1 (Three identical B’s). Here we have that only ONE favorable scenario is possible: that three visitors will buy - BBB.
BUT! The above case can be solved much easier: at least 1 visitor buys out of three is the opposite of NONE of three visitors will buy, B=0: so it’s better to solve it as below:
P(B>=1)=1-P(B=0, the same as N=3)=1-3!/3!*0.7^3=1-0.7^3.
3. The probability that a visitor at the mall buys a pack of candy is 30%. If five visitors come to the mall today, what is the probability that at exactly two visitors will buy a pack of candy?
P(B=2)=5!/2!3!*0.3^2*0.7^3
We want to count favorable scenarios possible for BBNNN (two bought the candy and three didn’t) --> 2 identical B-s and 3 identical N-s, total of five visitors --> 5!/2!3!=10 (BBNNN, BNBNN, BNNBN, BNNNB, NBNNB, NNBNB, NNNBB, NNBBN, NBBNN, NBNBN). And multiply this by the probability of occurring of 2 B-s=0.3^2 and 3 N-s=0.7^3.
Hope now it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 33103
Followers: 5782
Kudos [?]: 70969 [12] , given: 9857
### Show Tags
20 Oct 2009, 21:09
12
KUDOS
Expert's post
3
This post was
BOOKMARKED
Let’s move to the second problem:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.
What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000
I see no argument of multiplying this by two.
This problem can be solved in another way and maybe this way shows that no need of multiplication:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Let’s count favorable outcomes: 1 from 60=60C1=60
The pair of the one chosen=1C1=1
So total favorable outcomes=60C1*1C1=60
Probability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000
Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
The same way here: | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
The same way here:
What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000
Another way:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Favorable outcomes:
1 from 120=120C1=120
The pair of the one chosen=1C2=2
So total favorable outcomes=120C1*1C2=240
Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 33103
Followers: 5782
Kudos [?]: 70969 [8] , given: 9857
### Show Tags
20 Oct 2009, 21:10
8
KUDOS
Expert's post
1
This post was
BOOKMARKED
Third problem:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* 2/3
* 1/2
* 2/5
* 1/3
In this case we know that drawing has already happened and the sum of two cards was 8.
What combinations of two cards are possible to total 8?
(first card, second card)
(6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.
From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
If the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?
What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.
Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18
_________________
Senior Manager
Joined: 05 Oct 2008
Posts: 272
Followers: 3
Kudos [?]: 278 [0], given: 22
### Show Tags
21 Oct 2009, 06:44
Great Explanations and examples - couldn't have been more clear!!
Kudos given for all of them..
Thanks!!
Manager
Joined: 06 Oct 2009
Posts: 69
Followers: 2
Kudos [?]: 33 [1] , given: 5
### Show Tags
21 Oct 2009, 23:51
1
KUDOS
Problem 1:
As there are three visitors, scenarios can be as below
so Probability= (3/10*3/10*7/10)+(3/10*7/10*3/10)+(7/10*3/10*3/10) = 0.189
Problem 2:
Probability of selecting 1 sibling pair= (60/1000*60/800*1/60*1/60)= 1/800000
This can happen 60 times so probability= (1/800000)*60= 3/40000
Problem 3:
possible pairs are (2,6),(3,5),(4,4),(5,3),(6,2) so total 5 possibilities
two pairs are having 5
so probability = 2/5
Intern
Joined: 01 Dec 2008
Posts: 5
Followers: 0
Kudos [?]: 0 [0], given: 4
### Show Tags
08 May 2010, 03:32
Bunuel,
Excellent explanations. I really loved the way you explain with examples.
I am really glad that i started using gmatclub.. Thanks!
-Vinod
Manager
Joined: 05 Mar 2010
Posts: 221
Followers: 1
Kudos [?]: 28 [0], given: 8
### Show Tags
09 May 2010, 02:02
good questions and excellent explanations
_________________
Success is my Destiny
Current Student
Joined: 15 Jul 2010
Posts: 256
GMAT 1: 750 Q49 V42
Followers: 8
Kudos [?]: 143 [0], given: 65
### Show Tags
12 Nov 2010, 20:21
Thanks for the question and explanations.
Kudos for both!
_________________
Consider KUDOS if my post was helpful. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
Kudos for both!
_________________
Consider KUDOS if my post was helpful.
My Debrief: 750-q49v42-105591.html#p825487
Manager
Joined: 23 Jan 2011
Posts: 127
Followers: 1
Kudos [?]: 49 [0], given: 13
### Show Tags
24 May 2011, 10:30
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?
What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.
Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18
Can you please explain why number of outcomes will be 36?
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 920
Followers: 13
Kudos [?]: 294 [1] , given: 123
### Show Tags
24 May 2011, 22:12
1
KUDOS
Welcome to the club !
This interpretation is wrong. Probability = Favourables / Total. But this question total is constrained if you read it carefully - "the sum of two cards would be 8". Total is not the absolute ways in which you will pick the cards i.e. 6 * 6 = 36 absolute ways. Total ways to get a sum of 8 using two cards is 5. And that's why the ultimate probability is 2/5
Chetangupta wrote:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?
What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.
Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18
Can you please explain why number of outcomes will be 36?
Manager
Joined: 23 Jan 2011
Posts: 127
Followers: 1
Kudos [?]: 49 [0], given: 13
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
Kudos [?]: 49 [0], given: 13
### Show Tags
24 May 2011, 22:55
thank you. makes sense.
Manager
Joined: 13 May 2010
Posts: 124
Followers: 0
Kudos [?]: 10 [0], given: 4
### Show Tags
08 Jan 2012, 10:44
Regarding the below listed explanation from funnel -
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15
This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.
What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000
I see no argument of multiplying this by two.
-------------------------------------------------------------------------------------------------------------
I have a question regarding this one -
So basically you are considering the case when the first sibling selected is from the junior class and the second sibling is from the senior class.....but why does it have to be only in that order.....why would you not consider the case when first sibling is from senior class and the second is from the junior class?
Manager
Joined: 29 Jul 2011
Posts: 107
Location: United States
Followers: 5
Kudos [?]: 48 [0], given: 6
### Show Tags
08 Jan 2012, 20:47
1. (0.3)(0.3)(0.7) x 4C3 = 0.189
2. 60/1000 x 1/800 = 3/20000
_________________ | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
08 Jan 2012, 20:47
1. (0.3)(0.3)(0.7) x 4C3 = 0.189
2. 60/1000 x 1/800 = 3/20000
_________________
I am the master of my fate. I am the captain of my soul.
Please consider giving +1 Kudos if deserved!
DS - If negative answer only, still sufficient. No need to find exact solution.
PS - Always look at the answers first
CR - Read the question stem first, hunt for conclusion
SC - Meaning first, Grammar second
RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 9719
Followers: 466
Kudos [?]: 120 [0], given: 0
Re: The probability that a visitor at the mall buys a pack of [#permalink]
### Show Tags
21 Sep 2013, 12:02
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0
Kudos [?]: 38 [0], given: 23
### Show Tags
20 Apr 2014, 17:28
Bunuel wrote:
Third problem:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* 2/3
* 1/2
* 2/5
* 1/3
In this case we know that drawing has already happened and the sum of two cards was 8.
What combinations of two cards are possible to total 8?
(first card, second card)
(6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.
From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
If the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8?
What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.
Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18
Hi Bunuel,
Thanks for the great explanations. I'm stuck on #3 -- first two worked out great.
-In the scenario where we get 2/5, why isn't it 2/10 because isn't the favorable outcome(2) and the total outcomes is the total number of cards, which is 10? therefore 2/10?
-Can you please talk about the second scenario a little bit. I'm not catching the nuance that makes you multiply 6*6? Why wouldn't it be 6!?
Math Expert
Joined: 02 Sep 2009
Posts: 33103
Followers: 5782
Kudos [?]: 70969 [1] , given: 9857
### Show Tags
21 Apr 2014, 00:34
1
KUDOS
Expert's post
russ9 wrote:
Bunuel wrote:
Third problem:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?
* 1
* 2/3
* 1/2
* 2/5
* 1/3
In this case we know that drawing has already happened and the sum of two cards was 8.
What combinations of two cards are possible to total 8?
(first card, second card)
(6,2) (2,6) (5,3) (3,5) (44) – only 5 possible ways sum to be 8. One from this 5 has already happened.
From this five, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5=2/5.
If the problem were: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. What is the probability that one of the cards drawn would be 5 and the sum of two cards would be 8? | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
What combinations of two cards are possible to total 8 AND one of them to be 5?
(5,3) (3,5) – two favorable outcomes.
Total number of outcomes=6*6=36.
P(one card 5, sum 8)=2/36=1/18
Hi Bunuel,
Thanks for the great explanations. I'm stuck on #3 -- first two worked out great.
-In the scenario where we get 2/5, why isn't it 2/10 because isn't the favorable outcome(2) and the total outcomes is the total number of cards, which is 10? therefore 2/10?
-Can you please talk about the second scenario a little bit. I'm not catching the nuance that makes you multiply 6*6? Why wouldn't it be 6!?
The number of total outcomes is 5, not 10: (6,2) (2,6) (5,3) (3,5) (4, 4). Only 5 possible ways sum to be 8.
So, the probability that one of the cards drawn was a 5 is 2/5.
As for your second question: 6 options for the first card and 6 options for the second card, thus total 6*6=35 options.
_________________
Intern
Joined: 01 Aug 2010
Posts: 14
Followers: 0
Kudos [?]: 0 [0], given: 41
Re: The probability that a visitor at the mall buys a pack of [#permalink]
### Show Tags
14 Apr 2015, 22:08
Bunuel wrote:
Let’s move to the second problem:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A: 3/40000
B: 1/3600
C: 9/2000
D: 3/20000
E: 1/15
This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.
What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them). | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000
I see no argument of multiplying this by two.
This problem can be solved in another way and maybe this way shows that no need of multiplication:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Let’s count favorable outcomes: 1 from 60=60C1=60
The pair of the one chosen=1C1=1
So total favorable outcomes=60C1*1C1=60
Probability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000
Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
The same way here:
What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
So the probability of that the 2 students selected will be a sibling pair is: 120/1000*2/800=3/10000
Another way:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Favorable outcomes:
1 from 120=120C1=120
The pair of the one chosen=1C2=2
So total favorable outcomes=120C1*1C2=240
Probability=Favorable outcomes/Total # of outcomes=240/(1000*800)=3/10000
I understand your explanation but have a doubt. This is for the second question.
After 60/1000 * 1/800 = 3/40000. But shouldn't we add 59 to it, to include the different ways of choosing the first sibling pair? I don't think I am clear on when to count these different ways and when not to. For e.g., when we count the number of shoppers in the first question, we count the different arrangements of shoppers, then why shouldn't the different sibling pairs be counted here?
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 9719
Followers: 466
Kudos [?]: 120 [0], given: 0
Re: The probability that a visitor at the mall buys a pack of [#permalink]
### Show Tags
17 Apr 2016, 01:21
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: The probability that a visitor at the mall buys a pack of [#permalink] 17 Apr 2016, 01:21
Similar topics Replies Last post
Similar
Topics:
1 At the wholesale store you can buy an 8-pack of hot dogs for \$1.55, a 7 22 Dec 2015, 21:36
1 The probability that a visitor at the mall buys a pack of candy is 30% 4 03 Apr 2011, 16:14
Probability 9 23 Jan 2010, 15:49
24 The probability that a visitor at the mall buys a pack of 19 16 Nov 2007, 09:27
19 The probability that a visitor at the mall buys a pack of 23 13 Nov 2007, 09:20
Display posts from previous: Sort by | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972751232809,
"lm_q2_score": 0.8705972751232809,
"openwebmath_perplexity": 1810.2808087801413,
"openwebmath_score": 0.6559093594551086,
"tags": null,
"url": "http://gmatclub.com/forum/the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-85523.html"
} |
It is currently 23 Mar 2018, 17:45
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
### Show Tags
12 Dec 2012, 05:29
30
This post was
BOOKMARKED
00:00
Difficulty:
65% (hard)
Question Stats:
66% (01:53) correct 34% (01:59) wrong based on 1656 sessions
### HideShow timer Statistics
A pharmaceutical company received $3 million in royalties on the first$20 million in sales of the generic equivalent of one of its products and then $9 million in royalties on the next$108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first $20 million in sales to the next$108 million in sales?
(A) 8%
(B) 15%
(C) 45%
(D) 52%
(E) 56%
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 44419
### Show Tags
14 Dec 2012, 03:18
3
KUDOS
Ans:
for this kind of percentage change questions we apply the formula (change/original)x100, so here we have initial ratio=3/20 final ratio=1/12 . now change = 3/20-1/12=1/15 , putting these values in the formula we get the answer as (C).
_________________
www.mnemoniceducation.com
Intern
Joined: 04 Dec 2012
Posts: 2
### Show Tags
26 Feb 2013, 11:34
1
KUDOS
carloswn wrote:
To Brunuel:
Thanks for the sharing, I'm wondering whether your formula shouldn't be (9/108-3/20) / (3/20) ? This doesn't change the final answer anyway. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.870597270087091,
"lm_q2_score": 0.870597270087091,
"openwebmath_perplexity": 3873.8104959385996,
"openwebmath_score": 0.6309756636619568,
"tags": null,
"url": "https://gmatclub.com/forum/a-pharmaceutical-company-received-3-million-in-royalties-143985.html"
} |
I don't think it'll make a difference because in the numerator we are just looking for the difference between the two amounts. Just make sure that in the denominator you have the original value.
For better understanding:
$$Percent Change = \frac{Change in Value}{Original Value}$$
Hence, the "change in value" is just the difference. I wouldn't worry about a negative here.
_________________
If my post has contributed to your learning or teaching in any way, feel free to hit the kudos button ^_^
Manager
Joined: 07 Apr 2014
Posts: 130
### Show Tags
29 May 2015, 02:57
For me, percentage questions seem time consuming ..not sure if I am the only one feel this way..
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11313
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: A pharmaceutical company received $3 million in royalties [#permalink] ### Show Tags 29 May 2015, 15:53 3 This post received KUDOS Expert's post Hi katzzzz, Percent questions come from the broader family of 'ratio-based' questions and you're going to see a bunch of those on Test Day, so you have to make sure that you're ready for them. While some of these questions can be wordier/longer than average, the 'key' to answering these types of questions quicker is to organize information in the most effective way possible (for the question that is asked and for the answer choices that are given). For example, ALL of the following examples mean the same thing, so you have to decide which would be easiest to work with... Men/Women = 1 to 10 = 1:10 = 1/10 = 0.1 = 10% 10M = 1W GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/ | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.870597270087091,
"lm_q2_score": 0.870597270087091,
"openwebmath_perplexity": 3873.8104959385996,
"openwebmath_score": 0.6309756636619568,
"tags": null,
"url": "https://gmatclub.com/forum/a-pharmaceutical-company-received-3-million-in-royalties-143985.html"
} |
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 882
### Show Tags
23 Nov 2015, 15:55
1
KUDOS
1
This post was
BOOKMARKED
% change also equals :
$$\frac{new}{old}$$$$- 1$$
$$\frac{9}{108}$$*$$\frac{20}{3}$$$$- 1$$
= $$\frac{-4}{9}$$ ==> since 1/9 = 0,11 ==> -44%
_________________
New Application Tracker : update your school profiles instantly!
Manager
Joined: 30 Dec 2015
Posts: 90
GPA: 3.92
WE: Engineering (Aerospace and Defense)
### Show Tags
08 Mar 2016, 00:30
To Bunuel:
I tend to get confused when estimating in certain scenarios and select the wrong answer. In this case I took denominator 20 as 18 so that 3/20 becomes 3/18 = 1/6 and 9/108 = 1/12 so 1/6-1/12 =1/12 and now (1/12)/(1/6)*100 = 50% so I would have selected (D) - 52% which is the closest. In another case had I taken 20 as 21(more convenient) the fraction would have been 3/21 = 1/7 and to balance the round off I would have taken 12 as 10 (one increase other decrease) so (1/7-1/10)/(1/7) = (3/70)*7 = 30% its quite far apart from the options so either 15 or 45 although we are actually reducing the 108 mn to 90 mn which is a huge difference so the overall result should be higher so 45 should be the case. In the next scenario I took 20 mn as 21 mn and then ratio became (1/7) so ({1/7-1/12}/(1/7))*100 = 5/12*100 = 42% approx so again option C (45%). So in these cases the answers are different so then how to reach the correct answer in quick time if choices are quite close, there is a high chance of error. Kindly advise the extent to which we can approximate values of fractions as answers can be a huge difference in the answers arrived by taking multiple scenarios as if one scenario leads us to one choice we will quickly select it and move on as there is the time factor.
_________________
Give kudos and appreciate if you think its worthwhile | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.870597270087091,
"lm_q2_score": 0.870597270087091,
"openwebmath_perplexity": 3873.8104959385996,
"openwebmath_score": 0.6309756636619568,
"tags": null,
"url": "https://gmatclub.com/forum/a-pharmaceutical-company-received-3-million-in-royalties-143985.html"
} |
Give kudos and appreciate if you think its worthwhile
SVP
Joined: 12 Sep 2015
Posts: 2159
Re: A pharmaceutical company received $3 million in royalties [#permalink] ### Show Tags 10 Apr 2016, 10:43 2 This post received KUDOS Expert's post Walkabout wrote: A pharmaceutical company received$3 million in royalties on the first $20 million in sales of the generic equivalent of one of its products and then$9 million in royalties on the next $108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first$20 million in sales to the next $108 million in sales? (A) 8% (B) 15% (C) 45% (D) 52% (E) 56% First$20 million: royalties/sales ratio = 3/20 = 36/240
Next $108 million: royalties/sales ratio = 9/108 = 1/12 = 20/240 Noticed that I rewrote both with the SAME DENOMINATOR. So, now all we need to is determine the percent change from 36 to 20. To do so, we could use some more lengthy calculations [e.g., 100(36-20)/36] HOWEVER, notice that, if we start at 36, a 50% decrease would give us 18. So going from 36 to 20, must be a decrease that's LESS THAN 50% (but also pretty close to 50%) Only one answer choice works. Answer: C Related Resource The following free video covers the concepts/strategies that are useful for answering this question: Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2166 Re: A pharmaceutical company received$3 million in royalties [#permalink]
### Show Tags
03 May 2016, 09:27
2
KUDOS
Expert's post
A pharmaceutical company received $3 million in royalties on the first$20 million in sales of the generic equivalent of one of its products and then $9 million in royalties on the next$108 million in sales. By approximately what percent did the ratio of royalties to sales decrease from the first $20 million in sales to the next$108 million in sales? | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.870597270087091,
"lm_q2_score": 0.870597270087091,
"openwebmath_perplexity": 3873.8104959385996,
"openwebmath_score": 0.6309756636619568,
"tags": null,
"url": "https://gmatclub.com/forum/a-pharmaceutical-company-received-3-million-in-royalties-143985.html"
} |
(A) 8%
(B) 15%
(C) 45%
(D) 52%
(E) 56%
Solution:
This is a percent decrease problem. We will use the formula: percent change = (new – old)/old x 100 to calculate the final answer.
We first set up the ratios of royalties to sales. The first ratio will be for the first 20 million in sales, and the second ratio will be for the next 108 million in sales. Because all of the sales are in millions, we do not have to express all the trailing zeros in our ratios.
First 20 Million
royalties/sales = 3/20
Next 108 Million
royalties/sales = 9/108 = 1/12
Because each ratio is not an easy number to use, we can simplify each one by multiplying each by the LCM of the two denominators, which is 60. Keep in mind that we are able to do this only because our answer choices are expressed in percents.
First 20 Million
royalties/sales = (3/20) x 60 = 9
Next 108 Million
royalties/sales = 9/108 = (1/12) x 60 = 5
We can plug 9 and 5 into our percent change formula:
(new – old)/old x 100
[(5 – 9)/9] x 100
-4/9 x 100
At this point we can stop and consider the answer choices. Since we know that 4/9 is just a bit less than ½, we know that -4/9 x 100 is about a 45% decrease.
_________________
Jeffery Miller
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Manager
Joined: 03 Jan 2017
Posts: 191
### Show Tags
11 Jun 2017, 04:15
I did this by "Long hand", Of course it eat up more time!
Just sharing in case someone interested.
For the first 20 Million, $$\frac{Royalty}{sales}= \frac{3}{20}$$
For the next 108 Million, $$\frac{Royalty}{sales} = \frac{9}{108}$$
Required percentage
= $$(\frac{3}{20}- \frac{9}{108})/\frac{3}{20} * 100$$
= $$(\frac{3}{20} - \frac{9}{108})* \frac{20}{3} * 100$$
= $$3* (\frac{1}{20}-\frac{3}{108}) * \frac{20}{3} * 100$$
= $$(1 - \frac{5}{9}) * 100$$
= $$\frac{4}{9} * 100$$
= 44.44444
_________________
My Best is yet to come!
Intern
Joined: 03 Jun 2017
Posts: 15
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.870597270087091,
"lm_q2_score": 0.870597270087091,
"openwebmath_perplexity": 3873.8104959385996,
"openwebmath_score": 0.6309756636619568,
"tags": null,
"url": "https://gmatclub.com/forum/a-pharmaceutical-company-received-3-million-in-royalties-143985.html"
} |
_________________
My Best is yet to come!
Intern
Joined: 03 Jun 2017
Posts: 15
### Show Tags
24 Sep 2017, 22:33
Percentage change = (3/20*100 -9/108*100) / (3/20*100)
= only percentage change to near to 50 and less that 50 is 45% hence answer must be C
_________________
Regards,
Naveen
email: nkmungila@gmail.com
Please press kudos if you like this post
Study Buddy Forum Moderator
Joined: 04 Sep 2016
Posts: 780
Location: India
WE: Engineering (Other)
### Show Tags
14 Jan 2018, 00:22
1
KUDOS
Expert's post
Bunuel chetan2u amanvermagmat niks18
Quote:
$$=\frac{\frac{3}{20}-\frac{9}{108}}{\frac{3}{20}}*100\approx{44%}$$.
Is below approach the most efficient for simplification:
Taking 1/4 common after simplifying(3/20 - 1/12)
in numerator which finally simplifies to 2/3 and then multiplying by 20/3
which approx to 40/9. Now since denominator is slightly less than 10
and 40/10 is 4 so we shall get fraction as slightly more than 4.xx as a value.
I'd suggest another way:
$$\frac{\frac{3}{20}-\frac{9}{108}}{\frac{3}{20}}=(\frac{3}{20}-\frac{1}{12})*\frac{20}{3}=1 -\frac{1}{12}*\frac{20}{3}=1-\frac{5}{9}=\frac{4}{9}=0.444....$$
_________________ | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.870597270087091,
"lm_q2_score": 0.870597270087091,
"openwebmath_perplexity": 3873.8104959385996,
"openwebmath_score": 0.6309756636619568,
"tags": null,
"url": "https://gmatclub.com/forum/a-pharmaceutical-company-received-3-million-in-royalties-143985.html"
} |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 22 Feb 2019, 06:25
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT RC Webinar
February 23, 2019
February 23, 2019
07:00 AM PST
09:00 AM PST
Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
• ### FREE Quant Workshop by e-GMAT!
February 24, 2019
February 24, 2019
07:00 AM PST
09:00 AM PST
Get personalized insights on how to achieve your Target Quant Score.
# If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0,
Author Message
TAGS:
### Hide Tags
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2597
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 05:17
5
15
00:00
Difficulty:
75% (hard)
Question Stats:
59% (02:21) correct 41% (02:32) wrong based on 901 sessions
### HideShow timer Statistics
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?
(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above
This is Question 5 of
Provide your solution below. Kudos for participation. Happy Solving!
Best Regards
The e-GMAT Team
_________________
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Originally posted by EgmatQuantExpert on 20 May 2015, 09:28.
Last edited by EgmatQuantExpert on 07 Aug 2018, 05:17, edited 3 times in total.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2597
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
22 May 2015, 06:55
2
4
Official Explanation
|y + 3| ≤ 3
means that the distance of y from the point -3 on the number line is less than, or equal to, 3
This gives us, -6 ≤ y ≤ 0
So, possible values of y (keeping in mind the constraint that y is an integer): -6, -5, -4, -3, -2, -1, 0
We’re given 2y – 3x = -6
So, $$x= \frac{(2y+6)}{3}$$
For x to be an integer, y must be a multiple of 3.
This means, possible values of y: -6, -3, 0
Corresponding values of x: -2, 0, 2
Corresponding values of xy: 12, 0, 0
Thus, minimum possible value of xy: 0
_________________
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
##### General Discussion
Retired Moderator
Joined: 29 Apr 2015
Posts: 839
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
Updated on: 21 May 2015, 13:20
2
2
EgmatQuantExpert wrote:
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?
(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above
This is Question 5 of
Provide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 22nd May.
Till then, Happy Solving!
Best Regards
The e-GMAT Team | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
Till then, Happy Solving!
Best Regards
The e-GMAT Team
How to deal with inequalities involving absolute values? First example shows us the so called "number case"
In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:
y + 3 ≤ 3
y ≤ 0
So y is 0 or negative
Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:
y + 3 >= -3
y >= -6
Therefore we have a possible range for y: -6=<y<=0
Ok, so far so good, we're half way through. What about x?
Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x.
You can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers (-6, -3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value.
Hence, Answer Choice C is the one to go.
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!
PS Please send me PM if I do not respond to your question within 24 hours.
Originally posted by reto on 20 May 2015, 13:19.
Last edited by reto on 21 May 2015, 13:20, edited 2 times in total.
Manager
Joined: 18 Nov 2013
Posts: 78
Concentration: General Management, Technology
GMAT 1: 690 Q49 V34
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
20 May 2015, 21:43
3
Eq: 1 |y + 3| ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integers
Solve Eq: 1 |y + 3| ≤ 3
$$+(y + 3) ≤ 3$$
$$y≤ 0$$
$$-(y + 3) ≤ 3$$
$$y\geq{-6}$$
Solve Eq: 2 2y – 3x + 6 = 0
$$x = \frac{(2y + 6)}{3}$$
as range of y: $$-6=<y<=0$$
we get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}
lowest xy = 0
Ans : C
_________________
_______
- Cheers
+1 kudos if you like | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
lowest xy = 0
Ans : C
_________________
_______
- Cheers
+1 kudos if you like
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2597
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
21 May 2015, 03:39
2
reto wrote:
How to deal with inequalities involving absolute values? First example shows us the so called "number case"
In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:
y + 3 ≤ 3
y ≤ 0
So y is 0 or negative
Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:
y + 3 >= -3
y >= -6
Therefore we have a possible range for y: -6=<y<=0
Ok, so far so good, we're half way through. What about x?
Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.
Dear reto
Very well explained! Kudos for that.
Just want to ask 2 questions:
1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not?
2. Is the question asking you to find the value of the product xy?
Based on the answers to these 2 questions, you may feel a need to edit the last part of your solution
Best Regards
Japinder
_________________
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Manager
Joined: 09 Nov 2013
Posts: 70
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
21 May 2015, 05:23 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
### Show Tags
21 May 2015, 05:23
As per the equation |y+3| <=3, y can have value from -6 to 0.
Now as per the equation x=(2y+6)/3; x will have negative values for all y in {-6,-5,-4,-3} hence the xy will be (-x)*(-y) a positive value. for y in { -2,-1,0}, x will have { 2/3,4/3,2}. Hence x*y will have minimum at x=2/3,y=-2 or x=4/3,y=-1
Intern
Joined: 23 Nov 2014
Posts: 3
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
22 May 2015, 03:28
1
-6<=y<=0
To get the least value of xy, we have to take the least values of both the terms. Taking y=-6, x is -2. Xy is 12. Hence, none.
Note : It is clearly mentioned that both x and y are integers.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2597
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
22 May 2015, 07:03
csirishac wrote:
-6<=y<=0
To get the least value of xy, we have to take the least values of both the terms. Taking y=-6, x is -2. Xy is 12. Hence, none.
Note : It is clearly mentioned that both x and y are integers.
Dear csirishac
You bring up an interesting point by saying the part in red.
This part is not true. As you yourself saw, when you minimized the values of both x and y (both negative), you got a positive product.
However, if x = +2 and y = 0 (both are actually the maximum values of x and y in this question), the product = 0, far lesser than the product 12 that you got above.
So, be careful about what you need to minimize.
Best Regards
Japinder
_________________
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Senior Manager
Joined: 27 Jul 2014
Posts: 251
Schools: ISB '15
GMAT 1: 660 Q49 V30
GPA: 3.76
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
10 Nov 2015, 01:50
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy? | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above
solving |y+3|<=3
gives
-6<=Y<=0
further equation 2y-3x+6=0
In order to fiind least value of x we nned to maximize y i,e 0
this gives x =2
And least value of y=-6
So product will be -12
Bunuel @VeritasPrepKarsihma Engr2012
Experts please let me know where I am missing something?
Math Expert
Joined: 02 Sep 2009
Posts: 53069
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
10 Nov 2015, 02:07
3
kanigmat011 wrote:
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?
(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above
solving |y+3|<=3
gives
-6<=Y<=0
further equation 2y-3x+6=0
In order to fiind least value of x we nned to maximize y i,e 0
this gives x =2
And least value of y=-6
So product will be -12
Bunuel @VeritasPrepKarsihma Engr2012
Experts please let me know where I am missing something?
The relationship between the values of x and y are given by the formula 2y – 3x + 6 = 0. You cannot take the values of x and y separately, meaning that you cannot take x= 2 and y = -6, because if x = 2, then y = 0, and xy = 0.
_________________
Intern
Joined: 22 Nov 2012
Posts: 21
Location: United States
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
03 Dec 2015, 09:38
EgmatQuantExpert wrote:
reto wrote:
How to deal with inequalities involving absolute values? First example shows us the so called "number case"
In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:
y + 3 ≤ 3
y ≤ 0
So y is 0 or negative
Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:
y + 3 >= -3
y >= -6
Therefore we have a possible range for y: -6=<y<=0
Ok, so far so good, we're half way through. What about x? | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
Ok, so far so good, we're half way through. What about x?
Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x. Now it's all about finding extreme values by combining with -6=<y<=0! If you play with the numbers and plug in y = -6, x will be -2 and the product would be 12. That's not what we want... if you plug in -2 for y, x will be 2/3 and the product xy is -2*2/3 ... since with all trials i don't get any of the numbers in the question steam, I assume the ANSWER is E.
Dear reto
Very well explained! Kudos for that.
Just want to ask 2 questions:
1. Please refer to the orange part in your solution. Is x = 2/3 a valid value for x? Why/ why not?
2. Is the question asking you to find the value of the product xy?
Based on the answers to these 2 questions, you may feel a need to edit the last part of your solution
Best Regards
Japinder
Japinder,
I'm getting back to yours qs:
1) x=2/3 must be rejected as per costraints in the question stem.
2) we are asked to find the least product of xy. This happens when y=0 x=2 or y=-3 x=0.
Thanks.
_________________
GMAT,
It is not finished untill I win!!!
Manager
Joined: 03 Dec 2014
Posts: 100
Location: India
GMAT 1: 620 Q48 V27
GPA: 1.9
WE: Engineering (Energy and Utilities)
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
07 Dec 2015, 19:38
UJs wrote:
Eq: 1 |y + 3| ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integers
Solve Eq: 1 |y + 3| ≤ 3
$$+(y + 3) ≤ 3$$
$$y≤ 0$$
$$-(y + 3) ≤ 3$$
$$y\geq{-6}$$
Solve Eq: 2 2y – 3x + 6 = 0
$$x = \frac{(2y + 6)}{3}$$
as range of y: $$-6=<y<=0$$
we get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}
lowest xy = 0
Ans : C
is it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
is it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12
pls. clear the doubt. thank in advance.
Math Expert
Joined: 02 Aug 2009
Posts: 7334
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
07 Dec 2015, 20:17
robu wrote:
UJs wrote:
Eq: 1 |y + 3| ≤ 3 and Eq: 2 2y – 3x + 6 = 0, x & y are integers
Solve Eq: 1 |y + 3| ≤ 3
$$+(y + 3) ≤ 3$$
$$y≤ 0$$
$$-(y + 3) ≤ 3$$
$$y\geq{-6}$$
Solve Eq: 2 2y – 3x + 6 = 0
$$x = \frac{(2y + 6)}{3}$$
as range of y: $$-6=<y<=0$$
we get 3 integer value of x , as x & y are integers for y = {-6, -3, 0} x = {-2, 0, 0} & xy = {12, 0, 0}
lowest xy = 0
Ans : C
is it that we should multiply corresponding pair of XY or can we multiply y= -6 and x= 2 XY =-12
pls. clear the doubt. thank in advance.
Hi,
the eq ly+3l<=3 gives following values of y..
y=0 ,-1,-2,-3,-4,-5,-6..
for these values x=2,-,-,0,-,-,-2...as x has to be an integer..
you have to take the corresponding values only..
i)when y=-6, then x =-2
ii)and when y=0, x=2..
iii) when y=-3, x=0..
ofcourse x as 0 and y as 0 is wrong inputs
so you get xy as 12 in (i) case and 0 in (ii) and (iii) case..
hope it helps
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html
GMAT Expert
Board of Directors
Joined: 17 Jul 2014
Posts: 2587
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
### Show Tags
28 Oct 2016, 06:11
EgmatQuantExpert wrote:
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value of the product xy?
(A) -12
(B) -3
(C) 0
(D) 2
(E) None of the above
we are given that
-3<= y <= -1.
y can take 3 values:
-3, -2, -1, and 0.
2y-3x +6 = 0
2y+6 = 3x
we can test values, but we can clearly see that if y is not -3/0, then x can't be an integer, but we need it to be an integer.
therefore, the only cases that actually work are y=-3, x=0 or y=0, and x=2
xy = 0 in any case
Senior Manager
Joined: 15 Jan 2017
Posts: 353
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
26 Jul 2017, 04:33
A simple algebraic approach.
We have |y + 3| less than eq to 3 which gives us -6 <=y<=0. So we have two values which are definitely for y (-6 and 0). Now we find x.
2y - 3x + 6 = 0 => 2y = 3x - 6 => y = 3x - 6/2
If we take y = -6; then we have :
- 6 = 3x - 6/ 2
-12 = 3x - 6
-6 = 3x
There fore x = -2.
X * Y = -2*-6 = positive 12 --> not the least value
Now lets take y = 0
0 = 3x - 6/2
3x = -6 => 6 = 3x so x =2
X*Y = 2*0 = 0
Would this approach be right? Expert comment would be welcome!
Manager
Joined: 11 Feb 2017
Posts: 188
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
26 Jul 2017, 04:46
A simple algebraic approach.
We have |y + 3| less than eq to 3 which gives us -6 <=y<=0. So we have two values which are definitely for y (-6 and 0). Now we find x.
2y - 3x + 6 = 0 => 2y = 3x - 6 => y = 3x - 6/2
If we take y = -6; then we have :
- 6 = 3x - 6/ 2
-12 = 3x - 6
-6 = 3x
There fore x = -2.
X * Y = -2*-6 = positive 12 --> not the least value
Now lets take y = 0
0 = 3x - 6/2
3x = -6 => 6 = 3x so x =2
X*Y = 2*0 = 0
Would this approach be right? Expert comment would be welcome! | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
Would this approach be right? Expert comment would be welcome!
Yes Indeed....This approach is correct
Manager
Joined: 23 Oct 2017
Posts: 64
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
09 Dec 2017, 08:33
If you visualize |y+3|<=3 on the number line you will get to know that y can take values in the range [-6,0]
Next express the give relation in terms of x i.e. y= (2/3)*(y-3).
and since y is in the range of [-6,0] and not in (0,3) .. x will follow y's sign.
Thus, the product xy has to be >=0.
Minumum being 0 .. check whether one can be 0 .. y=0 falls in the range. min. value of the product =0
Non-Human User
Joined: 09 Sep 2013
Posts: 9889
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink]
### Show Tags
20 Jan 2019, 04:37
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, [#permalink] 20 Jan 2019, 04:37
Display posts from previous: Sort by | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 1847.3903594093272,
"openwebmath_score": 0.7057603597640991,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-and-y-are-integers-such-that-y-3-3-and-2y-3x-198567.html"
} |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 16 Jul 2018, 00:01
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Root(80)+root(125)
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 47006
### Show Tags
01 Oct 2012, 05:02
2
5
00:00
Difficulty:
5% (low)
Question Stats:
92% (00:38) correct 8% (01:14) wrong based on 1872 sessions
### HideShow timer Statistics
$$\sqrt{80}+\sqrt{125} =$$
(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100
Practice Questions
Question: 52
Page: 159
Difficulty: 600
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47006
### Show Tags
01 Oct 2012, 05:03
1
3
SOLUTION
$$\sqrt{80}+\sqrt{125} =$$
(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100
$$\sqrt{80}+\sqrt{125} =\sqrt{16*5}+\sqrt{25*5}=4\sqrt{5}+5\sqrt{5}=9\sqrt{5}$$.
_________________
Board of Directors
Joined: 01 Sep 2010
Posts: 3428
### Show Tags
01 Oct 2012, 07:43
2
Whe we deal with this problems may be the best thing to do is to break the number into factors
$$\sqrt{80}$$$$=$$$$\sqrt{2*5*2*2*2}$$ ---> $$\sqrt{4*4*5}$$ ---> $$4$$$$\sqrt{5}$$
$$\sqrt{125}$$ $$=$$ $$\sqrt{25 *5}$$ -----> $$5$$ $$\sqrt{5}$$
$$4$$$$\sqrt{5}$$ $$+ 5$$ $$\sqrt{5}$$ $$=$$ $$9$$$$\sqrt{5}$$
_________________
Intern
Joined: 30 Aug 2011
Posts: 17
Location: United States
Schools: ISB '15
GMAT 1: 680 Q46 V37
WE: Project Management (Computer Software)
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 13453.318723910088,
"openwebmath_score": 0.7343714833259583,
"tags": null,
"url": "https://gmatclub.com/forum/root-80-root-139868.html"
} |
### Show Tags
04 Oct 2012, 00:50
2
$$\sqrt{80} = \sqrt{16*5} = 4\sqrt{5}$$ ........(1)
$$\sqrt{125} = \sqrt{25*5} = 5\sqrt{5}$$ .......(2)
Thus $$4\sqrt{5} + 5\sqrt{5} = 9\sqrt{5}$$
Manager
Joined: 12 Jan 2013
Posts: 188
### Show Tags
18 Dec 2013, 14:04
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$
(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100
Practice Questions
Question: 52
Page: 159
Difficulty: 600
The only thing I know is that the answer is just under 24.. And it is definitely not D since D falls between 14 and 15.. And it's clearly not E. And out of the 3 left, B and C are waay too high above 24, and I figured sqrrt of 5 is around "2.something" so A made most sense.
Clearly this approach is very shaky but this time it worked.
Manager
Joined: 13 Feb 2011
Posts: 94
### Show Tags
20 Aug 2014, 23:32
$$\sqrt{80}$$ is approx. 9 and $$\sqrt{125}$$ is approx. 11 so the answer should be close to 20. Choice A is the closest.
Manager
Joined: 26 Feb 2015
Posts: 119
### Show Tags
05 Mar 2015, 07:51
I see this question is from a subcategory called "Practice Questions", where can I find that sub?
Math Expert
Joined: 02 Sep 2009
Posts: 47006
### Show Tags
05 Mar 2015, 07:58
erikvm wrote:
I see this question is from a subcategory called "Practice Questions", where can I find that sub?
Those are practice questions from OG13. You can find all questions HERE.
_________________
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2916
Location: United States (CA)
### Show Tags
04 May 2016, 10:03
1
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$
(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100
Solution: | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 13453.318723910088,
"openwebmath_score": 0.7343714833259583,
"tags": null,
"url": "https://gmatclub.com/forum/root-80-root-139868.html"
} |
(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100
Solution:
To solve the problem we must simplify the radicals. Radicals should be simplified whenever possible. Since the square root of a perfect square produces integers, it will often be helpful to locate and simplify perfect squares within a radical expression. Thus, we first locate the perfect squares that divide evenly into 80 and 125, making the simplification of each radical straightforward.
√80 = √16 x √5 = 4√5
√125 = √25 x √5 = 5√5
Now we can add these two results together. Remember to keep the value inside the radical constant and add together the values on the outside.
4√5 + 5√5 = 9√5
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Intern
Joined: 01 Jan 2015
Posts: 31
Location: India
GPA: 3.71
WE: Consulting (Retail Banking)
### Show Tags
04 May 2016, 10:06
A
80=4*4*5
125=5*5*5
Root(80)=4Root(5)
Similarly 125
hence 9Root(5)
_________________
_________________
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3622
Location: India
GPA: 3.5
### Show Tags
04 May 2016, 10:19
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$
(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100
Practice Questions
Question: 52
Page: 159
Difficulty: 600
$$\sqrt{80}+\sqrt{125} =$$ $$4\sqrt{5}+5\sqrt{5}$$ = $$9\sqrt{5}$$
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Manager
Status: On a 600-long battle
Joined: 22 Apr 2016
Posts: 138
Location: Hungary
Schools: Erasmus '19
GMAT 1: 410 Q18 V27
GMAT 2: 490 Q35 V23
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 13453.318723910088,
"openwebmath_score": 0.7343714833259583,
"tags": null,
"url": "https://gmatclub.com/forum/root-80-root-139868.html"
} |
### Show Tags
22 Mar 2017, 23:29
Prime factorization can really help you and so can converting the roots to fraction exponents.
$$\sqrt { 80 } +\sqrt { 125 } =\\ { ({ 2 }^{ 4 } }\times { 5 }^{ 1 })^{ \frac { 1 }{ 2 } }+{ ({ 5 }^{ 2 }\times { 5 }^{ 1 }) }^{ \frac { 1 }{ 2 } }=\\ ({ 2 }^{ \frac { 4 }{ 2 } }\times { 5 }^{ \frac { 1 }{ 2 } })+({ 5 }^{ \frac { 2 }{ 2 } }\times { 5 }^{ \frac { 1 }{ 2 } })\quad =\\ { 2 }^{ 2 }\sqrt { 5 } +5\sqrt { 5 } =\\ 4\sqrt { 5 } +5\sqrt { 5 } =\\ \Rightarrow 9\sqrt { 5 } \\$$
_________________
"When the going gets tough, the tough gets going!"
|Welcoming tips/suggestions/advices (you name it) to help me achieve a 600|
Manager
Joined: 07 Jun 2017
Posts: 177
Location: India
Concentration: Technology, General Management
GMAT 1: 660 Q46 V38
GPA: 3.6
WE: Information Technology (Computer Software)
### Show Tags
14 Sep 2017, 22:49
√80 ~~ 9
√ 125 ~~ 11
9+11 = 20
√5 ~~2.1
9*2.1~~20
all other answer not near to this
_________________
Regards,
Naveen
email: nkmungila@gmail.com
Please press kudos if you like this post
Re: Root(80)+root(125) [#permalink] 14 Sep 2017, 22:49
Display posts from previous: Sort by
# Events & Promotions
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8705972684083609,
"lm_q2_score": 0.8705972684083609,
"openwebmath_perplexity": 13453.318723910088,
"openwebmath_score": 0.7343714833259583,
"tags": null,
"url": "https://gmatclub.com/forum/root-80-root-139868.html"
} |
# Is there an idempotent element in a finite semigroup?
Let $(G,.)$ be a finite semigroup. Is there any $a\in G$ such that: $$a^2=a$$
It seems to be true in view of theorem 2.2.1 page 97 of this book (I'm not sure). But is there an elementary proof?
Theorem 2.2.1. [R. Ellis] Let $S$ be a compact right topological semigroup. Then there exists an idempotent in it.
-
"Pages 77 to 99 are not shown in this preview". Do you have another link? – 1015 Apr 6 '13 at 15:42
Note first that it suffices to prove that $a^k = a$ for some $k \geq 2$. If $k = 2$ we are done. Otherwise $k > 2$ and multiplying both sides by $a^{k-2}$ gives $(a^{k-1})^2 = a^{k-1}$.
Fix $x \in G$ and consider the sequence
$$x, x^2, x^4, x^8, x^{16}, \ldots$$
Since $G$ is finite, there is repetition in this sequence. That is, $x^{2^t} = x^{2^s}$ for some integers $t > s \geq 1$. Thus $x^{2^t} = (x^{2^s})^{2^{t-s}} = x^{2^s}$, so choosing $a = x^{2^{s}}$ and $k = 2^{t-s}$ gives $a^k = a$. Note that $k \geq 2$ since $t > s$.
-
Pick an arbitrary element and start iterating $x\mapsto x^2$. Since the semigroup is finite you will eventually hit a cycle. This gives you a $b$ such that $b^k=b$ for some $k\ge 2$. Now set $a=b^{k-1}$.
-
When I first saw the question, I remembered there was a proof on MO using Ramsey theory, but couldn't remember how the argument went, so I came up with the following, that I first posted as a comment:
A cute proof using Schur's theorem: Fix $a$ in your semigroup $S$, and color $n$ and $m$ with the same color whenever $a^n=a^m$. By Schur's theorem (and the fact that the semigroup is finite) there are $n\le m$ such that $n$, $m$, and $n+m$ have the same color. That is, $a^n=a^m=a^{n+m}=(a^n)^2$.
(Today I finally found the thread on MO with the Ramsey theory proof, using Ramsey's theorem directly rather than Schur's theorem.)
-
This is a very good point proved by E.H.Moore that says:
Some power of every element of a finite semigroup is idempotent. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717472321277,
"lm_q1q2_score": 0.870578406589985,
"lm_q2_score": 0.8824278710924296,
"openwebmath_perplexity": 389.4375717154165,
"openwebmath_score": 0.9014335870742798,
"tags": null,
"url": "http://math.stackexchange.com/questions/353028/is-there-an-idempotent-element-in-a-finite-semigroup"
} |
Some power of every element of a finite semigroup is idempotent.
Trans. Amer. Math. Soc. 3 1902
-
Good day, @Babak! How is your's going? ;-) – amWhy Apr 7 '13 at 12:13
Ahhh, maybe you need a nice, sweet nap? ;-) – amWhy Apr 7 '13 at 12:19
Oh yes Amy. Teaching Maths makes everyone tired. I am so glad I could see you at this time. ;-) – B. S. Apr 7 '13 at 12:21
8^) ${}{}{}{}{}{}{}{}$ – amWhy Apr 7 '13 at 12:23
You can find a proof $^1$ of the following theorem, from which your assertion follows, at Proof Wiki
Theorem
Let $(S,\circ)$ be a finite semigroup.
For every element in $(S,\circ),$ there is a power of that element which is idempotent. That is: $$\forall x \in S:\exists i \in \mathbb N:x^i=x^i\circ x^i$$
Essentially, then, for your purposes: you can simply set $a = x^i$,
and you have the existence of an idempotent element $a \in S$ such that $\;a^2 = a$.
$1.$ Source of proof: Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717472321277,
"lm_q1q2_score": 0.870578406589985,
"lm_q2_score": 0.8824278710924296,
"openwebmath_perplexity": 389.4375717154165,
"openwebmath_score": 0.9014335870742798,
"tags": null,
"url": "http://math.stackexchange.com/questions/353028/is-there-an-idempotent-element-in-a-finite-semigroup"
} |
$1.$ Source of proof: Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978).
-
N.B. Proof Wiki is a great resource when looking for a proof, or an alternative proof, to mathematical proofs. (Link is to "main page"). – amWhy Apr 6 '13 at 16:20
I do not agree with "stronger, and more general". Because it's a simple consequence of the question. – user59671 Apr 7 '13 at 17:53
@CutieKrait: Actually, no, this theorem talks about every element of a finite semigroup, not just the existence of some element such that... Your question and its answer follow from the theorem above. Besides, the point of my post was not to argue the merits of the theorem or your question, the point of my post was to simply offer help. – amWhy Apr 7 '13 at 18:06
assume we know every finite element in a semigroup $S$ has an idempotent. Then for each $a$ in $S$ , $\{a^n \mid n\in \Bbb N \}$ is a semigroup. So it has an idempotent, say $a^k$. $a^ka^k=a^k$ for some $k$. – user59671 Apr 7 '13 at 18:19
You asked about whether there exists an element in each semigroup...such that it is idempotent. We don't know that for each x $x \circ x = x$, we only know (from the theorem) that for each $x$, exists a power $i$ of $x$ such that $x^i\circ x^i = x^i$...that's what the theorem above states. From that we know that exists $y = x^i$ such that $y\circ y = y$, we don't know that for all $a$ in the semigroup, $a\circ a = a$. The assumption that "every finite element in a semigroup has [a power that is] an idempotent" is NOT to say ever finite element in a semigroup is idempotent. – amWhy Apr 7 '13 at 18:24 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717472321277,
"lm_q1q2_score": 0.870578406589985,
"lm_q2_score": 0.8824278710924296,
"openwebmath_perplexity": 389.4375717154165,
"openwebmath_score": 0.9014335870742798,
"tags": null,
"url": "http://math.stackexchange.com/questions/353028/is-there-an-idempotent-element-in-a-finite-semigroup"
} |
# The natural logarithmic function essay
The natural logarithmic function essay, The history of logarithms is the story of a correspondence around 1730, leonhard euler defined the exponential function and the natural logarithm by = →.
Home math, popular an intuitive guide to exponential functions & e the mathematical constant e is the base of the natural logarithm. 32 logarithmic functions and their graphs the definition above implies that the natural logarithmic function and the natural exponential function are inverse. Introduction to exponents and logarithms introduction to exponential and logarithmic functions the natural logarithm [latex]ln(x. 32 the natural logarithm function brian e veitch this means the value of lnx, x the the negative of the area shown above this should make sense from one of. Introduction to logarithms mathematicians use log (instead of ln) to mean the natural logarithm this can lead to confusion: example engineer thinks.
The common and natural logarithms (page since the log function is the inverse of the exponential function, the graph of the log is the flip of the graph of the. A summary of the number e and the natural log in 's inverse, exponential, and logarithmic functions learn exactly what happened in this chapter, scene, or section of. In fact, another approach to introduce the exponential and logarithmic functions in calculus is to present the natural logarithm first, defined exactly as we did l (x. Yes, exponential and logarithmic functions isn’t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong. | {
"domain": "allstarorchestra.info",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.986571742099477,
"lm_q1q2_score": 0.8705784035860106,
"lm_q2_score": 0.882427872638409,
"openwebmath_perplexity": 709.9886028109532,
"openwebmath_score": 0.8453141450881958,
"tags": null,
"url": "http://bstermpaperwcen.allstarorchestra.info/the-natural-logarithmic-function-essay.html"
} |
Natural logarithms used with base e logarithmic function: definition & examples related study materials essay prompts. College essay financial the number e and the natural logarithm function of an exponential function natural logarithms are special types of. A-1 natural exponential function in lesson 21, we explored the world of logarithms in base 10 the natural logarithm has a base of “e” “e” is approximately. 51 the natural logarithmic function and differentiation and the definition of the natural logarithmic function: to this point later in this essay.
The natural logarithmic function-integration name _____ (no trigonometry) homework find the indefinite integral of the following functions. Open document below is an essay on ma1310: week 1 exponential and logarithmic functions from anti essays, your source for research papers, essays, and term paper. The graph of an exponential function logarithmic and exponential logarithmic and exponential functions here is the graph of the natural logarithm. The integration of rational functions, resulting in logarithmic or arctangent functions where represents the natural (base e) logarithm of x and.
We have all seen the natural logarithm appear in why does the natural logarithm appear so much in its inverse function, the natural logarithm will crop up a. Read this essay on ma1310: week 1 exponential and logarithmic functions come browse our large digital warehouse of free sample essays get the knowledge you need in. | {
"domain": "allstarorchestra.info",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.986571742099477,
"lm_q1q2_score": 0.8705784035860106,
"lm_q2_score": 0.882427872638409,
"openwebmath_perplexity": 709.9886028109532,
"openwebmath_score": 0.8453141450881958,
"tags": null,
"url": "http://bstermpaperwcen.allstarorchestra.info/the-natural-logarithmic-function-essay.html"
} |
From the early 1600's there have been two kinds of logarithms—natural logarithms and common logarithms common logarithms are base 10, denoted [math]\log_{10. Natural logarithm functiongraph of natural logarithmalgebraic properties of ln(x) i applying the natural logarithm function to both sides of the equation ex 4. Evaluating natural logarithm with and so the reason why you wouldn't see log base e written this way is log base e is referred to as the natural logarithm. The natural logarithm of a number is its logarithm to the base of the plots of the natural logarithm function on the complex plane (principal branch.
The natural logarithmic function essay
Rated 4/5 based on 39 review
2017. bstermpaperwcen.allstarorchestra.info | {
"domain": "allstarorchestra.info",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.986571742099477,
"lm_q1q2_score": 0.8705784035860106,
"lm_q2_score": 0.882427872638409,
"openwebmath_perplexity": 709.9886028109532,
"openwebmath_score": 0.8453141450881958,
"tags": null,
"url": "http://bstermpaperwcen.allstarorchestra.info/the-natural-logarithmic-function-essay.html"
} |
# Can the inverse of a function be the same as the original function?
I was wondering if the inverse of a function can be the same function.
For example when I try to invert
$g(x) = 2 - x$
The inverse seems to be the same function. Am I doing something wrong here?
• No, you're not doing anything wrong. You can certainly have functions such that $f(f(x))=x$. Such functions are called involutions. – mjqxxxx Oct 27 '13 at 20:54
• Consider $f(x) = x$. – Thomas Oct 28 '13 at 1:35
• There is also a simple graphical interpretation: any function that when graphed is symmetrical over $y=x$ (that is, a diagonal line at 45° angle splits the graph in two mirror images of each other) will have this property. It's easy to see that you can draw boatloads of functions like that. – Euro Micelli Oct 28 '13 at 1:40
• In general, given an invertible function $g$, $f(x)=g^{-1}(-g(x))$ is an involution. – Workaholic Sep 23 '16 at 17:23
You're correct. A function that's its own inverse is called an involution.
Edit: Oh let's have some fun. :) What are some other functions that are easy to check are involutions? I've cherry picked some of my favorites in what follows, both from memory and also the references I provide below.
• First, note that there's an easy test to determine whether or not $f$ is an involution. Namely, since $f^{-1} = f$, you just need to double check that $f(f(x)) = x$ for all $x$ in the domain of $f$. This can be used to verify all three of the following examples are actually involutions.
• Your function $g(x)$ generalizes to a whole class of involutions! Namely, $$f(x) = a-x$$ is an involution for any real number $a$. In particular, $f(x) = 0 - x = -x$ is an involution (as is $f(x) = x$, of course). | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717484165853,
"lm_q1q2_score": 0.870578400009085,
"lm_q2_score": 0.8824278633625322,
"openwebmath_perplexity": 252.32056974090685,
"openwebmath_score": 0.8139545321464539,
"tags": null,
"url": "https://math.stackexchange.com/questions/541978/can-the-inverse-of-a-function-be-the-same-as-the-original-function?noredirect=1"
} |
• As someone already pointed out, $f(x) = 1/x$ (defined for all real $x \neq 0$) is also an involution. More generally, for any real $a$ and $b$ the function $$f(x) = a + \frac{b}{x-a} = \frac{ax + (b-a^2)}{x-a}$$ satisfies $$f(f(x)) = a + \frac{b}{a + \frac{b}{x-a} - a} = a + (x-a) = x$$ for all real $x \neq a$, and as such is also an involution on this domain.
• Here's a less obvious (but cool) example. Consider the function $f(x) = (a - x^3)^{1/3}$. You can check this directly that $f(f(x)) = (a - ((a-x^3)^{1/3})^3)^{1/3} = x$. This is an example of a large class of involutions generated by a special type of symmetric function $F(x,y)$ (as explained here).
• Fun Fact: The only continuous, odd ($f(-x) = -f(x)$ for all $x$) involutions with domain $(-\infty,\infty)$ are $f(x) = \pm x$. (A short proof of this fact is given here.)
• There are many, many, more of these functions, and they occur naturally/are useful tools in many branches of mathematics.
• For "fun fact", that is the only non-trivial involution, but the trivial involution is also odd and continuous. – Dietrich Epp Oct 28 '13 at 3:25
• @DietrichEpp: Good catch! I've edited the post. Thanks! – Dan Oct 28 '13 at 5:26
That's perfectly fine, and your answer is correct. For another function that is its own inverse, see: $$f(x) = \frac{1}{x} = f^{-1}(x)$$
$g(x): y=2-x$
$g^{-1}(x): x=2-y\implies x-2=-y\implies y=2-x$
So you're correct. It is possible that a function can be an inverse of itself.
Coming from theory of coding (LDPC codes decoding), you have another involution: $x\mapsto-\log(\tanh(x/2))$.
Yes, you are correct, a function can be it's own inverse. However, I noticed no one gave a graphical explanation for this.
The inverse for a function of $x$ is just the same function flipped over the diagonal line $x=y$ (where $y=f(x)$). So, if you graph a function, and it looks like it mirrors itself across the $x=y$ line, that function is an inverse of itself. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717484165853,
"lm_q1q2_score": 0.870578400009085,
"lm_q2_score": 0.8824278633625322,
"openwebmath_perplexity": 252.32056974090685,
"openwebmath_score": 0.8139545321464539,
"tags": null,
"url": "https://math.stackexchange.com/questions/541978/can-the-inverse-of-a-function-be-the-same-as-the-original-function?noredirect=1"
} |
This graphical representation can be used to understand a lot of really cool ideas about inverse functions.
For example, any straight line that's perpendicular to the line $x=y$ will of course be the exact same on both sides of the line $x=y$. Any function that is a straight line perpendicular to $x=y$ can be written in the form $f(x)=a−x$, which as Dan mentioned is an entire class of functions that your function just happens to be one of. Note that the $a$ just determines how high up on the $y$ axis the function starts, but it will always mirror itself across $x=y$.
You can also see some interesting facts about functions that are their own inverse that might be fun to take some time to prove. For example, since it needs to mirror itself over the line $x=y$, any function that is its own inverse that has any points where $f(x) = x$ must have a slope of either 1 or -1 at that point (or have an undefined slope).
Edit: I must have skipped over Euro Micelli's comment, which explained the graphical aspect of this literally years before I did.
There are many examples for such types of function's Y=1/x X^2+Y^2=1,2,3,4,5,6,7.....(any other positive number) Simply the fact behind this is that the graph of the function should be symmetric about line Y=X While calculating inverse what we actually calculate is image of that function with respect to line Y=X
• $x^2 + y^2 = 1$ is not a function. – mlc Aug 2 '17 at 6:01
• Welcome to the math.stachexchange. Try using mathjax for formulas to make your answers/questions more readable. Extensive how-to found on site. – Thanassis Aug 2 '17 at 6:04 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717484165853,
"lm_q1q2_score": 0.870578400009085,
"lm_q2_score": 0.8824278633625322,
"openwebmath_perplexity": 252.32056974090685,
"openwebmath_score": 0.8139545321464539,
"tags": null,
"url": "https://math.stackexchange.com/questions/541978/can-the-inverse-of-a-function-be-the-same-as-the-original-function?noredirect=1"
} |
# DFT: why are frequencies mirrored around the frequency axis?
I want to apply Mathematica's DFT procedure, Fourier[], to a Gaussian. Why a Gaussian? Because I know the analytical result.
So I define
a = 0.5;
h[t_] := Exp[-(t^2/a^2)];
The analytic FourierTransform of h using FourierParameters -> {0, -2 Pi} is:
HAnalytic[v] := Sqrt[Pi]*a*Exp[-a^2*Pi^2*v^2];
Then I create time values:
dt = 1/100;
tList = Table[t, {t, 0, 20000 dt, dt}];
and I create sample data:
inputData = h[tList]
I then create the frequency values:
dv = 1/((Length[tList] - 1)*dt);
vList = dv*Table[i, {i, 0, Length[tList] - 1}];
and carry out the DFT calculation:
HDFT = Re[Fourier[inputData]];
Now I plot it:
Show[
Plot[HAnalytic[v], {v, 0, 1000*dv}, PlotRange -> {{0, 1.5}, {-1, 1}},
PlotStyle -> Green],
]
So far no problem. Only the scaling is different.
BUT NOW MY QUESTION!
Why do I get
i.e. the Fourier-transformed values are mirrored around the frequency axis, when using
tListSym = Table[t, {t, -10000 dt, 10000 dt, dt}]
I mean h[tListSym] gives really kind of the whole function to the DFT, as h is symmetric around $t = 0$, whereas h[tList] reproduces only the right hand side of h.
If I look at the values of HDFT, I can see that the values are alternatively positive / negative.
Can someone explain to me what is going on? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.963779943094681,
"lm_q1q2_score": 0.8705768460347125,
"lm_q2_score": 0.9032942138630786,
"openwebmath_perplexity": 1364.7799658351746,
"openwebmath_score": 0.638904869556427,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/86126/dft-why-are-frequencies-mirrored-around-the-frequency-axis"
} |
Can someone explain to me what is going on?
• Does this help ? dsp.stackexchange.com/questions/431/… – image_doctor Jun 17 '15 at 11:07
• Not really... as I am not wondering about negative frequenies. I am wondering why the amplitude of the frequencies is alternating for t =ListSym = Table[t, {t, -10000 dt, 10000 dt, dt}]. – newandlost Jun 17 '15 at 12:24
• @Hugh has recently written up an excellent primer on DFT in Mathematica. Maybe something in there might be of help to you. In particular, I think you might want to plot the Abs[] of your DFT values, rather than the Re[]. – MarcoB Jun 17 '15 at 14:45
• Does not matter, as the fouriertransform of a real even function is a real even function. – newandlost Jun 17 '15 at 15:15
The answer is that Fourier does not know your time values and does not see an even function. Thus if we generate your new input data and plot we get
inputData2 = h[tListSym];
ListLinePlot[inputData2, PlotRange -> All]
The problem is thus how to tell Fourier that the first half of your input is for negative x values. We could change the phase of the values created by Fourier or we can put your input into the usual Fourier form of positive values then negative values - a shifted version. Thus
nn = Length[tListSym]
inputData3 =
Join[inputData2[[(nn - 1)/2 + 1 ;; -1]],
inputData2[[1 ;; (nn - 1)/2]]];
ListLinePlot[inputData3, PlotRange -> All]
Thus we are starting with the positive values and then putting the negative values after. Now when we do the Fourier transform we get what you want.
HDFT2 = Fourier[inputData2];
HDFT3 = Fourier[inputData3];
Here I have not taken the real parts as you did. We can compare the two versions
ListPlot[{Re[HDFT2][[1 ;; 500]], Re[HDFT3][[1 ;; 500]]},
PlotRange -> All]
ListPlot[{Im[HDFT2][[1 ;; 500]], Im[HDFT3][[1 ;; 500]]},
PlotRange -> All]
The imaginary part of the shifted version is zero because we have constructed an even function. We now get what you want. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.963779943094681,
"lm_q1q2_score": 0.8705768460347125,
"lm_q2_score": 0.9032942138630786,
"openwebmath_perplexity": 1364.7799658351746,
"openwebmath_score": 0.638904869556427,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/86126/dft-why-are-frequencies-mirrored-around-the-frequency-axis"
} |
Show[
Plot[HAnalytic[v], {v, 0, 1000*dv},
PlotRange -> {{0, 1.5}, {-1, 1}}, PlotStyle -> Green],
ListPlot[{Thread[{vList, Re[HDFT2]}][[1 ;; 500]]}, PlotStyle -> Blue],
ListPlot[{Thread[{vList, HDFT3}][[1 ;; 500]]}, PlotStyle -> Brown]
]
On a secondary point I think you are off by one point in defining your frequency values. The increment should be 1/(Length[] dt).
I suppose a deeper question is why does Fourier do positive values then negative values? There is an assumption of a periodic function so it does not matter. However, it is often a nuisance as you have found.
With regard to getting the scaling correct you need to work out your version of Parseval's Theorem using FourierParameters. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.963779943094681,
"lm_q1q2_score": 0.8705768460347125,
"lm_q2_score": 0.9032942138630786,
"openwebmath_perplexity": 1364.7799658351746,
"openwebmath_score": 0.638904869556427,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/86126/dft-why-are-frequencies-mirrored-around-the-frequency-axis"
} |
• I was hoping you might be induced to weigh in on this one :-) Let me also add a link to a post in the DSP stackexchange site dealing explicitly with the math behind the even/odd function problem, just for reference. (+1) – MarcoB Jun 17 '15 at 18:09
• @Hugh Thanks you :) "I suppose a deeper question is why does Fourier do positive values then negative values?" Yes ... But I think no I understand it. You have to think about it like this: It is a N periodic function, hence 0 = N, AND the first value given to the DFT defines 0. This is the same as sying the function lives on a ring. Therefore for getting a even function we have fist hand it the positive and then the negative values. I will make a small drawing: drive.google.com/file/d/0B1ecQW5sX1LJLWhfUUt6ckktY1U/edit – newandlost Jun 18 '15 at 11:15
• @newandlost Unfortunately I can't download or see your video -I will try when I get to another computer. However, I think I can see what is happening and I agree with your comment. There is also as solution that would involve adjusting the phase of the Fourier transform after the calculation i.e. making a transformation in time. Thus by writing Exp[- I 2 Pi f (t-t0)] in the Fourier integral choosing a t0 and then taking this phase outside the integral it would be possible to correct the spectrum. Glad I could help. – Hugh Jun 18 '15 at 14:07 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.963779943094681,
"lm_q1q2_score": 0.8705768460347125,
"lm_q2_score": 0.9032942138630786,
"openwebmath_perplexity": 1364.7799658351746,
"openwebmath_score": 0.638904869556427,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/86126/dft-why-are-frequencies-mirrored-around-the-frequency-axis"
} |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 20 Nov 2018, 19:59
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!
November 22, 2018
November 22, 2018
10:00 PM PST
11:00 PM PST
Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Free lesson on number properties
November 23, 2018
November 23, 2018
10:00 PM PST
11:00 PM PST
Practice the one most important Quant section - Integer properties, and rapidly improve your skills.
### Show Tags
Updated on: 22 Jan 2014, 06:15
3
00:00
Difficulty:
45% (medium)
Question Stats:
72% (01:55) correct 28% (01:51) wrong based on 128 sessions
### HideShow timer Statistics
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->
He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?
A)13,200
B)14,800
C)16,400
D)15,800
E)19,600 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799378929587,
"lm_q1q2_score": 0.8705768381997117,
"lm_q2_score": 0.9032942106088968,
"openwebmath_perplexity": 13958.469966398803,
"openwebmath_score": 0.45608147978782654,
"tags": null,
"url": "https://gmatclub.com/forum/paul-sells-encyclopedias-door-to-door-he-earns-160-on-ever-166375.html"
} |
A)13,200
B)14,800
C)16,400
D)15,800
E)19,600
Originally posted by guerrero25 on 20 Jan 2014, 21:16.
Last edited by guerrero25 on 22 Jan 2014, 06:15, edited 1 time in total.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8574
Location: Pune, India
Re: Paul sells encyclopedias door-to-door. [#permalink]
### Show Tags
20 Jan 2014, 22:00
1
1
guerrero25 wrote:
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->
He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?
A)13,200
B)14,800
C)16,400
D)15,800
E)19,600
I will post the OA soon
His pay check was $1320. Out of this,$160 was his fixed salary so the total commission he earned was $1320 -$160 = $1160 He earns 10% on the sales of first$10,000 which gives a commission of $1000. He earns 5% on every subsequent dollar. Since he earns another$160, he must have had sales of another 160*(100/5) = 3200
So his total sales must have been $10,000 +$3200 = $13,200 _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 04 Oct 2013 Posts: 152 Location: India GMAT Date: 05-23-2015 GPA: 3.45 Re: Paul sells encyclopedias door-to-door. He earns$160 on ever [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799378929587,
"lm_q1q2_score": 0.8705768381997117,
"lm_q2_score": 0.9032942106088968,
"openwebmath_perplexity": 13958.469966398803,
"openwebmath_score": 0.45608147978782654,
"tags": null,
"url": "https://gmatclub.com/forum/paul-sells-encyclopedias-door-to-door-he-earns-160-on-ever-166375.html"
} |
### Show Tags
22 Jan 2014, 00:11
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->
He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?
Using the equation y = mx + C, where, y = Paycheck amount; x= amount of sales exceeding $10000; m = 5% or 1/20; C = 160 + 10000(1/10) = 1160; $$1320 = \frac{x}{20} + 1160$$ Or,$$x = (1320-1160)(20) = 3200$$ So, Total sales $$= 10000+3200=13200$$ Answer: (A) Senior SC Moderator Joined: 22 May 2016 Posts: 2117 Paul sells encyclopedias door-to-door. He earns$160 on ever [#permalink]
### Show Tags
26 Aug 2018, 18:31
1
guerrero25 wrote:
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->
He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?
A)13,200
B)14,800
C)16,400
D)15,800
E)19,600
Paycheck = fixed pay + commissions
$$1,320 = 160$$ + commissions
$$(1,320 -160)=1,160$$ in commissions | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799378929587,
"lm_q1q2_score": 0.8705768381997117,
"lm_q2_score": 0.9032942106088968,
"openwebmath_perplexity": 13958.469966398803,
"openwebmath_score": 0.45608147978782654,
"tags": null,
"url": "https://gmatclub.com/forum/paul-sells-encyclopedias-door-to-door-he-earns-160-on-ever-166375.html"
} |
All of the answer choices are greater than $10,000 $$x=$$ first$10,000 in sales
$$y$$ = sales amount after the first $10,000 $$x+y=$$ total sales amount $$.10x + .05y=$$ total commissions amount $$.10x+.05y=1,160$$ $$.10(10,000)+.05(y)=1,160$$ $$1,000+0.05y=1,160$$ $$.05y=160$$ $$y=\frac{160}{.05}=\frac{16,000}{5}=3,200$$ Total sales = $$(x+y)=(10,000+3,200)=13,200$$ Answer A Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4170 Location: United States (CA) Re: Paul sells encyclopedias door-to-door. He earns$160 on ever [#permalink]
### Show Tags
12 Nov 2018, 07:22
guerrero25 wrote:
Paul sells encyclopedias door-to-door. He earns $160 on every paycheck, regardless of how many sets he sells. In addition, he earns commission as follows: Commission Sales 10%$0.00 - $10,000.00 5%$10,000.01 --->
He does not earn "double commission." That is, if his sales are $12,000, he earns 10% on the first$10,000 and 5% on the remaining $2,000. His largest paycheck of the year was$1,320. What were his sales for that pay period?
A)13,200
B)14,800
C)16,400
D)15,800
E)19,600
Since he earns $160 as base salary and 10% of$10,000 is $1,000, if he sells no more than$10,000 worth of encyclopedias, the maximum he earns is $1,160. However, since he earns$1,320, he must have sold more than $10,000 worth of encyclopedias. Therefore, if we let x = his sales for the paycheck of$1,320, we can create the equation:
160 + 0.1(10,000) + 0.05(x - 10,000) = 1,320
160 + 1,000 + 0.05x - 500 = 1,320
660 + 0.05x = 1,320
0.05x = 660
x = 13,200
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799378929587,
"lm_q1q2_score": 0.8705768381997117,
"lm_q2_score": 0.9032942106088968,
"openwebmath_perplexity": 13958.469966398803,
"openwebmath_score": 0.45608147978782654,
"tags": null,
"url": "https://gmatclub.com/forum/paul-sells-encyclopedias-door-to-door-he-earns-160-on-ever-166375.html"
} |
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12891
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Paul sells encyclopedias door-to-door. He earns $160 on ever [#permalink] ### Show Tags 13 Nov 2018, 11:15 Hi All, We're told that Paul sells encyclopedias door-to-door. He earns$160 on every paycheck, regardless of how many sets he sells and a commission as follows:
10% on the first $10,000.00 and 5% for any additional sales beyond that. His largest paycheck of the year was$1,320. We're asked for his sales for that pay period. This question can be approached in a couple of different ways. Based on the given answer choices, we can take advantage of a couple of built-in patterns and avoid some of the extra 'math steps.'
To start, we can see that all of the answers are ABOVE $10,000, so we know that part of his commissions was 10% of$10,000 = $1,000. Paul received$160 on top of that piece of his commission, so we have accounted for $1,160. Now, we just have to account for the extra$1320 - $1160 =$160 he earned...
That extra commission was based on 5% of the sales ABOVE $10,000. 5% of$1,000 = $50$160 is a little more than (3)($50), so the extra sales had to be a little more than (3)($1000). Thus, his total sales for that period were a LITTLE MORE than $10,000 +$3000 = $13,000... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9637799378929587,
"lm_q1q2_score": 0.8705768381997117,
"lm_q2_score": 0.9032942106088968,
"openwebmath_perplexity": 13958.469966398803,
"openwebmath_score": 0.45608147978782654,
"tags": null,
"url": "https://gmatclub.com/forum/paul-sells-encyclopedias-door-to-door-he-earns-160-on-ever-166375.html"
} |
prove that opposite angles of a parallelogram are equal | {
"domain": "eventfullevents.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9697854146791214,
"lm_q1q2_score": 0.8705718084111486,
"lm_q2_score": 0.8976952996340946,
"openwebmath_perplexity": 507.28567752899374,
"openwebmath_score": 0.6664319038391113,
"tags": null,
"url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal"
} |
Strategy: how to prove that opposite sides of a parallelogram are equal. Always […] if angle AOB=118 degree, find i) angle ABO, ii) angle ADO, iii) angle … The reflexive property refers to a number that is always equal to itself. Join now. Opposite Angles of a Parallelogram. Most questions answered within 4 hours. Prove that. The congruence of opposite sides and opposite angles is a direct consequence of the Euclidean parallel postulate and neither condition can be proven without appealing to the Euclidean parallel postulate or one of its equivalent formulations. (iv) In quadrilateral ACFD, AD || CF and AD = CF | From (iii) ∴ quadrilateral ACFD is a parallelogram. A good way to begin a proof is to think through a game plan that summarizes your basic argument or chain of logic. Answer Save. Each pair of co-interior angles are supplementary, because two right angles add to a straight angle, so the opposite sides of a rectangle are parallel. 12 Answers. Relevance? There is a theorem in a parallelogram where opposite angles are equal. ... angle ABD≅angle BDC --> Alternate Interior Angles Theorem. One of the properties of parallelograms is that the opposite angles are congruent, as we will now show. To Prove that the Opposite Angles of a Parallelogram are equal in measure 00:06:23 undefined Related concepts Properties of a Parallelogram - Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram. Property: The Opposite Angles of a Parallelogram Are of Equal Measure. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram. In a parallelogram , 1. prove that in a parallelogram opposite angles are equal. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. how_to_reg Follow . For example, z = z or 1000 = 1000 are examples of the reflexive property. Class-IX . Properties of Parallelogram: A parallelogram is a special type of | {
"domain": "eventfullevents.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9697854146791214,
"lm_q1q2_score": 0.8705718084111486,
"lm_q2_score": 0.8976952996340946,
"openwebmath_perplexity": 507.28567752899374,
"openwebmath_score": 0.6664319038391113,
"tags": null,
"url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal"
} |
the reflexive property. Class-IX . Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! Answer: 3 question Given: SV || TU and SVX = UTX Prove: VUTS is a parallelogram. 1. Log in. In the adjoining figure abcd is a rectangle and diagonals intersect at O point. | ∵ A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of equal length (v) ∵ ACFD is a parallelogram | … State the given then use CPCTC to say all their parts match and say what specifically makes it a parallelogram. 1 decade ago. Favourite answer. Prove that opposite sides of a parallelogram are congruent.? if one pair of opposite sides are parallel and one pair of opposite angles are congruent? This video is highly rated by Class 8 students and has been viewed 322 times. Since each pair of alternate interior angles are congruent, the sum of them must be as well. Why are the opposite angles of a parallelogram equal? Its diagonals bisect each other.. or i need both pairs for it to be a parallelogram? For the other opposite angles, we can prove that the angles are equal by drawing another diagonal line and proving that the triangles are congruent. Parallel Lines Transversals Angle. - Get the answer to this question and access a vast question bank that is tailored for students. Prove that A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel. In any case, in a parallelogram, the opposite angles are always equal. Yes if diagonals of a parallelogram are equal then it is a rectangle. Given, opposite angles of a quadrilateral are equal. Opposite angles of a parallelogram are equal (or congruent) Consecutive angles are supplementary angles to each other (that means they add up to 180 degrees) Read more: Area of Parallelogram. There can be more than one community in a | {
"domain": "eventfullevents.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9697854146791214,
"lm_q1q2_score": 0.8705718084111486,
"lm_q2_score": 0.8976952996340946,
"openwebmath_perplexity": 507.28567752899374,
"openwebmath_score": 0.6664319038391113,
"tags": null,
"url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal"
} |
add up to 180 degrees) Read more: Area of Parallelogram. There can be more than one community in a society. Click hereto get an answer to your question ️ Prove that in a parallelogram the opposite angles are equal . Area = base $$\times$$ height = b $$\times$$ h. It includes every relationship which established among the people. A parallelogram has two parallel pairs of opposite sides. The diagonals of a parallelogram are not of equal length. 2. asked Jan 19, 2019 in Mathematics by Bhavyak ( 67.3k points) quadrilaterals So as you can see parallelogram has opposite sides parallel. - the answers to estudyassistant.com So adjacent angles must sum up to 180°(because if the angle is more or less the lines would converge on either side) Is it possible to have a regular polygon with measure of each interior angle of 45 degree 3. Angles "a" and "b" add up to 180°, so they are supplementary angles. Update: simple geometry proof required. Lv 6. Construction: Draw a parallelogram ABCD. Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram. don't use any co-ordinate fundamentals. Prove theorem: If a quadrilateral is a parallelogram, then opposite angles are congruent. Things that you need to keep in mind when you prove that opposite sides of a parallelogram are congruent. Say m alternate interior angles are equal prove: If each pair of opposite sides of a.... Its opposite sides parallel and one pair of opposite sides of a are... The given then use CPCTC to say all their parts match and say specifically... Need at least one side, in addition to the angles, to show congruency,! Each pair of opposite angles are equal aware of about the proof above following examples of the properties parallelogram. Proof above this polygon good way to begin a proof is to through! In Mathematics by Golu ( 105k points ) in a parallelogram, the sum of them be! Hereto Get an answer to a quick problem a game plan outlining how thinking. Their parts match and | {
"domain": "eventfullevents.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9697854146791214,
"lm_q1q2_score": 0.8705718084111486,
"lm_q2_score": 0.8976952996340946,
"openwebmath_perplexity": 507.28567752899374,
"openwebmath_score": 0.6664319038391113,
"tags": null,
"url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal"
} |
Hereto Get an answer to a quick problem a game plan outlining how thinking. Their parts match and say what specifically makes it a parallelogram where opposite angles is equal, then is. As we will now show think through a game plan that summarizes your basic argument or chain of logic CF... 8.6K ) edit answer of equal measure supplementary angles are congruent. is think... Of equal measure all their parts match and say what specifically makes it a parallelogram opposite angles are?... Community in a parallelogram, so: its opposite sides of a quadrilateral is a.. Show congruency are supplementary angles b '' add up to 180°, so are! 2-D geometry visibility Views ( 8.6K ) edit answer Notice the congruent triangles to the angles to! Formal proofs is defined as a quadrilateral, each pair of opposite angles of a parallelogram opposite... To a quick problem, the opposite angles are congruent, the opposite sides are equal Class students... Good way to begin a proof is to think through a game plan how... The four sides of a parallelogram are of equal length and the opposite angles of a quadrilateral the! Is equal, the opposite angles are equal makes it a parallelogram equal parallelogram bisect other. It possible to have a regular polygon exit exterior angle by 100 degree 322 times pairs for it to a. Of the following examples of parallelogram proofs show game plans followed by the resulting formal.. B '' add up to 180°, so: its opposite sides of a parallelogram are of equal...., then it is a rhombus theorem: If each pair of opposite sides of a parallelogram are and! The proof in line 6 when you prove that in a parallelogram opposite parallel! A quadrilateral, each pair of opposite sides of a quadrilateral, each pair of opposite sides prove that opposite angles of a parallelogram are equal pairs it. Pairs of opposite sides are equal opposite angles are equal and its diagonal bisects the 2... Been viewed 322 times to the angles, to show congruency angles to... I need both | {
"domain": "eventfullevents.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9697854146791214,
"lm_q1q2_score": 0.8705718084111486,
"lm_q2_score": 0.8976952996340946,
"openwebmath_perplexity": 507.28567752899374,
"openwebmath_score": 0.6664319038391113,
"tags": null,
"url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal"
} |
bisects the 2... Been viewed 322 times to the angles, to show congruency angles to... I need both pairs for it to be a parallelogram opposite angles are equal the following examples of reflexive. Need both pairs for it to be a parallelogram figure ABCD is a parallelogram, sum. Area = base \ ( \times\ ) h. it includes every relationship which established among people. '' and b '' add up to 180°, so: its opposite sides parallel each other O... Has opposite sides are equal and parallel opposite angles are equal and parallel think a... We will now show community in a parallelogram the opposite angles are equal opposite angles are equal property refers a. Z or 1000 = 1000 are examples of the following reasons would complete the in. Why are the opposite angles are equal quadrilateral are equal answers you need keep... In line 6 equal and its diagonal bisects the parallelogram 2 its diagonals bisect each other the properties of is. 1000 are examples of the properties of parallelogram proofs show game plans followed by the resulting proofs! Click hereto Get an answer to a number that is always equal for example, z z... Ad = CF need to keep in mind when you prove that opposite sides of a parallelogram are,. From ( 1 ) and ( 2 ), we obtain AD || CF AD... Why are the opposite angles are congruent. is always equal to.... One community in a quadrilateral is a parallelogram where opposite angles of a quadrilateral each! Parallelogram are of equal measure the interior angle of 45 degree 3 we... Side, in addition to the angles, to show congruency interior angle of parallelogram... As well your basic argument or chain of logic should be aware of about the proof above proves. So they are supplementary angles is the number of sides of a parallelogram not... A question for free Get a free answer to a quick problem ( 2,! To prove that opposite sides of a quadrilateral is equal, then it is a parallelogram equal... This question and access a vast question bank that is always equal to | {
"domain": "eventfullevents.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9697854146791214,
"lm_q1q2_score": 0.8705718084111486,
"lm_q2_score": 0.8976952996340946,
"openwebmath_perplexity": 507.28567752899374,
"openwebmath_score": 0.6664319038391113,
"tags": null,
"url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal"
} |
it is a parallelogram equal... This question and access a vast question bank that is always equal to itself not of equal.... … prove that in a quadrilateral is equal, then opposite angles in a parallelogram If one pair opposite!, as we will now show of 45 degree 3 ( \times\ ) h. it includes every which. And say what specifically makes it a parallelogram opposite sides are equal and.! All their parts match and say what specifically makes it a parallelogram If one of! Pairs for it to be a parallelogram are equal is it possible to have regular... Get the answer to your question ️ prove that in a quadrilateral is parallelogram... Think through a game plan that summarizes your basic argument or chain of logic, in a parallelogram show... Get the answer to your question ️ prove that in a parallelogram and we know,! Intersect at O point to have a regular polygon exit exterior angle by 100.! Pairs for it to be a parallelogram and we know that, a. Parallelogram has opposite sides parallel and one pair of opposite sides are also equal diagonal bisects the 2! ( 105k points ) in a society sides are equal parallel and equal, then is... Diagonals of a parallelogram need to keep in mind when you prove that in a parallelogram equal! Use CPCTC to say all their parts match and say what specifically makes it a parallelogram, the opposite of! Get a free answer to a quick problem or 1000 = 1000 examples... Has two parallel pairs of opposite angles of a parallelogram, so they supplementary... Angles is equal, then it is a flat shape with opposite sides a. Good way to begin a proof is to think through a game plan how! | {
"domain": "eventfullevents.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9697854146791214,
"lm_q1q2_score": 0.8705718084111486,
"lm_q2_score": 0.8976952996340946,
"openwebmath_perplexity": 507.28567752899374,
"openwebmath_score": 0.6664319038391113,
"tags": null,
"url": "http://eventfullevents.com/swamp-animals-ujouar/de8a57-prove-that-opposite-angles-of-a-parallelogram-are-equal"
} |
# Direct Proof instead of an Indirect Proof: If $B$ is the set of lower bounds of $A$, then $\sup B \in B$
The author proved the following exercise using a proof by contradiction, but I think it can be accomplished using a direct proof, so I was wondering if someone would verify that my argument is sound.
Exercise
Suppose $A$ is a nonempty set of reals that is bounded below. Let $B$ be the set of lower bounds for $A$, and assume further that $B$ is not empty and bounded above.
Prove that $\sup B \in B$.
Before proving the statement, I present the following lemma.
Lemma Let $x,y \in \mathbb{R}$ such that $x \leq y + \epsilon$ for every $\epsilon > 0$. Then $x \leq y$.
Proof
Since $A$ is not empty and bounded below, the completeness axiom implies that there exists a real number $\psi$ such that $\psi = \inf A$. Further, because $B$ is not empty and bounded above, the completeness axiom entails the existence of a real number $\phi$ such that $\phi = \sup B$.
Now choose any $\epsilon > 0$. Because $\phi$ is the least upper bound of B, it follows that $\phi - \epsilon$ is not an upper bound of B. Thus there exists an $x \in B$ such that
$$\phi - \epsilon < x$$
Since $x \in B$, it follows by definition of $B$ that $x$ is a lower bound of A. Due to the fact that $\psi$ is the greatest lower bound of A, it follows that
$$x \leq \psi$$
By transitivity, we deduce that $\phi - \epsilon < \psi$, which implies that
$$\phi < \psi + \epsilon$$
From the lemma we conclude that
$$\phi \leq \psi$$ From the last inequality, we conclude that $\phi$ is a lower bound of A, i.e. $\phi = \sup B \in B$, as desired. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127430370156,
"lm_q1q2_score": 0.8705029715565851,
"lm_q2_score": 0.8807970732843033,
"openwebmath_perplexity": 112.7183002591206,
"openwebmath_score": 0.9573087692260742,
"tags": null,
"url": "https://math.stackexchange.com/questions/1461604/direct-proof-instead-of-an-indirect-proof-if-b-is-the-set-of-lower-bounds-of"
} |
• Note: I understand that $\phi$ and $\psi$ are typically used to denote either angle degrees or as functions due to convention, but I like using them because of how they look haha. Oct 2 '15 at 21:18
• "Let B be the set of lower bounds for A, and assume further that B is not empty and bounded above." You can not make that assumption you must demonstrate that, which is relatively easy to do. (Anything less than inf A is a lower bound and anything in A is greater than all lower bounds). Your lemma is a direct consequence of axioms of inequalities and doesn't need stating. Otherwise your proof seems sound and valid. (Not sure why your professor choice a proof by contradiction as your proof is the standard and quite direct.) Oct 2 '15 at 21:41
• @fleablood The exercise that is in my post is the second part of a three part exercise, where part one is to prove that $B \neq \emptyset$ and $B$ is bounded above. So, I included the assumption without proof for convenience. Oct 2 '15 at 21:56
• Votaire. Never mind then :) Oct 2 '15 at 22:58
• Instead of saying $\phi - \epsilon$ just say "for any y < $\phi$". We know R isn't bounded below so we know y exists. This is simpler, more direct, easier to read and means you don't need to present the lemma. Oct 2 '15 at 23:05
The proof is correct except a possible minor modification. So, $x\in B$ which happens to be the set of all lower bounds of $A$ and $\psi$ is the greatest lower bound of $A$. This implies $x\leq \psi$ (contrary to strict inequality in your proof.) However, this won't change your proof because we have $\phi -\epsilon <x\leq \psi \implies \phi-\epsilon<\psi$.
• ah, good point. Thank you! Oct 2 '15 at 21:34
• Also, you can use a stronger lemma which states that x=y with the same hypothesis. This will also prove that sup (B) =inf (A)! Oct 2 '15 at 21:40 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127430370156,
"lm_q1q2_score": 0.8705029715565851,
"lm_q2_score": 0.8807970732843033,
"openwebmath_perplexity": 112.7183002591206,
"openwebmath_score": 0.9573087692260742,
"tags": null,
"url": "https://math.stackexchange.com/questions/1461604/direct-proof-instead-of-an-indirect-proof-if-b-is-the-set-of-lower-bounds-of"
} |
# Thread: discs in a bag
1. ## discs in a bag
a bag contains 20 discs marked with numbers from 1 to 20 inclusive. if two discs are picked at random from the bag whats the probability of getting one even numbered disc and one odd numbered disc.
there are 10 even and odd numbers so i was thinking the answer would be 10/20 *10/20 =1/4 the answer at the back of the book is 10/19
2. ## Re: discs in a bag
Hey markosheehan.
You should think about the order of picking things.
If you pick one even one then the next even one has a lower chance.
Try listing the different combinations of frequencies for picking even and odd in different orders and add them up.
3. ## Re: discs in a bag
sorry yes so the answer would be 10/20*10/19=5/19 this is not the right answer either
4. ## Re: discs in a bag
The first disc chosen is even or odd ... probability of choosing one or the other is 1. Choosing the first leaves 19 in the bag.
You're looking for the second disc to be different than the first ... odd if the first was even or even if the first was odd. In either case, there are 10 discs of the remaining 19 in the bag that are different than the first choice. The probability of pulling out odd/even disc different than the first is 10/19.
the probability that the fisrt two discs drawn are odd/even or even/odd is 1 times 10/19
5. ## Re: discs in a bag
There are initially 20 disks in the bag, 10 of them even. The probability the first disk drawn is even is 10/20= 1/2. There are then 19 disks in the bag, 10 of them odd. The probability the second disk drawn is odd is 10/19. The probability the first disk drawn is even and the second odd, in that order, is (1/2)(10/19)= 5/19. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127406273587,
"lm_q1q2_score": 0.8705029709804077,
"lm_q2_score": 0.8807970748488297,
"openwebmath_perplexity": 593.9153418589925,
"openwebmath_score": 0.8258749842643738,
"tags": null,
"url": "http://mathhelpforum.com/statistics/274308-discs-bag.html"
} |
Similarly, since there were also 10 odd disks in the bag initially, the probability the first disk drawn is odd is 1/2. Then there are 19 disks left in the bag, 10 of them even. The probability the second disk drawn is even is 10/19. The probability the first disk drawn is odd and the second even, in that order, is (1/2)(10/19)= 5/19.
The probability one disk is even and the other odd, in either order, is 5/19+ 5/19= 10/19.
(It is always the case, in problems like this, that the probabilities of the two orders are the same. We could have just calculated the first probability, 5/19, then multiplied by 2 for the two orders.)
6. ## Re: discs in a bag
another way of doing it is using combinations.
choose 2 from 20 = 20ncr2=190 total number
(choose 1 from 20 even)and(chose 1 from 10 odd)
10ncr1*10ncr1=100
100/190=10/19
i do not like this method as i am finding it hard to see why it works even though it gives the right answer | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127406273587,
"lm_q1q2_score": 0.8705029709804077,
"lm_q2_score": 0.8807970748488297,
"openwebmath_perplexity": 593.9153418589925,
"openwebmath_score": 0.8258749842643738,
"tags": null,
"url": "http://mathhelpforum.com/statistics/274308-discs-bag.html"
} |
# How to prove that gcd(m+1, n+1) divides (mn-1)
I'm learning about divisors and the gcd, but now I'm stuck at proving:
gcd(m+1, n+1) divides (mn-1) for all m,n in the set of Integers
Help is appreciated on how to prove this! Thanks
• $mn-1=(m+1)(n+1)-(m+1)-(n+1)$ – Wojowu Jan 8 '17 at 20:29
A common strategy to solve this type of problem is using the following simple fact:
If $d$ divides $a$ and $b$, then d divides $$a\cdot r+b\cdot s$$ for all $r,s \in \mathbb{Z}$.
Thus if $d=\gcd(m+1,n+1)$, then obviously $d$ divides $m+1$ and $n+1$, and so d divides $$(m+1)\cdot r+(n+1)\cdot s$$ for all $r,s \in \mathbb{Z}$. In particular, for $r=n$ and $s=-1$, we have that $d$ divides $$(m+1)\cdot n+(n+1)\cdot (-1)=mn-1.$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668701095653,
"lm_q1q2_score": 0.8704957558273562,
"lm_q2_score": 0.8872045952083047,
"openwebmath_perplexity": 504.38805445444433,
"openwebmath_score": 0.9780576825141907,
"tags": null,
"url": "https://math.stackexchange.com/questions/2089393/how-to-prove-that-gcdm1-n1-divides-mn-1"
} |
• Thanks for this clarification! One of my attempts stranded at (m+1)⋅r + (n+1)⋅s. Could you maybe explain why in particular r = n and s = -1? Or do you simply take these to get to mn-1 as we can state that (m+1)⋅r+(n+1)⋅s for all r and s in Z? – Superkuuk Jan 8 '17 at 21:10
• Exactly! The suitable choices you make for $r$ and $s$ is completely based on your target "$mn-1$". – MathChat Jan 8 '17 at 21:13
• @Superkuuk You hit the nail on the head.The problem with this approach is that it pulls the answer out of a hat, i.e. there is no motivation for the choice of $\,r,s.\,$ But if you use the remainder-based approach as in my answer it is much more more natural, i.e. it is just a special case of the Polynomial Remainder Theorem, i.e that $\,P(x) \equiv P(c)\pmod{x-c}.\$ Good luck trying as above to find those coefficients for more complex polynomials $\,P(m,n).\,$ OTOH, evaluating $\,P(m,n)\,$ at $\,n,m = -1\,$ is utterly trivial. – Bill Dubuque Jan 8 '17 at 21:36
• @BillDubuque - I disagree that it's "pulling the answer out of a hat". Mathematics is, first and foremost, a pursuit of pattern-matching. As neither $m^2$ nor $m^2$ appear in the target term, we can see that, for the simplest solution, $r$ does not contain $m$ and $s$ does not contain $n$. So we have $(m+1)(an+b)+(n+1)(cm+d)=mn-1$. Expanding the left and equating, we get $a+c=1$, $b+c=0$, $a+d=0$, and $b+d=-1$. It can easily be seen that one equation is redundant, and so we can set, for example, $b=0$, which gives $a=1$, $c=0$, $d=-1$, which is MathChat's solution. – Glen O Jan 9 '17 at 8:45
• @GlenO But no such derivation is given in the answer. Further, as I emphasized above, ad hoc methods like that do not generalize, but the remainder based methods do. In fact they lead to very very powerful algorithms such as the Grobner basis algorithm. Ad-hoc attempts at "pattern matching" cannot compete with algorithms. – Bill Dubuque Jan 9 '17 at 14:57 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668701095653,
"lm_q1q2_score": 0.8704957558273562,
"lm_q2_score": 0.8872045952083047,
"openwebmath_perplexity": 504.38805445444433,
"openwebmath_score": 0.9780576825141907,
"tags": null,
"url": "https://math.stackexchange.com/questions/2089393/how-to-prove-that-gcdm1-n1-divides-mn-1"
} |
Hint $\$ mod \,\gcd(m\!+\!1,n\!+\!1)\!:\ \ \begin{align} &m\!+\!1\equiv0\equiv n\!+\!1\\ \Rightarrow\ &\ \ \,m\equiv -1\equiv n\\ \Rightarrow\ &mn\equiv (-1)(-1)\equiv 1\end{align}
Remark $\$ More generally, as above, using the Polynomial Congruence Rule we deduce
$$P(m,n)\equiv P(a,b)\,\ \pmod{\gcd(m\!-\!a,m\!-\!b)}$$
for any polynomial $\,P(x,y)\,$ with integer coefficients, and for any integers $\,m,n,a,b.$
OP is special case $\, a,b = -1\$ and $\ P(m,n) = mn\$ (or $\ mn-1)$
Although it's less elegant than the other approaches, you can also prove this through direct substitution. First observe that \begin{align} \gcd(m+1, n+1)=d & \implies m+1=pd,n+1=qd \\ & \implies m=pd-1,n=qd-1 \end{align} for some $p,q\in\mathbb{Z}$. Then \begin{align} mn-1 & = (pd-1)(qd-1)-1 \\ & = (pqd^2-(p+q)d+1)-1 \\ & = (pqd-(p+q))d, \end{align} which is an integer multiple of $d$.
Use the fact that
$$(m+1)(n+1) = mn + m + n + 1 = (mn - 1) + (m+1) + (n+1)$$
THen it is straightforward | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668701095653,
"lm_q1q2_score": 0.8704957558273562,
"lm_q2_score": 0.8872045952083047,
"openwebmath_perplexity": 504.38805445444433,
"openwebmath_score": 0.9780576825141907,
"tags": null,
"url": "https://math.stackexchange.com/questions/2089393/how-to-prove-that-gcdm1-n1-divides-mn-1"
} |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 20 Nov 2018, 20:01
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!
November 22, 2018
November 22, 2018
10:00 PM PST
11:00 PM PST
Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Free lesson on number properties
November 23, 2018
November 23, 2018
10:00 PM PST
11:00 PM PST
Practice the one most important Quant section - Integer properties, and rapidly improve your skills.
# A lottery is played by selecting X distinct single digit numbers from
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
### Hide Tags
Manager
Joined: 11 Aug 2012
Posts: 116
Schools: HBS '16, Stanford '16
A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
05 Feb 2013, 07:40
2
1
23
00:00
Difficulty:
25% (medium)
Question Stats:
69% (01:00) correct 31% (01:05) wrong based on 923 sessions
### HideShow timer Statistics | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
69% (01:00) correct 31% (01:05) wrong based on 923 sessions
### HideShow timer Statistics
A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?
(1) Players must match at least two numbers with machine to win.
(2) X = 4
##### Most Helpful Expert Reply
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
30 Jun 2017, 12:07
6
2
TheMastermind wrote:
Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.
Thank you.
Dear TheMastermind,
I'm happy to respond.
Just so we are clear, here would be the PS version of the problem:
A lottery is played by selecting 4 distinct single digit numbers from 0 to 9 at once such that order does not matter. The lottery machine also picks four digits in the same way. Players must match at least two numbers with the machine to win. What is the probability that a player will win playing the lottery?
Think about it this way. I pick four of the digits at random. That's fixed. Now, the lottery machine is going to pick four digits at random. Here, I will use "at least" logic in the calculation. That is, for ease and convenient, we will calculate the probabilities of all the cases not included in the solution.
Case I: the machine's four choices don't overlap at all with mine.
There are four digits, my chosen digits, that the machine can't pick. The machine has to pick from the remaining six digits:
first choice P = 6/10 = 3/5
second choice P = 5/9
third choice P = 4/8 = 1/2
fourth choice = 3/7
$$P_1 = \dfrac{3}{5} \times \dfrac{5}{9} \times \dfrac{1}{2} \times \dfrac{3}{7}$$ | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
$$P_1 = \dfrac{3}{5} \times \dfrac{5}{9} \times \dfrac{1}{2} \times \dfrac{3}{7}$$
$$P_1 = \dfrac{1}{1} \times \dfrac{1}{1} \times \dfrac{1}{2} \times \dfrac{1}{7} = \dfrac{1}{14}$$
That's the Case I probability.
Case II: the machine's four choices overlap with exactly one digit of mine.
First, there's the one digit that matches one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc.
$$P_2 = \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$$
$$P_2 = \dfrac{1}{1} \times \dfrac{2}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}$$
$$+ \dfrac{2}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}$$
$$+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}$$
$$+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}$$
$$P_2 = \dfrac{4*2}{3*7} = \dfrac{8}{21}$$
That's the Case II probability.
Add those two:
$$P_{1,2} = \dfrac{1}{14} + \dfrac{8}{21} = \dfrac{3}{42} + \dfrac{16}{42} = \dfrac{19}{42}$$
That's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question.
$$P = 1 - \dfrac{19}{42} = \dfrac{23}{42}$$
That's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown.
Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939) | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
##### General Discussion
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
05 Feb 2013, 10:12
5
danzig wrote:
A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?
(1) Players must match at least two numbers to win.
(2) X = 4
I'm happy to help.
This is a somewhat offbeat question, but then again, that's just what the GMAT will throw at you.
So, from the prompt, we know we are picking X different single digit numbers: X must be greater than two and less than 9 (or 10, if we are counting zero as a "single digit number" ----- let's ignore that complication). Order doesn't matter. We know nothing about what constitutes winning.
Statement #1: Players must match at least two numbers to win.
Now, at least we know what constitutes winning. The trouble is --- we don't know the how many digits are picked. If X = 9 ---- the lottery picks all the digits from 1-9, then I also pick all the digits from 1-9 --- then I have 100% chance of matching at least two digits and winning. That wouldn't be much of a lottery. If X = 3 --- the lottery picks three, and then I pick three --- well, that's harder. Clearly the probability of winning depends on the value of X, and we don't know that in Statement #1. This statement, alone and by itself, is insufficient.
Statement #2: X = 4
Now, we know how many digits are picked ---- lottery picks 4, then I pick 4 --- but now I have no idea what constitutes "winning". (This is an example of a DS question in which it's crucially important to forget all about Statement #1 when we are analyzing Statement #2 on its own.) In Statement #2, we know how many digits are picked, but we have absolutely no idea what constitutes winning. This statement, alone and by itself, is insufficient. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
Combined statements:
Now, we know --- the lottery picks 4 digits, then I pick 4 digits, and if at least two of my digits match two of the lottery's cards, I win. This is now a well defined math problem, and if we wanted, we could calculate the numerical value of the probability. Of course, since this is DS, it would be a big mistake to waste time with that calculation. We have enough information now. Combined, the statements are sufficient.
Answer = C
Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Manager
Joined: 27 Jan 2013
Posts: 228
GMAT 1: 780 Q49 V51
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
05 Feb 2013, 11:49
1
Hi,
I think that Mike's explanation is great. I'm just wondering where this question came from? The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9
What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone?
I guess that if you assume that the question includes the extra decimals then Mike's solution still works although I'm curious about whether this is an official question.
HG.
_________________
"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land
GMAT vs GRE Comparison
If you found my post useful KUDOS are much appreciated.
IMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE: | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
IMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE:
Here is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html
Intern
Joined: 09 Oct 2012
Posts: 7
Location: United States
Concentration: Strategy, Finance
GMAT 1: 590 Q34 V38
GPA: 3.56
WE: Analyst (Consumer Products)
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
07 Feb 2013, 08:22
HerrGrau wrote:
Hi,
I think that Mike's explanation is great. I'm just wondering where this question came from? The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9
What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone?
I guess that if you assume that the question includes the extra decimals then Mike's solution still works although I'm curious about whether this is an official question.
HG.
It's tagged as a grockit problem, so I would guess that this is not an official question. It is also tagged as a 600-700 lvl question, which seems high to me for this question...I don't have any experience with Grockit, but MGMAT questions that are in the 600-700 range tend to be more difficult for me to solve.
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
### Show Tags
07 Feb 2013, 10:09
HerrGrau wrote:
Hi,
The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9 What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone?
Dear HerrGrau
Thank you for your kind words.
In my experience, I have never heard "single digit numbers" apply to anything other than non-negative integers. I have never seen any book or hear anyone ever refer to, say, -5 or 0.008 or 3 x 10^8 as a "single digit number. Yes, technically, each of these is written with a single non-zero digit, but I have never heard the term "single digit number" used for them. (We could say that each of these has "one significant figure", but that carries us far afield into measurement theory, well beyond GMAT territory.)
In my mind, the only ambiguity in this question is whether zero is included --- I guess I was assuming, whatever their understanding of the term, this understanding would be fixed and wouldn't change as we moved through the statements of the DS ---- therefore, with both statements, and with whatever convention they are following, the question can be definitely answered.
I agree with you ---- this question is not written with the tight precision so characteristic of official GMAT questions. The very fact that there's any ambiguity at all makes this a woefully substandard question.
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Manager
Joined: 27 Jan 2013
Posts: 228
GMAT 1: 780 Q49 V51
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
07 Feb 2013, 14:47
Hi Mike, | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
### Show Tags
07 Feb 2013, 14:47
Hi Mike,
In general I agree with everything that you're saying. Just to clarify, I think that .008 is a three digit number with 2 unique digits and so would be excluded. The numbers in question are .1, .2, .3, .4, .5, .6, .7, .8, and .9. They are numbers that have only one digit and hence are "single digit numbers". I have never seen this distinction be an actual issue on a GMAT question but am curious as to what the general consensus is.
HG.
_________________
"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land
GMAT vs GRE Comparison
If you found my post useful KUDOS are much appreciated.
IMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE:
Here is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html
Manager
Joined: 30 Mar 2013
Posts: 109
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
16 Oct 2014, 11:04
just for the sake of understanding, how would we solve this question?
1- probability of matching none - prob of matching 1?
none= (5/9)*4/9*3/9*2/9
probability of matching one: 1*5/9*4/9*3/9
is this the way?
Thank you!
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6531
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
05 Dec 2015, 11:47
2
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?
1) Players must match at least two numbers to win.
(2) X = 4
There are 2 variables (x and how many to be win) and 2 equations are given, so there is high chance (C) will be the answer.
Looking at the conditions together,
the probability to win is 1-(10C4/10^4), which is unique and makes the condition sufficient,
and the answer becomes (C).
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Manager
Joined: 02 Feb 2016
Posts: 89
GMAT 1: 690 Q43 V41
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
30 Jun 2017, 04:06
mikemcgarry wrote:
danzig wrote:
A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?
(1) Players must match at least two numbers to win.
(2) X = 4
I'm happy to help. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
I'm happy to help.
This is a somewhat offbeat question, but then again, that's just what the GMAT will throw at you.
So, from the prompt, we know we are picking X different single digit numbers: X must be greater than two and less than 9 (or 10, if we are counting zero as a "single digit number" ----- let's ignore that complication). Order doesn't matter. We know nothing about what constitutes winning.
Statement #1: Players must match at least two numbers to win.
Now, at least we know what constitutes winning. The trouble is --- we don't know the how many digits are picked. If X = 9 ---- the lottery picks all the digits from 1-9, then I also pick all the digits from 1-9 --- then I have 100% chance of matching at least two digits and winning. That wouldn't be much of a lottery. If X = 3 --- the lottery picks three, and then I pick three --- well, that's harder. Clearly the probability of winning depends on the value of X, and we don't know that in Statement #1. This statement, alone and by itself, is insufficient.
Statement #2: X = 4
Now, we know how many digits are picked ---- lottery picks 4, then I pick 4 --- but now I have no idea what constitutes "winning". (This is an example of a DS question in which it's crucially important to forget all about Statement #1 when we are analyzing Statement #2 on its own.) In Statement #2, we know how many digits are picked, but we have absolutely no idea what constitutes winning. This statement, alone and by itself, is insufficient.
Combined statements:
Now, we know --- the lottery picks 4 digits, then I pick 4 digits, and if at least two of my digits match two of the lottery's cards, I win. This is now a well defined math problem, and if we wanted, we could calculate the numerical value of the probability. Of course, since this is DS, it would be a big mistake to waste time with that calculation. We have enough information now. Combined, the statements are sufficient.
Answer = C
Does all this make sense?
Mike | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
Answer = C
Does all this make sense?
Mike
Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.
Thank you.
Intern
Joined: 12 Sep 2018
Posts: 1
Re: A lottery is played by selecting X distinct single digit numbers from [#permalink]
### Show Tags
14 Nov 2018, 17:47
mikemcgarry wrote:
TheMastermind wrote:
Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.
Thank you.
Dear TheMastermind,
I'm happy to respond.
Just so we are clear, here would be the PS version of the problem:
A lottery is played by selecting 4 distinct single digit numbers from 0 to 9 at once such that order does not matter. The lottery machine also picks four digits in the same way. Players must match at least two numbers with the machine to win. What is the probability that a player will win playing the lottery?
Think about it this way. I pick four of the digits at random. That's fixed. Now, the lottery machine is going to pick four digits at random. Here, I will use "at least" logic in the calculation. That is, for ease and convenient, we will calculate the probabilities of all the cases not included in the solution.
Case I: the machine's four choices don't overlap at all with mine.
There are four digits, my chosen digits, that the machine can't pick. The machine has to pick from the remaining six digits:
first choice P = 6/10 = 3/5
second choice P = 5/9
third choice P = 4/8 = 1/2
fourth choice = 3/7
$$P_1 = \dfrac{3}{5} \times \dfrac{5}{9} \times \dfrac{1}{2} \times \dfrac{3}{7}$$
$$P_1 = \dfrac{1}{1} \times \dfrac{1}{1} \times \dfrac{1}{2} \times \dfrac{1}{7} = \dfrac{1}{14}$$
That's the Case I probability. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
That's the Case I probability.
Case II: the machine's four choices overlap with exactly one digit of mine.
First, there's the one digit that matches one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc.
$$P_2 = \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$$
$$P_2 = \dfrac{1}{1} \times \dfrac{2}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}$$
$$+ \dfrac{2}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}$$
$$+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}$$
$$+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}$$
$$P_2 = \dfrac{4*2}{3*7} = \dfrac{8}{21}$$
That's the Case II probability.
Add those two:
$$P_{1,2} = \dfrac{1}{14} + \dfrac{8}{21} = \dfrac{3}{42} + \dfrac{16}{42} = \dfrac{19}{42}$$
That's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question.
$$P = 1 - \dfrac{19}{42} = \dfrac{23}{42}$$
That's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown.
Does all this make sense?
Mike
Hi Mike,
I know this is DS, so we do not need to solve, but I was able to come up with the same solution, using the following method - is there anything done incorrectly?
1. Total possible machine selections: 10C4 = 210
2. Total possible human selections where no numbers match (choose any 4 from the remaining 6 numbers the machine did not select): 6C4 = 15 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
3. Total possible human selections where 1 number matches (choose one of the 4 numbers selected by machine x 3 of the unselected numbers): 4 x 6C3 = 4 x 20 = 80
Probability that human does not win: 1- (combined losing selections/total possible machine selections) = 1 - ((15+80)/210) = 1 - 19/42 = 23/42
Re: A lottery is played by selecting X distinct single digit numbers from &nbs [#permalink] 14 Nov 2018, 17:47
Display posts from previous: Sort by
# A lottery is played by selecting X distinct single digit numbers from
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9811668679067631,
"lm_q1q2_score": 0.8704957494836778,
"lm_q2_score": 0.8872045907347108,
"openwebmath_perplexity": 1063.8596555375786,
"openwebmath_score": 0.585705041885376,
"tags": null,
"url": "https://gmatclub.com/forum/a-lottery-is-played-by-selecting-x-distinct-single-digit-numbers-from-146802.html"
} |
# Math Help - Question from Cambridge STEP Paper.
1. ## Question from Cambridge STEP Paper.
Hi everyone, I was looking through STEP Papers because I'm considering doing Mathematics at University. On the University of Cambridge website, they have past STEP papers, I tried a question and I was wondering if anyone could confirm the method for doing this. The question was:
How many integers between 10 000 and 100 000 (inclusive) contain exactly 2 different digits? (23 332 contains exactly 2 different digits but neither of 33 333 or 12 331 does.)
My calculation gave me an answer of 1 217.
Can someone try this question and let me know what they got and how they got it?
I'll explain my method too after there has been confirmation of whether my answer is right or wrong.
2. Originally Posted by AAKhan07
Hi everyone, I was looking through STEP Papers because I'm considering doing Mathematics at University. On the University of Cambridge website, they have past STEP papers, I tried a question and I was wondering if anyone could confirm the method for doing this. The question was:
How many integers between 10 000 and 100 000 (inclusive) contain exactly 2 different digits? (23 332 contains exactly 2 different digits but neither of 33 333 or 12 331 does.)
My calculation gave me an answer of 1 217.
Can someone try this question and let me know what they got and how they got it?
I'll explain my method too after there has been confirmation of whether my answer is right or wrong.
Here's my naive counting method for doing this. Start by counting only the five-digit numbers, but then we must remember to add 1 at the end to allow for the number 100000. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9728307700397331,
"lm_q1q2_score": 0.8704787246496275,
"lm_q2_score": 0.8947894654011352,
"openwebmath_perplexity": 556.3250619187047,
"openwebmath_score": 0.6888198852539062,
"tags": null,
"url": "http://mathhelpforum.com/discrete-math/35974-question-cambridge-step-paper.html"
} |
If both digits are nonzero then one of them, say x, occurs once or twice, and the other one, y, occurs 3 or 4 times. If x occurs once, there are 5 positions where it can occur. If it occurs twice, there are $\textstyle{5\choose2}=10$ possible positions. In each case, there are 9 ways to choose x and 8 ways to choose y. This gives a total of (5+10)×9×8 = 1080 possible numbers.
To those we must add the numbers containing one or more zeros. But a zero cannot occur as the first digit. The zeros can occur in any of the four other positions, and there can be one, two, three or four of them. There are 4×9 numbers with one zero, 6×9 with two zeros, 4×9 with three zeros, and 9 with four zeros. Total: 15×9 = 135.
My final answer is thus 1080 + 135 + 1 = 1216 (the "1" is for the number 100000 that we had to remember to account for).
So, where did number 1217 come from?
The papers are brilliant they cover come really good topics, Cambrdige do set high offers based on the papers though, I have my test in two months
Let me know if you want more resources.
Bobak
4. I agree with the 1216. Here is my approach.
$\binom{9}{2}\sum\limits_{k = 1}^4 \binom{5}{k}+ 9\sum\limits_{k = 1}^4 \binom{4}{k} +1$
5. Originally Posted by bobak
The papers are brilliant they cover come really good topics, Cambrdige do set high offers based on the papers though, I have my test in two months
You are probably going to do well.
6. ## My mistake, it was 1,216
Yes, sorry, you were all right, I meant to say 1,216. I remember my initial calculation gave me 1,215 so I added 1 for 100,000 but I think I added one again because I forgot I had already done it.
My method was 9(4C1+4C2+4C3+4C4)+36(5C1+5C2+5C3+5C4), same as Plato's I think although I could not express it as elegantly as Plato. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9728307700397331,
"lm_q1q2_score": 0.8704787246496275,
"lm_q2_score": 0.8947894654011352,
"openwebmath_perplexity": 556.3250619187047,
"openwebmath_score": 0.6888198852539062,
"tags": null,
"url": "http://mathhelpforum.com/discrete-math/35974-question-cambridge-step-paper.html"
} |
Thanks to bobak for the link, I'll be happy to hear about any more Cambridge STEP resources you have, and good luck for your test. Are you hoping to get into Cambridge? Which college? Are you doing AS Mathematics now?
7. ## my crap approach :)
This is my approach to the problem:
(I found the total possible values of "x" and "y", excluding 0 for now, by finding the sum of one to 8 which is 36)
Then calculating the number of posible possitions of "x" and "y" by doing 2^5 -2 = 30 (the -2 is because the digits can not all be the same)
So the number of combinations of the digits excluding 0 is therefore 36 * 30 = 1080
The total posible values of "x" and "y", when "x" = 0 is 9, and as the first digit can't be 0, you half the number of posible possitions making it 15, 15 * 9 = 135
so 1080 + 135 + 1 = 1216 .... | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9728307700397331,
"lm_q1q2_score": 0.8704787246496275,
"lm_q2_score": 0.8947894654011352,
"openwebmath_perplexity": 556.3250619187047,
"openwebmath_score": 0.6888198852539062,
"tags": null,
"url": "http://mathhelpforum.com/discrete-math/35974-question-cambridge-step-paper.html"
} |
# Math Help - general solution to trig problem
1. ## general solution to trig problem
Hi,
I'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated!
Find the general solution to
$\sin 2x + \sin 4x = \cos 2x + \cos 4x$
Regards,
Stonehambey
2. Originally Posted by Stonehambey
Hi,
I'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated!
Find the general solution to
$\sin 2x + \sin 4x = \cos 2x + \cos 4x$
Regards,
Stonehambey
Use the sum to product identities
List of trigonometric identities - Wikipedia, the free encyclopedia
$2\sin\left(\frac{2x+4x}{2} \right)\cos\left(\frac{2x-4x}{2} \right)=2\cos\left(\frac{2x+4x}{2} \right)\cos\left(\frac{2x-4x}{2} \right)$
$2\sin(3x)\cos(-2x)=2\cos(3x)\cos(-2x)$
$2\sin(3x)\cos(2x)=2\cos(3x)\cos(2x)$
$2\sin(3x)\cos(2x)-2\cos(3x)\cos(2x)=0$
$2\sin(3x)\cos(2x)-2\cos(3x)\cos(2x)=0$
$2\cos(2x)[\sin(3x)-\cos(3x)]=0$
You should be able to finish from here
3. sum to product! Now it's obvious :P
Thanks!
4. I'm assuming you are looking for all solutions on interval [0,2pi)
I include thr graph just to show we can expect 8 solutions
I've included the solutions and method in the attachment .
But the basic method i used was to rearrange as
sin(2x)-cos(2x) = cos(4x)-sin(4x)
Then I squared both sides and got 2 fairly simple equations
however there are 16 possibilities of which 8 are solutions--this is the tedious part.
If someone has an easier way I'd like to know it.
should have waited | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9728307668889048,
"lm_q1q2_score": 0.8704787177361966,
"lm_q2_score": 0.8947894611926921,
"openwebmath_perplexity": 380.09546312716003,
"openwebmath_score": 0.9131166338920593,
"tags": null,
"url": "http://mathhelpforum.com/trigonometry/89376-general-solution-trig-problem.html"
} |
If someone has an easier way I'd like to know it.
should have waited
5. hmm, I tried the way theEmptySet suggested and got a little stuck again
I applied the formulas and got to the same bit that you did. When I solved for
$2\cos(2x)=0$
I get $x=\frac{\pi}{4} + n\pi$, which doesn't work when I sub back into the original equation, so where did I go wrong?
6. Originally Posted by Stonehambey
hmm, I tried the way theEmptySet suggested and got a little stuck again
I applied the formulas and got to the same bit that you did. When I solved for
$2\cos(2x)=0$
I get $x=\frac{\pi}{2} + n\pi$, which doesn't work when I sub back into the original equation, so where did I go wrong?
You are really close you should have gotten
$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$
7. Originally Posted by TheEmptySet
You are really close you should have gotten
$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$
eek, sorry that was a typo on my part What I meant to type was
$\cos(2x) = 0$
$2x = \pm \frac{\pi}{2} + 2\pi n$
$\implies x = \pm \frac{\pi}{4} + n \pi$
If we let n = 0 and sub back into the original equation we have
$\sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)$
$1 = -1$
which is obviously wrong but I can't see where the error is
EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P
8. Originally Posted by Stonehambey
eek, sorry that was a typo on my part What I meant to type was
$\cos(2x) = 0$
$2x = \pm \frac{\pi}{2} + 2\pi n$
$\implies x = \pm \frac{\pi}{4} + n \pi$
If we let n = 0 and sub back into the original equation we have
$\sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)$
$1 = -1$
which is obviously wrong but I can't see where the error is | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9728307668889048,
"lm_q1q2_score": 0.8704787177361966,
"lm_q2_score": 0.8947894611926921,
"openwebmath_perplexity": 380.09546312716003,
"openwebmath_score": 0.9131166338920593,
"tags": null,
"url": "http://mathhelpforum.com/trigonometry/89376-general-solution-trig-problem.html"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.