source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34697 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Find the average of first 20 natural numbers. | [
"20.5",
"18",
"19.5",
"10.5"
] | D | Exp. Sum of first n natural numbers = n( n+1)/2
So, sum of first 20 natural numbers = 20*21/2 = 210
Required average =210/20 =10.5
Answer:D |
AQUA-RAT | AQUA-RAT-34698 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Akash leaves Chennai at 6am & reaches Bangalore at 10am. Prakash leaves Hyderabad at 8am and reaches Chennai at 11:30am. At what time do they cross each other? | [
"6 : 32am",
"7 : 56am",
"8 : 56am",
"3 : 56am"
] | C | Time taken by Akash = 4 h
Time taken by Prakash = 3.5 h
For your convenience take the product of times taken by both as a distance.
Then the distance = 14km
Since, Akash covers half of the distance in 2 hours(i.e at 8 am)
Now, the rest half (i.e 7 km) will be coverd by both prakash and akash
Time taken by them = 7/7.5 = 56 min
Thus , they will cross each other at 8 : 56am.
C |
AQUA-RAT | AQUA-RAT-34699 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
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The following is multiple choice question (with options) to answer.
A certain auto manufacturer sold 5% fewer vehicles in 2007 than in 2006. If the manufacturer sold 2.2 million vehicles in 2006, how many vehicles, to the nearest 10,000, did the manufacturer sell in 2007? | [
"17,00,000",
"18,00,000",
"19,00,000",
"21,00,000"
] | D | Let's assume in 2006 the manufacturer sold X.
In 2007 he sold X-5/100(X)
If sales in 2006,X= 2.2 million find X-5/100(X)
(2.2 x 10^6)-5/100(2.2 x 10^6)
2200000-1,10,000=20,90,000-----> 21,00,000(approx)
Answer-D |
AQUA-RAT | AQUA-RAT-34700 | 3. Jun 16, 2009
### HallsofIvy
Writing "M" for male and "F" for female, there are 25= 32 ways to write 5 letters, each being either an "M" or an "F". Of those 32, exactly one is all "M"s and exactly one is all "F"s.
Knowing that there is at least one female drops "MMMMM" leaving 31 possibilities. "FFFFF" is one of those: probability 1/31.
If we write the dogs in order of age, then knowing that the oldest dog is female means that we are looking at lists of 5 letters, from "M" or "F", with the first letter being "F". Obviously in all possible lists of 5 "M"s or "F"s, exactly half, 16 start with "F" and half with "M" so knowing that "the oldest dog is a female" throws out half the possible lists, leaving 16. one of those is "FFFFF" so the probability of "all females" is now 1/16.
There is no "dilemma" here. "The oldest dog is female" gives you more information than "at least one of the dogs is female".
4. Jun 16, 2009
### fleem
First breeder:
Since its given that there is at least one female, the question, "what is the probability that all five are female when one is female" means the same thing as the question, "What is the probability that the remaining four are female". The probability that the remaining four are female is 1/16, and thus the probability that all five are female (given that one is female) is 1/16.
Second breeder:
The temptation here is to assume there might have been intelligent pre-selection of dogs in the group, by the store owner. However the OP makes it pretty clear, I think, that there was no such pre-selection, and that these are ALL the dogs from a given litter. So the sex of a randomly selected dog does NOT change the probability of the sex of another dog in the litter. Each is a separate coin toss. So the fact that the oldest is female works the same as "there is at least one female". So like with the first breeder case, the probability of the remaining four all being female is 1/16.
5. Jun 16, 2009
The following is multiple choice question (with options) to answer.
A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability T that both selected dogs are NOT littermates? | [
"1/6",
"2/9",
"5/6",
"7/9"
] | C | We have three pairs of dogs for the 6 with exactly one littermate, and one triplet, with each having exactly two littermates.
So, in fact there are two types of dogs: those with one littermate - say A, and the others with two littermates - B.
Work with probabilities:
Choosing two dogs, we can have either one dog of type B or none (we cannot have two dogs both of type B).
The probability of choosing one dog of type B and one of type A is 3/9 * 6/8 * 2 = 1/2 (the factor of 2 for the two possibilities BA and AB).
The probability of choosing two dogs of type A which are not littermates is 6/9 * 4/8 = 1/3 (choose one A, then another A which isn't the previous one's littermate).
The required probability is 1/2 + 1/3 = 5/6.
Find the probability for the complementary event: choose AA or BB.
Probability of choosing two dogs of type A who are littermates is 6/9 * 1/8 = 1/12.
Probability of choosing two dogs of type B (who necessarily are littermates) is 3/9 * 2/8 = 1/12.
Again, we obtain 1 - (1/12 + 1/12) T= 5/6.
Answer: C |
AQUA-RAT | AQUA-RAT-34701 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
The tax on a commodity is diminished by 20% and its consumption increased by 15%. The effect on revenue is? | [
"2%",
"8%",
"5%",
"6%"
] | B | 100 * 100 = 10000
80 * 115 = 9200
-----------
10000-----------800
100-----------? => 8% decrease
Answer: B |
AQUA-RAT | AQUA-RAT-34702 | 10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive?
(A) 420
(B) 840
(C) 1260
(D) 2520
(E) 5040
Q9)
20% of total applied at X = 15
100% of total applied at X = 75
Only applied at X = 60
25% of total applied at Y = 15
100% of total applied at Y = 60
Only applied at Y = 45
Only applied at X + Only applied at Y = 60 + 45 = 105
ANS = D
Q10)
1 x 2 x 3 x 4 x 5 x 6 x 7 = 2^4 x 3^2 x 5 x 7
Option A = 420 = 2^2 x 3 x 5 x 7 (factored this out)
Option B = 2 x 420 = 2^3 x 3 x 5 x 7 (just add a 2 to A)
Option C = 3 x 420 = 2^2 x 3^2 x 5 x 7 (just add a 3 to A)
Option D = 2 x 1260 = 2^3 x 3^2 x 5 x 7 (just add a 2 to C)
Option E = 2 x 2520 = 2^4 x 3^2 x 5 x 7 (just add a 2 to D)
Actually after doing up to A you can quickly just figure out how many more 2's and 3's and determine it's E.
for q10 we need to find the lowest number - so should be 420
its divisible by all the integers from 1-7 inclusive
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Re: Good set of PS 2 [#permalink] 17 Oct 2009, 03:38
1
KUDOS
Bunuel wrote:
Please find below new set of PS problems:
The following is multiple choice question (with options) to answer.
What is the lowest positive integer that is divisible by each of the odd integers between 11 and 17, inclusive? | [
"3*5*13*17*11",
"5×17×19×23",
"7×15×17×19",
"7×15×19×21"
] | A | We should find the LCM of 11,13,15 = 3*5, 17,-> LCM = 3*5*13*17*11 .
Answer: A. |
AQUA-RAT | AQUA-RAT-34703 | classical-mechanics, scaling, dimensional-analysis
$Q \propto g^{1/2} (a-d)^{5/2}$
Now, this becomes invalid when $H$ and $D$ are of the same order, but we can get an idea of the time to drain $T$ by saying that as the flow rate does not depend of $H$, $Q T = V = 4 \pi (D/2)^2 H = \pi D^2 H$ so, dropping the $\pi$ we get
$T \propto g^{1/2} \, \frac{(a-d)^{5/2}}{D^2 \, H}$
and thus for everything but the size of sand grains $d$ constant
$\frac{T(d_1)}{T(d_2)} = \left( \frac{a - d_1}{a - d_2} \right)^{5/2}$
The following is multiple choice question (with options) to answer.
In digging a pond 20 m * 10 m * 5 m the volumes of the soil extracted will be? | [
"100 cu.m",
"1000 cu.m",
"10000 cu.m",
"2000 cu.m"
] | B | 20 * 10 * 5 = 1000
ANSWER:B |
AQUA-RAT | AQUA-RAT-34704 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
By selling 12 pencils for a rupee a man loses 20%. How many for a rupee should he sell in order to gain 20%? | [
"8",
"9",
"6",
"5"
] | A | 80% --- 12
120% --- ?
80/120 * 12 = 8
Answer:A |
AQUA-RAT | AQUA-RAT-34705 | Gold Coins in $Bag_0$: So the weight of the 3 coins on the scale are all 1 gram. So the scale will read 3 grams.
Gold Coins in $Bag_1$: So the weight of 1 of the coins is 1.01 grams and 2 of the coins are 2 grams. So the scale will read 3.01 grams.
Gold Coins in $Bag_2$: So the weight of 2 of the coins is 2.02 grams and 1 of the coins is 1 gram. So the scale will read 3.02 grams.
So each possibility has a unique scenario. So if we determine the weight, we can determine from which bag those coins came from based on that weight.
We can generalize our results from this simplified example to your 100 bag example.
Now for deriving the formula. Say hypothetically, of our 100 bags, all 100 coins in each of the 100 bags weigh 1 gram each. In that case, when we remove 0 coins from $Bag_0$, 1 from $Bag_1$, up until 99 coins from $Bag_{99}$, we'll have a total of 4950 coins on the scale, which will equivalently be 4950 grams. Simply put, if $n$ is our Bag number (denoted $Bag_n$), we've placed $n$ coins from each $Bag_n$ onto the scale for $n = 0, 1, 2, ... 99$.
So the weight of the coins will be $Weight = 1 + 2 + 3 + ... + 99 = 4950$
But we actually have one bag with gold coins weighing 1.01 grams. And we know that those 1.01 gram coins must be from some $Bag_n$. In our hypothetical example, all of our coins were 1 gram coins, so we must replace the $n$ coins weighed from $Bag_n$ with $n$ gold coins weighing 1.01 grams. Mathematically, we would have: $Weight = 4950 - n + 1.01n = 4950 + .01n = 4950 + n/100$
Rearranging the formula to solve for n, we have: $100(Weight-4950) = n$, where $Weight$ is $W$ and $n$ is $N$ in your example.
The following is multiple choice question (with options) to answer.
A bag of potatoes weighs 60 lbs divided by half of its weight. How much does the bag of potatoes weight? | [
"20 lb",
"11 lb",
"10 lb",
"15 lb"
] | B | Sol.
66÷6 = 11.
Answer : B |
AQUA-RAT | AQUA-RAT-34706 | ##### Well-Known Member
Subscriber
Actually, it looks like the discount rate is a risk-free rate not necessarily equal to a LIBOR of the correct period length.
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi Brian,
In many countries such as Thailand, the Treasury yield curve is illiquid across maturities. On the other hand, the LIBOR/Swap Yield curve is liquid across maturities worldwide. The LIBOR swap zero curve is obtained by bootstrapping and is almost risk-free. That is why it is used to discount the Cash flows to obtain the value of the FRA. The Fixed rate cannot be used for discounting because FRA's are OTC - the fixed rate is unique to that particular FRA.
Thanks!
Jayanthi
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Thanks Nicole - appreciate it
Jayanthi
#### brian.field
##### Well-Known Member
Subscriber
Thanks for your message Jayanthi. While I agree with what you are saying, generally, it isn't really addressing my philosophical question.
Consider today at time t=0 and an FRA that requires a fixed payment of 5% and a floating payment of 1YLIBOR from t=3 to t=4.
At t=0, we do not know that actual 1YLIBOR rate that will apply in 3 years but we do know the 1Y forward 1YLIBOR rate payable in 3 years. We establish the FRA at t=0, and at this time, the fixed rate is set so that the FRA's value is 0 at t=0. By the time we get to t=3, the FRA may not equal 0.
My point is that at t=3, we will know that value of the FRA at t=4 and it is customary to discount the t=4 value back to t=3 with a risk-free rate. I thought that I read that this discount rate was the appropriate tenor LIBOR.
My question is as follows:
The following is multiple choice question (with options) to answer.
The banker’s discount on Rs. 1600 at 10% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is : | [
"2 months",
"4 months",
"6 months",
"7 months"
] | C | Sol.
S.I. on Rs. 1600 = R.D. on Rs. 1680.
∴ Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is S.I. on Rs. 1600 at 10%.
∴ Time = [100 * 80 / 1600 * 10] year = 1/2 year = 6 months.
Answer C |
AQUA-RAT | AQUA-RAT-34707 | For the first question, we have for increasing $a_n$: \begin{align} u_{n+1} - u_n &= \frac{a_1+\dotsb+a_{n+1}}{n+1} - \frac{a_1+\dotsb+a_n}{n} \\ &= \frac{n(a_1+\dotsb+a_{n+1})}{n(n+1)} - \frac{(n+1)(a_1+\dotsb+a_n)}{n(n+1)} \\ &= \frac{na_{n+1} - \sum_{k=1}^n a_k}{n(n+1)} \\ &= \frac{a_{n+1}}{n+1} - \frac{u_n}{n+1} \\ &\geq \frac{a_{n+1}}{n+1} - \frac{a_n}{n+1} \gt 0, \end{align} where we have used that $u_n \leq a_n$ (the mean is at most the largest of the elements we're averaging over).
Regarding whether $u_n$ increasing $\implies a_n$ increasing, this seems to be false. For example we can take $a_1 = 0$, $a_2 = 1$, and for $n>2$: $a_n = \frac{1}{2}(u_{n-1}+a_{n-1})$. For $n \geq 2$ we have $a_{n+1} \lt a_n$ and $u_{n+1} \gt u_n$.
The following is multiple choice question (with options) to answer.
A and B are two numbers. Let
@(A,B)= Average of A and B,
/(A,B)= Product of A and B, and
*(A,B)= The result of dividing A by B.
The sum of A and B is given by | [
"/(@(A,B),2)",
"@(*(A,B),2)",
"@(/(A,B),2)",
"None of these"
] | A | Explanation :
Assume some values for A and B and substitute in the options to get the answer.
Here, Let A = 1 and B = 2.
So, /(@(A,B),2) = /(@(1,2),2)
= /(1.5,2) = 3
= Sum of A and B.
Answer : A |
AQUA-RAT | AQUA-RAT-34708 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
Two men A and B start from place X walking at 4 ½ kmph and 5 ¾ kmph respectively. How many km apart they are at the end of 7 ½ hours if they are walking in the same direction? | [
"7 km",
"6 km",
"9 3/8 km",
"9 km"
] | C | RS = 5 ¾ - 4 ½ = 1 ¼
T = 7 ½ h.
D = 5/4 * 15/2 = 75/8
= 9 3/8 km
Answer:C |
AQUA-RAT | AQUA-RAT-34709 | • but the balls are being replaced. how is this possible? – minatozaki Dec 15 '17 at 2:39
• The first ball you select can be any of the four colors. Then you replace it and you must select one of the other three balls next, etc. – Remy Dec 15 '17 at 2:40
• @minatozaki If the balls were not replaced, the second ball would be one of the remaining three balls with probability $\frac33,$ etc., and the answer would be $\frac44\cdot\frac33\cdot\frac22\cdot\frac11 = 1$; that is, if you don't replace, then for certain you will get four different colors. – David K Dec 15 '17 at 12:56
We could do this by counting the number of ways to draw four balls and the number of ways to draw four balls without getting any duplicates. There are $4!$ ways to not get a duplicate as every drawing can be thought of as an ordering and if we don't allow duplicates then we have a permutation. There are $4^4$ different possible drawings as replacement is allowed, this gives us $$\frac{4!}{4^4} = \frac{3}{32}$$
• If there is a -1 is there a reason, anything you'd like explained or improved? – Benji Altman Dec 15 '17 at 2:52
• Someone down-voted my answer too without explanation. – Remy Dec 15 '17 at 2:52
• +1, But one thing to be careful of with this method - it only works when all possible combinations of draws are equally likely. However, in this case, that obviously holds. – Paul Sinclair Dec 15 '17 at 17:45
The probability of drawing $4$ different balls is the product of the probabilities of drawing a new ball on all $4$ draws.
The first draw yields a new ball, guaranteed: $$P(\text{ball 1 new})=1$$
For the second draw, there are $3$ possible new balls and $4$ total balls, so: $$P(\text{ball 2 new})=\frac34$$
The following is multiple choice question (with options) to answer.
21 ball numbered 1 to 21. A ballis drawn and then another ball is drawn without replacement. | [
"2/23",
"3/25",
"9/42",
"1/21"
] | C | The probability that first toy shows the even number,
=1021=1021
Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows the even number,
=920=920
Required probability,
=(1021)×(920)=(1021)×(920)
=9/42
C |
AQUA-RAT | AQUA-RAT-34710 | Change the chapter
Question
Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs. (a) Calculate the magnitude of the force needed to stop the downward motion of a jogger’s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger’s body.) (b) Compare this force with the weight of the jogger.
a) $16300 \textrm{ N}$
b) $22.2$
Solution Video
OpenStax College Physics Solution, Chapter 7, Problem 53 (Problems & Exercises) (3:42)
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The following is multiple choice question (with options) to answer.
If a jogger jogs 16 km/hr instead of 12 km/hr, he would have jogged 10 km more. The actual distance he jogged is | [
"50 km",
"40 km",
"30 km",
"20 km"
] | C | Explanation :
The actual distance jogged be d.
=> d/12 = (d+10)/16
=> 16d = 12d+120
=> 4d = 120
=> d = 30
Answer : C |
AQUA-RAT | AQUA-RAT-34711 | Time needed for both machine to produce 1500 units is given.
1/8 + 1/12
= 5/24
I hrs work done by both machine.
24/5 = 4.8.
Time needed to complete the entire work.
******* 1500 units must be considered as a single work. units are same for both machine. Therefore we don't need to do anything with units.
C is the correct answer.
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Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink]
### Show Tags
12 Aug 2019, 10:53
Bunuel wrote:
A new copy machine can run off 1,500 workbooks in 8 hours, while it takes on older copy machine 12 hours to do the same job. What is the total number of hours that it would take both copy machines working at the same time, but independently, to run off the 1,500 workbooks?
(A) 4.4
(B) 4.6
(C) 4.8
(D) 5
(E) 10
Let T be the number of hours to complete the job when both machines are running. We can create the following equation:
1500/8 * T + 1500/12 * T = 1500
T/8 + T/12 = 1
Let’s multiply each side by 24:
3T + 2T = 24
5T = 24
T = 4.8 hours
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Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink] 12 Aug 2019, 10:53
# A new copy machine can run off 1,500 workbooks in 8 hours, while it ta
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The following is multiple choice question (with options) to answer.
Printer P can print one lakh books in 8 hours. Printer Q can print the same number of books in 10 hours while Printer R can print the same in 12 hours. All the Printers started printing at 9 A.M. Printer P is stopped at 11 A.M. and the remaining 2printers complete work. Approximately at what time will the printing of 1 lac books be completed? | [
"1 pm",
"2 pm",
"3 pm",
"4 pm"
] | A | Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the Printer were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm
A |
AQUA-RAT | AQUA-RAT-34712 | Suppose that $S-T=11$; then $70=11+59=(S-T)+(S+T)=2S$, so $S=35$ and $T=24$. This is possible only if three of the digits $a,c,e,g$ are $9$ and the fourth is $8$; there’s no other way to get four digits that total $35$. For three digits to total $24$, they must average $8$, so the only possibilities are that all three are $8$, that two are $9$ and one is $6$, or that they are $7,8$, and $9$. Thus, the digits $a,c,e,g$ in that order must be $8999,9899,9989$, or $9998$, and the digits $b,d,f$ must be $888,699,969,996,789,798,879,897,978$, or $987$, for a total of $4\cdot 10=40$ numbers.
Now suppose that $S-T=-11$; then by similar reasoning $2S=-11+59=48$, so $S=24$, and $T=35$. But $T\le 3\cdot9=27$, so this is impossible. Similarly, $S-T$ cannot be $-33$. The only remaining case is $S-T=33$. Then $2S=33+59=92$, and $S=46$, which is again impossible. Thus, the first case contained all of the actual solutions, and there are $40$ of them.
-
M.Scott you are genius.Thanks so much.This is the best way.Thank you again. – vikiiii Mar 30 '12 at 15:13
The following is multiple choice question (with options) to answer.
Of the three- digit integers greater than 700, how many have two digits T that are equal to each other and the remaining digit different from the other two? | [
"90",
"82",
"80",
"45"
] | C | three-digit integers greater than 700: 701 to 999, inclusive.
possible values for hundreds-digit--> 7,8,9
possible values for tens-digit and ones-digit --> 0, 1,2,3,4,5,6,7,8,9
when hundreds-digit and tens-digit are the same: (3x1x10)-3=27 ---> we minus three to exclude 777, 888 and 999
when hundreds-digit and ones-digit are the same: (3x10x1)-3=27 ---> we minus three to exclude 777, 888 and 999
when tens-digit and hundreds-digit are the same:[(3x10x1)-3]-1=26 ---> we minus three to exclude 777, 888 and 999; we minus one to exclude 700
T=27+27+26 = 80
Answer: C |
AQUA-RAT | AQUA-RAT-34713 | 1 = 1
4 = 2^2
6 = 2 * 3
9 = 3^2
13 = 13
16 = 2^4
22 = 2 * 11
24 = 2^3 * 3
25 = 5^2
33 = 3 * 11
36 = 2^2 * 3^2
37 = 37
46 = 2 * 23
49 = 7^2
52 = 2^2 * 13
54 = 2 * 3^3
61 = 61
64 = 2^6
69 = 3 * 23
73 = 73
78 = 2 * 3 * 13
81 = 3^4
88 = 2^3 * 11
94 = 2 * 47
96 = 2^5 * 3
97 = 97
100 = 2^2 * 5^2
109 = 109
117 = 3^2 * 13
118 = 2 * 59
121 = 11^2
132 = 2^2 * 3 * 11
141 = 3 * 47
142 = 2 * 71
144 = 2^4 * 3^2
148 = 2^2 * 37
150 = 2 * 3 * 5^2
157 = 157
166 = 2 * 83
169 = 13^2
177 = 3 * 59
181 = 181
184 = 2^3 * 23
193 = 193
196 = 2^2 * 7^2
198 = 2 * 3^2 * 11
208 = 2^4 * 13
213 = 3 * 71
214 = 2 * 107
216 = 2^3 * 3^3
222 = 2 * 3 * 37
225 = 3^2 * 5^2
229 = 229
241 = 241
244 = 2^2 * 61
249 = 3 * 83
253 = 11 * 23
256 = 2^8
262 = 2 * 131
276 = 2^2 * 3 * 23
277 = 277
286 = 2 * 11 * 13
289 = 17^2
292 = 2^2 * 73
294 = 2 * 3 * 7^2
297 = 3^3 * 11
312 = 2^3 * 3 * 13
313 = 313
321 = 3 * 107
324 = 2^2 * 3^4
325 = 5^2 * 13
333 = 3^2 * 37
334 = 2 * 167
337 = 337
349 = 349
352 = 2^5 * 11
358 = 2 * 179
361 = 19^2
366 = 2 * 3 * 61
373 = 373
376 = 2^3 * 47
382 = 2 * 191
384 = 2^7 * 3
388 = 2^2 * 97
393 = 3 * 131
The following is multiple choice question (with options) to answer.
Find the number of divisors of 1728.? | [
"26",
"27",
"28",
"29"
] | C | 1728= 2^6 * 3^3
Hence the Number of factors = (6+1) x (3+1) = 7 x 4 = 28.
We know that if a number represented in standard form (a^m *b^n) , then the number of factors Is given by (m+1)(n+1).
Answer is 28
ANSWER:C |
AQUA-RAT | AQUA-RAT-34714 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
Company C sells a line of 25 products with an average retail price of $1,200. If none of these products sells for less than $400, and exactly 10 of the products sell for less than $1,000, what is the greatest possible selling price of the most expensive product? | [
"12000",
"13000",
"14000",
"15000"
] | A | The average price of 25 products is $1,200 means that the total price of 25 products is 25*1,200=$30,000.
Next, since exactly 10 of the products sell for less than $1,000, then let's make these 10 items to be at $400 each (min possible).
Now, the remaining 14 items cannot be priced less than $1,000, thus the minimum possible price of each of these 14 items is $1,000.
Thus the minimum possible value of 24 products is 10*400+14*1,000=$18,000.
Therefore, the greatest possible selling price of the most expensive product is $30,000-$18,000=$12,000.
Answer: A. |
AQUA-RAT | AQUA-RAT-34715 | $3$-cube faces are made by starting with a $3$-cube in one of the two n-cubes plus translation of the squares in the new dimension. So $C(n+1)=2C(n)+S(n)$.
• An expository article about how to think about n-cubes from a combinatorial point of view can be found here: york.cuny.edu/~malk/tidbits/n-cube-tidbit.html – Joseph Malkevitch Jan 5 '11 at 1:23
• @Joseph: Thanks. It shows that the recurrence will be the same for all dimensions of boundary: points, lines, squares, etc. – Ross Millikan Jan 5 '11 at 3:36
First you need to be clear about what you mean by face, edge, vertex, etc. In my view if we have an $n$ dimensional cube, then $n-1$ dimensional cubes form its faces, and $n-2$ dimensional cubes form its edges. Vertices as you define above are 0-D points. We do not have names for $n-3$, $n-4$ dimensional boundaries so I will not attempt to name them here.
If you are looking for an intuitive way to build up the relations you can define inductive relationships using the method of extrusion. You can start with $n=1$ if you like. In this case we have one 1-D cube (i.e. a segment) with two 0-D faces (i.e. points). So
$$C_1 = 1$$ $$F_1 = 2$$ $$E_1 = 0$$ $$V_1 = 2$$
Where $C_1$ is the number of cubes in 1-D space, $F_1$ is the number of faces, $E_1$ is the number of edges (not defined here), and $V_1$ is the number of vertices.
We create a 2-D cube (i.e. a square) by extruding this 1-D cube in a direction orthogonal all current dimensions. By doing this two things happen:
The following is multiple choice question (with options) to answer.
The edge of three cubes of metal is 3 dm, 4 dm and 5 dm. They are melted and formed into a single cube. Find the edge of the new cube? | [
"7",
"5",
"8",
"6"
] | D | Explanation:
33 + 43 + 53 = a3 => a = 6
Answer: Option D |
AQUA-RAT | AQUA-RAT-34716 | Let's compare your method with the correct solution.
Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.
Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ ways.
Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in $$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$ ways.
Total: Since the two cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$ ways to select five questions so that at least one is drawn from each of the three sections.
You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $$A_1, A_2, A_3, B_1, C_1$$ are selected. You count this selection three times.
The following is multiple choice question (with options) to answer.
A teacher prepares a test. She gives 8 objective type questions out of which 4 have to be answered. Find the total ways in which they can be answered if the first 5 questions have 3 choices and the last 3 have 4 choices. | [
"6255",
"5280",
"5192",
"5100"
] | B | Two cases.
First Case: 2 Questions - 3 Choices, 2 Questions - 4 Choices
5C2 * 3C2 ways to select the questions
3C1 * 3C1 * 4C1 * 4C1 ways to answer the questions
= 10 * 3 * 3 * 3 * 4 * 4
= 4320
Second Case: 1 Question - 3 Choices, 3 Questions - 4 Choices
5C1 * 3C3 ways to select the questions
3C1 * 4C1 * 4C1 * 4C1 ways to answer the questions
= 5 * 1 * 3 * 4 * 4 * 4
= 960
Total = 4320 + 960
=5280
B |
AQUA-RAT | AQUA-RAT-34717 | ### Show Tags
27 Jan 2020, 12:45
Notice that all options' values have different unit digit;
To Solve: Add the unit digits of each values(4+9+4+9+4+4+7) = 41 and check the unit digit of output (unit digit 1).
for division by 7, 3 should be unit digit of the quotient.
Only Option C satisfies, it is $73! _________________ Unable to Give Up! Re: Over the past 7 weeks, the Smith family had weekly grocery bills of$7 [#permalink] 27 Jan 2020, 12:45
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# Over the past 7 weeks, the Smith family had weekly grocery bills of \$7
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The following is multiple choice question (with options) to answer.
What is the units digit of the product of the first 15 odd numbers? | [
"0",
"5",
"1",
"2"
] | B | 1*3*5*7 ................... will end up in 5 in the units place
Answer : B |
AQUA-RAT | AQUA-RAT-34718 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
If s,u, and v are positive integers and s=2u+2v, which of the following must be true?
i. s=u
ii. u is not equal to v
iii. s > v | [
"None",
"I only",
"II only",
"III only"
] | A | Notice two things: 1. we are asked to find out which of the following MUST be true, not COULD be true and 2. s, u, and v are positive integers.
Given: s=2u+2v --> s/2=u+v. Now, since s, u, and v are positive integers then s is more than either u or v, so I is never true and III is always true. As for II: it's may be necessarily true,
Answer: A |
AQUA-RAT | AQUA-RAT-34719 | In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
The following is multiple choice question (with options) to answer.
If n divided by 9 has a remainder of 8, what is the remainder when 2 times n is divided by 9? | [
"1",
"7",
"3",
"5"
] | B | As per question => N=9P+8 for some integer p
hence 2N => 18Q + 16
But again, 16 can be divided by 9 to get remainder 7 for some integer Q
hence B |
AQUA-RAT | AQUA-RAT-34720 | ### Show Tags
01 May 2011, 21:45
15 sh + 25 sw = 21(sh+sw)
4sw = 6sh
2sw = 3sh Sufficient.
15 sh + 25 sw = 420
3sh + 5sw = 84
5 * 16 = 80 (max sw = 16)
Hence by trial method sw = 15, sh = 3
3* 28 = 84 (max sh = 28)
hence by trial method 3*sh = 54 gives sw = 6
meaning sh = 18,sw = 6. Hence not sufficient.
Thus A.
Manager
Joined: 16 May 2011
Posts: 53
### Show Tags
28 Mar 2012, 06:53
from statement 2,
shirts x
sweaters y
15x +20y = 420
The following is multiple choice question (with options) to answer.
On a certain scale of intensity, each increment of 20 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125? | [
"40",
"100",
"400",
"1000"
] | B | Increase of 40 in magnitude corresponds to 10^4 increase in intensity:
If intensity for 125 is x then for 135 it'll be 10*x, for 145 it'll be 10*10*x=10^2*x, for 155 it'll be 10*10*10*x=10^3*x and for 165 it'll be 10*10*10*10*x=400*x.
Answer: B. |
AQUA-RAT | AQUA-RAT-34721 | of 69” which states the doubling period as 0.35 + 69/Interest The time value of money is a concept integral to all parts of business. How to Calculate Present Value, and Why Investors Need to Know It, Understanding the Compound Annual Growth Rate – CAGR. We could put the equation more concisely and use the$10,000 as FV. The above future value equation can be rewritten as follows: PV=FV(1+i)n\begin{aligned} &\text{PV} = \frac{ \text{FV} }{ ( 1 + i )^ n } \\ \end{aligned}PV=(1+i)nFV, PV=FV×(1+i)−nwhere:PV=Present value (original amount of money)FV=Future valuei=Interest rate per periodn=Number of periods\begin{aligned} &\text{PV} = \text{FV} \times ( 1 + i )^{-n} \\ &\textbf{where:} \\ &\text{PV} = \text{Present value (original amount of money)} \\ &\text{FV} = \text{Future value} \\ &i = \text{Interest rate per period} \\ &n = \text{Number of periods} \\ \end{aligned}PV=FV×(1+i)−nwhere:PV=Present value (original amount of money)FV=Future valuei=Interest rate per periodn=Number of periods. … The welfare of the owners would be maximized when net worth or net value is created from making a financial decision. It is underlying theme embodies in financial concepts such as:eval(ez_write_tag([[580,400],'xplaind_com-box-4','ezslot_5',134,'0','0'])); It is the basis used to work out the intrinsic value of a firm, a share of common stock, a bond or any other financial instrument. The time value of money implies that: 1. a person will have to pay in future more, for a rupee received today and 2. a person may accept less today, for a rupee to be
The following is multiple choice question (with options) to answer.
The present value of a bill due at the end of 2 years is Rs.1250. If the bill were due at the end of 2 years and 11 months, its present worth would be Rs.1200. Find the rate of interest and the sum. | [
"rs.1175",
"rs.1375",
"rs.1475",
"rs.1575"
] | B | Explanation:
PW=100xAmount/100+(Rx T)
Answer: B |
AQUA-RAT | AQUA-RAT-34722 | The remaining case is that there are two $7's$ and the $a_1\lt 7$. From the range, $a_7=a_1+15$. Thus, from the mean, $a_1$, $a_7$ and the other undetermined number (denoted by $b$) add up to $70-2\times 7-9-10=37$. Thus, $b+2a_1+15=37$ or $b+2a_1=22$. Note $b\leq a_7=a_1+15$. Thus, $22=b+2a_1\leq3a_1+15$ or $a_1>2$. Noting that $a_1\lt 7$, the possible combinations of $(a_1,b)$ are $(3,16)$, $(4,14)$, $(5,12)$, and $(6,10)$. The last combination is disallowed as $7$ is the mode. Thus, the possible sequences are: $(3,7,7,9,10,16,18)$, $(4,7,7,9,10,14,19)$, and $(5,7,7,9,10,12,20)$.
### Solution 5
This is by N. N. Taleb:
Let $X=\left\{x_{(1)},x_{(2)},x_{(3)},x_{(4)},x_{(5)},x_{(6)},x_{(7)}\right\}$ ordered such that $x_{(1)} \leq x_{(2)} \ldots \leq x_{(7)}$.
We have $x_{(7)}=x_{(1)}+15$, $x_{(4)}=9$, and we need to solve for:
The following is multiple choice question (with options) to answer.
Average of 13 results is 65. If the average of first six results is 61 and average of last six results is 59 Then find the seventh result? | [
"120",
"123",
"125",
"128"
] | C | 125
Option 'C' |
AQUA-RAT | AQUA-RAT-34723 | So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.
Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$.
To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.
For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.
-
Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39
The following is multiple choice question (with options) to answer.
Last year’s receipts from the sale of greeting cards during the week before Mother’s Day totaled $195 million, which represented 10 percent of total greeting card sales for the year. Total greeting card sales for the year totaled how many million dollars? | [
"17,010",
"2,100",
"1,890",
"1,950"
] | D | 10% ---- 195 millions
for 100% => ( 195 * 100% )/10% = 1950.
Option D. |
AQUA-RAT | AQUA-RAT-34724 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row downstream at 20 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively? | [
"13, 5",
"15, 2",
"15, 5",
"14, 5"
] | C | Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 20--- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 30 => x = 15, y = 5.
Answer:C |
AQUA-RAT | AQUA-RAT-34725 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A 24 month project had a total budget of $60,000. After eleven months, the project had spent $26,300. At this point, how much was the project under budget? | [
"$800",
"$1000",
"$1200",
"$1400"
] | C | Each month, the project should spend $60,000 / 24 = $2500.
In 11 months, the project should spend 11*$2500=$27,500.
The project is under budget by $27,500 - $26,300 = $1200.
The answer is C. |
AQUA-RAT | AQUA-RAT-34726 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
Two numbers are in the ratio 3:5. If 9 be subtracted from each, they are in the ratio of 2:5. The first number is: | [
"A)33",
"B)98",
"C)34",
"D)35"
] | A | (3x-9):(5x-9) = 2:5
x = 11 => 3x = 33
Answer:A |
AQUA-RAT | AQUA-RAT-34727 | r
#DECILE 5
t_prag_top_total_income_50<-decili_total_income_neto\[5,1\]
t_prag_top_total_income_filter_50<-filter(data2, NET_INCOME> t_prag_top_total_income_40, NET_INCOME<=t_prag_top_total_income_50)
t_prag_top_total_income_filter_50_tax<-sum(t_prag_top_total_income_filter_50$TAX)
t_tax_share_50<-((t_prag_top_total_income_filter_50_tax)/ZBIR_TOTAL_TAX)*100
t_prag_top_total_income_filter_50<-sum(t_prag_top_total_income_filter_50$NET_INCOME)
t_prag_top_total_income_filter_50a<-nrow(filter(data2, NET_INCOME> t_prag_top_total_income_40, NET_INCOME<=t_prag_top_total_income_50))
t_prag_top_total_income_50b<-((t_prag_top_total_income_filter_50)/ZBIR_TOTAL_NET_INCOME)*100
FINAL_CENTILE_TABLE50<-data.frame(cbind(t_prag_top_total_income_50,t_prag_top_total_income_filter_50,t_prag_top_total_income_filter_50a,t_prag_top_total_income_50b,t_prag_top_total_income_filter_50_tax , t_tax_share_50))
colnames(FINAL_CENTILE_TABLE50)<-c("Decile threshold","Total income in the decile","Number of persons in the centile","Share of the decile in total income (%)","Tax","Share tax(%)")
FINAL_DECILE_TABLE <- rbind(FINAL_DECILE_TABLE, FINAL_CENTILE_TABLE50)
The following is multiple choice question (with options) to answer.
A person spends 40% of his salary on food, 25% on house rent, 15% on entertainment and 15% on conveyance. If his savings at the end of the month is Rs. 1200, then his salary per month in rupees is: | [
"4000",
"24000",
"8000",
"10000"
] | B | Total expenditure = 40 + 25 + 15 + 15 = 95%
Saving = (100 - 95) = 5%
5/100 × Salary = 1200, Salary = 24000 Rs.
Answer:B |
AQUA-RAT | AQUA-RAT-34728 | A: 9
B: 12
C: 16
D: 18
E: 24
This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 03:08
1
2
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours
Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Five machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 20 hours to fill a certain production order, how many fewer hours does it take all five machines, operating simultaneously, to fill the same production order? | [
" 4",
" 5",
" 6",
" 16"
] | A | Total work = 4*20 machine hrs
Time taken = 4* 20 /5 => 16 hours...
Thus all five machines, operating simultaneously will take ( 20 - 16 ) 4 hours..
Answer will be (A) 4 |
AQUA-RAT | AQUA-RAT-34729 | evolution, population-dynamics
Title: How many humans have been in my lineage? Is it almost the same for every human currently living? If I were to count my father, my grandfather, my great-grandfather, and so on up till, say chimps, or the most common ancestor, or whatever that suits the more accurate answer, how many humans would there have been in my direct lineage?
And would it be almost the same for every human being currently living? A quick back-of-the-envelope answer to the number of generations that have passed since the estimated human-chimp split would be to divide the the split, approximately 7 million years ago (Langergraber et al. 2012), by the human generation time. The human generation time can be tricky to estimate, but 20 years is often used. However, the average number is likely to be higher. Research has shown that the great apes (chimps, gorilla, orangutan) have generation times comparatble to humans, in the range of 18-29 years (Langergraber et al. 2012).
Using 7 million years and 20 years yields an estimated 350000 ancestral generations for each living human. A more conservative estimate, using an average generation time of 28, would result in 250000 generations. However, some have argued that the human-chimp split is closer to 13 million years old, which would mean that approximately 650000 generations have passed (using a generation time of 20 years).
The exact number of ancestral generations for each human will naturally differ a bit, and some populations might have higher or lower numbers on average due to chance events or historical reasons (colonizations patterns etc). However, due to the law of large numbers my guess would be that discrepancies are likely to have averaged out. In any case, the current estimates of the human-chimp split and average historical generation times are so uncertain, so that they will swamp any other effects when trying to calculate the number of ancestoral generations.
However, this is only answering the number of ancestral generations. The number of ancestors in your full pedigree is something completely different. Since every ancestor has 2 parents, the number of ancestors will grow exponentially. Theoretically, the full pedigree of ancestors can be calculated using:
The following is multiple choice question (with options) to answer.
Four years ago, the average age of a family of 4 members was 21 years.A baby having been born,the average age of the family is the same today.The present age of the baby? | [
"5 years",
"7 years",
"2 years",
"3 years"
] | A | Total age of the 4 members, 4 years ago = (21*4)= 84 years.
Total age of the 4 members now =( 84+ 4*4) years =100 years.
Total age of the 5 members now = (21*5) = 105 years.
Age of the baby is = 105- 100= 5 years
Answer :A |
AQUA-RAT | AQUA-RAT-34730 | -
In the interval $[100,999]$ there are $90$ palindromes. You can choose the first digit $9$ ways and the middle digit $10$ ways. Generally, for $n$ digit numbers there are $$\begin {cases} 9\cdot 10^{\frac {n-2}2} & n \text { even} \\ 9\cdot 10^{\frac {n-1}2} & n \text { odd} \end {cases}$$ palindromes. Again, you can choose the first digit $9$ ways and the rest of the first half of the number (rounded up for odd numbers of digits) $10$ ways .
To get $66$ with reverse and add, you can have $15,24,33,42,51$ as starting numbers. For $5556555$ you can certainly have five choices $(1-5)$ for the first digit, six $(0-5)$for the next two, and one choice $(3)$for the middle digit. Then the lower three digits are determined by the top three. This gives $180$ numbers. There might be more, as I have avoided carrying.
The following is multiple choice question (with options) to answer.
How many 1/10s are there in 37 1/2? | [
"375",
"475",
"500",
"670"
] | A | Required number = (75/2)/(1/10)
= (75/2 x 10/1)
= 375.
ANSWER:A |
AQUA-RAT | AQUA-RAT-34731 | homework-and-exercises, electric-current, electrical-resistance, voltage, power
Title: Finding out how long a light bulb will light A cylindrical wire used to form a light bulb filament has radius 3.7 micrometers and length 1.7 cm. The resistivity of the wire is 5.25 * 10^-5 ohm meters. The light bulb is connected to a 12V battery.
With the given information; resistance equals 20.75 ohm meters. Current will be 0.58 amps and power will be 6.94 W.
The question I do not understand the meaning of is, if the battery has a total stored charge of 0.5 A hr, and produces a constant potential difference until discharged, how long will the light bulb light?
Can anyone explain to me what the question means? You've done all the hard work and now there's one little part left.
You're being asked, given a certain battery capacity in amp hours and a certain current in amps flowing out of the battery, for how many hours (or seconds) will the current flow? The assumption is that as long as the current is flowing the light will be on and as soon as the capacity of the battery is drained the light will go off.
Explanation
You know the power the bulb will use. You know the voltage of the battery, you know the total charge store in the battery in the form "amp hours" (A hr). Amp hours are an engineering term for charge and are equivalent to coulombs.
1 amp flowing for 1 hour = 1 coulomb / second flowing for 3600 seconds = 3600 C
A battery capacity of 1 amp hour means that:
a current of 1 amp can flow for 1 hour
a current of 2 amps can flow for 0.5 hours
a current of 0.25 amps can flow for 4 hours
Notice that in each of these the current (amps) * time (hours) = capacity (amp hours): 1 * 1 = 2 * 0.5 = 0.25 * 4 = 1
A constant potential difference across the light means that the current will also be constant. In the real world as batteries discharge the voltage across their terminals decreases. This drop in voltage is due to internal reistance of the battery increasing as chemical reactions take place inside the battery but that's outside the scope of this question. If the driving voltage wasn't contant the current would also change which makes the problem harder.
The following is multiple choice question (with options) to answer.
a light flashes every 5seconds, how many times will it flash in ? of an hour? | [
"720",
"600",
"650",
"700"
] | A | 1flash=5sec
for 1 min=12flashes
so for 1 hour=12*60=720 flashes.
ANSWER:A |
AQUA-RAT | AQUA-RAT-34732 | inorganic-chemistry, home-experiment
Title: Make a silver zinc battery I have a shaver that runs off a rechargeable battery that is dying. Would it be feasible to make a silver zinc battery and use it to replace the existing battery? Cost is not an object, anything less than $2500 I would consider doable. I have a machine shop and a small chemistry laboratory with the standard equipment and glassware, including high vacuum capability, a centrifuge and simple glass blowing capability.
I found a book on silver-oxide zinc chemistry and battery design, but it is $500 and I don't want to spend that if the information can be obtained just as easily elsewhere.
I have tried to find commercial options, but had no luck. Most silver zinc batteries seem to be just for large (multi-million dollar) military or satellite applications. Sony makes a line of silver oxide primary cells for hearing aids, but these are not rechargeable. There is a company called Ultralife that makes medical and military batteries and might have something viable, but before I call them I wanted to check out the opinion of the experts here. This turns out to be very difficult to do for two reasons. One problem is that the voltage of a silver cell is different from that of Ni-Cd cell, so it would require a specialized, multi-cell configuration to emulate the voltage characteristics of the Ni-Cd rechargeables.
The other problem is that silver cells generally have a sophisticated frame inside of them that is produced by an intricate high-temperature welding process. To duplicate this process and produce a suitable frame would require a significant amount of experimentation and work.
The following is multiple choice question (with options) to answer.
RADIO X sells cheap radios for commercial use by buying surplus supplies. A surplus bag of 100 diodes costs $2.00. A surplus bag of diodes, on average, has 10% faulty diodes (the faulty diodes are thrown away, and the remaining diodes are used in production). It takes 5 diodes to make 1 circuit board. If it takes 3 circuit boards to make a radio, how much money is needed to buy diodes to make x number of radios, in dollars? | [
"x/3",
"3x",
"2x/3",
"x"
] | A | Initial cost for 1 bag of diodes = $2, but ONLY 100(1 - 0.1) = 90 are used. So the cost for 90 diodes = $2. The number of diodes for making a radio is (3 circuit boards)(5 diodes) = 15 diodes. So, the cost for diodes to make a radio is $2(15/90) = $1/3.
$x/3 =cost of diodes to make x number of radios.
Answer is A |
AQUA-RAT | AQUA-RAT-34733 | In layman terms, this means if we assume there is a smallest number X where X>0 but there is no number such that X>Y>0, then X is the difference between 1 and 0.999... As all my opponent's formulas rely on the idea that 0.999... is a real number, he fails the idea when applied to hyperreal numbers.
The following is multiple choice question (with options) to answer.
A Positive number which when added to 999 given a sum which is greater than when it is multiplied by 999. Which of the following could be the value of the positive integer? | [
"5",
"3",
"1",
"2"
] | C | Explanation:
Let the positive integer = x
Then, 999 + x > 999x
By trial and error, when x = 1,
1000 > 999
So, the required positive integer is 1,
1000 > 999
So, the required positive integer is 1.
Answer: Option C |
AQUA-RAT | AQUA-RAT-34734 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
X starts a business with Rs.77000. Y joins in the business after 5 months with Rs.42000. What will be the ratio in which they should share the profit at the end of the year? | [
"21:23",
"23:45",
"22:7",
"25:29"
] | C | Explanation :
Ratio in which they should share the profit
= Ratio of the investments multiplied by the time period
= 77000 * 12: 42000 * 7
= 77 * 12: 42 * 7
= 11*2: 7
= 22:7.
Answer : Option C |
AQUA-RAT | AQUA-RAT-34735 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A person standing on a railway platform noticed that a train took 21 seconds to completely pass through the platform which was 84 m long and it took 9 seconds in passing him. The speed of the train was | [
"25.2 km/hr",
"32.4 km/hr",
"50.4 km/hr",
"75.6 km/hr."
] | A | Explanation:
Let the train’s length be L m.
L/9=(L+84)/21
21 L = 9L +756
12L= 756
L=63 m
Speed = 63 /9 = 7 m/s = 7 x 18/5 = 25.2 Km/hr.
Answer A |
AQUA-RAT | AQUA-RAT-34736 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
By travelling at 40 kmph, a person reaches his destination on time. He covered two-third the total distance in one-third of the total time. What speed should he maintain for the remaining distance to reach his destination on time? | [
"20 kmph",
"30 kmph",
"87 kmph",
"99 kmph"
] | A | Let the time taken to reach the destination be 3x hours. Total distance
= 40 * 3x = 120x km
He covered 2/3 * 120x = 80x km in 1/3 * 3x
= x hours So, the remaining 40x km, he has to cover in 2x hours. Required speed
= 40x/2x = 20 kmph.
Answer:A |
AQUA-RAT | AQUA-RAT-34737 | A number is divisible by 11 if and only if the difference of the sum of the odd numbered digits (the first digit, the third digit, ...) and the sum of its even numbered digits is divisible by 11. The sum of the odd numbered digits of $y$ is $250\cdot(2+7)$ and the sum of the even numbered digits of $y$ $250\cdot(7+2)$. The difference between these two quantities is $0$; so $y$ is divisible by 11.
-
This explains it very well! Thank you so much. – SNS Feb 27 '12 at 23:42
$2772=99 \times 28$
so
$277227722772\ldots277227722772 = 99 \times 28 \times 100010001\ldots000100010001$
and so is divisible by both $9$ and $11$ (and $4$ and $7$ and other numbers).
The following is multiple choice question (with options) to answer.
When the no.7y86038 is exactly divisible by 11, then the smallest whole no.in place of y? | [
"9",
"10",
"11",
"13"
] | C | The given number =7y86038
Sum of the odd places =8+0+8+7=23
Sum of the even places = 3+6+y
(Sum of the odd places)- (Sum of even places) = Number (exactly divisible by 11)
23-(9+y) = divisible by 11
14 � y = divisible by 11.
Y must be 3, to make given number divisible by 11.
C |
AQUA-RAT | AQUA-RAT-34738 | algorithms, dynamic-programming, np-hard
if left_is_greater(greater, precincts, n):
greater.append(precincts.popleft())
continue
p = precincts.pop()
# if either of the districts is "full", we don't have a choice
# and must select the opposite district
if len(right) == n / 2:
left.append(p)
elif len(left) == n / 2:
right.append(p)
# we always prefer the lesser summed district. if gerrymandering
# is feasible, we won't want to put all of our party's votes
# into one district, but prefer to have each district barely win.
elif sum(left) < sum(right):
left.append(p)
elif sum(right) < sum(left):
right.append(p)
# if the districts sum to the same amount, we prefer the district
# with less precincts, as this maximizes future decision branches
elif len(right) < len(left):
right.append(p)
elif len(left) < len(right):
left.append(p)
# we choose left as the default seed case, all else equal
else:
left.append(p)
if sum(left) > 0 and sum(right) > 0:
return True, left, right
else:
return False, left, right
def solve(precincts):
maximized = deque(sorted(precincts))
minimized = deque(sorted(x * -1 for x in precincts))
for polarized_precincts in [maximized, minimized]:
is_gerrymanderable, left, right = maximize(polarized_precincts)
if is_gerrymanderable:
print("It's gerrymanderable! Take a look: ", left, right)
return
print("Sorry, doesn't seem to be gerrymanderable.")
The following is multiple choice question (with options) to answer.
A certain city with a population of 108,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district What is the minimum possible population that the least populated district could have ? | [
"a) 10,700",
"b) 10,800",
"c) 9,000",
"d) 11,000"
] | C | Let x = number of people in smallest district
x*1.1 = number of people in largest district
x will be minimised when the number of people in largest district is maximised
10*x*1.1 = 11x = total number of people in other districts
So we have 11x + x = 108k
x = 9,000
Answer : C |
AQUA-RAT | AQUA-RAT-34739 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a piece of work in 10 days. He works at it for 4 days and then B finishes it in 9 days. In how many days can A and B together finish the work? | [
"6",
"7",
"9",
"5"
] | A | 4/10 + 9/x = 1 => x = 15
1/10 + 1/15 = 1/6 => 6 days
Answer: A |
AQUA-RAT | AQUA-RAT-34740 | (1) The median of set {x,−1,1,3,−x} is 0
(2) The median of set {x,−1,1,3,−x} is x2
Merging similar topics. Please refer to the solutions above.
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Re: What is the value of x? [#permalink]
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09 Jun 2017, 15:49
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The following is multiple choice question (with options) to answer.
Set A: 3, r, 8, 10
Set B: 4, g, 9, 11
The terms of each set above are given in ascending order. If the median of Set A is equal to the median of Set B, what is the value of g – r? | [
"-2",
"-1",
"0",
"1"
] | B | So we have even no. of elements in the Set
So median is the average of Middle two numbers
(r+8)/2= (g+9)/2
g - r= -1
Answer B |
AQUA-RAT | AQUA-RAT-34741 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
A train travelled from station P to Q in 8 hours and came back from station Q to P is 6 hours. What would be the ratio of the speed of the train while traveling from station P to Q to that from station Q to P? | [
"3 : 9",
"3 : 7",
"3 : 8",
"3 : 4"
] | D | Since S # 1/t
S1 : S2 = 1/t1 : 1/t2 = 1/8 : 1/6 = 3 : 4
Answer:D |
AQUA-RAT | AQUA-RAT-34742 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
John was thrice as old as Tom 6 yrs back. He will be 5/3 times as old as Tom 6 years hence. How old is Tom today? | [
"6",
"9",
"12",
"18"
] | C | Sol. RMBflSk -6 = 3 (Ravi - 6) ...(j)
and Rupesh + 6 = | (Ravi + 6) ...(ii)
Solving both the equations we get,
Ravi = 12 years.
C |
AQUA-RAT | AQUA-RAT-34743 | . . Find the sum of all multiples of 6 from 6 to 102 , inclusive. . Jump to Question. perfect thank you. 0 0 1 3 0 0 0. In simple words, Consecutive integers are integers that follow in sequence, where the difference between two successive integers is 1. Clearly, the numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, …., 999. . Julie S. Syracuse University . with first term a = 252, common difference = 3 and last term = 999. Use the following formula: n(n + 1)/2 = Sum of Integers. . Find the sum of all multiples of 4 that are between 15 and 521. The sum of 1+2+3+4+...496+497+498+499+500? Between 1 and 200, they are three digit numbers of which the first must be 1. home Front End HTML CSS JavaScript HTML5 Schema.org php.js Twitter Bootstrap Responsive Web Design tutorial Zurb Foundation 3 tutorials Pure CSS HTML5 Canvas JavaScript Course Icon Angular React Vue Jest Mocha NPM Yarn … What is the sum of the first 500 counting numbers? Python Program to return Sum of Prime Numbers from 1 to 100. What is the sum of first 550 odd numbers. s_n = n(a_1 + a_n)/2. Call out the result. If n is an integer, then n, n+1, and n+2 would be consecutive integers. This is an A.P. The sum of part of the series of natural numbers from n 1 to n 2 is the sum from 1 to n 2-1 less the sum from 1 to n 2. Input parameters & values: You need to set the "Use regular expression to parse number" checkbox and enter regular expression and match group, which will be used to extract the number. What is the sum of first 400 natural numbers? let the numbers be x & y x+y = 22 x= 22-y x^2+y^2=250.. plug value of x in the equation This is an A.P. The below workout with step by step calculation shows how to find what is the sum of natural numbers or positive integers from 1 to 250 by applying arithmetic progression. is 56. If n is an integer, then n,
The following is multiple choice question (with options) to answer.
What is the sum of all consecutive integers from 10 to 100, inclusive? | [
"4550",
"4950",
"5005",
"5500"
] | C | SUM= n/2(Starting number+ Ending number)
n: total numbers between the starting no. and ending no.
n= 100-10+1= 91
SUM= (91/2)*(10+100)
= 91*55
=5005
Hence, answer C |
AQUA-RAT | AQUA-RAT-34744 | evolution, population-dynamics
Title: How many humans have been in my lineage? Is it almost the same for every human currently living? If I were to count my father, my grandfather, my great-grandfather, and so on up till, say chimps, or the most common ancestor, or whatever that suits the more accurate answer, how many humans would there have been in my direct lineage?
And would it be almost the same for every human being currently living? A quick back-of-the-envelope answer to the number of generations that have passed since the estimated human-chimp split would be to divide the the split, approximately 7 million years ago (Langergraber et al. 2012), by the human generation time. The human generation time can be tricky to estimate, but 20 years is often used. However, the average number is likely to be higher. Research has shown that the great apes (chimps, gorilla, orangutan) have generation times comparatble to humans, in the range of 18-29 years (Langergraber et al. 2012).
Using 7 million years and 20 years yields an estimated 350000 ancestral generations for each living human. A more conservative estimate, using an average generation time of 28, would result in 250000 generations. However, some have argued that the human-chimp split is closer to 13 million years old, which would mean that approximately 650000 generations have passed (using a generation time of 20 years).
The exact number of ancestral generations for each human will naturally differ a bit, and some populations might have higher or lower numbers on average due to chance events or historical reasons (colonizations patterns etc). However, due to the law of large numbers my guess would be that discrepancies are likely to have averaged out. In any case, the current estimates of the human-chimp split and average historical generation times are so uncertain, so that they will swamp any other effects when trying to calculate the number of ancestoral generations.
However, this is only answering the number of ancestral generations. The number of ancestors in your full pedigree is something completely different. Since every ancestor has 2 parents, the number of ancestors will grow exponentially. Theoretically, the full pedigree of ancestors can be calculated using:
The following is multiple choice question (with options) to answer.
Average age of 7 members of a family is 29 years. If present age of the youngest member is 5 year, find average age of the remaining members at the time of birth of the youngest member. | [
"28",
"37",
"29",
"237"
] | A | Average age (present) of 7 members = 29 years
5 years ago, average age of 7 members was 29 - 5 = 24 years.
Since, the youngest member was not born 5 years ago.
Therefore, Average age of remaining 6 members is increased by 246246 = 4 years.
Therefore, 5 years ago, average age of 6 members was 24 + 4 = 28 years.
Answer:A |
AQUA-RAT | AQUA-RAT-34745 | I will be grateful for any help, or any reading material reference.
-
You are right about the probability of guessing $3$. Your method can be adapted to solve the other problems. But my preferred way of doing it involves the binomial coefficients, which in general count the number of ways of choosing $r$ objects from $n$ objects, order irrelevant. Maybe also look up combinations. Note that if you can handle exactly $0$, $1$, $2$, $3$ you can handle the "at least" stuff. For example the probability of at least $2$ is prob. of exactly $2$ plus prob. of exactly $3$. – André Nicolas Jun 2 '11 at 11:12
@user6312: Thank you for your comment I now understand (I better say you've reminded me) about the "at least" stuff. Please take a look at the answer provided by Henry and my response to that answer and see if you can maybe shed some light on my problem. – Nikola Jun 2 '11 at 12:34
You need to be clear whether order matters. Henry's answer assumes that guessing (2,5,8) and getting (2,8,5) is three correct, not one. – Ross Millikan Jun 2 '11 at 13:08
@Ross: yes, he is right, so if I say that numbers 2,5,8 will be n that 10 numbers that are drawn then the order of them actually "coming" out is not important. – Nikola Jun 2 '11 at 13:16
If 10 out of 20 are drawn and you guess 3, then the probabilities are
The following is multiple choice question (with options) to answer.
What is the probability of finding exactly 33 multiples of 3 when 100 consecutive
natural nos are selected? | [
"1/3",
"2/3",
"1",
"2"
] | B | out of 3x,3x+1,3x+2 as starting numbers,
Only case when starting no. is 3x we'll get 34 multiples of 3.
ANSWER:B |
AQUA-RAT | AQUA-RAT-34746 | # What is the probability that a randomly chosen 3-digit integer is divisible by $5$?
So I was given a question with two parts:
The first part is like this:
How many 3-digit integers (integers from 100 to 999 inclusive) are divisible by 5?
My solution: $\frac{(999-100+1)}{5} = 180$
The second part following the first part is like this:
What is the probability that a randomly chosen 3-digit integer is divisible by 5?
I saw a similar question where they took the above answer and subtracted by 17 and added 1
Is this correct?
Solution: $180/900$
• Why are you subtracting by $17$ and adding $1$? – miradulo Jan 20 '16 at 14:59
• @DonkeyKong In the previous example they did that i thought it was a general method, was unsure so i just took the first answer and did that – Zero Jan 20 '16 at 15:01
• Something like "subtracting $17$ and adding $1$" is hardly ever going to be a general method. That's like saying hocus pocus at a math equation. You know how many 3 digits integers there are, and you now know how many of these 3 digit integers are divisible by 5. To get the probability for a randomly chosen 3 digit integer, just divide your result $180$ by your sample space of $900$. – miradulo Jan 20 '16 at 15:05
• @DonkeyKong Thank you – Zero Jan 20 '16 at 15:07
Number of $3$ digit numbers which are divisible by $5$ :
To calculate this, note that there are $9$ choices for putting digits in the hundreds' place $[1-9]$, $10$ choices for putting digits in the tens' place $[0-9]$ and $2$ choices for putting digits in the units' place $[0,5]$. Hence total number of $3$ digit numbers which are divisible by $5$ is $9\times 10 \times 2 =180$.
However total number of $3$-digit numbers is $900$.
So the probability is $\large\frac{180}{900}=\frac{1}{5}$
The following is multiple choice question (with options) to answer.
What is the probability that a three digit number is divisible by 7? | [
"1/7",
"127/900",
"32/225",
"129/900"
] | C | Total 3 digit integers which are divisible by 7 :
(994-105)/7+1 = 128
The total number of 3 digit integers = (999-100)+1 = 900
The required probability : 128/900 = 32/225
answer:C. |
AQUA-RAT | AQUA-RAT-34747 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 46, the dividend is: | [
"5324",
"5334",
"5336",
"5356"
] | C | Divisor = (5 * 46) = 230
= 10 * Quotient = Divisor
=> Quotient = 230/10 = 23
Dividend = (Divisor * Quotient) + Remainder
Dividend = (230 * 23) + 46 = 5336.
C |
AQUA-RAT | AQUA-RAT-34748 | So if we are looking for the rational roots of $16x^2-24x-7=0$, the only possible candidates are $\pm\frac{1}{1}$, $\pm\frac{1}{2}$, $\pm\frac{1}{4}$, $\pm\frac{1}{8}$, $\pm\frac{1}{16}$, $\pm\frac{7}{1}$, $\pm\frac{7}{2}$, $\pm\frac{7}{4}$, $\pm\frac{7}{8}$, and $\pm\frac{7}{16}$. We can just try all these, to see which, if any, work. It isn't quite as unpleasant as it looks. However, it is not really a suitable approach to quadratic equations. After all, the roots might not be rational. And the Quadratic Formula is simple to use.
-
A full, very concise and understandable explanation. I need to practice a few more problems just to make sure but the concept seems clear. There are no words to express my gratitude. Thank you. – luclabs Nov 1 '11 at 3:21
The following is multiple choice question (with options) to answer.
If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5, then what is the value of m? | [
"-22.5",
"16",
"-10.5",
"-17.5"
] | D | Explanatory Answer
We know that the product of the roots of a quadratic equation ax2 + bx + c = 0 is
In the given equation, x2 + mx + 24 = 0, the product of the roots = = 24.
The question states that one of the roots of this equation = 1.5.
If x1 and x2 are the roots of the given quadratic equation and let x1 = 1.5.
Therefore, x2 == 16.
In the given equation, m is the co-efficient of the x term.
We know that the sum of the roots of the quadratic equation ax2 + bx + c = 0 is = -m
Sum of the roots = 16 + 1. 5 = 17 = -17.5.
Therefore, the value of m = -17.5
Answer D |
AQUA-RAT | AQUA-RAT-34749 | # In how many ways can I select 4 groups from 10 indistinguishable items where each selected group only contains either 0 or an odd number of items
If this question asked instead to select groups of even number of items, then I know how to approach it.
The even approach is: there is a bijection to a sequence of 8 bits (10/2+4-1), where each 0 correspond to $$two$$ items and 1's are partition lines between groups. For example, 01100001 would be groups of 2, 0, 8, and 0 items. The number of ways to select these $$even$$ groups is $${8 \choose 3}$$ because there are 8 bits, and 3 partitions.
But I'm confused what to do about the odd case. It's not correct to now assume that each 0 correspond to an $$odd$$ number of item, because then a multiple of 0's could be $$even$$ or $$odd$$.
How should this problem be approached?
• Are the groups labeled? Sep 30 '18 at 23:03
• groups are not labeled Oct 1 '18 at 0:36
First: Give one item to each group. So every group has one item, and we have last 6 items. Then by using your even approach, use three $$0$$ (every zero means two items) and three $$1$$, get $${6 \choose 3}$$. ( for example: in 010110 situation we distribute 2+1 items to 1st group, 2+1 items to 2nd group, 0+1 item the third one, and 2+1 items the last one. Obviously every group has odd many items)
What we do in the above is not enough since we do not calculate one of the groups has zero item stuation. So we will continue.
Second: Now we consider to give zero item to one group. So we take care the last three groups. Again we give one item each of one of the three groups. Then we have 7 items. So we can not distribute the items by using $$0$$ approach. It means the situation is impossible.
The following is multiple choice question (with options) to answer.
When toys are grouped into 12 and 18, always 3 toys remain ungrouped. What is the minimum number of toys to be added so that when they are grouped into 7, no toys will remain ungrouped ? | [
"0",
"1",
"2",
"3"
] | D | T= 12a+3 = 18b + 3
2a=3b --> min (a,b) = (3,2)
So minimum number of toys --> 12*3+3 = 39
When toys are grouped into 7, 4 will remain ungrouped so in order to make another group of 7 toys -->
7-4 = 3 toys at least to be added.
ANSWER: D |
AQUA-RAT | AQUA-RAT-34750 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
The expenditure of a businessman increase consistently by 10% per year. If his present expenditure is Rs. 20000 then what will his expenditure be after 2 years ? | [
"24500",
"24400",
"24300",
"24200"
] | D | Explanation :
Expenditure = 20000 x 1.1 x 1.1 = 24200
Answer : Option D |
AQUA-RAT | AQUA-RAT-34751 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
In a 500 m race,the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m.Then,A wins by | [
"60 m",
"40 m",
"20 m",
"10 m"
] | C | Solution
To reach the winning post A will have to cover a distance of ( 500 -140)m, i.e,360m.
While A covers 3 m, B covers 4 m.
While A covers 360 m, B covers (4/3 x 360)m =480m
Thus,when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.
∴ A wins by 20 m.
Answer C |
AQUA-RAT | AQUA-RAT-34752 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
Two pipes A and B can separately fill a cistern in 45 minutes and 60 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 40 minutes. In how much time, the third pipe alone can empty the cistern? | [
"90 min",
"100 min",
"72 min",
"75 min"
] | C | 1/40-(1/45+1/60)=-1/72
third pipe can empty in 72 minutes
ANSWER:C |
AQUA-RAT | AQUA-RAT-34753 | # Inequality with four positive integers looking for upper bound
Umm. This comes from Diophantine quartic equation in four variables and will finish the most important part if it can be done.
Four positive integers $w,x,y,z.$ One equation and two inequalities $$wxyz = (w+x+y+z)^2,$$ $$w \geq x \geq y \geq z \geq 1,$$ $$xyz \geq 2(w+x+y+z).$$
I am hoping for an upper bound. Since i made $w$ biggest, it would be an UPPER BOUND on $w.$ For example, I am running a computer program to find all such quadruples with $w \leq 1000.$
Sample question: is it true that $w \leq 1000?$
This is the method of Hurwitz 1907. I have a pdf. His techniques are almost right for this problem, to the point where i am already convinced that the answer to the question by hardmath comes out the exact same way.
EDIT, these imply easily that $$\color{green}{ x+y+z \geq w}.$$ Could be useful.
EDIT: almost forgot, these are what I believe to be all such quadruples:
w x y z xyz 2(w+x+y+z)
4 4 4 4 64 32
6 6 3 3 52 36
8 5 5 2 50 40
10 10 9 1 90 60
12 6 4 2 48 48
15 10 3 2 60 60
18 9 8 1 72 72
21 14 6 1 84 84
30 24 5 1 120 120
The following is multiple choice question (with options) to answer.
Which of the following CANNOT be the median W of the four consecutive positive integers w, x, y, and z, where w < x < y < z ? | [
"(w+x)/2 - 1",
"(w+z)/2",
"(x+y)/2",
"(y+z)/2 -1"
] | A | Using the properties of consecutive positive integers we could right away eliminate B,C and E.
(B) - The average of first and last terms is the median
(C) - For even number of integers, the median is the average of middle terms
(E) - The average is equal to median if the integers are consecutive
Remaining answer choices are A and D.
For A, the average of first two consecutive numbers (w,x) will definitely be less than x making it impossible to be a median W.
Answer (A) |
AQUA-RAT | AQUA-RAT-34754 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
The time it took car P to travel 600 miles was 2 hours less than the time it took car R to travel the same distance. If car P’s average speed was 10 miles per hour greater than that of car R, what was car R’s average speed, in miles per hour? | [
"40",
"50",
"60",
"70"
] | B | Let speed of car R be=x
Then speed of car P= x+10
A/Q,
(600/x)-(600/(x+10))=2
Solving for x=50 miles\hr
ANSWER:B |
AQUA-RAT | AQUA-RAT-34755 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Carrie likes to buy t-shirts at the local clothing store. They cost $8.75 each. One day, she bought 18 t-shirts. How much money did she spend? | [
"$150",
"$70",
"$200",
"$171.6"
] | B | $8.75*18=$70. Answer is B. |
AQUA-RAT | AQUA-RAT-34756 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
Sides of a rectangular park are in the ratio 3: 2 and its area is 3750 sq m, the cost of fencing it at 50 ps per meter is? | [
"Rs.122",
"Rs.129",
"Rs.125",
"Rs.120"
] | C | 3x * 2x = 3750 => x = 25
2(75 + 50) = 250 m
250 * 1/2 = Rs.125
Answer:C |
AQUA-RAT | AQUA-RAT-34757 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
The compound and the simple interests on a certain sum at the same rate of interest for two years are Rs.11730 and Rs.10200 respectively. Find the sum. | [
"Rs.17277",
"Rs.17000",
"Rs.17287",
"Rs.172827"
] | B | Explanation:
The simple interest for the first year is 10200/2 is Rs.5100 and compound interest for first year also is Rs.5100. The compound interest for second year on Rs.5100 for one year
So rate of the interest = (100 * 1530)/ (5100 * 1) = 30% p.a.
So P = (100 * 10200)/ (30 * 2) = Rs.17000
Answer: B |
AQUA-RAT | AQUA-RAT-34758 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B? | [
"1 years",
"20 years",
"10 years",
"15 years"
] | C | Explanation:
Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence, B's age = 2x = 10 years.
Answer: C |
AQUA-RAT | AQUA-RAT-34759 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are 40 students in a certain geometry class. If two thirds of the students are boys and three fourths of the boys are under six feet tall, how many boys in the class are under six feet tall? | [
"6",
"12",
"20",
"24"
] | C | 40*2/3 *3/4= 20
Answer: C |
AQUA-RAT | AQUA-RAT-34760 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A dishonest dealer professes to sell goods at the cost price but uses a weight of 900 grams per kg, what is his percent? | [
"28%",
"25%",
"11.11%",
"28%"
] | C | 900 --- 100
100 --- ? => 11.11%
Answer: C |
AQUA-RAT | AQUA-RAT-34761 | To fill it completely we need 2 (jugs needed) + 5 (remaining jugs) = 7 (one carton's capacity)
Hence (B) 2 jugs
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Re: A worker carries jugs of liquid soap from a production line to a packi [#permalink]
### Show Tags
08 Apr 2018, 06:05
1
Since the worker, carried jugs of liquid soap(4 jugs/trip)
In 17 trips, he would carry 68 jugs of liquid soap.
Since he can fill 7 jugs in a carton, he will need to add 2 more jugs
such that he has 70 jugs and he can fill the jugs in 10 cartons.
Hence he needs 2 jugs(Option B) such that the last partially filled carton is full.
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A worker carries jugs of liquid soap from a production line to a packi [#permalink]
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The following is multiple choice question (with options) to answer.
A grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each soap box is 7 inches by 6 inches by 6 inches, then what is the maximum number of soap boxes that can be placed in each carton? | [
"210",
"252",
"250",
"300"
] | C | however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers
Dimensions of the box =25*42*60
Dimensions of the soap = 6*6*7
placing the 7 inch side along 42 inch side we get 6 soaps in a line and in a similar way 5 along 25 and 6 along 60
we get = 5x5x10 = 250
so the question is why this particular arrangement, in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number
the answer is (C) |
AQUA-RAT | AQUA-RAT-34762 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
If x is an odd negative integer and y is an even integer, which of the following statements must be true?
I. (3x - 4y) is odd
II. xy^2 is an even negative integer
III. (y^2 - x) is an odd negative integer | [
"(3X-4Y) IS ODD TRUE",
"II only",
"I and II",
"I and III"
] | A | If x is an odd negative integer and y is an even integer, which of the following statements must be true?
I. (3x - 4y) is odd
3x - is always negative and odd
2y - can be 0, when y=0, or always even integer (positivie or negative)
odd integer +/- even integer = always odd
I statement is always true
II. xy^2 is an even negative integer
y can be 0 => xy^2 = 0 (non-negative even) => II statement is not true
III. (y^2 - x) is an odd negative integer
y can be 0 => -x will be positive => III statement is not true
basing on above, only I statement will always be true
answer is A |
AQUA-RAT | AQUA-RAT-34763 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is? | [
"100",
"150",
"140",
"130"
] | B | Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m.
Answer: Option B |
AQUA-RAT | AQUA-RAT-34764 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are moving at 50 kmph and 70 kmph in opposite directions. Their lengths are 150 m and 100 m respectively. The time they will take to pass each other completely is? | [
"7 1/2 sec",
"2 sec",
"3 sec",
"5 sec"
] | A | 70 + 50 = 120 * 5/18 = 100/3 mps
D = 150 + 100 = 250 m
T = 250 * 3/100 = 15/2 = 7 ½ sec
ANSWER A |
AQUA-RAT | AQUA-RAT-34765 | $\Rightarrow AC+CB=4(AB)$
$\Rightarrow (AB+BC)+CB=4(AB)$
$\Rightarrow BC+CB = 3(AB)$
$\Rightarrow BC=\dfrac{3}{2}(AB)$ -------- $(1)$
Similarly by the time Dinesh reaches point $D$ from $C$ walking, Mukesh and Suresh reach $D$ riding bike.
Here also distance travelled by bike $(=BD)$ is 4 times the distance travelled on foot $(=CD)$
$\Rightarrow CB+(BD)=4(CD)$
$\Rightarrow BC+(BC+CD)=4(CD)$
$\Rightarrow 2(BC)= 3(CD)$
$\Rightarrow CD= \dfrac{2}{3}(BC)$ -------- $(2)$
Now, it is given that total distance is given as $300 \text{ km}$
$\Rightarrow AB+BC+CD=300$
Using values from, equations $(1)$ and $(2)$,
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}(BC)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}\times \dfrac{3}{2}(AB)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+AB=300$
$\Rightarrow \dfrac{7}{2}(AB)=300$
$\Rightarrow AB=\dfrac{600}{7}$
So, $BC$
$=\dfrac{3}{2}(AB)$
$=\dfrac{3}{2}\times \dfrac{600}{7}$
$=\dfrac{900}{7}$
Similarly, $CD$
$=\dfrac{2}{3}(BC)$
The following is multiple choice question (with options) to answer.
The jogging track in a sports complex is 900 m in circumference. Sunil and Sachin start from the same point and walk in opposite directions at 5.4 km/hr and 3.6km/hr respectively. They will meet for the first time in? | [
"2.98 min",
"5.7 min",
"6.0 min",
"6.3 min"
] | C | Clearly, the two will meet when they are 900m apart.
To be (5.4+ 3.6) = 9.0 km apart, they take 1 hour.
To be 900 m apart, they take
= (900/9000 * 60) min
= 6.00 min.
Answer: C |
AQUA-RAT | AQUA-RAT-34766 | Your are trying to understand important instructions from someone who is soft-spoken. From your distance $r_1$ her voice sounds like an average whisper of $20\, {\rm dB}$. So you move to a position where you are a distance $r_2$ from her and the sound level is at $60\, {\rm dB}$.
(a) If the intensity is $I_1$ at distance $r_1$ and $I_2$ at distance $r_2$, calculate the ratio $I_2/I_1$.
(b) Use your answer to part (a) to calculate the ratio $r_2/r_1$.
(a) The intensity level ($\beta$) of the sound waves is measured in ${\rm dB}$ by the following equation
$\beta=10\,{\log \frac{I}{I_0}\ }$
Where $I$ is the intensity of the source of sound at distance $r$ and $I_0$ is the hearing threshold. So find the ratio $I_2/I_1$ as follows
$\beta_1=10\,{\log \frac{I_1}{I_0}\ }\Rightarrow 20=10\,{\log \frac{I_1}{I_0}\ }\Rightarrow \frac{I_1}{I_0}={10}^2$
$\beta_2=10\,{\log \frac{I_2}{I_0}\ }\Rightarrow 60=10\,{\log \frac{I_2}{I_0}\ }\Rightarrow \frac{I_2}{I_0}={10}^6$
We have used the definition of the logarithm i.e. $y={\log x\ }\Rightarrow x={10}^y$. Dividing the two relation, we obtain
$\frac{I_2}{I_1}=\frac{{10}^6}{{10}^2}={10}^4$
The following is multiple choice question (with options) to answer.
A trainer is standing in one corner of a square ground of side 25m. His voice can be heard upto 140m. Find the area of the ground in which his voice can be heard? | [
"12300",
"14500",
"15400",
"16700"
] | C | Area covered by goat = Pi*r^2/4 ( here we divide by 4 because the trainer is standing in a corner of the ground and only in 1/4 part , the voice can be heard)
where r= 14 m = length reaching the voice
So area = (22/7)*140*140/4 = 15400 sq m
answer :C |
AQUA-RAT | AQUA-RAT-34767 | Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Originally posted by EgmatQuantExpert on 13 Jul 2018, 06:28.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:04, edited 2 times in total.
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Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
### Show Tags
Updated on: 18 Jul 2018, 12:09
3
EgmatQuantExpert wrote:
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
Let the tank = 60 liters.
Since P takes 3 hours to fill the 60-liter tank, P's rate $$= \frac{60}{3} = 20$$ liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate $$= \frac{60}{4} = 15$$ liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate $$= \frac{60}{-5} = -12$$ liters per hour.
Since R works to EMPTY the tank, R's rate is negative.
The following is multiple choice question (with options) to answer.
Pipe A can fill a tank in 6 hours. Due to a leak at the bottom, it takes 9 hours for the pipe A to fill the tank. In what time can the leak alone empty the full tank? | [
"17",
"16",
"18",
"77"
] | C | Explanation:
Let the leak can empty the full tank in x hours 1/6 - 1/x = 1/9
=> 1/x = 1/6 - 1/9 = (3 - 2)/18 = 1/18
=> x = 18.
Answer: C |
AQUA-RAT | AQUA-RAT-34768 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
What is the difference between the compound interests on Rs. 5000 for 11â„2 years at 4% per annum compounded yearly and half-yearly? | [
"2.04",
"4.8",
"3.06",
"8.3"
] | A | compunded yearly
= 5000*(1+4/100)*(1+1/2*4/100)
=5304
compound interest for 1 1/2 years when interest=>(5304-5000)
=>P(1+(R/2)/100)^2T = 5000(1+(4/2)/100)2*3/2 = 5000(1+2/100)^3
=5000(102/100)^3 = 5306.04
1 1/2 Years half yearly = (5306.04 - 5000)=> 2.04
ANSWER A |
AQUA-RAT | AQUA-RAT-34769 | homework-and-exercises, pressure, fluid-statics
Title: Which tank fills up first? Which tank would fill first. My first guess was 3 and 4 simultaneously due to Pascal's Law of pressure distribution. Then tank 2 and then 1. Could you please help? This is my first question ever on Stack Exchange. Tank 1 will to the level of the pipe. Then water will flow into 2.
If the pipe is blocked, 2 will fill. When the water in 2 reaches the level of the upper pipe, tanks 1 and 2 will stay even with each other. When tank 2 reaches the top, water will spill out. It ends there.
If the pipe to 2 is open, tank 2 will fill to the level of the lower pipe. Then water will flow into 3. Water in tank 3 will stay even with the level in the pipe to 4.
It looks like the level of the upper part of both pipes from 3 are the same. When the level in 3 rises to the pipes, water will begin to spill into 4. When 4 is full up to the pipe, the level will rise in 2, 3, and 4 until it spills over the top of 3 and 4.
The following is multiple choice question (with options) to answer.
A water tank is four-fifth full.Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes.If both the pipes are open,how long will it take to empty or fill the tank completely? | [
"12 min.to empty",
"9 min.to empty",
"5 min.to empty",
"4 min.to empty"
] | A | Explanation:
Clearly,pipe B is faster than pipe A and so,the tank will be emptied.
part to be emptied = 4/5
part emptied by (A+B) in 1 minute=(1/6-1/10)=1/15
so, the tank will be emptied in 12 min
Answer: A |
AQUA-RAT | AQUA-RAT-34770 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A garrison of 2000 men has provisions for 54 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement? | [
"1977",
"1893",
"1979",
"1900"
] | D | 2000 ---- 54
2000 ---- 39
x ----- 20
x*20 = 2000*39
x = 3900
2000
-------
1900
Answer: D |
AQUA-RAT | AQUA-RAT-34771 | Then, $$\sum_{k = 0}^{n}{n + k \choose k}2^{n - k} = 2^{n}\,\mrm{f}\pars{1 \over 2} = \bbx{\large 4^{n}}$$
The following is multiple choice question (with options) to answer.
If 2^k = 4, then 2^(2K+2) = | [
"29",
"45",
"81",
"64"
] | D | 2^k=4
2^2k=4^2
2^2k=16
2^(2k+2)
=2^2k*2^2
=16*4
=64
Answer: D |
AQUA-RAT | AQUA-RAT-34772 | This series of cash flows will yield exactly 10 % is $56.07 scenario... Stan also wants his son to be paid or received in the future amount that you expect receive! Is extremely important in many financial calculations 510.68 ; discount rate. discount rate is investment! Money received in the discussion above, we looked at one investment over the course of year! Now knows all three variables for the first offer suggests the value of$ 100 today or can... For the four discount rates trend, the amount $100 being$... Would not have realized a future sum the applicable discount rate. a present value formula shown! Costs, inflation will cause the price they pay for an investment might earn example, future... ; discount rate or the interest rate ” is used in the future cashflows expected from an investment might.... That with an initial investment of exactly $100 today or I can pay you back 100! Periods interest rates rise and the CoStar product suite often used as the present value provides a basis assessing... Today, you can buy goods at today 's prices, i.e must... Earned on the funds over the next five years time refers to future value and a... If you receive money today, you can buy goods at today 's prices, i.e from now 1 4... Aone-Size-Fits-All approach to determining the appropriate discount rate is used when referring to present. Discount lost business profits to a present value, the future value of cash flows will yield exactly %... Than$ 1,000 five years time value takes the future receiving $1,000 five years discount rate present value not. The concept that states an amount for any timeframe other than one.! More Answers money worth in today ’ s lost earnings the idea of net present value money...: present value becomes equal to the present value of money that is expected to arrive at a time. This Table are from partnerships from which investopedia receives compensation states that an amount of money that expected! Financial planning formula given below PV = CF / ( 1 + r ) t 1 of. Can pay you back$ 100 today or I can pay you $110 year. Bob knows the future amount that you expect to earn a rate return... 5,000 lump sum payment in five years from now % ) 3 2 earn. Bob gets up and says, “ I
The following is multiple choice question (with options) to answer.
The true discount on Rs.2562 due 4 months hence is Rs.122.The rate % is? | [
"12%",
"13 (1/3)%",
"15%",
"14%"
] | A | Explanation:
P.W = Rs.( 2562 - 122) = Rs.2440
S.I on Rs.2440 for 4 months is Rs.122
Rate = [100X 122/2440X1/3 ] % = 15%
Answer: A |
AQUA-RAT | AQUA-RAT-34773 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
Find the compound interest on $1200 for 5 years at 20% p.a. if CI is component yearly? | [
"$120",
"$150",
"$1786",
"$250"
] | C | A = P(1 + R/100)^T
= 1200(1 + 20/100)^5
=$2986
CI = $1786
Answer is C |
AQUA-RAT | AQUA-RAT-34774 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
An article is bought for Rs.675 and sold for Rs.700, find the gain percent? | [
"2%",
"3%",
"4%",
"5%"
] | C | 675 ---- 25
100 ---- ? =>
=4%
Answer:C |
AQUA-RAT | AQUA-RAT-34775 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
0, 180, 12, 50, 100, 200, ?, 3, 50, 4, 25, 2, 6, 30, 3 | [
"150",
"175",
"225",
"250"
] | A | 90, 180, 12, 50, 100, 200, ?, 3, 50, 4, 25, 2, 6, 30, 3
I think 1st term will be 90
two series
90, 180, 12, 50, 100, 200, ?,
and 3, 50, 4, 25, 2, 6, 30, 3
3*50=150
50*4=200
4*25=100
25*2=100
2*6=12
6*30=180
30*3=90
so ans is 150
ANSWER:A |
AQUA-RAT | AQUA-RAT-34776 | # Random Gift Giving at a Party - Combinatorics Problem
Each of $$10$$ employees brings one (distinct) present to an office party. Each present is given to a randomly selected employee by Santa (an employee can get more than one present). What is the probability that at least two employees receive no presents?
Firstly, there are $$10^{10}$$ total ways to give the $$10$$ employees the $$10$$ presents. So this is our denominator.
My attempt was to consider the complement and consider the number of ways that either $$0$$ employees receive no presents (every employee gets a present) or $$1$$ employee receives no present.
Case 1: $$0$$ employees
There are $$10$$ employees and $$10$$ presents. So there are $$10^{10}$$ ways to give the presents.
Case 2: $$1$$ employee
Step 1: Decide which employee receives no presents: $$10$$ possibilities.
Step 2: Distribute the $$10$$ presents to the remaining $$9$$ employees: $$9^{10}$$ ways.
So the number of ways in which at least $$2$$ employees receive no presents is: $$1-(10^{10}+9^{10}$$).
So my final answer is: $$1-\displaystyle\frac{(10^{10}+9^{10})}{10^{10}}$$.
However, this answer does not match the answer in my textbook. Which is: $$1-\displaystyle\frac{10!-10\times 9 \times \frac{10!}{2!}}{10^{10}}$$
Where did my attempt go wrong and how can I correct it?
The following is multiple choice question (with options) to answer.
At a particular graduation party with 220 guests, 70% of the guests brought gifts, and 40% of the female guests brought gifts. If 12 males did not bring gifts to the party, how many females did bring gifts? | [
"18",
"36",
"42",
"68"
] | B | the correct method
total = 220..
70% of 220= 154 got gifts..
66 did not get gift, out of which 12 are males, so remaining 60-12=54 are females..
But 40% females brought gift, so 60% did not get it..
so 60% = 54, 100%= 54*100/60= 90..
ans 40% of 90 =36
B |
AQUA-RAT | AQUA-RAT-34777 | algorithms
Title: In the Gas-Up problem, how is the amount of gas the same up to a cyclic shift regardless of starting city? I'm working through Elements of Programming Interviews as practice for finding a job. I've spent a ridiculous amount of time on Problem 18.7.
In the gas-up problem, $n$ cities are arranged on a circular road. You
need to visit all of the $n$ cities and come back to the starting
city. A certain amount of gas is available at each city. The total
amount of gas is equal to the amount of gas required to go around the
road once. Your gas tank has unlimited capacity. Call a city $c$
ample if you can begin at $c$ with an empty tank, refill at it, then travel through each of the remaining cities, refilling at each, and
return to $c$, without running out of gas at any point. See Figure
18.3 for an example. [You can see Figure 18.3 on the Google Books result if you search for "Elements of Programming Interviews Problem
18.7.]
Given an instance of the gas-up problem, how would you efficiently [i.e. in $\mathcal{O}(n)$ time or better]
compute an ample city, if one exists?
I couldn't figure it out. I read the hint, and I still couldn't figure it out. I gave up, and read the answer in the back of the book. I still couldn't figure it out. I searched Google and this site for a more complete solution, but didn't find one.
This is the part that's getting me:
On closer inspection, it becomes apparent that the graph of the amount
of gas as we perform the traversal is the same up to a cyclic shift
regardless of the starting city.
The following is multiple choice question (with options) to answer.
A car traveled 480 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city? | [
"14",
"16",
"21",
"22"
] | A | Let the speed in highway be h mpg and in city be c mpg.
h = c+6
h miles are covered in 1 gallon
462 miles will be covered in 462/h.
Similarly c miles are covered in 1 gallon
336 miles will be covered in 336/c.
Both should be same (as car's fuel capacity does not change with speed)
=> 336/c = 480/h
=> 336/c = 480/(c+6)
=> 336c+336*6=480c
=>c=336*6/144 =14
Answer A. |
AQUA-RAT | AQUA-RAT-34778 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
Ramesh has solved 108 questions in an examination. If he
got only ‘0’ marks, then how many questions were wrong
when one mark is given for each one correct answer and 1/3
mark is subtracted on each wrong answer. | [
"78",
"79",
"80",
"81"
] | D | If Ramesh attempts 'x' questions correct and 'y' questions wrong, then
x + y=108 ---(i) &
x - (1/3)y=0 ---(ii)
On solving x=27, y=81
ANSWER:D |
AQUA-RAT | AQUA-RAT-34779 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
Sara bought both German chocolate and Swiss chocolate for some cakes she was baking. The Swiss chocolate cost $2.5 per pound, and German chocolate cost $1.5 per pound. if the total the she spent on chocolate was $30 and both types of chocolate were purchased in whole number of pounds, how many total pounds of chocolate she purchased? | [
"7",
"18",
"10",
"12"
] | B | If there were all the expensive ones 2.5....
There would be 30/2.5 or 12 of them but since 1.5$ ones are also there, answer has to be >12....
If all were 1.5$ ones, there will be 30/1.5 or 20...
So ONLY 18 is left
Ans B.. |
AQUA-RAT | AQUA-RAT-34780 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec. What is the length of the platform? | [
"340",
"350",
"360",
"370"
] | B | Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/39 = 50/3
3x + 900 = 1950 => x = 350 m.
Answer: Option B |
AQUA-RAT | AQUA-RAT-34781 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
In a river flowing at 2 km/hr, a boat travels 72 km upstream and then returns downstream to the starting point. If its speed in still water be 6 km/hr, find the total journey time. | [
"10 hours",
"12 hours",
"14 hours",
"27 hours"
] | D | Explanation :
speed of the boat = 6 km/hr
Speed downstream = (6+2) = 8 km/hr
Speed upstream = (6-2) = 4 km/hr
Distance travelled downstream = Distance travelled upstream = 72 km
Total time taken = Time taken downstream + Time taken upstream
= (72/8) + (72/4) = 27 hr. Answer : Option D |
AQUA-RAT | AQUA-RAT-34782 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Each week a restaurant serving Mexican food uses the same volume of chili paste, which comes in either 25-ounce cans or 15-ounce cans of chili paste. If the restaurant must order 30 more of the smaller cans than the larger cans to fulfill its weekly needs, then how manysmallercans are required to fulfill its weekly needs? | [
"60",
"70",
"75",
"100"
] | C | let x be the number of 25 ounce cans. Therefore (x+30) is the number of 15 ounce cans.
Total volume is same, therefore
25x=15(x+30)
10x=450
x=45
Therefore, number of 15 ounce cans=45+30=75
Ans - C |
AQUA-RAT | AQUA-RAT-34783 | Square [1]: . $(x+y)^2 \:=\:1^2 \quad\Rightarrow\quad x^2+2xy + y^2 \:=\:1$
$\text{We have: }\;x^2 + y^2 + 2(xy) \:=\:1$
. . . . . . . . . . . . . . $\uparrow$
. . . . . . . . . . . . $^{\text{This is }\text{-}6}$
Therefore: . $x^2+y^2 - 12 \:=\:1 \quad\Rightarrow\quad \boxed{x^2+y^2 \:=\:13}$
This method is guarenteed to impress/surprise/terrify your teacher.
.
4. Originally Posted by Soroban
Hello, Joel!
Welcome aboard!
I have a back-door approach . . .
I like your solution. I wish I had thought about the problem a bit more before going down the obvious path.
The following is multiple choice question (with options) to answer.
If x^2 + y^2 = 2xy, then (x+y)^2 = | [
"x^2",
"3x^2",
"4xy",
"2y^2"
] | C | Its B
x^2 +y^2 = 2xy
Add 2xy on both sides we get
x^2 +y^2+2xy = 2xy+2xy
(x+y)^2=4xy
Correct Option : C |
AQUA-RAT | AQUA-RAT-34784 | a) 22
b) 35
c) 56
d) 70
e) 105
Select 3 out of 5 = 5c3 = 5
Select 1 out of 7 = 7c1 = 7
Total = 5c3 x 7c1 = 5x7 = 35
_________________
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
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Math Expert
Joined: 02 Sep 2009
Posts: 42280
Kudos [?]: 132893 [4], given: 12391
### Show Tags
27 Dec 2009, 17:54
4
KUDOS
Expert's post
GMAT TIGER wrote:
dk94588 wrote:
This is a section I have had a lot of trouble with, and I was hoping someone could show me how to do this.
Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?
a) 22
b) 35
c) 56
d) 70
e) 105
Select 3 out of 5 = 5c3 = 5
Select 1 out of 7 = 7c1 = 7
Total = 5c3 x 7c1 = 5x7 = 35
Just a little correction 5C3=10 not 5. So, 5C3*7C1=10*7=70. (5C3 # of selections of 3 women out of 5 and 7C1 # of selections of 1 man out of 7)
_________________
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Posts: 64
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### Show Tags
28 Dec 2009, 00:02
5c3*7c1=70
but come on 75% jobs for females? where is the right to equality
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The following is multiple choice question (with options) to answer.
A company D has 30 percent of the employees are secretaries and 20 percent are salespeople. If there are 60 other employees of Company D, how many employees does Company D have? | [
"120",
"162",
"180",
"152"
] | A | Let the total number of employees in the company be x
% of secretaries = 30%
% of salespeople = 20%
% of of employees other than secretaries and salespeople = 100 - 50 = 50%
But this number is given as 60
so 50% of x = 60
x = 120
Therefore there a total of 120 employees in the company D
Correct answer - A |
AQUA-RAT | AQUA-RAT-34785 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A tradesman by means of his false balance defrauds to the extent of 20%? in buying goods as well as by selling the goods. What percent does he gain on his outlay? | [
"74%",
"49%",
"44%",
"74%"
] | C | g% = 20 + 20 + (20*20)/100
= 44%
Answer:C |
AQUA-RAT | AQUA-RAT-34786 | $P(-5) = 100,000 * (1.0594631)^{-5} \approx 74,915$.
=====
Alternatively
$(1 + rate)^{12} = 2$. What is $(1 + rate)^5$?(1+rate)^{12} = 2 \implies rate = .0594631$so$(1+rate)^5 = 1.594631^5 = 1.3348398541700343648308318811845$So from$1995$to$2000$the population grew by a factor of$1.3348398541700343648308318811845$. Or$x*1.3348398541700343648308318811845 = 100,000$so$x= \frac {100,000}{1.3348398541700343648308318811845} = 74,915\$
• Thank you very much. I'm sorry but I don't have enough reputation to declare your answer as useful but of course, it is. – Bachir Messaouri Dec 11 '17 at 22:16
The following is multiple choice question (with options) to answer.
Find the value of p: 5p^2-12p+1 = 20.25 | [
"4.5",
"3.5",
"20.25",
"6.2"
] | B | Using the elimination method substitute options for x and find the correct option.
Answer: B |
AQUA-RAT | AQUA-RAT-34787 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
The price of a shirt is increased by 15% and then reduced by 15%. The final price of the shirt is? | [
"$97.75",
"$90",
"$91.56",
"$95.62"
] | A | Let the original price be $100
New final price = 85% of (115% of $100) = 85/100 * 115/100 * 100 = $97.75
Answer is A |
AQUA-RAT | AQUA-RAT-34788 | # an exam has 50 multiple choice questions with 5 options
An exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that "a student" has no prior knowledge and randomly guess on all questions exam,
1. Compute the expected mean for the student score
2. Compute the standard deviation for the student score
3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?
4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam : What is the expected success rate? How do you expect the proportion of students who will score less or equal to 20?
If you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?
a. What is the expectation of this student's degree?
b. what is the standard deviation of this student's grade?
b. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?
1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct
• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz May 21 at 0:59
• thanks I did it – Nidal May 21 at 1:29
It seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.
The following is multiple choice question (with options) to answer.
The average marks in mathematics scored by the pupils of a school at the public examination were 39. If four of these pupils who actually scored 5, 12, 15 and 19 marks at the examination had not been sent up, the average marks for the school would have been 44. Find the number of pupils sent up for examination from the school? | [
"33",
"25",
"66",
"72"
] | B | 39x = 5 + 12 + 15 + 19 + (x – 4)44
x = 25
Answer: B |
AQUA-RAT | AQUA-RAT-34789 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph, find his average speed for the total journey? | [
"3",
"4",
"8",
"9"
] | B | M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4
Answer:B |
AQUA-RAT | AQUA-RAT-34790 | homework-and-exercises, kinematics
Title: Train problem question in kinematics engine of a train moving with uniform acceleration passes an electric pole with velocity u and the last
compartment with velocity v. The middle point of the train passes past the same pole with a velocity of.
My thinking:
Q1 Will the values of v and u be constant since they can’t pass it with different velocities because they are one thing.
Q2 Shouldn’t the middle part also cover it with u velocity.?
Will the values of v and u be constant since they can’t pass it with different velocities because they are one thing.
You are right in saying that two ends of a same rigid body can't have different values of speed but you should note that your line is correct only if you are talking about the speed of engine and the last compartment at a given point of time.
In your question , the engine passes the pole with speed $u$ and note that since the train can't elongate or compress , the speed of the middle part as well as the last compartment at that instant is $u$ .
When the last compartment reaches the pole, in that time interval the train has accelerated (first line of your question). So, at that instant, the speed of the engine , the middle part and the last compartment is $v$.
So your first question is just a mere confusion.
For the center of the train to pass the pole , the train has to travel for some time and since it is an accelerated motion , the speed with which it passes the pole is different than $u$.
Note: The value of speed with which the center passes the pole can be calculated using the three equation of motion involving constant acceleration :
$$ v = u + at$$
$$s = ut + \frac{1}{2} a t^2$$
$$v^2 = u^2 + 2as$$
Hope it helps .
The following is multiple choice question (with options) to answer.
In what time will a train 100 m long cross an electric pole, it its speed be 144 km/hr? | [
"2.5 sec",
"2.9 sec",
"3.5 sec",
"7.5 sec"
] | A | Speed = 144 * 5/18 = 40 m/sec
Time taken = 100/40
= 2.5 sec.
Answer: A |
AQUA-RAT | AQUA-RAT-34791 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
A man bought 20 shares of Rs. 50 at 5 discount, the rate of dividend being 13. The rate of interest obtained is: | [
"12\t1/2%",
"13\t1/2%",
"15%",
"16\t2/3%"
] | C | there is a small correction in this question. the rate of dividend is 13 1/2.
replace 13 in the solution by 13 1/2.
the ans is 15%
ANSWER:C |
AQUA-RAT | AQUA-RAT-34792 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
The ratio of the present age of Sandy to that of Molly is 9:2. Sixteen years from now, the ratio of the ages of Sandy to Molly will be 5:2. What was Sandy's age 6 years ago? | [
"30",
"36",
"42",
"48"
] | D | Let the present age of Sandy be 9x years and that of Molly be 2x years.
(9x + 16) / (2x + 16) = 5 / 2
8x = 48
x = 6
Six years ago, Sandy's age was 9(6) - 6 = 48
The answer is D. |
AQUA-RAT | AQUA-RAT-34793 | We see that the last digits form a 4-step cycle: . $7, 9, 3, 1$
We find that: . $7^7 \:=\:823,543 \:=\:4(205,885) + 3$
Therefore, the last digit of $7^{7^7}$ is the same as $7^3$: .3
Thanks for this answer, the only thing is how do I get the deduction that $7^{7^7}$ is the same as $7^3$?
7. ## Re: The last digit of 7^7^7
Really? We've done this twice!
The last digit ("last digit" means mod 10) of powers of 7:
7
9
3
1
7
9
3
1
.
.
.
7^3 has last digit 3
7^7 has last digit 3
7^11
7^15
7^19
.
.
.
7^10000000000000000000000000000000003 will also have last digit equal to three, since the exponent is congruent to 3 (mod4).
8. ## Re: The last digit of 7^7^7
This problem is not hard.
We see $7 \equiv -1 \pmod{4} \Rightarrow 7^7 \equiv -1 \pmod{4}$.
Then let $7^7=4k+3$.
So $7^{7^7}=7^{4k+3}=(7^4)^k.7^3$.
From here it is easy to find next.
,
,
,
,
,
,
,
,
,
,
,
,
,
,
# find last two digits of 7^7^7
Click on a term to search for related topics.
The following is multiple choice question (with options) to answer.
What is the tens digit of 7^1213? | [
"0",
"1",
"2",
"3"
] | A | 7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649
We should see this as pattern recognition . We have a cycle of 4 . (We can multiply the last 2 digits only as we care about ten's digit )
0 , 4 , 4 , 0 .
1213= 4*304 + 1
The ten's digit will be 0 .
Answer A |
AQUA-RAT | AQUA-RAT-34794 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
Director
Joined: 25 Jul 2018
Posts: 731
A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
Posted from my mobile device
Stern School Moderator
Joined: 26 May 2020
Posts: 268
Concentration: General Management, Technology
WE: Analyst (Computer Software)
Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
A car engine is half filled and holds 15 litres of petrol.what fraction of the engine is full if it contains 18 litres of petrol? | [
"1/4",
"2/3",
"2/5",
"3/5"
] | D | half of the car engine filed with 15 ltrs means full engine filed with 30 ltrs so the fraction will be=18/30=6/10=3/5
ANSWER:D |
AQUA-RAT | AQUA-RAT-34795 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
How many pairs of letters in the word 'CHAIRS' have as many letters between them in the word as in the alphabet? | [
"2",
"1",
"3",
"5"
] | A | Letters in the word Letters in the alphabet
C H A C B A
R S R S
ANSWER A |
AQUA-RAT | AQUA-RAT-34796 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
For integers x and y if x^2+x/y is always an odd integer then which of the following must be true? | [
"x is even ; y is odd",
"x is even ; y is even",
"x is odd ; y is odd",
"x is odd ; y is even"
] | B | Given x and y are integers...NOTE : While taking values, x/y must be properly divisible and the result has to integer itself..
x^2+x/y = odd always...
lets take x even (2) and y even (2)
4 + 2/2 = odd...
lets take x odd(3) and y odd(3)
9+3/3 = even...
So only x even and y even satisfies the equation..
If we take one even and other odd then we get x/y as non integer..which shouldn't be the case as per our question stem.
Option B is correct... |
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