mathewhe/aqua_rat huggingface dataset:
Question: In a group of ducks and cows, the total number of legs are 28 more than twice the no. of heads. Find the total no.of buffaloes.
(A) 10
(B) 12
(C) 13
(D) 15
(E) 14
Rationale: Let the number of buffaloes be x and the number of ducks be y
=> 4x + 2y = 2 (x + y) + 28
=> 2x = 28 => x = 14
E
Answer: (E) 14
Question: A type R machine can complete a job in 5 hours and a type B machine can complete the job in 7 hours. How many hours will it take 2 type R machines and 3 type B machines working together and independently to complete the job?
(A) 1/5
(B) 29/35
(C) 5/6
(D) 35/29
(E) 35/12
Rationale: now D should be the answer.
R need 5 hours to complete
and B needs 7 hours to compete
so 2R + 3B will complete 2/5 + 3/7 or 29/35 portion of the job in 1 hour
so the whole job will take 35/29 hours....=D
Answer: (D) 35/29
Question: 0.0008?=0.02
(A) 40
(B) 4
(C) 0.4
(D) None of these
(E) 0.04
Rationale: Explanation :
Required Answer = 0.0008/0.02 = 0.08/2 = 0.04. Answer : Option E
Answer: (E) 0.04
Question: 3 pumps working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?
(A) 34
(B) 77
(C) 36
(D) 49
(E) 38
Rationale: Explanation:
Let the required length be x meters
More men, More length built (Direct proportion)
Less days, Less length built (Direct Proportion)
\inline \fn_jvn \left.\begin{matrix} Men\: \: \: 20:35\\ Days\: 6:3 \end{matrix}\right\}::56:x
\inline \fn_jvn \therefore (20 x 6 x X)=(35 x 3 x 56) \inline \fn_jvn \Leftrightarrow \inline \fn_jvn x=\frac{35\times 3\times 56}{120} =49
Hence, the required length is 49 m.
Answer: D
Answer: (D) 49
Question: A started a bussiness with a sum of Rs. 14000. After 4 months, B entered with an amount of Rs. 12000 and A added another 2,000 in the business. After 6 months, C entered with Rs. 20000 and 'B' withdrew Rs. 2000 from the business. What will be C's share of profit out of a total profit of Rs. 94000 at the end of 2 years.
(A) Rs. 376000
(B) Rs. 204000
(C) Rs. 36000
(D) Rs. 18000
(E) Rs. 260000
Rationale: Investments--
A=14000x4+16000x20=376000
B=12000x2+10000x18=204000
C=20000x18=360000
So C's Share = 94000xC/(A+B+C)=36000
ANSWER:C
Answer: (C) Rs. 36000
Question: A box contains nine slips that are each labeled with one number: 1, 3, 5, 8, 13, 21, 34 and 55. Two of the slips are drawn at random from the box without replacement. What is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box?
(A) 7/72
(B) 1/6
(C) 5/28
(D) 15/36
(E) 21/36
Rationale: Probability = No: of desired outcomes / Total No: of outcomes.
You are picking two slips out of 8 slips. So
Total No: of outcomes = 8C2 = 28
Desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box. How many such outcomes are there?
If you look at the numbers closely, you will see that the following pair of numbers will give you the desired outcome.
(3,5) (5,8) (8,13) (13,21) (21,34) . There are 7 such pairs. If the two numbers which I pick is from any of these 5 pairs, then I get my desired outcome.
So No: of desired outcomes = 5
Probability = 5/28
Answer :C
Answer: (C) 5/28
Question: Given that there are 11 cricket players per team, how many ways can you select 10 cricket players from 2 teams if no more than 2 players can be selected from each team?
(A) 3025
(B) 1500
(C) 2600
(D) 3500
(E) 4200
Rationale: Now from each group we have to select 5 persons only: 11C2 * 11C2= 55 * 55 = 3025
Total ways=313444 ways.
Answer A
Answer: (A) 3025
Question: The game 'Chunk-a-Luck' is played at carnivals in some parts of Europe. Its rules are as follows:
If you pick a number from 1 to 6 and the operator rolls three dice.
If the number you picked comes up on all three dice, the operator pays you Rs. 3.
If it comes up on two dice, you are paid Rs. 2.
And it comes up on just one dice, you are paid Rs. 1.
Only if the number you picked does not come up at all, you pay the operator Rs. 1.
The probability that you will win money playing in this game is:
(A) 0.52
(B) 0.753
(C) 0.42
(D) 0.68
(E) None of the above
Rationale: Explanation :
If one picks up a number out of six numbers then the only case in which he/she will lose money if none of the three dice shows the picked number on the top surface.
Required probability of losing the game:
=>(5/6) x (5/6) x (5/6).
=>125/216.
Hence, the required probability of winning the game is :-
=>1-(125/216).
=>91/216.
=>0.42.
Answer : C
Answer: (C) 0.42
Question: Jack, Jill, and Sandy each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/6, 1/7, and 1/8 respectively, what is the probability that Jill and Sandy will make a basket but Jack will miss?
(A) 3/168
(B) 5/168
(C) 7/168
(D) 5/336
(E) 7/336
Rationale: The probability that Jill and Sandy will make a basket but Jack will miss is 5/6*1/7*1/8=5/336.
The answer is D.
Answer: (D) 5/336
Question: If A = 625 * 7776 * 16 / 1296, How many digits are in A?
(A) 5
(B) 2
(C) 3
(D) 4
(E) 0
Rationale: A = 625 * 7776 * 16 / 1296
A = 5^4 * 6^5 * 2^4 / 6^4
A = 10^4*6 = 60000
Option: 5
Answer: (A) 5
Question: 34.95+240.016+23.98=?
(A) 298.0946
(B) 298.111
(C) 298.946
(D) 299.09
(E) None of them
Rationale: 34.95
240.016
23.98
----------
298.946
Answer is C
Answer: (C) 298.946
Question: a,b,c,d,e,f,g sitting in a row what is the probability that a,b & c sitting together.
(A) 3/7
(B) 5/7
(C) 6/7
(D) 2/7
(E) 1/7
Rationale: total number of arrangement is= 7!= 5040
Favorable event i.e a,b and c can be arranged together in 3! and the remaining can be arranged in5! since abc can be in any of the five positions.
so 5!*3!/7! = 2/5
ANSWER:E
Answer: (E) 1/7
Question: If A is a positive integer, which of the following must be divisible by 6?
(A) A(A^2−2)
(B) A^2*(A−5)
(C) A(A+2)^3
(D) A^2*(A−5)
(E) (A−1)*A*(A+1)
Rationale: option E) (A−1)*A*(A+1) which represents the product of three consecutive integers
the product of 3 consecutive integers is divisible by both 2 and 3 and so also divisible by 6.
Answer is E
Answer: (E) (A−1)*A*(A+1)
Question: The circulation for magazine P in 1971 was 4 times the average (arithmetic mean) yearly circulation for magazine P for the years 1972-1980. What is the ratio of the circulation in 1971 to the total circulation during 1971-1980 for magazine P ?
(A) 4/7
(B) 5/7
(C) 3/7
(D) 2/7
(E) 1/7
Rationale: There are 9 years from 1972-1980, inclusive. Let's say the average circulation every year between 1972-1980 inclusive is x.
So the total circulation is 9x from 1972-1980, inclusive.
In 1971, the circulation is 4x.
So total circulation for 1971-1980 is 4x + 9x = 13x.
Ratio of circulation in 1971 to total circulation during 1971-1980 is 4x to 14x or 4/14=2/7
Answer : D
Answer: (D) 2/7
Question: A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 6, 4 and 3, which can be repeated. If the product of the digits in the area code must be even, how many different codes can be created?
(A) 20
(B) 22
(C) 26
(D) 24
(E) 30
Rationale: Total # of codes possible is 3*3*3 = 27. Oit of those 27 codes only the product of 333 and will be odd, the remaining 26 will have either 2 or 4 in them, which ensures that their product will be even. Therefore the number of codes where the product of the digits is even = (total) - (restriction) = 27 - 1 = 26.
Answer: C
Answer: (C) 26
Question: Which is greater than 16
(A) 1.6
(B) 66
(C) 6
(D) -6
(E) 6.1
Rationale: 66 greater than 16.
Answer: B
Answer: (B) 66
Question: A clothing store originally bought 100 jackets at a total cost of j dollars. If each jacket was sold for 30 percent more than its original cost, what was the individual price, in terms of j, of each jacket sold?
(A) j/77
(B) 7j/500
(C) 140j
(D) j/100+40
(E) 140j/100
Rationale: lets take j=1000, as decided by you..
now 30% increase makes it 1300..
this is the price for 100 jackets, so selling price for each jacket=1300/100=$13..
so ans is correct as A..
Answer: (A) j/77
Question: John spent a total of $135 on baseball tickets. The price of a ticket is either $12 for the field seats or $5 for the balcony. What is the minimum amount W of dollars he could have spent for the $12 tickets?
(A) $48
(B) $60
(C) $84
(D) $96
(E) $120
Rationale: 60 ..... 12 *5 W= 60. (min). Thus answer .B
Answer: (B) $60
Question: A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
(A) 2951
(B) 8125
(C) 15600
(D) 16302
(E) 18278
Rationale: Remember that Numbers can be repeated.
Number of 1 Digit Codes = 26
Number of 2 Digit Codes = 26 * 26 = 676
Number of 2 Digit Codes = 26 * 26 * 26 = 17576
Total = 26 + 676 + 17576 = 18278
ANS:E
Answer: (E) 18278
Question: Out of three given numbers, the first one is twice the second and three times the third. If the average of these numbers is 88, then the difference between first and third is.
(A) 92
(B) 39
(C) 87
(D) 96
(E) None
Rationale: Sum of three number is=88*3=264
Let three numbers are a, b,c and a is the highest and c is the lowest
Then,
2b=a so b=a/2 and 3c=a so c=a/3
we can write,
a+b+c=264
a+a/2+a/3=264
11a/6=264
a=144
So, c=144/3=48
So there difference is=144-48=96
Answer D
Answer: (D) 96
Question: If x is the interest on y and y is the interest on z, the rate and time is the same on both the cases. What is the relation between x, y and z?
(A) Y2 = xz
(B) Y2 = xy
(C) Y2 = Yz
(D) Y3 = xz
(E) Y5 = xY
Rationale: X = (Y*NR)/100 Y = (Z*NR)/100
X/Y = NR/100 Y/Z = NR/100
X/Y = Y/Z
Y2 = XZ.Answer:A
Answer: (A) Y2 = xz
Question: H(x, y) is the intersection point between the circle (x^2 + y^2 = 4) and the line (y = x +2). Which of the following can be the point H?
(A) (1, 2).
(B) (2, 0).
(C) (0, -2).
(D) (-2, 0).
(E) (2, 2).
Rationale: A quick glance at the options, H(x, y)and we can see that A and E do not satisfy the circle equation. Now we are left with B, C and D. Now B and C do not satisfy the line equation. So we are just left with D. And thats the answer=D
Answer: (D) (-2, 0).
Question: Of the films Empty Set Studios released last year, 60% were comedies and the rest were horror films. 75% of the comedies were profitable, but 75% of the horror moves were unprofitable. If the studio made a total of 45 films, and broke even on none of them, how many of their films were profitable?
(A) 18
(B) 19
(C) 20
(D) 21
(E) 44
Rationale: After fiddling some more, this is how I got to answer choice E...but did I get here using voodoo math or is it sound?
(.60*.75) + (.40*.25) = .44 + .10 = .55
55/100 = 11/20
Then subtract from 45:
45 - 11/20 = 44 (answer E)
Answer: (E) 44
Question: Find the principle on a certain sum of money at 5% per annum for 2 2/5 years if the amount being Rs.1344?
(A) 1000
(B) 2217
(C) 1200
(D) 2777
(E) 2991
Rationale: 1344 = P [1 + (5*12/5)/100]
P= 1200
Answer: C
Answer: (C) 1200
Question: If k is the greatest positive integer such that 4^k is a divisor of 16! then k =
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: 16/4=4
16/16=1
4+1=5
k=5
Answer: C
Answer: (C) 5
Question: How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?
(A) 577
(B) 567
(C) 360
(D) 376
(E) None of these
Rationale: Explanation :
The smallest number in the series is 1000, a 4-digit number.The largest number in the series is 4000, the only 4-digit number to start with 4.The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.Hence, there are 3×5×5×5 or 375 numbers from 1000 to 3999.Including 4000, there will be 376
Answer : D
Answer: (D) 376
Question: If m^(-1) = -1/4 then m^(-2) is equal to
(A) -9
(B) -3
(C) -1/9
(D) 1/16
(E) 9
Rationale: You made a mistake here.
You're right till
m^-2 = 1/m^2 and m^-1 = 1/m^1 = 1/m
given m^-1 = -(1/4) so, 1/m = -(1/4) solving this, m = -4
Now, m^-2 = 1/m^2 = 1/(-4)^2 = 1/16 ANS D
Answer: (D) 1/16
Question: If the cost price of 50 articles is equal to the selling price of 30 articles, then the gain or loss percent is?
(A) 16
(B) 127
(C) 12
(D) 18
(E) 67
Rationale: Percentage of profit = 20/30 * 100 = 67%
Answer: E
Answer: (E) 67
Question: A package contains 8 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
(A) 3/5
(B) 1/2
(C) 6/8
(D) 7/9
(E) 1/3
Rationale: Let number of balls = (8+8) = 16.
Number of white balls = 8.
P (drawing a white ball) =	8/16	= 1/2
B
Answer: (B) 1/2
Question: How many different 5 letter combinations can be created from the letters A,B,C,D and E?
(A) 625
(B) 125
(C) 3125
(D) 25
(E) 5
Rationale: This is a case of number of arrangements possible with repetition.
Thus you have 5 options each for the 1st,2nd,3rd,4th and 5th letter giving you a total of 5*5*5*5*5 = 3125 number of possible combinations.
Answer : C
Answer: (C) 3125
Question: Find the odd man out. 7,13,19,25,29,37,43
(A) 19
(B) 29
(C) 25
(D) 43
(E) 37
Rationale: Explanation :
All given numbers except 25 are prime numbers.
Answer : Option C
Answer: (C) 25
Question: When Greenville State University decided to move its fine arts collection to a new library, it had to package the collection in 20-inch by 20-inch by 15-inch boxes. If the university pays $0.60 for every box, and if the university needs 3.06 million cubic inches to package the collection, what is the minimum amount the university must spend on boxes?
(A) $306
(B) $275
(C) $510
(D) $1,250
(E) $2,550
Rationale: Total no. Of boxes = 3060000/(20×20×15) = 510
Total cost = 510×$0.6=$306
Answer A
Answer: (A) $306
Question: Nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw. If the total weight of one screw and one nail is half that of 6 screws and one nail, then the total weight of 3 screws, and 3 nails is how many times that of 4.5 screws and 1.5 nails?
(A) a. 4/3
(B) b. 5/2
(C) c. 2
(D) d. 3/2
(E) e. 5/4
Rationale: let the weight of nail be n and that of screw be s..
so s+w=1/2 * (6s+1n)...
or 1n = 4s..
lets see the weight of 3 s and 3n= 3s+3*4s=15s..
and weight of 4.5s and 1.5n= 4.5s+1.5*4.5s=11.25s..
ratio= 15s/11.25s=15/11.25=4/3
A
Answer: (A) a. 4/3
Question: Find the area of circle whose radius is 4m?
(A) 50.28
(B) 51.54
(C) 51.77
(D) 52.88
(E) 52.98
Rationale: 22/7 * 4 * 4
= 50.28
Answer:A
Answer: (A) 50.28
Question: If a 4-digit integer is selected at random from the integers 6000 through 6999, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to three times the tens digit and two times the hundreds digit?
(A)  1/999
(B)  3/6000
(C)  1/1000
(D)  3/2000
(E)  2/3000
Rationale: Thousand digit has to be 6 . So units digit needs to 6 as well
Both Hundreds and Units digit needs to be 3 times the tens digit
6/3 = 2
The tens digit is 2
Both Hundreds and Units digit needs to be 2 times the hundreds digit
6/2 = 3
The hundreds digit is 3
The number is 6326
Total numbers between 6000 and 6999 = (6999-6000)+1 = 1000
Probability = 1/1000
Answer C
Answer: (C)  1/1000
Question: A vessel of capacity 2 litre has 25% of alcohol and another vessel of capacity 6 litre had 50% alcohol. The total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water. What is the new concentration of Mixture?
(A) 31%.
(B) 71%.
(C) 49%.
(D) 29%.
(E) 35%.
Rationale: 25% of 2 litres = 0.5 litres
50% of 6 litres = 3.0 litres
Therefore, total quantity of alcohol is 3.5 litres.
This mixture is in a 10 litre vessel. Hence, the concentration of alcohol in this 10 litre vessel is 35%
ANSWER:E
Answer: (E) 35%.
Question: If 7 and 6 are factors of 60n, what is the minimum value of n?
(A)  2
(B)  3
(C)  7
(D)  14
(E)  56
Rationale: 60n/7*6 should be integer => 2*2*3*5*n/7*2*3 = 2*5*n/7 must be an integer
for this to be true n must multiple of 7, thus min of n = 7
hence C
Answer: (C)  7
Question: A retailer sold 2 houses at each $10000. On 1st house he got 30% profit and on 2nd house he loss 10% then find his net profit or loss?
(A) 15%
(B) 17%
(C) 21%
(D) 19%
(E) 25%
Rationale: A = 30%
B = -10%
profit% = 30-10 + (-300/100) = 17%
Answer is B
Answer: (B) 17%
Question: Bill drives from city A to city B. After two hours of driving he noticed that he covered 80 km and calculated that, if he continued driving at the same speed, he would end up been 15 minutes late. So he increased his speed by 10 km/hr and he arrived at city B 36 minutes earlier than he planned.
Find the distance between cities A and B.
(A) 132 km
(B) 250 km
(C) 390 km
(D) 440 km
(E) 578 km
Rationale: Let x be the distance between A and B. Since Bill covered 80 km in 2 hours, his speed was V=80/2=40 km/hr.
If he continued at the same speed he would be 15 minutes late, i.e. the planned time on the road is x/40−15/60x hr. The rest of the distance is (x−80)) km. V=40+10=50 km/hr.
So, he covered the distance between A and B in 2+x−80/50 hr, and it was 36 min less than planned. Therefore, the planned time was 2+x−80/50+36/60
When we equalize the expressions for the scheduled time, we get the equation:
x/40−15/60=2+x−80/50+36/60
x−10/40=100+x−80+30/50
x−10/4=x+50/5
5x−50=4x+200
x=250
So, the distance between cities A and B is 250 km.
correct answer is B)250km
Answer: (B) 250 km
Question: Sum of 3 consecutive even no.'s is 24 more than the 1st no. of the series. Find the middle no.?
(A) 11
(B) 12
(C) 14
(D) 16
(E) 18
Rationale: Let the numbers be x, x+2 and x+4 then
x + x+2 + x+4 = x+24
3x + 6 = x + 24
2x = 18
\inline \therefore x = 9
\inline \therefore Middle number = x + 2 = 9+2 = 11
A
Answer: (A) 11
Question: If a,b are positive integers and a/b=85.75 which of the following could be the value of b?
(A) 1. 25
(B) 4.24
(C) 3. 27
(D) 4. 27
(E) 5. 30
Rationale: Given a/b = 85.75.
this is 85and 3/4 = 343/4
Since a and b are integers we know that whatever is the value of b, it must be divisible by 4 completely(to make a an integer)
Ans:B
Answer: (B) 4.24
Question: On Sunday morning, Pugsley and Wednesday are trading pet spiders. If Pugsley were to give Wednesday two of his spiders, Wednesday would then have seven times as many spiders as Pugsley does. But, if Wednesday were to give Pugsley four of her spiders, Pugsley would now have four fewer spiders than Wednesday had before they traded. How many pet spiders does Pugsley have before the trading game commences?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 10
Rationale: If Pugsley were to give Wednesday two of his spiders, Wednesday would then have seven times as many spiders as Pugsley does:
(w + 2) = 7(p - 2)
If Wednesday were to give Pugsley four of her spiders, Pugsley would now have four fewer spiders than Wednesday had before they traded:
p + 4 = w - 4
Solving gives p = 4 and w = 12.
Answer: B.
Answer: (B) 4
Question: A watch which gains uniformly is 2 minutes low at noon on Monday and is 4 min. 48 sec fast at 2 p.m. on the following Monday. When was it correct
(A) 2 p.m. on Monday
(B) 2 p.m. on Tuesday
(C) 2 p.m. on Wednesday
(D) 3 p.m. on Thursday
(E) None
Rationale: Sol.
Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours.
∴ The watch gains [2 + 4 4/5] min. or 34/5 min. in 170 hrs.
Now, 34/5 min. are gained in 170 hrs.
∴ 2 min. are gained in [170 * 5/34 * 2] hrs = 50 hrs.
∴ Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e. it will be correct at 2 p.m. on Wednesday.
Answer C
Answer: (C) 2 p.m. on Wednesday
Question: A can finish a piece of work in 15 days. B can do it in 30 days. They work together for 6 days and then A goes away. In how many days will B finish the work?
(A) 11 days
(B) 12 days
(C) 14 days
(D) 10 days
(E) 15 days
Rationale: 6/15 + (6 + x)/30 = 1 => x = 12 days
Answer: B
Answer: (B) 12 days
Question: Find the average of first 40 natural numbers.
(A) 15.5
(B) 20.5
(C) 40.5
(D) 25
(E) 30.5
Rationale: Explanation:
Sum of first n natural numbers = n(n+1)/2
Hence, sum of first 40 natural numbers = (40 x 41)/2 = 820
Therefore, required average of = 820/40 = 20.5
ANSWER: B
Answer: (B) 20.5
Question: A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included in a particular student?
(A) 6
(B) 8
(C) 10
(D) 12
(E) 15
Rationale: If students are A, B, C, D, E and F; we can have 6C3 groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get 5C2 = 10 ways of doing it.
ANSWER:C
Answer: (C) 10
Question: In a graduate physics course, 70 percent of the students are male and 40 percent of the students are married. If two-sevenths of the male students are married, what fraction of the female students is single?
(A) 2/7
(B) 5/3
(C) 1/2
(D) 2/3
(E) 1/3
Rationale: let assume there are 100 students of which 70 are male and 30 are females
if 40 are married then 60 will be single.
now its given that two-sevenths of the male students are married that means 2/7 of 70 = 20 males are married
if 40 is the total number of students who are married and out of that 20 are males then the remaining 20 will be females who are married.
total females = 30
married females = 20
then single females = 30-20 = 10
we need to find the fraction of female students who are single i.e single female students / total female student
= 10/30 = 1/3 [E]
Answer: (E) 1/3
Question: A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 17 seconds. What is the length of the platform in meters?
(A) 240 meters
(B) 260 meters
(C) 420 meters
(D) 600 meters
(E) Cannot be determined
Rationale: Speed of the train in metres/sec = 72000/3600 = 20
Distance travelled by train to cross the platform = 30 * 20 = 600 = Length of train + Length of platform
Distance travelled by train to cross the man = 17 * 20 = 340 = Length of train
Length of platform = 600 - 340 = 260
Answer: B
Answer: (B) 260 meters
Question: If the cost price of 121 chocolates is equal to the selling price of 77 chocolates, the gain percent is :
(A) 56.14%
(B) 57.44%
(C) 57.14%
(D) 56.34%
(E) 57.87%
Rationale: Explanation :
Solution: Let C.P. of each chocolate be Re. 1. then, C.P. of 77 chocolates = Rs. 77;
S.P. of 77 chocolates =Rs. 121.
.'. Gain% = 44*100/77 = 57.14%
Answer : C
Answer: (C) 57.14%
Question: Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill cistern. How much time will be taken by A to fill the cistern separately?
(A) 7
(B) 9
(C) 6
(D) 2
(E) 3
Rationale: Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
1/x + 1/(x + 6) = 1/4
x2 - 2x - 24 = 0
(x - 6)(x + 4) = 0 => x = 6.
Answer: C
Answer: (C) 6
Question: From a group of 6boys&4girls a Committee of 4 persons is to be formed. In how many different ways can it be done, so that the committee has at least 2 boys?
(A) 120
(B) 140
(C) 185
(D) 190
(E) 210
Rationale: The committee of 4 persons is to be so formed that it has at least 2 men. The different ways that we can choose to form such a committee are:
(i) 2m. 2w in 6C2 x 4 C1=6X5 - 2X1 X 3X3 2X1 =n an (ii) 3m. 1w in 6C3 x 4C1 =6X5X4 3X2X1 X4 _n - a ''''
(iii) 4m in 6C4 -2x1 = 15 Total no. of different ways in which a committee of 4 persons can be formed so that it has at least 2 men. = 90 + 18 + 15 = 185
C
Answer: (C) 185