source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34897 | I'll have a go and answer this the maths-lite way (though there are a number of answers with more mathematic rigor and .. dare I say it vigor posted here already).
Note that there is:
• 1 result with a face value 1
• 3 results with a face value 2,
• 5 results with a face value 3,
• 7 results with a face value 4,
• 9 results with a face value 5, and
• 11 results with a face value 6
The Average is defined to be: $$\text{Average} = \frac{\text{Sum of the Results}}{\text{Total number of Results}}$$
The Sum of the Results is: $$\begin{eqnarray} \text{Sum} &=& (1 \times 1) + (3 \times 2) + (5 \times 3) + (7 \times 4) + (9 \times 5) + (11 \times 6) \nonumber \\ &=& 1 + 6 + 15 + 28 + 45 + 66 \nonumber \\ &=& 161 \nonumber \end{eqnarray}$$
The Total number of Results is: $6 \times 6 = 36$
So the Average is: $$\text{Average} = \frac{161}{36} \approx 4.472$$
-
This is very much delayed, but consider the case with an $n$-sided die. As has already been observed, the expected value of the maximum of two $n$-sided die is
$${1 \over n^2} \sum_{k=1}^n (2k^2-k)$$
and we can write out this sum explicitly. In particular, we can expand to get
$${1 \over n^2} \left( \left( 2 \sum_{k=1}^n k^2 \right) - \sum_{k=1}^n k \right)$$ and recalling the formulas for those sums, this is
$${1 \over n^2} \left( {2n(n+1)(2n+1) \over 6} - {n(n+1) \over 2} \right)$$
or after some rearrangement
The following is multiple choice question (with options) to answer.
The average of 25 results is 24. The average of first 12 of those is 14 and the average of last 12 is 17. What is the 13th result? | [
"740",
"228",
"690",
"780"
] | B | Solution: Sum of 1st 12 results = 12*14
Sum of last 12 results = 12*17
13th result = x (let)
Now,
12*14+12*17+x = 25*24
Or, x = 228.
Answer: Option B |
AQUA-RAT | AQUA-RAT-34898 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
In a consumer survey, 65% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? | [
"30",
"10",
"15",
"20"
] | A | pick 100 and x= all set with exactly two items
65=30+50+20 -(X)-10
X=25
so the answer is 25+5/100= 30%
Answer A |
AQUA-RAT | AQUA-RAT-34899 | Here, since counting the number of total fish in the pond is hard, they tagged 50 and let them disperse evenly in the population. Then they caught 50 and found 2 to be tagged. So approximately 4% of the fish were tagged. So 50 is 4% of the entire fish population of the pond. Note that the method uses huge approximation because of the small sample number. If 1 more tagged fish were caught among the 50, it would change the approximated fish population number by a huge amount. But they have given us that "the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond" so we can make this approximation.
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
### Show Tags
27 Mar 2016, 12:49
Is it a 600 level question or more?
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
### Show Tags
21 Feb 2017, 02:30
Hi can you help me with this problem? The last part of the problem It is not clear for me. I don't know if I have to use 50 fish or 48 fish still in the pond
- 2 fish tagged / 50 in the second catch
- 48 fish are still in the pond
So the equation is:
48 fish tagged in the pond/Tot in the pond = 2 fish tagged out/50 fish catched
Tot in the pond = 48*2/50 = 1200 (aprox. 1250) --> answer is C
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
### Show Tags
30 Apr 2017, 11:41
Sachin9 wrote:
so, x * 4% = 50
The following is multiple choice question (with options) to answer.
At the moment there are 54,210 tagged birds in a certain wildlife refuge. If exactly 30 percent of all birds in the refuge are tagged, what percent of the untagged birds must be tagged so that half of all birds in the refuge are tagged? | [
"66.67",
"66",
"67",
"65.99"
] | A | All birds = 54210
Currently Tagged = 54210 * 30/100 = 16263
So Untagged birds = 54210 - 16263 = 37947
Half of all birds = 27105
The number of birds to be tagged to make half of all birds tagged = 27105 - 16263 = 10842
So now the question remains - 10842 is how much percentage of untagged birds (16263)
= 10842*100/16263
= 66.67
Answer (A) |
AQUA-RAT | AQUA-RAT-34900 | 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.
Renuka got married 8 years ago. Today her age is 4/3 times of marriage. Her daughter age is 1/8 times her age. her daughter age is | [
"3 years",
"4 years",
"6 years",
"8 years"
] | B | Renuka got married 8 years ago. Today her age is 4/3 times of marriage.
If her age at time of marriage was x, then
x*4/3 = x+8x/3=8
x=24 yrs
Her present age = 24+8=32 yrs
Daughter's age = 32/8= 4 years
ANSWER:B |
AQUA-RAT | AQUA-RAT-34901 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
A cricket player whose bowling average was 24.50 runs per wicket, takes 5 wicket for 52 runs in a match. Due to this his average decreases by 0.50. What will be the number of wickets taken by him till the last match? | [
"64",
"72",
"141",
"96"
] | C | Average = Total Runs / Total wickets
Total runs after last match = 24.50w + 52
Total wickets after last match = w + 5
(24.50w + 52) / (w + 5) = 24.5 - 0.50 = 24
w = 136
so total wickets aftr last match = w+5 = 141
ANSWER:C |
AQUA-RAT | AQUA-RAT-34902 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
The average of first 18 even numbers is? | [
"19",
"18",
"16",
"11"
] | A | Sum of 18 even numbers = 18 * 19 = 342
Average = 342/18 = 19
Answer:A |
AQUA-RAT | AQUA-RAT-34903 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
Mr. RIDDLE invested in Fund X and Fund Y. The total amount she invested, in both funds combined, was $200,000. In one year, Fund X paid 43% and Fund Y paid 28%. The interest earned in Fund Y was exactly $200 greater than the interest earned in Fund X. How much did Ms. Tom invest in Fund X? | [
"43,870.34",
"56,876.45",
"77,789.45",
"78,450.70"
] | D | X+Y=200,000
0.28Y=0.43A+300
Take away decimals first: 28Y+43X+20,000
Isolate first equation to solve for X (your goal): Y=200,000-Y
Plug in for B:
28(200,000-X)=43X+30,000
5,600,000-28X=43X+30,000
5,570,000=71X
5,570,000/71=X
5,570,000/71=X
X=78,450.70=Answer Choice D |
AQUA-RAT | AQUA-RAT-34904 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100? | [
"22",
"88",
"25",
"99"
] | C | A scores 60 while B score 40 and C scores 30.
The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75.
In a game of 100 points, B gives (100 - 75) = 25 points to C.
Answer: C |
AQUA-RAT | AQUA-RAT-34905 | Probability in getting two numbers exactly the same after rolling three dices?
What is the probability of getting two numbers the same having three dices? I had this on my exam, it sounded super easy but isn't so.
Solution: I know that there are $$6^3$$ combinations for all three dices. Then I just wrote the following:
And the combinations that I want are: $$first-second:(1,1,x),(2,2,x),(3,3,x),(4,4,x),(5,5,x),(6,6,x)$$
$$second-third:(x,1,1),(x,2,2),(x,3,3),(x,4,4),(x,5,5),(x,6,6)$$ $$first-third:(1,x,1),(2,x,2),(3,x,3),(4,x,4),(5,x,5),(6,x,6)$$ wich gives me the combination of $$6+6+6=18$$ and final result $$18/216$$, which is incorrect. What am I doing wrong?
• Exactly two the same, or at least two the same? – tilper Sep 7 '16 at 14:27
• Exactly the same – eugene_sunic Sep 7 '16 at 14:28
• Refer to Lovsovs' answer-hint. What you've done so far doesn't exclude, for example, (1,1,1). – tilper Sep 7 '16 at 14:29
• Still unclear. Do you mean exactly 2 the same? – drhab Sep 7 '16 at 14:29
• Yes exactly to the same... – eugene_sunic Sep 7 '16 at 14:29
Hint: For each of your eighteen cases, what can $x$ be?
The following is multiple choice question (with options) to answer.
Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is? | [
"5/19",
"5/12",
"5/18",
"2/12"
] | B | Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9)
= 5/12
Answer: B |
AQUA-RAT | AQUA-RAT-34906 | 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.
On dividing 171 by a number, the quotient is 8 and the remainder is 3. Find the divisor. | [
"18",
"19",
"20",
"21"
] | D | d = (D-R)/Q
= (171 - 3)/8
= 168/8 = 21
D |
AQUA-RAT | AQUA-RAT-34907 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
If 0 < a < b and k = (2a + 8b)/b , which of the following must be true? | [
"k < 2",
"k < 7",
"k < 10",
"k > 10"
] | C | Here's another approach:
k = (2a + 8b)/b
= 2a/b + 8b/b
= 2(a/b) + 8
Since 0 < a < b, we know that a/b is less than 1, which means that 2(a/b) is some number less than 2.
So, we get k = (some number less than 2) + 8
From here, we can see that k must be less than 10
Answer: C |
AQUA-RAT | AQUA-RAT-34908 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
The area of a square field 3136 sq m, if the length of cost of drawing barbed wire 3 m around the field at the rate of Rs.1.50 per meter. Two gates of 1 m width each are to be left for entrance. What is the total cost? | [
"278",
"279",
"999",
"287"
] | C | a2 = 3136 => a = 56
56 * 4 * 3 = 672 – 6 = 666 * 1.5
= 999
Answer:C |
AQUA-RAT | AQUA-RAT-34909 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
How many seconds will it take for a car that is traveling at a constant rate of 14 miles per hour to travel a distance of 96 yards? (1 mile = 1,160 yards) | [
"11",
"12",
"13",
"14"
] | D | speed = 14 miles/hr = 6.84 yard/s
distance = 96 yards
time = distance/speed
=96/6.84
=14 sec
ans - D |
AQUA-RAT | AQUA-RAT-34910 | 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=100*Amount / 100+(R+T)
ANSWER IS B |
AQUA-RAT | AQUA-RAT-34911 | Pre-checking the number $A$
Very quickly, 7 is found to be a factor of $A$. We have $A=7 \times B$ where $B$ is:
$B=$ 9636514689378269172619627962314350702683338048951
We then try to find small factors of $B$. The number 283 is found to be a factor of $B$. Then we have $A=7 \times 283 \times C$ where $C$ is the following 47-digit number:
$C=$ 34051288655046887535758402693690285168492360597
By tagging a 6 in front of the 49-digit factor of the 9th Fermat number, we obtain a composite number. The prime or composite status for the number $A$ is settled. Now the next question: is the remaining factor $C$ prime or composite?
___________________________________________________________________
Pre-checking the number $C$
We perform the two pre-checks as described above. In searching for small factors, we find that $C$ is not divisible by all prime numbers less than 200,000. So if $C$ has a factor, it will have to be a large one. Or it could be that $C$ is prime. Instead of continuing to search for factors, we can do the second check. We calculate $2^{C-1} \ (\text{mod} \ C)$ and find that:
$2^{C-1} \equiv t \ (\text{mod} \ C)$
where $t=$ 7426390354305013563302537374271875139618265902
Since $t$ clearly is not 1, this tells us that $C$ is composite.
The following is multiple choice question (with options) to answer.
The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 10 and 11. The larger of the two numbers is: | [
"276",
"253",
"322",
"345"
] | B | Clearly, the numbers are (23 x 10) and (23 x 11).
Larger number = (23 x 11) = 253.
Answer: Option B |
AQUA-RAT | AQUA-RAT-34912 | 5C2 * 4!(5C2 for selecting 2 out of 5 places for E & rest can be arranged in 4! ways.)
Ans with this approach 240 which is correct in this case.
Now the question discussed above:
2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?
Going as per the above approach it comes
8C4 * (7!/4!2!) (8C4 for filling 4 I's in 8 blank spaces and arranging rest 7 alphabets in 7!/4!2!)
But in this case answer doesn't match.
Please if you could explain the flaw in my approach will be very helpful.
Cheers.
_________________
Sailing through rough waters. Stars & few kudos will steer me past.
Re: In how many ways can the letters of the word PERMUTATIONS be &nbs [#permalink] 09 May 2018, 01:12
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
How many different 6-letter sequences are there that consist of 1A, 4B's and 5C's ? | [
"6",
"60",
"100",
"360"
] | C | How many different 6-letter sequences are there that consist of 1 A, 4B's, and 5 C's?
(A) 6
(B) 60
(C) 100
(D) 360
(E) 720 (ANS C) |
AQUA-RAT | AQUA-RAT-34913 | acid-base, titration
I got that I need $\pu{0.8 g}$ to make the solution.
Then I did the lab and got this data:
\begin{array}{lrrrr}
\text{Drink}
& \text{Vol. of Drink}
& \text{Init. Vol. NaOH}
& \text{Final Vol. NaOH}
& \text{Vol.-Difference}
\\\hline
\text{Lemonade} & \pu{25 mL} & \pu{6.0 mL} & \pu{12.0 mL} & \pu{6.0 mL}\\
\text{Gatorade} & \pu{25 mL} & \pu{16.5 mL} & \pu{19.5 mL} & \pu{3.0 mL}\\
\text{Ginger Ale} & \pu{25 mL} & \pu{12.0 mL} & \pu{16.5 mL} & \pu{4.5 mL}
\end{array}
With this information I need to answer these questions:
The following is multiple choice question (with options) to answer.
How many litres of 6% acid solution must be mixed with 120 gallons of 20% acid solution to obtain an acid solution that is 16%? | [
"48 litres",
"98 litres",
"28 litres",
"49 litres"
] | A | QUESTION : Rs. 2250 is divided among three boys named Rama, Krishna and Kalki in such a way that 1/6th of Rama’s share, 1/4th of Krishna’s share and 2/5th of Kalki’s share are equal. Find Rama’s share.
ANSWER : A |
AQUA-RAT | AQUA-RAT-34914 | # 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 woman can row upstream 5 kmph and downstream 10 kmph. Find the woman rate in still water and rate of the stream. | [
"2.5,2.5",
"7.5,2.5",
"5.5,2.5",
"5,2.5"
] | B | Explanation:
Please remember,
If a is rate downstream and b is rate upstream
Rate in still water = 1/2(a+b)
Rate of current = 1/2(a-b)
=> Rate in still water = 1/2(10+5) = 7.5 kmph
=> Rate of current = 1/2(10-5) = 2.5 kmph
Option B |
AQUA-RAT | AQUA-RAT-34915 | Now let's go to Righty:
If R were 11 or more, then L and M would have to be 0 or negative, so R is at most 10. If R is 10, then L+M must be 3, which only leaves 1 and 2 as possible values, but Righty couldn't deduce it, so R is 9 or less. Also, as a perfect logician, Righty would know that L is 1 or 2, because if it were 3, then Lefty would know the answer.
If R is 9, then L+M must be 4, leaving only 1 and 3 as possible values for L and M. So R must be 8 or lower.
If R is 8, then L+M = 5, with possible values of: 1/4, 2/3
If R is 7, then L+M = 6, possible values: 1/5, 2/4
If R is 6, then L+M = 7, possible values: 2/5
If R is 5, then L+M = 8, no possible values
So L is [1,2] and R is [7,8]
Do the same for Middly:
As a perfect logician he would know than L is 1 or 2 and R is 7 or 8.
So M must be between 2 and 7, L can only be 1 or 2 and R can only be 7 or 8:
If M is 2, then L+R = 11, possible values: none
If M is 3, then L+R = 10, possible values: 2/8
If M is 4, then L+R = 9, possible values: 1/8, 2/7
If M is 5, then L+R = 8, possible values: 1/7
If M is 6, then L+R = 7, possible values: none
If M is 7, then L+R = 6, possible values: none
From here we can deduce that M is not 3 or 5. And because Middly could not deduce it, we can see that M must be 4
Possible solutions:
L M R
1 4 8
2 4 7
I agree with Novarg's answer, but it's a bit hard to follow. Here's how I solved it:
We can know the possible values are these:
The following is multiple choice question (with options) to answer.
If 6^r is a factor of (10!)^2, What is the greatest possible value of r ? | [
"2",
"4",
"6",
"8"
] | D | r 10!= 10*9*8*7*6*5*4*3*2*1 = 2*5*3*3*2*2*2*7*2*3*5*2*2*3*2 = 2^8*3^4*5^2*7
6= 2*3
Therefore only the exponents of 2 and 3 are relevant, 2^8 or 3^4 -> higher number counts = 8 -> Answer Choice D |
AQUA-RAT | AQUA-RAT-34916 | Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13
The following is multiple choice question (with options) to answer.
In the following equation what value would come in place of question mark (?)?
5798 – ? = 7385 – 4632 | [
"3225",
"2595",
"2775",
"3045"
] | D | ? = 5798 + 4632 – 7385 = 3045
Answer D |
AQUA-RAT | AQUA-RAT-34917 | Kudos [?]: 13 [0], given: 21
GMAT 1: 570 Q46 V24
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
26 Dec 2013, 23:04
Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played?
Then is 28*2 the correct approach?
Kudos [?]: 13 [0], given: 21
Math Expert
Joined: 02 Sep 2009
Posts: 42302
Kudos [?]: 133018 [1], given: 12402
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
27 Dec 2013, 03:08
1
KUDOS
Expert's post
4
This post was
BOOKMARKED
Kudos [?]: 133018 [1], given: 12402
Manager
Joined: 07 Apr 2014
Posts: 138
Kudos [?]: 31 [0], given: 81
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
12 Sep 2014, 06:36
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
total 8 teams & each game by 2 pair then 8C2
Kudos [?]: 31 [0], given: 81
Intern
Joined: 20 Sep 2014
Posts: 9
Kudos [?]: 4 [1], given: 49
There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
26 Nov 2014, 02:14
1
KUDOS
1
This post was
BOOKMARKED
SreeViji wrote:
Hi Bunnel,
I would also like to learn this approach. Can u help me?
Sree
Hey SreeViji,
The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2}
The following is multiple choice question (with options) to answer.
There are, in a certain league, 14 teams, and each team face another team for a total of 5 times. How many games are played in the season? | [
"455",
"525",
"1400",
"180"
] | A | By using the formula, t[n(n - 1) / 2], where t = no. of games between two teams and n = total no. of teams, we get: 455
Option A. |
AQUA-RAT | AQUA-RAT-34918 | java, beginner, homework, finance
System.out.println("How many nickles do you have?");
int nickles = scanner.nextInt();
double total = new CoinAdder()
.addCoins(Coin.QUARTER, quarters)
.addCoins(Coin.DIME, dimes)
.addCoins(Coin.NICKLE, nickles)
.getValue();
System.out.println("You have: " + moneyFormat.format(total));
}
private static void calculateCompoundInterest(Scanner scanner, NumberFormat moneyFormat) {
System.out.println("What is the initial investment?");
double investment = scanner.nextDouble();
System.out.println("At what intrest rate is the intrest compounded annually?");
double intrestRate = scanner.nextDouble();
double futureValueFive = investment * Math.pow(1 + intrestRate, 5.);
System.out.println("In five years the investment will be worth : " + moneyFormat.format(futureValueFive));
double futureValueTen = investment * Math.pow(1 + intrestRate, 10.);
System.out.println("In ten years the investment will be worth : " + moneyFormat.format(futureValueTen));
double futureValueTwenty = investment * Math.pow(1 + intrestRate, 20.);
System.out.println("In twenty years the investment will be worth : " + moneyFormat.format(futureValueTwenty));
}
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
NumberFormat moneyFormat = NumberFormat.getCurrencyInstance();
calculateDollarValueOfCoins(scanner, moneyFormat);
calculateCompoundInterest(scanner, moneyFormat);
}
}
The following is multiple choice question (with options) to answer.
If John invested $ 1 at 4 percent interest compounded annually, the total value of the investment, in dollars, at the end of 4 years would be | [
"(1.5)^4",
"4(1.5)",
"(1.05)^4",
"1 + (0.04)^4"
] | C | In this prompt, the answer choices show that we don't actually have to calculate anything - we just need to 'format' the calculation correctly. We're told that the interest is COMPOUNDED ANNUALLY, so we need the Compound Interest Formula, with a Principal of $1, an R of 5% (which would 'translate' as .05) and a T of 4. We'd have....
Answer :C |
AQUA-RAT | AQUA-RAT-34919 | Hint: Start your proof like this:
Let $k$ be any arbitrary odd integer. Then by the definition of an odd integer, we have $k=2a+1$ for some integer $a$. Thus...
Then consider $-k=-(2a+1)$ and perform some algebraic manipulations. Your final step should look something like:
...hence, since $-k=2b+1$ where $b$ is an integer, it follows by definition that $-k$ is also an odd integer, as desired.
-
One could prove this inductively:
Assume that the $n^\text{th}$ odd positive integer, $2n-1$, has an odd negation. Then $-2n+1$ is odd, so $$\underbrace{-2n+1}_{\text{odd}}-\underbrace{2}_{\text{even}}=\underbrace{-2n-1}_{\text{odd}}.$$ Thus the statement holds for the $(n+1)^\text{th}$ positive integer, $2n+1$.
-
I like the simplicity of this approach. Pedantic: you still also need to prove the base case. – Lie Ryan Jun 23 '13 at 17:12
Here's my proof of the equivalent statement for even integers. I'll let you figure out what to do for odds.
Let $n$ be an even integer. By definition of "even", $\exists m \in \mathbb{Z}: n=2m$. Then $-n = -2m = 2(-m)$. Because $\mathbb{Z}$ is closed under additive inverse, $-m$ is an integer. By the definition of "even", $2(-m) = -n$ is also even, Q.E.D.
-
Let our odd number be $o$. Since $-1\equiv 1 \pmod{2}$, $-o\equiv o \pmod{2}$. By the definition of odd numbers, $o\equiv 1 \pmod{2}$. Combining this with the earlier statement, one gets
The following is multiple choice question (with options) to answer.
If a and b are odd integers, which of the following is an even integer? | [
"2a+b",
"3a+b",
"4a+b",
"2a-b"
] | B | If a is odd, then 3a is odd.
Then 3a+b is even.
The answer is B. |
AQUA-RAT | AQUA-RAT-34920 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
A man walks at a speed of 7 km/hr and runs at a speed of 7 km/hr. How much time will the man require to cover a distance of 10 1/2 km, if he completes half of the distance, i.e., (5 1/4) km on foot and the other half by running? | [
"1 hours",
"1 1/2 hours",
"2 1/6 hours",
"2 1/1 hours"
] | B | Required time
= (5 1/4)/7 + (5 1/4)/7
= 1 1/2 hours.
Answer:B |
AQUA-RAT | AQUA-RAT-34921 | 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.
In a competitive examination in state A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each state? | [
"8200",
"7500",
"7000",
"8000"
] | D | Equal number of candidates appeared in each state.
In state A, 6% candidates got selected.
In state B, 7% candidates got selected.
Given that 80 more candidates got selected in state B than A.
Therefore, 1% of candidates appeared in each state =80
Answer: Option D
= 100% of candidates appeared in each state
=80×100=8000
i.e., number of candidates appeared from each state = 8000 |
AQUA-RAT | AQUA-RAT-34922 | 2. Mike says:
of course i meant 5 buckets, one for each student ;).
3. meichenl says:
Let there be $c$ cookies and $n$ students. Denote the number of ways to distribute $c$ cookies between $n$ students by $S_n(c)$. This is claiming that for a given number of students, the number of ways to distribute the cookies is a function of the number of cookies.
The last student can receive anywhere from $0$ to $c$ cookies. Suppose he receives $k$. Then there are $c-k$ cookies to distribute between $n-1$ students. This can be done $S_{n-1}(c-k)$ ways. Summing over all possible values of $k$ we obtain
$S_n(c) = \sum_{k=0}^c S_{n-1}(c-k)$.
We know
$S_1(c) = 1$
because if there is one student, he must get all the cookies (AWESOME!).
Then
$S_2(c) = \sum_{k=0}^c 1 = 1+c$
$S_3(c) = \sum_{k=0}^c c-k+1 = \frac{1}{2}c(3+c)$
$S_4(c) = \sum_{k=0}^c (c-k)(c-k+3)/2 = \frac{1}{6} c(1+c)(5+c)$
\begin{align*} S_5(c) &= \sum_{k=0}^c \frac{1}{6} (c-k)(1+c-k)(5+c-k) \\ &= \frac{1}{24} c(1+c)(2+c)(7+c) \end{align*}
$S_5(10) = 935$.
So the answer is 935. For larger $n$ I conjecture,
The following is multiple choice question (with options) to answer.
If F(x, n) be the number of ways of distributing "x" toys to "n" children so that each child receives at the most 2 toys then F(4, 3) = | [
"3",
"4",
"6",
"5"
] | C | Explanation :
We have to find the number of ways in which 4 toys can be distributed to 3 children so that each child receives at the most 2 toys.There are two possible cases:Case 1: Two of them receive 2 toys each and one of them doesn’t get any toy.There are 3 possible ways to distribute the toys in this case i.e., the three possible ways of selecting the child who will not get any toy.Case 2: Two of them receive 1 toy each and one of them receives 2 toys.Again there are 3 possible ways to distribute the toys in this case i.e., the three possible ways of selecting the child who will get 2 toys.So there are a total of 6 possible ways.
Answer : C |
AQUA-RAT | AQUA-RAT-34923 | # Help me understand events/sample space
## Homework Statement
1.
Suppose that A, B, and C are 3 independent events such that Pr(A)=1/4, Pr(B)=1/3 and Pr(C)=1/2.
a. Determine the probability that none of these events will occur.
Is it just:
(1-P(a))(1-P(b))(1-P(c)) = 3/4 * 2/3 * 1/2 = 1/4
## The Attempt at a Solution
I tried to do 1. another way:
The probability that all theses events will occur: 1/4 * 1/3 * 1/2 = 1/24
1-(1/24) = 23/24
Obviously this is wrong. Is the reason it is wrong, because: the complement of "all of these events will occur" is that "not all of these events will occur," meaning, it is not "none of these events will occur."
None of these events will occur is included in the compliment 1-(1/24), but so is that 1 of the events occur, and that 2 of the events occur, etc.
Am I right in my reasoning?
PeroK
the (1) is correct.
for reference, in general the answer can be found by calculating multinomial distribution.
in (3), the 23/24 probability is sum of "no events", "A only", "B only", "C only", "A&B", "A&C", "B&C".
PeroK
Homework Helper
Gold Member
2020 Award
## Homework Statement
1.
Suppose that A, B, and C are 3 independent events such that Pr(A)=1/4, Pr(B)=1/3 and Pr(C)=1/2.
a. Determine the probability that none of these events will occur.
Is it just:
(1-P(a))(1-P(b))(1-P(c)) = 3/4 * 2/3 * 1/2 = 1/4
## The Attempt at a Solution
I tried to do 1. another way:
The probability that all theses events will occur: 1/4 * 1/3 * 1/2 = 1/24
The following is multiple choice question (with options) to answer.
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability E that all three secretary are assigned at least one report? | [
"8/9",
"64/81",
"4/9",
"16/81"
] | C | I got the same answer: Here's my reasoning
First Report you have 3 choices
Second Report you have 2 choices
Third Report you have 1 choice
Fourth report 3 choices again
Then total number of ways is: 3*2*1*3=3^2*2
This is not correct. You have assumed that the 4th report must go to someone who already has a report. There is no such constraint. You can easily give the 1st and 2nd reports to secretary 1, 3rd report to secretary 2 and 4th report to secretary 3. But you have ignored all such cases.
The number of ways of ensuring at least one report goes to each secretary is 4C2 (select 2 reports out of 4 which go to the same person)*3C1 (select the person who must type 2 reports)*2! (since you have 2 reports left which you must distribute to the 2 remaining people such that each person gets one) = 36
Required probability E= 36/81.C |
AQUA-RAT | AQUA-RAT-34924 | 20. mathslover
And.,.... Hence proved.. Right @joemath314159 ?
21. joemath314159
very interesting :)
22. mathslover
:) yes .
23. hartnn
finally u are doing only AM>=HM only
24. hartnn
but other way round......
25. mathslover
yep :)
The following is multiple choice question (with options) to answer.
Of the 200 math majors at a certain university, 30 percent will be assigned to advanced calculus on Monday and Wednesday and the remaining 80 percent will be assigned to advanced calculus on Tuesday and Thursday. However, 40 percent of the math majors prefer Tuesday and Thursday and 60 percent prefer Monday and Wednesday. What is the lowest possible number of math majors who will NOT be assigned to the days they prefer? | [
" 15",
" 20",
" 30",
" 45"
] | D | 30% of students are assigned to Monday and Wednesday = 0.3*200 = 60
70% of students are assigned to Tuesday and Thursday = 0.7*200 = 140
60% of students prefer Monday and Wednesday = 0.6*200 = 120
40% of students prefer Tuesday and Thursday = 0.4*200 = 80
Since we need theleastnumber of students who will not be assigned to their preferred days, we shall fit in themaximumpossible students as per their preference:
All of the 40% of students who prefer Tuesday and Thursday can be accommodated to their preferred days, whereas only 60 out of 120 students who prefer Monday and Wednesday can be accommodated to their preferred days.
The remaining 45 students who prefer Monday and Wednesday will be allocated to Tuesday and Thursday by default, which is not their choice.
Answer is D. |
AQUA-RAT | AQUA-RAT-34925 | 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.
Frances can complete a job in 8 hours, and Joan can complete the same job in 12 hours. Frances starts the job at 9 a.m., and stops working at 3 p.m. If Joan starts working at 4 p.m. to complete the job, at what time is the job finished? | [
"6 p.m.",
"7 p.m.",
"8 p.m.",
"10 p.m."
] | B | Let the total work be 24 units.
Frances can complete 3 units/hr
Joan can complete 2 units/hr
Work done by Frances in 6 hours is 18 units ; work left for Joan is 6 units ( 24 - 18)
Time required by Jones to complete the remaining work will be 6/2 => 3 hours
Given Jones starts at 4PM , so she would complete the work at 7 PM
Answer will be (B) |
AQUA-RAT | AQUA-RAT-34926 | 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.
Amar takes as much time in running 15 meters as a car takes in covering 40 meters. What will be the distance covered by Amar during the time the car covers 1.9 km? | [
"700 m",
"720 m",
"712 m",
"600 m"
] | C | Distance covered by Amar
= 15/40 (1.9km)
= 3/8(1900) = 712 m
Answer: C |
AQUA-RAT | AQUA-RAT-34927 | # 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.
Ms. Mary sold two properties, X and Y, for $25,000 each. She sold property X for 20 % more than she paid for it and sold property Y for 20% less than she paid for it. If expenses are disregarded , what was her total net gain or loss, if any, on the two properties ? | [
"$2100",
"$2222",
"$2320",
"$2083.33"
] | D | There is a property to solve such questions withCommon Selling PriceandCommon %gain and loss. such cases always result in a loss and...
Total %Loss = (Common Gain% or Loss%/10)^2
Hence Here Loss% = (20/10)^2 = 4%
which means he recovered only 96% of his investment which amount to a total revenue = 25000 + 25000 = 50000
i.e. 96% of cost = 40000
therefore, 4% of cost (Loss) = $2083.33
Answer: D |
AQUA-RAT | AQUA-RAT-34928 | # 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.
The ratio of the selling prices of three articles is 5:6:9 and the ratio of their cost prices is 4:5:8 respectively. What is the ratio of their respective percentages of profit, if the profit on the first and the last articles is the same? | [
"4:5:6",
"10:8:5",
"5:6:9",
"7:6:9"
] | B | Given that the selling prices of three articles,
S.P1 =5X, S.P2 =6X, S.P3 =9X,
And their cost prices are C.P1 =4Y, C.P 2 =5Y, C.P3 =8Y
Given that , S.P1 –C.P1 =S.P2 –C.P2, 5X-4Y =9X-8Y, X =Y,
Their profit percentages are, p1 =(5-4)/4*100 = 25% ,
p2 = (6-5)/5*100 = 20%, p3 = (9-8)/8*100 = 12 1/2 %
Ratio of the percentages is 25:20:12 1/2 = 10:8:5
ANSWER:B |
AQUA-RAT | AQUA-RAT-34929 | 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.
42 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work? | [
"10",
"16",
"13",
"14"
] | D | Let the required number of days be x.
Less persons, More days (Indirect Proportion)
More working hours per day, Less days (Indirect Proportion)
Persons 30 : 42 :: 12 : x
Working hours/day 6 : 5
30 x 6 x x = 42 x 5 x 12
x = (42 x 5 x 12)/(30 x 6)
x = 14
ANSWER D |
AQUA-RAT | AQUA-RAT-34930 | # Difference between revisions of "2009 AMC 10B Problems/Problem 21"
## Problem
What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$
## Solution
### Solution 1
The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore
$(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})$
is divisible by 8. So the required remainder is $3^0 + 3^1 = \boxed {4}$. The answer is $\mathrm{(D)}$.
### Solution 2
We have $3^2 = 9 \equiv 1 \pmod 8$. Hence for any $k$ we have $3^{2k}\equiv 1^k = 1 \pmod 8$, and then $3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3 \pmod 8$.
Therefore our sum gives the same remainder modulo $8$ as $1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3$. There are $2010$ terms in the sum, hence there are $2010/2 = 1005$ pairs $1+3$, and thus the sum is $1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8$.
### Solution 3
The following is multiple choice question (with options) to answer.
What is the remainder when 8^1 + 8^2 + 8^3 +...+ 8^9 is divided by 8? | [
"4",
"0",
"2",
"5"
] | B | Notice that in the brackets we have the sum of 9 even multiples of 8, which yields remainder of 0 upon division by 8.
Answer : B |
AQUA-RAT | AQUA-RAT-34931 | # Difference between revisions of "1992 AHSME Problems/Problem 11"
## Problem
$[asy] draw(circle((0,0),18),black+linewidth(.75)); draw(circle((0,0),6),black+linewidth(.75)); draw((-18,0)--(18,0)--(-14,8*sqrt(2))--cycle,black+linewidth(.75)); dot((-18,0));dot((18,0));dot((-14,8*sqrt(2))); MP("A",(-18,0),W);MP("C",(18,0),E);MP("B",(-14,8*sqrt(2)),W); [/asy]$
The ratio of the radii of two concentric circles is $1:3$. If $\overline{AC}$ is a diameter of the larger circle, $\overline{BC}$ is a chord of the larger circle that is tangent to the smaller circle, and $AB=12$, then the radius of the larger circle is
$\text{(A) } 13\quad \text{(B) } 18\quad \text{(C) } 21\quad \text{(D) } 24\quad \text{(E) } 26$
## Solution (Similarity)
The following is multiple choice question (with options) to answer.
A circular logo is enlarged to fit the lid of a jar. The new diameter is 40 per cent larger than the original. By what percentage has the area of the logo increased? | [
"50",
"96",
"100",
"125"
] | B | Let old diameter be 4 , so radius is 2
Old area = 4π
New diameter is 5.6 , so radius is 2.8
New area = 7.84π
Increase in area is 3.84π
% increase in area = 3.84/4*100
So, % increase is 96%
Answer will be (B) |
AQUA-RAT | AQUA-RAT-34932 | The book has 300 pages, 200 of which are in long chapters and 100 of which are in short chapters. If you pick a random page it is $\frac 23$ to be in a long chapter. So strategy 1 gives you a long chapter $\frac 23$ of the time. Strategy 2 gives you the chapter after a long one $\frac 23$ of the time. If the chapters alternate, strategy 2 will give you a short chapter $\frac 23$ of the time.
Thanks, but doesn't the book still have 200 pages from long and 100 pages from short chapters both ways? How come the probability of picking long chapters isn't $\frac{2}{3}$ for both strategies? In other words, doesn't that logic for strategy 1 also also apply to 2? – David Faux May 5 '12 at 0:04
The initial page selection is in fact $\frac 23$ to be in a long chapter either way. But the selected chapter in strategy 2 is not the chapter the selected page is in. Think of a 2 chapter book with the first chapter 200 pages and the second 100. $\frac 23$ of the time the page chosen will be in chapter 1. In strategy 1 you then read chapter 1. In strategy 2 you then read chapter 2. – Ross Millikan May 5 '12 at 0:08
Ah thanks. So if short and long chapters alternate, then a reader using strategy 2 flips forward to a short chapter if he/she lands in a long chapter. However, what if the chapters don't alternate and are ordered randomly? Would the probability of reading a short chapter for strategy 2 be less than $\frac{2}{3}$ because the reader will not always move forward to a short chapter after landing in a long chapter? – David Faux May 5 '12 at 0:23
It will depend upon the ordering of the chapters. If chapters 1-5 are long and 2-6 are short, you will have $\frac 2{15}$ to read any of 2-6 and $\frac 1{15}$ to read 1 or 7-10. So that is $\frac 9{15}$ to read a long and $\frac 6{15}$ to read a short. – Ross Millikan May 5 '12 at 0:34
The following is multiple choice question (with options) to answer.
Ron begins reading a book at 2: 30 p.m. and reads at a steady pace of 30 pages per hour. Michelle begins reading a copy of the same book at 6: 00 p.m. If Michelle started 5 pages behind the page that Ron started on and reads at an average pace of 50 pages per hour, at what time would Ron and Michelle be reading the same page? | [
"7: 00 p.m.",
"7: 30 p.m.",
"10: 45 p.m.",
"11: 30 p.m."
] | D | M gains 50-30=20 pages per hour on R
at 6pm R has read 105 pages and M is 105+5=110 pages behind him
time=110/20=5.5 hours
they are on the same page at 11:30pm
D |
AQUA-RAT | AQUA-RAT-34933 | # 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 boat travels 80 km downstream in 8 hours and 60 km upstream in 12 hours. Find the speed of the boat in still water and the speed of the water current. | [
"7.5 km/hr, 2.5 km/hr",
"5 km/hr, 4 km/hr",
"1 km/hr, 6 km/hr",
"2 km/hr, 1 km/hr"
] | A | Downstream speed=80km/8hrs
=10kmph
upstream speed=60km/12hrs
=5kmph
speed of boat=avg of downstream and upstream speeds
speed of boat=(10+5)/2kmph
=7.5kmph.
current speed=half of the difference of downstream and upstream speeds
currend speed=(10-5)/2kmph
=2.5kmph
ans is 7.5km/hr, 2.5km/hr
ANSWER:A |
AQUA-RAT | AQUA-RAT-34934 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter? | [
"287",
"132",
"156",
"158"
] | B | 2 * 22/7 * 14 = 88
88 * 1 1/2 = Rs.132
Answer: B |
AQUA-RAT | AQUA-RAT-34935 | • +1 though irritated that you just scooped me! Apr 20, 2016 at 17:16
• Thanks for the answer. I have made an appropriate system of equations and solved it but my results differ from those from the answers in my textbook. I got $3x^3-2x^2+3x+1$ (confirmed by WolframAlpha: m.wolframalpha.com/input/?i=%28x^2-x-3%29%28ax%2Bb%29%2B13x-2%3D%28x^2-2x%2B5%29%28cx%2Bd%29-1-7x&x=0&y=0 ) but according to the textbook the answer is $3x^3-5x^2+6x+4$ could you please verify if I have done this correctly? Apr 20, 2016 at 18:03
• It's easy enough to check. The remainders on dividing $3x^3 - 2x^2 + 3 x + 1$ by $x^2-x-3$ and $x^2 -2x+5$ are $13x+4$ and $-4x-19$ respectively. So you must have done something wrong. The textbook's answers are correct. Apr 20, 2016 at 19:46
The following is multiple choice question (with options) to answer.
What no. should be subtracted from x^3+4x^2−7x+12x^3+4x^2−7x+12, if it is to be perfectly divisible by x+3x+3? | [
"37",
"42",
"49",
"52"
] | B | According to remainder theorem when dfracf(x)x+adfracf(x)x+a, then the remainder is f(−a)f(−a).
In this case, as x+3x+3 divides x3+4x2−7x+12–kx3+4x2−7x+12–k perfectly (kk being the number to be subtracted), the remainder is 0 when the value of xx is substituted by -3.
i.e., (−3)3+4(−3)2−7(−3)+12−k=0(−3)3+4(−3)2−7(−3)+12−k=0
or −27+36+21+12=k−27+36+21+12=k
or k=k= 42
B |
AQUA-RAT | AQUA-RAT-34936 | 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 running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train. | [
"150 meter",
"286 meter",
"186 meter",
"288 meter"
] | A | Speed = 60*(5/18) m/sec
= 50/3 m/sec
Length of Train (Distance)
= Speed * Time
(50/3) * 9 = 150 meter
Answer:A |
AQUA-RAT | AQUA-RAT-34937 | Substituting $$e$$ with $$a-d$$ gives $$b^{2}c=4d(a-d)$$. So $$b^{2}c=4ad-4d^{2}$$. Rearranging the terms gives us $$4d^{2}-4ad+b^{2}c=0$$
Using the Quadratic Equation, we have $$d=\frac {a\pm\sqrt{a^{2}-b^{2}c}}{2}$$
And since $$a=d+e$$, $$e$$ is the conjugate of $$d$$. So $$e=\frac {a-\sqrt{a^{2}-b^{2}c}}{2}$$ and $$d=\frac {a+\sqrt{a^{2}-b^{2}c}}{2}.\,$$ Thus
$$\sqrt{a+b\sqrt{c}}\,=\, \sqrt{\frac {a+\sqrt{a^{2}-b^{2}c}}{2}} +\sqrt{\frac {a-\sqrt{a^{2}-b^{2}c}}{2}}$$
(I will assume $b$ is not a square, since otherwise it would not be a nested radical.)
A nested radical can be denested if and only if there exist $u,v\in\mathbb{N}$ such that the nested radical is of the form $\sqrt{u^2+v\pm2u\sqrt{v}}$ in which case it is also equal to $|u\pm\sqrt{v}|$.
It's not hard to show that those expressions are equal, which means that all nested radicals of that form can indeed be denested.
The following is multiple choice question (with options) to answer.
Which of the following equations is NOT equivalent to 4c^2 = d^2 - 9 ? | [
" 4c^2 + 9 = d^2",
" 4c^2 - d^2 = -9",
" 4c^2= (d + 3)(d - 3)",
" 2c = d - 3"
] | D | 4c^2 = d^2 - 9
The basic rule when dealing with equations is that you can do ANYTHING to both sides of an equation as long as you do it EQUALLY to BOTH sides. Everything gets a bit more complex when you're dealing with variables in the denominator of a fraction and/or inequalities, but neither of those subjects is a factor in this prompt.
Looking at Answer D, we have....
2c = d - 3
While you COULD take the square root of both sides of the original equation, it's important to note that the square root of d^2 - 9 is NOT (d-3).
d^2 - 9 can be factored into (d-3)(d+3), but neither of these parentheses is the square root of d^2 - 9.
You can see the proof when you square either of the two parentheses:
(d-3)^2 = d^2 - 6c+ 9
(d+3)^2 = d^2 + 6c + 9
Thus, D is NOT equivalent to the prompt. |
AQUA-RAT | AQUA-RAT-34938 | # 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.
Virat buys an old car for Rs. 50000 and spends Rs. 4000 on its repairs. If he sells the scooter for Rs. 60000, his gain percent is: | [
"3 (4/7) %",
"11 (1/9) %",
"12 (2/3) %",
"12%"
] | B | Explanation:
C.P. = Rs. (50000 + 4000) = Rs. 54000
S.P. = Rs. 60000.
Gain% = ( (6000 / 54000) * 100 ) % = 11 (1/9)%
ANSWER B |
AQUA-RAT | AQUA-RAT-34939 | 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.
A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. what is the rate of interest ? | [
"3 %",
"4 %",
"5 %",
"6 %"
] | D | Solution
S.I= Rs.(15500-12500) =Rs.3000
Rate=(100*3000/12500*4)% =6%
Answer D |
AQUA-RAT | AQUA-RAT-34940 | Quick way
Use Smart Numbers
Give 100 for the initial amount
Then you will have 50-0.25x = 30
x = 80
So % is 80/100 is 80%
Hope it helps
Cheers!
J
SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance
Re: If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
29 May 2014, 11:41
Or one can use differentials to slve
Initially 50% alcohol
Then 25% alcohol
Resulting mixture 30% alcohol
Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4
Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.
Hope this helps
Cheers
J
Senior Manager
Status: Verbal Forum Moderator
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If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
02 Jul 2014, 17:12
Bunuel wrote:
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
The following is multiple choice question (with options) to answer.
How much is 85% of 40 is greater than 4/5 of 25? | [
"22",
"14",
"88",
"12"
] | B | (85/100) * 40 – (4/5) * 25
= 14
Answer: B |
AQUA-RAT | AQUA-RAT-34941 | • In the problem stated above, there are two towers of different heights. The distance of the tower with height $h_1$ to the horizon is $d_1=\sqrt{2rh_1}$. The distance of the tower with height $h_2$ to the horizon is $d_2=\sqrt{2rh_2}$. Thus, the distance between the towers is $d_1+d_2=\sqrt{2rh_1}+\sqrt{2rh_2}$. If $h_1=h_2=h$, then this reduces to $2\sqrt{2rh}=\sqrt{8rh}$. – robjohn May 9 '13 at 1:52
The following is multiple choice question (with options) to answer.
On a map the distance between two mountains is 312 inches. The actual distance between the mountains is 136 km. Ram is camped at a location that on the map is 28 inch from the base of the mountain. How many km is he from the base of the mountain? | [
"14.83",
"12.2",
"14.8",
"14.82"
] | B | Explanation:
Since 312 inch = 136 km
So 1 inch = 136/312 km
So 28 inch = (136 × 28)/ 312 = 12.2 km
Answer: B |
AQUA-RAT | AQUA-RAT-34942 | The number of time periods to achieve three rings is two. The number of time periods to achieve eleven rings is ten.
So $5\times 6=30$
• Okay, but. Mind if you explain it a bit slowly for someone who is not familiar with any of the terms which you had just used. In other words in layman terms. What is the thing that I should do first?, then what I shouldn't do? and more importantly, what are the limitations that should be considering when working with time periods, although I believe that this would had been about other things, maybe lengths. – Chris Steinbeck Bell Apr 2 '18 at 9:44
• as I'm sure you are aware by now, the total time for a set of rings is about counting the three-second gaps between the rings, which is one less than the number of rings. – David Quinn Apr 2 '18 at 10:58
$$\textrm{11 hour}\times\frac{\textrm{6 sec}}{\textrm{3 hour}}=22\,\textrm{sec}$$
you did succumb to the flagpole error, because you used the total number of flagpoles ($11$ and $3$) rather than the number of flagpoles minus one.
Indeed, when you later corrected for the flagpole error, you got:
$$\textrm{(11-1) hour}\times\frac{\textrm{6 sec}}{\textrm{(3-1) hour}}=30\,\textrm{sec}$$
and now it works.
You need to do the latter, since you are asked the amount of time it takes to announce the hour, and assuming that each individual chime is instantaneous, you shouldn't focus on the number of chimes, but rather on the intervals between the chimes. But, as the title of your question reveals, you did focus on the "total number of bells tollings", and thus succumbed to the flagpole error.
You ask about what happens when there are larger number involved. I am not sure why that would change the math, really, other than that you are dealing with larger numbers. But the formulas should all remain exactly the same.
The following is multiple choice question (with options) to answer.
Five bells ring together and ring at intervals of 2, 6, 8, 12 and 15 seconds respectively. How many times will they ring together in 60 minutes? | [
"34",
"33",
"32",
"31"
] | D | Explanation:
LCM of 2, 6, 8, 12 and 15 is 120
120 second = 2 minutes
Six bells rings in 2 min intervals
Therefore, Number of times they ring together in 60 min = 1 + (60/2) = 31
Answer: Option D |
AQUA-RAT | AQUA-RAT-34943 | newtonian-mechanics, fluid-dynamics, momentum, propulsion
Title: Can a submarine be powered by a wind turbine? This is not intended as a realistic way for propulsion of submarines, it is just an interesting thought experiment.
Say I want to travel directly towards the wind, only powered by the wind. This is obviously possible on land, as I can just stop, anchor to the ground for some time, and charge a battery. Then I can use the stored energy to drive for a while.
But is this also possible if it is not possible to anchor to the medium I am travelling through? I assume that I can have a net forward movement as I can gear down from a small amount of quickly moving air to a large amount of slowly moving water, but the actual relationship seems kind of tricky.
Is it possible for a submarine powered by a wind turbine to travel directly against the air-flow? Whether it's a submarine, boat, car, whatever, if there is a propeller being driven by the wind it can provide forward motion, even directly upwind.
Here's an example.
If you're a high-speed wind sailor, you know why.
You can sail upwind close-hauled, and the only thing that limits your speed is your drag in the water.
Now suppose you tack back and forth - your progress is directly upwind.
An easy way to tack is to mount your sail on an axle so it can just go around and around.
Then you can go upwind, downwind, any way you like, speed limited only by drag.
The following is multiple choice question (with options) to answer.
Submarine M and Submarine N are equipped with sonar devices that can operate within a 5,000 yard range. Submarine M remains in place while Submarine N moves 4,700 yards south from Submarine M. Submarine N then changes course and moves due east, stopping at the maximum range of the sonar devices. In which of the following directions can Submarine N continue to move and still be within the sonar range of Submarine M? I. North II. South III. West | [
"I only",
"II only",
"I and III only",
"III only"
] | C | It's evident if we make a diagram for the Movement of N respect to M.
From its South position to the east position, it has covered the maximum radius for the south direction. Similarly, it's east movement further will also make it out of range.
Thus it can only move north or west or anything in that curve.
Answer : C |
AQUA-RAT | AQUA-RAT-34944 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
GMAT assassins aren't born, they're made,
Rich
_________________
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
### Show Tags
03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
In a group of 100 cars, 37 cars do not have air conditioning. If at least 41 cars have racing stripes, what is the greatest number of cars that could have air conditioning but not racing stripes? | [
"45",
"47",
"59",
"51"
] | C | Lets assume AC=63(includesonly AC carsandcars with AC and racing stripes)
lets assume RS(racing stripes)>=41(includescars with AC and racing stripesandonly racing stripes).
Now since we want to maximize(only AC) we have to see to it thatcars with AC and racing stripesis minimal(assume 0) But since RS>=41.. we have to assign atleast 4 tocars with AC and racing stripes.
Hence AC=63-4=59.
The answer is |
AQUA-RAT | AQUA-RAT-34945 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
The difference between simple and compound interest on Rs. 1700 for one year at 10% per annum reckoned half-yearly is? | [
"8.25",
"4.25",
"9.25",
"3.25"
] | B | S.I. = (1700 * 10 * 1)/100 = Rs. 170
C.I. = [1700 * (1 + 5/100)2 - 1700]
= Rs. 174.25 Difference = (174.25 - 174)
= Rs.4.25.
Answer:B |
AQUA-RAT | AQUA-RAT-34946 | - kante314 - ~ minor edit by junokim1011
## Solution 10(not rigorous)
An ideal solution come to mind is where they intersect at the $x$-axis at the same time, which is $(a, 0)$ and $(-a, 0).$ Take the root of our $y=x^2-a$ we get $x=\sqrt{a}$, set them equal we get $a=\sqrt{a}.$ The only answer is $1$ so it only left us with the answer choice $\boxed{a>\frac{1}{2}}$.
~@azure123456
~ dolphin7
The following is multiple choice question (with options) to answer.
Lines y=√5·x+1 and y=2√5·x−1 intersect at what height above the x axis? | [
"3",
"√5",
"3√5",
"0"
] | A | Solve two equations for y
y=√5·x+1
y=2√5·x−1
Set the two equations equal to each other and solve for x
√5·x+1=2√5·x−1
2 = √5·x
2/(√5) = x
Plug x = 2/(√5) into any one of the original equations to find the y value (height above x axis) where the two lines intersect
y=√5·x+1
y=√5·(2/(√5))+1
y = 2 + 1
y= 3
Final answer:
A) 3 |
AQUA-RAT | AQUA-RAT-34947 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
A number when successively divided by 3,5 and 8 leaves remainders 1,4 and 7 respectively. Find the respective remainders E when the order of the divisors is reversed. | [
"8,5,3",
"4,2,1",
"E=3,2,1",
"E=6,4,2"
] | D | My approach would be
Three Divisors: 3, 5, 8
Three Remain: 1, 4, 7
Start from the bottom of the last column i.e. from the third remainder:
Go up diagonally and multiply by the second divisor: 5*7 = 35
Go down and add the second remainder: 35 + 4 = 39
Go up diagonally and multiply by the first divisor: 39* 3 = 117
Go down and add the first remainder: 117 + 1 = 118
Divide 118 by 8, 5, 3
118/8 gives quotient = 14 and remainder = 6
14/5 gives quotient = 2 and remainder = 4
2/3 gives quotient = 0 and remainder = 2
Ans is D i.e. 6,4,2 |
AQUA-RAT | AQUA-RAT-34948 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
Of the 100 athletes at a soccer club, 40 play defense and 60 play midfield. If at least 20 of the athletes play neither midfield nor defense, the number of athletes that play both midfield and defense could be any number between | [
"10 to 20",
"10 to 40",
"20 to 40",
"30 to 70"
] | C | First of all notice that since only 40 athletes play defense, then the number of athletes that play both midfield and defense cannot possibly be greater than 40.Eliminate D and E.
{Total} = {defense} + {midfield} - {both} + {neither}
100 = 40 + 60 - {both} + {neither}
{both} = {neither} + 0
Since the least value of {neither} is given to be 20, then the least value of {both} is 20+0=20.Eliminate A and B.
Answer: C. |
AQUA-RAT | AQUA-RAT-34949 | \frac{\sqrt{137147}}{2 {{3}^{\frac{3}{2}}}}+\frac{15}{2}\right) }^{\frac{1}{3}}}-\frac{32}{3 {{\left( \frac{\sqrt{137147}}{2 {{3}^{\frac{3}{2}}}}+\frac{15}{2}\right) }^{\frac{1}{3}}}}+2,1],[3,1],[2,3,1]]$
The following is multiple choice question (with options) to answer.
What is the units digit of (12^3)(15^4)(31^7)? | [
"0",
"2",
"4",
"6"
] | A | The units digit of 12^3 is the units digit of 2^3 which is 8.
The units digit of 15^4 is the units digit of 5^4 which is 5.
The units digit of 31^7 is the units digit of 1^7 which is 1.
The units digit of 8*5*1 is 0.
The answer is A. |
AQUA-RAT | AQUA-RAT-34950 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election between two candidates A and B, the number of valid votes received by A exceeds those received by B by 15% of the total number of votes polled. If 20% of the votes polled were invalid and a total of 7720 votes were polled, then how many valid votes did B get? | [
"1888",
"2999",
"2834",
"2509"
] | D | Let the total number of votes polled in the election be 100k.
Number of valid votes = 100k - 20% (100k) = 80k
Let the number of votes polled in favour of A and B be a and b respectively.
a - b = 15% (100k) => a = b + 15k
=> a + b = b + 15k + b
Now, 2b + 15k = 80k and hence b = 32.5k
It is given that 100k = 7720
32.5k = 32.5k/100k * 7720 = 2509
The number of valid votes polled in favour of B is 2834.
Answer: D |
AQUA-RAT | AQUA-RAT-34951 | (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.
We bought a total of 80 books at the store. Math books cost $4 and history books cost $5. The total price was $368. How many math books did we buy? | [
"28",
"32",
"36",
"40"
] | B | M + H = 80
H = 80 - M
4M + 5H = 368
4M + 5*(80 - M) = 368
M = 32
The answer is B. |
AQUA-RAT | AQUA-RAT-34952 | 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 90 meters long completely crosses a 270-meter long bridge in 30 seconds. What is the speed of the train in km/h? | [
"34.2 km/h",
"37.2 km/h",
"40.2 km/h",
"43.2 km/h"
] | D | speed = distance / time = (90 + 270)/30 = 360/30 (m/s) * 3600/1000 = 43.2 km/h
The answer is D. |
AQUA-RAT | AQUA-RAT-34953 | # Faulty Combinatorial Reasoning?
I have 10 books, 4 of which are biographies while the remaining 6 are novels. Suppose I have to choose 4 total books with AT LEAST 2 of the 4 books being biographies. How many different combinations of choosing 4 books in such a way are there?
The following line of reasoning is faulty, but I can't figure out why:
First we figure out how many ways there are of choosing 2 biographies from 4. Then we multiply this by the number of ways there are of choosing 2 of any of the remaining books from 8. This way we will ensure that we get at least two biographies (perhaps more) when we enumerate the choices. Then we have:
1. BIOGRAPHIES: There are (4*3)/2! choices for the two biographies (we divide by 2! since the order in which the two biographies are chosen doesn't matter).
2. REMAINING BOOKS: There are now 8 books left (6 novels, 2 biographies), which can be chosen in any order. This leaves us with (8*7)/2! choices.
3. Overall we have [(4*3)/2!]*[(8*7)/2!] = 168 total choices.
Where did I go wrong?
-
How could I adjust for the over-counting I did here? (Rather than constructing the answer of 115 by adding together the discrete cases of choosing 2 bios, 3 bios, and 4 bios)? – George Sep 4 '12 at 22:15
In your reasoning, you are counting some cases several times. For example, if you take the biographies $B_1$ and $B_2$ as your mandatory biographies and take $B_3$ and $B_4$ as the two other ones, or if you take $B_£$ and $B_4$ as the mandatory ones and $B_1$ and $B_2$ as the other books, it is the same choice of $4$ books, but it will be counted twice.
To solve the problem:
The following is multiple choice question (with options) to answer.
The county-mandated guidelines at a certain community college specify that for the introductory English class, the professor may choose one of three specified novels, and choose two from a list of 5 specified plays. Thus, the reading list for this introductory class is guaranteed to have one novel and two plays. How many different reading lists could a professor create within these parameters? | [
"15",
"30",
"90",
"150"
] | B | There are three possibilities for the novel. With the plays, we are taken a combination of 2 from a set of 5 right n = 5, r = 2, n – r = 3
# of combinations = 5!/2!3! = (5)(4)(3)(2)(1)/(2)(1)(3)(2)(1) = (5)(4)/2 = 10
If the plays are P, Q, R, S, and T, then the 10 sets of two are PQ, PR, PS, PT, QR, QS, QT, RS, RT, & ST.
Any of the three novels can be grouped with any of the 10 possible pairs of plays, for a total of 30 possible reading lists.
Answer: B. |
AQUA-RAT | AQUA-RAT-34954 | Of course, since all four triangles have the same base lengths, the given area ratios are arithmetically equivalent to corresponding height ratios. I used that to write a second equation.
$\displaystyle \frac{\Delta BCP}{\Delta CDP} = \frac{y}{12-x} = \frac{1}{3}$
Simplifying terms and clearing denominators leads to $4x=36-3y$ and $3y=12-x$, respectively.
A VERY INTERESTING insight at this point is that there is an infinite number of locations within the square at which each ratio is true. Specifically, the $\Delta ABP : \Delta DAP = 3:4$ ratio is true everywhere along the line 4x=36-3y. This problem constrains us to only the points within the square with vertices (0,0), (12,0), (12,12), and (0,12), but setting that aside, anywhere along the line 4x=36-3y would satisfy the first constraint. The same is true for the second line and constraint.
I think it would be very interesting for students to construct this on dynamic geometry software (e.g., GeoGebra or the TI-Nspire) and see the ratio remain constant everywhere along either line even though the triangle areas vary throughout.
Together, these lines form a 2×2 system of linear equations with the solution to both ratios being the intersection point of the two lines. There are lots of ways to do this; I wonder how a typical 6th grader would tackle them. Assuming they have the algebraic expertise, I’d have work them by hand and confirm with a CAS.
The question asks for the area of $\Delta ABP = \frac{1}{2}*12*x = 6*8 = 48$.
PROBLEM VARIATIONS
Just two extensions this time. Other suggestions are welcome.
The following is multiple choice question (with options) to answer.
Triangle ATriangle B are similar triangles with areas 2023 units square and 1792 units square respectively. The ratio of there corresponding height would be | [
"9:10",
"17:16",
"23:27",
"13:17"
] | B | Let x be the height of triangle A and y be the height of triangle of B.
since triangles are similar, ratio of area of A and B is in the ratio of x^2/y^2
Therefore, (x^2/y^2)=2023/1792
(x^2/y^2)=(17*17*7)/(16*16*7)
(x^2/y^2)=17^2/16^2
x/y=17/16
Ans=B
You need to take a hint from given options to find factors of 2013 and 2527. |
AQUA-RAT | AQUA-RAT-34955 | # computing probability that at least one letter matches with envelope
Four letters to different insurers are prepared along with accompanying envelopes. The letters are put into the envelopes randomly. Calculate the probability that at least one letter ends up in its accompanying envelope.
### Attempt
Since this is tedioues, we can do
$$P(at \; least \; one ) = 1 - P( no \; match )$$
We notice that sample space is $4!$ since for letter 1 it has 4 choices but letter 2 has 3 choices and so on. Now, we wanna count in how many ways we get no match.
Let start with first one, we only have $3$ choices for this since it can go to either 2,3,4 envelope.
Now, as for the second one, we have to possibilities. If the first letter went to the second envelope, then the second letter now will have 3 different choices, but if the first letter didnt go to second letter, then the second letter will have 2 choices. Assume the former. Then we have 3 choices for this stage.
Now, for the third one (envelope 2 is taken already and assume letter 2 went to letter 1) then it would have 1 choice only and
last one must go to envelope 3.
Thus, we have $3 \times 3 \times 1 \times 1 = 9$ choices in total
Thus,
$$P(at \; least \; 1 \; letter) = 1 - \frac{9}{24}$$
IS this correct? I still feel as is something wrong because I assumed the letter 2 went to 1 and letter 1 went to envelope 2. Can we do that?
The following is multiple choice question (with options) to answer.
7 letters have to be put into their 7 respective envelopes. If the letters are inserted randomly into the envelopes, what is the probability that all the letters go into their respective envelopes? | [
"1/30",
"1/720",
"1/5040",
"1/240"
] | C | A-B-C-D-E (envelopes)
a-b-c-d-e (letters)
7 letters (a, b, c, d, and e) can be assigned to 7 envelopes (arranged) in 7! ways, out of which only one will be correct, thus the probability is 1/7! = 1/5040.
Answer: C. |
AQUA-RAT | AQUA-RAT-34956 | # 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.
If an article is sold at 18% profit instead of 11% profit, then the profit would be $63 more. What is the cost price? | [
"$600",
"$900",
"$1200",
"$1500"
] | B | 7% * cost price = $63
1% * cost price = $63 / 7 = $9
The cost price is $900.
The answer is B. |
AQUA-RAT | AQUA-RAT-34957 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work? | [
"12",
"18",
"22",
"24"
] | D | Let the required number of days be x.
Less men, More days (Indirect Proportion)
27 : 36 :: 18 : x 27 * x = 36 * 18
x = (36 * 18)/27
x = 24.
Answer is D. |
AQUA-RAT | AQUA-RAT-34958 | 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.
What is the present age of Kim, given that the ratio between her age and her sisters age 7 years ago was 4:3, and her sister is presently 31 years old. | [
"23",
"19",
"22",
"21"
] | D | Let the present ages of Kim's sister and Kim be 4x years and 3x years respectively
4x+7 = 35
4x = 28
x = 7
Kim's age = 3x = 21 years
Answer is D |
AQUA-RAT | AQUA-RAT-34959 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
What is the smallest positive perfect square that is divisible by 4, 9, and 18? | [
"36",
"81",
"144",
"625"
] | A | The number needs to be divisible by 2^2, 3^2, and 2*3^2.
The smallest such perfect square is 2^2*3^2 = 36
The answer is A. |
AQUA-RAT | AQUA-RAT-34960 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
In a cinema hall for the matinee show which started at 2.30 pm and ended at 5.30 pm the vehicle parking lot had only three wheelers (auto rickshaws) and four wheelers. Today the wheels of the vehicles parked was counted and found to be 240. What could be the number of three wheelers parked therein ? | [
"48",
"37",
"26",
"98"
] | A | Explanation:
3X + 4y = 240
by substitute through options 48 is correct
Answer: A |
AQUA-RAT | AQUA-RAT-34961 | algorithms
Title: How many times can you pour milk among 3 buckets? This is a problem from usaco, which I solved, but I don't know how to estimate the number of times you can pour milk among the jugs, my solutions just uses a big enough number.
Here's the problem statement:
Farmer John has three milking buckets of capacity A, B, and C liters. Each of the numbers A, B, and C is an integer from 1 through 20, inclusive. Initially, buckets A and B are empty while bucket C is full of milk. Sometimes, FJ pours milk from one bucket to another until the second bucket is filled or the first bucket is empty. Once begun, a pour must be completed, of course. Being thrifty, no milk may be tossed out.
Write a program to help FJ determine what amounts of milk he can leave in bucket C when he begins with three buckets as above, pours milk among the buckets for a while, and then notes that bucket A is empty.
For example, if the input is 8, 9, 10 then the output is 1, 2, 8, 9, 10.
Another example: if the input is 2, 5, 10 then the output is 5, 6, 7, 8, 9, 10.
My approach is to recursively try all possible move sequences, keeping track of the contents of bucket C each time bucket A is empty. Currently I stop the recursion at the arbitrary depth 200. What is the correct depth to stop the recursion at? Since the total amount of water is C liters (C being the capacity of bucket C), the number of possible states is at most the number of non-negative integer solutions of $x+y+z = C$, which is $\binom{C+2}{2}$. Any sequence of moves longer than that must repeat a state. Therefore it is enough to recurse up to level $\binom{C+2}{2}$.
The following is multiple choice question (with options) to answer.
A dishonest milk man tops up his bucket which is only 4/5 th full of milk with water. He again removes 5 liters of this mixture from the bucket and adds an equal quantity of water. If milk is now 60% of the mixture, what is the capacity of the bucket ? | [
"15 lit",
"20 lit",
"22.5 lit",
"25 lit"
] | B | Solution:
Let x be the capacity of the bucket.then, initial mixture has 80% milk.
then after 5 lt replaced with water---->(80/100)∗(x−5)=(60/100)∗x(80/100)∗(x−5)=(60/100)∗x
x=20 lt.
Ans:- B. |
AQUA-RAT | AQUA-RAT-34962 | Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5
Original Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1-x) of the original. So volume of acid left is (1-x)50.
in the new solution.. you added x of 30% solution. i.e 30x acid
(1-x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40%
[(1-x)50 + 30x] / 100 = 40/100
solve to get x=1/2
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Joined: 30 Apr 2008
Posts: 1835
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
### Show Tags
12 Aug 2008, 15:16
The reason that 1/2 is correct is this:
What would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).
The result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.
Here we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.
chrissy28 wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
The following is multiple choice question (with options) to answer.
How many ounces of a 60% salt solution must be added to 40 ounces of a 20 percent salt solution so that the resulting mixture is 40% salt? | [
"16.67",
"40",
"50",
"60.33"
] | B | let x=ounces of 60% salt solution to be added
.2*40+.6x=.4(40+x)
x=40
Amswer B |
AQUA-RAT | AQUA-RAT-34963 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B start a business jointly. A invests Rs. 16,000 for 8 months and B remains in the business for 4 months, Out of total profit, B claims - of the profit. How much money was contributed by B ? | [
"12822",
"12800",
"12819",
"12129"
] | B | Let the total profit be Rs. x. Then, B = 2x/7 and A = (x - 2x /7) = 5x / 7.
So, A : B = 5x/7 : 2x/7 = 5 : 2.
Let B s capital be Rs. y. Then, (16000 * 8) / ( y * 4) = 5/2 <=> (16000 * 8 * 2) / (5 * 4) = 12800..
Answer: B |
AQUA-RAT | AQUA-RAT-34964 | Ans.: A
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Posts: 776
Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink]
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19 Apr 2017, 23:24
dynamo wrote:
Triathlete Dan runs along a 2-mile stretch of river and then swims back along the same route. If Dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour, what is his average rate for the entire trip in miles per minute?
A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8
We can approach this problem using conversion factors
If the total distance is 2 miles and Dan runs at a rate of 10 miles per hour then he will cover in 1/5 of an hour or 1 mile in 6 minutes
1 mile/6 minutes = 2 miles/ 20 minutes
cross multiply add and then divide by 2
(16 miles /2) / 60 minutes
(16 miles/2)/ 1 minute
8 miles/ 1 minute
1 minute/ 8 miles
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Joined: 09 Oct 2016
Posts: 34
Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink]
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01 Aug 2017, 02:38
Dienekes wrote:
If distance traveled is same but at varying rates, the average rate can be found through the shortcut $$\frac{2ab}{a+b}$$ where a and b are individual rates.
In this problem $$a=10$$ and $$b=6$$ therefore average speed = $$\frac{15}{2}$$ miles per hour, i.e. $$\frac{1}{8}$$ miles per minute.
P.S. Added tags as saw this question in an MGMAT CAT.
Thank you for the explanation. I picked answer choice "B" because I did not pay attention to the unit (minutes) in the question.
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Posts: 2298
Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Tabby is training for a triathlon. She swims at a speed of 1 mile per hour. She runs at a speed of 7 miles per hour. She wants to figure out her average speed for these two events. What is the correct answer for her? | [
"8 mph",
"5.25 mph",
"3.5 mph",
"4 mph"
] | D | (1 mph +7 mph)/2 = 4 mph
Correct option is: D |
AQUA-RAT | AQUA-RAT-34965 | A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit
Total cost 60*($250/1.2)=50*250; # of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D. this must be kind of awkward question but i got no option other than asking you. 20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;"
(Cost per unit) + 0.2*(Cost per unit) = $250 1.2*(Cost per unit) =$250
(Cost per unit) = $250/1.2 Total cost for 60 units = 60*(Cost per unit) = 60*($250/1.2) = 50*250.
Hope it's clear.
_________________
Intern
Joined: 02 Mar 2010
Posts: 19
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 10 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? | [
"7% loss",
"13% loss",
"5% profit",
"13% profit"
] | C | Total cost 60*($250/1.1)=54.54*250;
# of cameras sold is 60-5=55 total revenue is 55*250;
# of cameras returned is 6 total refund 6*(250/1.1)*0.5;
So, total income 55*250+ 6*(250/1.1)*0.5
The dealer's approximate profit is (55*250+ 6*(250/1.1)*0.5-54.54*250)/(54.54*250)*100=5%
Answer: C. |
AQUA-RAT | AQUA-RAT-34966 | ##### 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 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is: | [
"7",
"4",
"5",
"3"
] | B | S.I. on Rs. 1600 = T.D. on Rs. 1680.
Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%.
Time = 100 x 80 year = 1 year = 4 months.
1600 x 15 3
Answer:B |
AQUA-RAT | AQUA-RAT-34967 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election between two candidates, one got 55% of the total valid votes. 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got? | [
"2800",
"2700",
"2100",
"2500"
] | B | Total no of votes = 7500
20 % invalid
=>80% valid
tot valid votes = 7500*80/100
=>valid for 2nd candidate got
7500*80/100*45/100
=2700
ANSWER B |
AQUA-RAT | AQUA-RAT-34968 | 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 of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 72 kmph and 90 kmph. In how much time will the trains cross each other? | [
"80/8",
"80/9",
"80/5",
"80/3"
] | B | Relative speed = (72 + 90)* 5/18 = 9* 5 = 45 mps.
The time required = d/s = (100 + 100 + 200)/45
= 400/45 = 80/9 sec.
Answer: B |
AQUA-RAT | AQUA-RAT-34969 | Any particular sequence made up of $3$ A's, $2$ B's, and $1$ C has probability $a^3b^2c^1$. There are $\dbinom{6}{3}\dbinom{3}{2}$ such sequences, so our probability is $$\binom{6}{3}\binom{3}{2}a^3b^2c^1.$$ In the case of your question, we have $a=b=c=\dfrac{1}{3}$.
The same basic strategy can be used for any similar problem. For example, if we modify your problem slightly to have $4$ equally likely outcomes A, B, C, D, then the probability of $4$ A's, $2$ B's, $1$ C and $3$ D's in $10$ trials is $\dbinom{10}{4}\dbinom{6}{2}\dbinom{4}{1}$ divided by $4^{10}$.
-
The following is multiple choice question (with options) to answer.
The probabilities of solving a question by 3students A, B & C are 1/2,1/3 &1/4, respectively. The probability that the problem will be solved is? | [
"3/4",
"3/5",
"4/7",
"4/9"
] | B | Sol. First, we find the probability of not solving the problem x PTO x = (1 — D x - D x - D
1 2 3 1 =iXiX71=4;
1 3 Required probability 3/4
B |
AQUA-RAT | AQUA-RAT-34970 | To find the final answer, we need to find a number $y$ such that $y \equiv 1 \pmod 5$ and $y \equiv 9 \pmod{19}$ (where $0 \leq y \leq 95$). By trying different values of $y' \in \mathbb{N}$ in the equation $x = 19y' + 9$, and seeing which have a last digit of 1 or 6, we get that $y' = 3$ and $y = 66$. So the least residue modulo 95 of $x$ is 66.
3.2Accuracy and discussion¶
This problem was worked through by the TA during the tutorial session on Friday, October 19. Also worked through by Darmon on Monday, October 22.
4Question 4¶
Using only the basic properties of the gcd proved in class, (and not the fundamental theorem of arithmetic) show that if a $p$ is a prime and $p$ divides a product $ab$ of two integers, then $p$ necessarily divides either $a$ or $b$.
4.1Solution¶
Since $p\mid ab$ and $p \mid p$ (by the reflexivity of $\mid$), then $\gcd(p, ab) = p$ (since nothing greater than $p$ can divide $p$). We assume that $p \nmid a$ and $p \nmid b$. So $\gcd(p, a) = 1$ and $\gcd(p, b) = 1$. By theorem 9.1.2 in the notes (also known as Bézout's identity), there exist integers $x$ and $y$ such that $xp + ya = 1$ (from the fact that $\gcd(p, a) = 1$). If we multiply both sides of the identity by $b$, we get:
$$xbp + yab = b$$
The following is multiple choice question (with options) to answer.
42573y is exactly divisible by 72, then what is the minimum value of y? | [
"3",
"4",
"6",
"8"
] | C | 72 = 9 x8, where 9 and 8 are co-prime.
The minimum value of x for which 73x for which 73x is divisible by 8 is, x = 6.
Sum of digits in 425736 = (4 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 9.
Required value of y is 6
C |
AQUA-RAT | AQUA-RAT-34971 | ) the! Two groups of students are within three standard deviations from a mean value, I! Measure this spread around the mean © 2021 Stack Exchange Inc ; user licensed! Reconceptualized in terms of distance from the mean exposition on a magic system when no character has an or. Have the same average but very different spreads since it is 496.9-86.6 = 410.3, a. 1 that 140 children had a urinary lead concentration of 2.18 µmol24hr, standard... For FEV1 graph there is much more variation among scores ) is a risky! Side note: these conversions and comparisons are complicated and sometimes misleading calculate the mean become... 5:00 pm ET a property looks like this, AVM: $200,000 scores are directly! Suppose mutual fund a has a 10 % standard deviation and population standard is... We have a group of 1,000 students who take a test ( scored 0-20 ) speaking, in a distribution. Is one s.d. ) 84th percentile a test ( scored 0-20 ) the data! By context. * * is greater than 3.89 and so falls in the second,... Positions of asi or any related entity or organization student Growth percentiles use... Blog offers informal commentary on the dates in their names very different spreads informal on! We say that the deviation of the children had a urinary lead concentration of over. T telling you the same thing about a given score is one s.d..! That 140 children had a mean urinary lead concentration of just over µmol24hr... For help, clarification, or responding to other answers reported plus or minus of a mean urinary concentration. Small, and over 99 % are within three to build crewed rockets/spacecraft able to reach escape?... On your business needs all close to 7 deviations ( this is in of. Other snow has melted ( or decrease ) in percentile varies depending on questions. 6-Pound melon is 1 s.d. ) compactly, whereas a score of 13 is the. Limits: three-sigma limit ( 3-sigma limits ) is a statistical calculation that refers to data within.. Its weight is equal to the work of the scale, and controversies related to the or... Makes no warranties, either express
The following is multiple choice question (with options) to answer.
For a certain exam,a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean.What was the mean score R for the exam? | [
"74",
"76",
"78",
"80"
] | A | A score of 58 was 2 standard deviations below the mean --> 58 = Mean - 2d
A score of 98 was 3 standard deviations above the mean --> 98 = Mean + 3d
Solving above for Mean R= 74.
Answer: A. |
AQUA-RAT | AQUA-RAT-34972 | 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.
5 years ago, Paula was 3 times as old as Karl. In 6 years, Paula will be twice as old as Karl. What is the sum of their ages now? | [
"42",
"46",
"50",
"54"
] | D | P-5 = 3(K-5) and so P = 3K-10
P+6 = 2(K+6)
(3K-10)+6 = 2K+12
K = 16
P = 38
P+K = 54
The answer is D. |
AQUA-RAT | AQUA-RAT-34973 | Note: The author(s) of your text are calculating the probability that the top-ranked woman is in the $k$th position. For that to occur, we must select $k - 1$ of the five men to be ranked ahead of her and one of the five women to fill the $k$th position while choosing $k$ of the $10$ people. Hence, the answer in the text is equivalent to \begin{align*} P(X = 1) & = \frac{\binom{5}{0}\binom{5}{1}}{\binom{10}{1}} = \frac{1}{2}\\ P(X = 2) & = \frac{\binom{5}{1}\binom{5}{1}}{\binom{10}{2}} = \frac{5}{18}\\ P(X = 3) & = \frac{\binom{5}{2}\binom{5}{1}}{\binom{10}{3}} = \frac{5}{36}\\ P(X = 4) & = \frac{\binom{5}{3}\binom{5}{1}}{\binom{10}{4}} = \frac{5}{84}\\ P(X = 5) & = \frac{\binom{5}{4}\binom{5}{1}}{\binom{10}{5}} = \frac{5}{252}\\ P(X = 6) & = \frac{\binom{5}{5}\binom{5}{1}}{\binom{10}{6}} = \frac{1}{252} \end{align*}
The following is multiple choice question (with options) to answer.
A five-person team is to be formed from a pool of 6 East All Stars and 6 West All Stars. What is the probability that the team will contain at least 2 East All Stars? | [
"29/33",
"81/169",
"57/120",
"47/144"
] | A | Teams can be formed as 2E-3W , 3E-2W , 4E-W and 5E
Possible ways to make a team 2E-3W = 6c2 * 6c3 = 300
Possible ways to make a team 3E-2W = 6c3 * 6c2 = 300
Possible ways to make a team 4E-W = 6c4 * 6c1 = 90
Possible ways to make a team 5E = 6c5 = 6
Number of ways to select a team of 5 out of 12 people = 12c5
=12!/(7!*5!)
= 792
Probability that the team will contain at least 2 East All Stars
= (300+300+90+6) / 792
=696/792
= 29/33
Answer A
Alternatively we can also calculate the number of possible ways to make a team of 5W and E-4W .
Possible ways to make a team 5W = 6c5 = 6
Possible ways to make a team E-4W = 6c1 * 6c4 = 90
Probability that the team will contain at least 2 East All Stars = 1 - (96/792)
= 29/33 |
AQUA-RAT | AQUA-RAT-34974 | • #### 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.
A rectangular park is 110 m long and 65 m wide.in this park there is 2.5m wide path all four side.on this path on the rate of 7.20 per sq. m what will be cost of spreading sand? | [
"6120",
"7350",
"6250",
"8285"
] | A | Area of park=110*65=7150sq.m
Area of park without path=[(110-2.5*2)×(65-2.5*2)]sq.m
(105*60) sq.m=6300 sq.m
Area of path=(7150-6300)=850 sq.m
Cost of spreading sand=[850×720/100]=6120
Answer A |
AQUA-RAT | AQUA-RAT-34975 | (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.
9 men went to a hotel. Eight of them spent Rs.3 each over their meals and the ninth spent Rs.2 more than the average expenditure of all the nine. Determine the total money spent by them? | [
"29.25",
"30",
"31.2",
"34"
] | A | Average of 9 = x
9x = 8 * 3 + x * 2 x = 3.25
Total = 9 * 3.25 = 29.25
A |
AQUA-RAT | AQUA-RAT-34976 | # Generalized repetitions of letters with limited amount of adjacent letters
Say I have the first $$x$$ letters of the alphabet, and I want to generate a sequence of length $$y$$, such that there are no more than $$z$$ adjacent repeated letters. For example, if $$x = 2$$, $$y = 3$$ and $$z = 2$$, here are all the valid sequences:
AAB ABA ABB BAA BAB BBA
How many total possible sequences are there?
Without the restriction of z, the question devolves to $$x^y$$, but I am stuck as to how to proceed further.
The case I have described earlier is also not too hard, I think it is $$x^y-x$$ since $$z = y-1$$.
Would it be better to count all cases and subtract (like I did for the $$z = y-1$$ case) or should I try and count upwards?
To be clear, you are allowed to repeat letters as many times as you want so long as they never lie next to each other more than $$z$$ times.
Thanks for the help!
The following is multiple choice question (with options) to answer.
There are 6 distinct letters of the English alphabet and 4 distinct digits. all possible 6 character apha-numero codes are generated using any 4 letters of the alphabet and any 2 available digits. If in any given code, the characters are all distinct, then what is the maximum number of such codes that can be generated? | [
"4320",
"64800",
"8800",
"22000"
] | B | From the six letters, we need four distinct letters
# of combinations = 6C4 = 15
From the four numerals, we need 2 distinct digits
4C2 = 6
Once we have our unique four letter and our unique 2 digits, we can put them in any order: each new order would be a different code. The number of orders would be
6! = 720
Total number of codes equals = 15*6*720
Without a calculator, first do 15*6 = 90. Then, 90 = 100 - 10, so
720*90 = 720*(100 - 10) = 72000 - 7200 = 64800
Answer = (B) |
AQUA-RAT | AQUA-RAT-34977 | electricity
Title: Why does this automatic night light advertise money savings? I just bought a night light to light the hallway since it receives no outside light. The one I bought advertises that it saves money because it turns on only during the night or (dark hours). I did the calculations and I am confused as to why it is even advertised because I did the calculations to find the cost per month to run it and as it turns out the money savings are absolutely miniscule.
The power information on the night light is 120v, 60hz, and 0.15w.
To find the cost to run it per month:
0.15w * 1kw/1000w * 24 hours * 30 days = 0.108 kwh per month.
At a cost of 9 cents per kwh, the total cost comes to less than 1 cent per month. The fact that the night light is automatic does not make a big difference on my wallet--why with a dollar I could run it for 8.3 years. Am I missing something or is this just an example of advertising? This is pure marketing, but not a lie... 0.036cents/mo is less than 0.108 cents/mo.
The following is multiple choice question (with options) to answer.
A certain electric-company plan offers customers reduced rates for electricity used between 8 p.m. and 8 a.m. weekdays and 24 hours a day Saturdays and Sundays. Under this plan, the reduced rates Z apply to what fraction of a week? | [
" 1/2",
" 5/8",
" 9/14",
" 16/21"
] | C | Number of hours between 8 PM to 8 AM = 12
Number of hours with reduced rates = (12*5)+(24*2)
Hours with reduced rates Z/Total number of hours in a week = (12*5)+(24*2)/(24*7) = 108/(24*7) = 9/14
Answer: C |
AQUA-RAT | AQUA-RAT-34978 | solubility, alloy, teaching-lab
What other such situations are there is chemistry; when would one actually need to do such simultaneous calculations? You may also give the weight of a sample made of an alloy $\ce{Zn-Mg}$. Then you dip it into some diluted $\ce{HCl}$. Both metals will react simultaneously according to $$\ce{Zn + 2 HCl -> H_2 + ZnCl_2}$$ and $$\ce{Mg + 2 HCl -> H_2 + MgCl_2}$$ This will produce an important amount of $\ce{H_2}$ gas. You can measure its volume. So you have two measured data, the original mass and the volume of gas. It is enough for calculating the proportion of $\ce{Zn}$ and $\ce{Mg}$ in the original sample.
Example : Take a mixture of $0.1$ mol $\ce{Zn}$ + $0.3$ mol $\ce{Mg}$. Of course the molar masses of $\ce{Zn}$ and $\ce{Mg}$ are respectively $65.39$ g/mol and $24.32$ g/mol. The total mass is : $6.54$ g + $7.29$ g = $13.83$ g. Suppose the temperature and the pressure of the gas is such that $1$ mole gas occupies $24$ L. So $0.1$ mol $\ce{Zn}$ will produce $0.1$ mol $\ce{H_2}$ which occupies $2.4$ L gas. Then $0.3$ mol $\ce{Mg}$ will produce $0.3$ mol $\ce{H_2}$, or $7.2$ L gas. The total volume of $\ce{H_2}$ is $2.4$ L + $7.2$ L = $9.6$ L.
The following is multiple choice question (with options) to answer.
Zinc and copper are melted together in the ratio 9 : 11. What is the weight of melted mixture, if 27 kg of zinc has been consumed in it? | [
"58 kg",
"60 kg",
"64 kg",
"70 kg"
] | B | Sol.
For 9 kg Zinc, mixture melted = (9+11) kg.
For 27 kg Zinc, mixture, melted = [20/9 x 27] kg
= 60 kg.
Answer B |
AQUA-RAT | AQUA-RAT-34979 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
A certain party consists of four different group of people 30 students, 35 politicians, 20 actors and 27 leaders. On a particular function day, the total cost spent on party members was Rs.9000. It was found that 6 students spent as much as 7 politicians, 15 politicians spent as much as 12 actors and 10 actors spent as much as 9 leaders. How much did students spent? | [
"2191",
"2200",
"2291",
"2391"
] | C | Let S be the amount spent for all students
Let P be the amount spent for all politicians
Let A be the amount spent for all actors and
Let L be the amount spent for all leaders.
given that 6 students spent as much as 7 politicians
i.e Amount spent per student x 6 = Amount spent per politician x 7
Or (Total amount spent on students / Total Students) x 6 = (Total amount spent on politicians / Total politicians)x 7
Or., (S/30)x 6 = (P/35)x7
=> S/5 = P/5
=> S = P ..... eqn 1
15 politicians spent as much as 12 actors :
Therefore, (P/35)x15 = (A/20)x12
=> 3P/7 = 3A/5
=> 5P = 7P
=> 5S = 7A (since S = P)
=> A = 5S/7 ........eqn2
and 10 actors spent as much as 9 leaders :
Therefore, (A/20)x10 = (L/27)x9
=> 3A = 2L
=> A = 2L/3
=> 5S/7 = 2L/3 (since A=5S/7)
=> L = 15S/14 ......eqn3
the total amount spent for the party is Rs.9000
i.e., S+P+A+L = 9000
from eqn1, eqn2, eqn3 we have
S + S + 5S/7 + 15S/14 = 9000
53S/14 = 9000
s=9000*14 / 53
s=2290.90 ~ 2291
Hence the answer is Rs.2291
ANSWER:C |
AQUA-RAT | AQUA-RAT-34980 | (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.
Anna and Carol buy CDs and tapes at a music store that sells each of its CDs for a certain price and each of its tapes for a certain price. Anna spends twice as much as Carol spends, buying three times as many CDs and the same number of tapes. If Carol spends $70.00 on four CDs and five tapes, how much does one tape cost? | [
"$5.00",
"$6.25",
"$7.00",
"$25.00"
] | C | CD => C
Type => T
Carol: 4C+5T=70
Anna: 12C+5T=140
12C-4C=70
8C=70
C=8.75
4C+5T=70
35+5T=70
5T=35
T=7
Answer is C |
AQUA-RAT | AQUA-RAT-34981 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
Anne, Markov, Peter line up at a ticket window. In how many ways can they arrange themselves so that Peter is first in line from the window? | [
"a 24",
"b 12",
"c 9",
"d 2"
] | D | D
Peter's position is fixed. Now we need to arrange all others around Peter.
There are 2x1= 2 ways to arrange two people. |
AQUA-RAT | AQUA-RAT-34982 | # 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.
When a merchant imported a certain item, he paid a 9 percent import tax on the portion of the total value of the item in excess of $1,000. If the amount of the import tax that the merchant paid was $76.50, what was the total value of the item? | [
"$1,250",
"$1,450",
"$1,650",
"$1,850"
] | D | Let x be the value in excess of $1,000.
0.09x = 76.5
x = $850
The total value was $850 + $1,000 = $1,850.
The answer is D. |
AQUA-RAT | AQUA-RAT-34983 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election only two candidates contested. A candidate secured 70% of the valid votes and won by a majority of 172 votes. Find the total number of valid votes? | [
"430",
"436",
"477",
"439"
] | A | Let the total number of valid votes be x.
70% of x = 70/100 * x = 7x/10
Number of votes secured by the other candidate = x - 7x/100 = 3x/10
Given, 7x/10 - 3x/10 = 172 => 4x/10 = 172
=> 4x = 1720 => x = 430.
Answer:A |
AQUA-RAT | AQUA-RAT-34984 | +0
# What is the greatest common divisor of $2^{1001}-1$ and $2^{1012}-1$?
0
234
2
+474
What is the greatest common divisor of $$2^{1001}-1$$ and $$2^{1012}-1$$?
Sep 22, 2018
#2
+4032
+2
The theorem being used here is that
$$gcd(a^n -1,~a^m-1) = a^{gcd(m,n)}-1$$
in this case $$n=1001,~m=1012$$
$$1001 = 7 \cdot 11 \cdot 13 \\ 1012 = 2 \cdot 11 \cdot 23 \\ \text{so clearly }gcd(1001,~1012)=11\\ 2^{11}-1 = 2047 \text{ as guest correctly computed}$$
The proof of this theorem can be found online
Sep 22, 2018
#1
+1
I think the GCD of ((2^1001) - 1), ((2^1012) - 1) =2^(1012 - 1001) - 1 =2^11 - 1 =2,047
Sep 22, 2018
#2
+4032
+2
The theorem being used here is that
$$gcd(a^n -1,~a^m-1) = a^{gcd(m,n)}-1$$
in this case $$n=1001,~m=1012$$
$$1001 = 7 \cdot 11 \cdot 13 \\ 1012 = 2 \cdot 11 \cdot 23 \\ \text{so clearly }gcd(1001,~1012)=11\\ 2^{11}-1 = 2047 \text{ as guest correctly computed}$$
The proof of this theorem can be found online
Rom Sep 22, 2018
The following is multiple choice question (with options) to answer.
The greatest common divisor of (b-2)!,(b+1)!, and (b+4)! is 5040. what is the value of b? | [
"8",
"10",
"12",
"9"
] | D | GCD of (b-2)!,(b+1)!, and (b+4)! is 5040. Find x.
Best way to find n is substitute each of the answer and find out the correct one.
Only option D gave the correct answer.
Option D: 9 and find GCD or 7!, 10! and 13!. 7! itself is 5040 and 7! which is contained in the other two factorials. This satisfies the condition, hence answer is 9.
Answer is D. |
AQUA-RAT | AQUA-RAT-34985 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
By walking at 3/4th of his usual speed, a man reaches office 18 minutes later than usual. What is his usual time? | [
"43min",
"54min",
"65min",
"60min"
] | B | let T be the usual time.
time spent=4T/3
therefore, 4T/3=T+18
4T=3T+54...therefor T=54min
ANSWER:B |
AQUA-RAT | AQUA-RAT-34986 | 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 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train ? | [
"230",
"777",
"267",
"288"
] | A | Explanation:
Relative speed = (120 + 80) km/hr
==
Let the length of the other train be x metres.
Then,
x + 270 = 500
x = 230.
Answer: A) 230 m |
AQUA-RAT | AQUA-RAT-34987 | • Possible duplicate of Chess Master Problem – Aqua Dec 21 '17 at 20:01
• – Aqua Dec 21 '17 at 20:04
• – Aqua Dec 21 '17 at 20:05
We define a sequence $a_1,a_2,\ldots$, by letting $a_i$ be the amount of games played on day $i$. It is given that for all $n\in\mathbb{N}$, we have: $$a_{7n+1}+a_{7n+2}+a_{7n+3}+a_{7n+4}+a_{7n+5}+a_{7n+6}+a_{7n+7}\le 12$$ Now, define $$b_k=\sum_{i=1}^{k}a_i$$ for all $k=1,2\ldots$. By the pigeonhole principle, we can find some $l>m$ such that: $$b_l\equiv b_m\pmod {20}$$ and $l-m\le 20$. This means that: $$\sum_{i=m+1}^{l}a_i\equiv 0\pmod {20}$$ and $$1\le\sum_{i=m+1}^{l}a_i<3\cdot 12=36$$ Since the chess player plays at least one game per day and $l>m$. We conclude that: $$\sum_{i=m+1}^{l}a_i=20$$ and we are done.
The following is multiple choice question (with options) to answer.
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants. | [
"15",
"16",
"17",
"18"
] | D | If the number of participants is 3 (say A, B, C) the number of games played will be 2 (A plays against B and C) + 1 (B plays against C) = 3
Using the same logic, if the number of participants is n, the number of games played will be (n-1) + (n - 2) + (n - 3) + ... 3 + 2 + 1
Given that this sum = 153 = 1 + 2 + 3 + ... ( n - 1)
Sum of first m positive integers is given by m(m+1)/2. So sum of first (n-1) positive integers is (n-1)*n/2
153 = (n-1)*n/2
(n-1)*n = 306
17*18 = 306 (We know that 15^2 = 225 so the two consecutive numbers must be greater than 15. Also, 20^2 = 400 so the two numbers must be less than 20. The pair of numbers in between 15 and 20 whose product ends with 6 is 17 and 18)
So n = 18
Answer (D) |
AQUA-RAT | AQUA-RAT-34988 | python, python-3.x, calculator
Title: Taxi fare calculator, trip recorder, and fuel calculator I solved this practice problem in preparation for school exams.
The owner of a taxi company wants a system that calculates how much money his taxis take in one day.
Write and test a program for the owner.
Your program must include appropriate prompts for the entry of data.
Error messages and other output need to be set out clearly and understandably.
All variables, constants and other identifiers must have meaningful names.
You will need to complete these three tasks. Each task must be fully tested.
TASK 1 – calculate the money owed for one trip
The cost of a trip in a taxi is calculated based on the numbers of KMs traveled and the type of taxi that you are traveling in. The taxi company has three different types of taxi available:
a saloon car, that seats up to 4,
a people carrier, that seats up 8,
a mini-van, that seats up to 12.
For a trip in a saloon car the base amount is RM2.50 and then a charge of RM1.00 per KM. For a trip in a people carrier the base amount is RM4.00 and a charge of RM1.25 per KM. For a trip in a mini- van the base amount is RM5.00 and a charge of RM1.50 per KM. The minimum trip length is 3KM and the maximum trip length is 50KM. Once the trip is complete a 6% service tax is added.
TASK 2 – record what happens in a day
The owner of the taxi company wants to keep a record of the journeys done by each one of his taxis in a day. For each one of his three taxis record the length of each trip and the number of people carried. Your program should store a maximum of 24 trips or 350km worth of trips, whichever comes first. Your program should be able to output a list of the jobs done by each one of the three taxis.
TASK 3 – calculate the money taken for all the taxis at the end of the day.
At the end of the day use the data stored for each taxi to calculate the total amount of money taken and the total number of people carried by each one of the three taxis. Using the average price of RM2.79 per litre use the information in the table below to calculate the fuel cost for each taxi:
The following is multiple choice question (with options) to answer.
A cab driver 5days income was $30, $50, $60, $65 , $70. Then his average income is? | [
"A)$55",
"B)$42",
"C)$65",
"D)$48"
] | A | avg = sum of observations/number of observations
avg income=(30+50+60+65+70)/5 = 55
Answer is A |
AQUA-RAT | AQUA-RAT-34989 | Circular motion question
1. Nov 10, 2005
donjt81
This is the question...
A small wheel of radius 1.4cm drives a large wheel of radius 15cm by having their circumferences pressed together. If the small wheel turns at 407 rad/s, how fast does the larger one turn? Answer in rad/s
This is what I was thinking...
radius of smaller wheel = .014m
radius of larger wheel = .15m
circumference of smaller wheel = 2*pi*r = 2*3.14*.014 = .08792
angular velocity of smaller wheel (given) = 407 rad/s
angular velocity = circumference/time
time = circumference/angular velocity
=.08792/407 = .000216s
circumference of larger wheel = 2*pi*r = 2*3.14*.15 = .942
angular velocity = circumference/time
=.942/.000216 = 4361.11 rad/s
Does this approach look right?
2. Nov 10, 2005
BerryBoy
I disagree... By intuition, you can predict that the larger wheel is going to turn more slowly.. Try another approach..
Hint: Consider the fact that the speeds of circumferences are equal.
Eq: speed = w x r
w = angular velocity
r = radius
Does this help?
Sam
Last edited: Nov 10, 2005
3. Nov 10, 2005
donjt81
You are right... the larger wheel should go slower.
but since the speed of smaller wheel is 407 rad/s wont the larger wheel speed be the same?
so is the answer to the problem 407 rad/s for the larger wheel? but that doesnt make sense because the larger wheel is supposed to go slower...
I am confused...
4. Nov 10, 2005
BerryBoy
OK, so if the speed at the circumfrence is:
v = w x r (as I stated above).
If the wheels are in contact this speed is equal on both wheels (not the angular velocity). Therefore:
wsmall x rsmall = wlarge x rlarge
I can't give you anymore hints without doing it now.
The following is multiple choice question (with options) to answer.
A wheel rotates 15 times every minute and moves 35 cm during each rotation. How many metres does the wheel move in one hour? | [
"6 metre",
"12 metre",
"120 metre",
"315 metre"
] | D | Number of times wheel moves in 1 hour = 15 * 60 = 900
:. Distance moves = (900 *35) cms = 31500 cms
In metres = 315 metre
ANSWER:D |
AQUA-RAT | AQUA-RAT-34990 | # Clock losing time puzzle
The question goes as:
A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?
My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720$$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM}$.
Is this correct?
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
The following is multiple choice question (with options) to answer.
A clock loses a minute every three hours for 4 days and gains 1% in the subsequent 6 days. If it was set right on Friday at 8 AM, what will be the time at the end of 10 days? | [
"08:54:40 AM",
"08:56:24 AM",
"08:16:40 PM",
"08:54:24 AM"
] | D | Loses 8 minutes each day.
so total loss = 4×8 = 32min
subsequent 6 days = 6×24×60 minutes
1% gain = ( 6×24×60)/100 = 86.4 minutes
so, total gain = 86.4 - 32 = 54.4 min = 54 + 0.4 min = 54 minutes + 0.4 × 60 seconds = 54min + 24seconds
10 days later the clock should be 54 min and 24seconds fast.
so time , 09:54:24 am (Answer D) |
AQUA-RAT | AQUA-RAT-34991 | Could you please verify that my 1st and 3rd prize probabilities are well calculated and help me calculate the 2nd prize probability?
Your logic for case number (3) does not seem correct.
First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $$C(6,5)\times36$$. So the probability is:
$$p_1=\frac{36\times{6\choose 5}}{42\choose 6}$$
But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:
$$p_2=\frac{35}{36}$$
The total proability is:
$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{35}{36}=\frac{35\times{6\choose 5}}{42\choose 6}=\frac{15}{374699}$$
You can use a similar logic for case (2).
The probability $$p_1$$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $$p_2$$ is:
$$p_2=\frac{1}{36}$$
...and the final probability for the second prize is:
$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{1}{36}=\frac{{6\choose 5}}{42\choose 6}=\frac{3}{2622893}$$
(35 times smaller than the probability for the third prize)
The following is multiple choice question (with options) to answer.
Judges will select 6 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prize-winners are possible? | [
"140",
"21",
"210",
"420"
] | A | First Select 5 from 7 contestants and then select top 3 from those 5 =
7C6*6C3 = 7 * 20 = 140
ANS:A |
AQUA-RAT | AQUA-RAT-34992 | 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.
What will come in place of the x in the following Number series? 68, 36, 20, 12, 8, x | [
"6",
"4",
"2",
"1"
] | A | 68
68 / 2 + 2 = 36
36 / 2 + 2 = 20
20 / 2 + 2 = 12
12 / 2 + 2 = 8
8 / 2 + 2 = 6 A |
AQUA-RAT | AQUA-RAT-34993 | javascript, datetime
// a simple helper function
function nextDay(date) {
return new Date(date.getFullYear(), date.getMonth(), date.getDate() + 1);
}
// as long we're in the same year, keep adding 1 day,
// and store the ones that match the weekday we're looking for
while(current.getFullYear() === year) {
if(current.getDay() === weekday) {
dates.push(current);
}
current = nextDay(current);
}
return dates;
}
No need to build ranges, map, and filter them. Just a while loop and push.
Incidentally, if you want to learn more about why programming calendar and time things are just hideously complex, check out this video on youtube
1) Happy Guy Fawkes day everyone
The following is multiple choice question (with options) to answer.
Today is Thursday. The day after 54 days will be? | [
"Monday",
"Tuesday",
"Saturday",
"Sunday"
] | B | Explanation:
54 days = 7 weeks 3 days = 5 odd days
Hence if today is Thursday, After 54 days, it will be = (Thursday + 5 odd days)
= Tuesday
Answer: Option B |
AQUA-RAT | AQUA-RAT-34994 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
Amit and Ananthu can do a work in 15 days and 90 days respectively. Amit started the work and left after 3 days. Ananthu took over and completed the work. In how many days was the total work completed? | [
"68 days",
"70 days",
"73 days",
"75 days"
] | D | Amit’s one day’s work= 1/15
Amit’s 3 day’s work = 1/15 *3 = 1/5
Work left= 1-1/5 = 4/5
Ananthu’s one day’s work= 1/90
Ananthu can do work in = 4/5 *90 = 72 days
So total days = 72+3 = 75 days
ANSWER:D |
AQUA-RAT | AQUA-RAT-34995 | # 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.
The ratio of three numbers is 4:2:5 and their sum is 121. The second number of the three numbers is? | [
"22",
"26",
"27",
"29"
] | A | 4:2:5
Total parts = 11
11 parts --> 121
1 part ---->11
The second number of the three numbers is = 2 * 11= 22
Answer : A |
AQUA-RAT | AQUA-RAT-34996 | # Integers and Number-theory
The positive integers from $1$ to $n$ (where $n>1$) are arranged in a line such that the sum of any two adjacent numbers is a square. What is the minimum value of $n$?
• Is this your own question or you heard it somewhere because n=1 – Abhishek May 13 '18 at 8:28
• I found one for n=15 but don't know about maximum value of n $8,1,15,10,6,3,13,12,4,5,11,14,2,7,9$ and this one for n=32 on internet $1,8,28,21,4,32,17 ,19$,$30,6,3, 13,12,24,25,11,5,31$,$18,7,29,20 ,16,9,27,22,14,2,23,26,10,15$ – Love Invariants May 13 '18 at 8:39
• @LoveInvariants, it is lovely to get it for two numbers – Satish Ramanathan May 13 '18 at 8:54
• @Satish Ramanathan-Thx – Love Invariants May 13 '18 at 9:08
It is conjectured that there is not a maximum value of $n$. See this post. It is a question of mine regarding exactly this, and today, it still remains an open problem such that most people I know refer to it as the "Square-Sum Problem."
An Australian mathematician by the name of Matt Parker perhaps introduced this problem or popularised it in one of his books, Things to Make and Do in the Fourth Dimension $(2014)$. In the book, he conjectured that there always exists such a sequence for $n > 89$, and then $n > 91$.
However, in a Numberphile YouTube video, Matt Parker and his friend talks about how they tested this conjecture for more values, and as far as I know, the conjecture is true thus far and we now want to test for values $n > 300$.
The video has two parts, Part 1 and Part 2.
The following is multiple choice question (with options) to answer.
The product of the squares of two positive integers is 81. How many pairs of positive integers satisfy this condition? | [
"0",
"1",
"2",
"3"
] | C | Ans: C - 2 pairs
(xˆ2)(yˆ2) = 81 [square root both sides]
xy = 9
9 = 1x9, 9x1, 3x3
Cancel the repeats
This leaves us with exactly 2 options.
Hence, C |
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