mathewhe/aqua_rat huggingface dataset:
Question: A number whose fifth part increased by 3 is equal to its fourth part diminished by 3 is ?
(A) 120
(B) 180
(C) 200
(D) 220
(E) None
Rationale: Answer
Let the number be N.
Then, (N/5) + 3 = (N/4) - 3
â‡’ (N/4) - (N/5) = 6
â‡’ (5N - 4N)/20 = 6
âˆ´ N = 120
Option: A
Answer: (A) 120
Question: One-sixth of the attendees at a certain convention are female students, two-thirds of the attendees are female, and one-third of the attendees are students. If 50 of the attendees are neither female nor students, what is the total number of attendees at the convention?
(A) 600
(B) 300
(C) 400
(D) 500
(E) 700
Rationale: Let no. of Attendee are A.
Now As per the condition of the problem stated above .
We have following population in the convention as Attendee.
Total no. of females = [2/3]*A
Total no. of females as student : [1/6]*A
Total no. of students = [1/3]*A
Total no. of male as student = [1/6]*A
Total no. of males = A - [2/3]*A = [1/3]A
No. of males which are not student = [1/3]A - [1/6]*A = 50
Hence A =300
Total no of males who are not student will be the answer as it states it should be neither female nor student
So Answer is B
Answer: (B) 300
Question: A doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 90 pounds. If the typical dosage is 2 cubic centimeters per 15 pounds of the body weight, by what percent was the prescribed dosage greater than the typical dosage?
(A) 8%
(B) 9%
(C) 11%
(D) 12.5%
(E) 16.7%
Rationale: Typical dosage is Dose : weight :: 2: 15.
Now if weight is 90 (Multiplying factor is 6: (90/15)) then typical dosage would be 2 *6= 12 cc.
Dosage = 18 cc. Dosage is greater by 2 cc.
% dosage is greater: (2 / 12)*100 = 16.67 %
E is the answer.
Answer: (E) 16.7%
Question: Tickets numbered 1 to 20 are mixed up and then a ticked is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3?
(A) 3/10
(B) 3/20
(C) 2/5
(D) 1/2
(E) 1/3
Rationale: S={1,2,3,….20}
E={3, 6,9,12,15,18}
P(E)=6/20
=3/10
Option A
Answer: (A) 3/10
Question: A cicketer bas a certain average for 10 innings, In the eleventh inning, he scorod 128 runs, thereby increasing his average by 6 runs. His new average is
(A) 18 runs
(B) 28 runs
(C) 48 runs
(D) 42 runs
(E) 68 runs
Rationale: Explanation:
Let average for 10 innings be x. Then,
(10x + 108)/11 = x + 6 => 11x + 66 = 10x + 128 => x = 62.
New average = (x + 6) = 68 runs.
Answer: E
Answer: (E) 68 runs
Question: Twelve people are planning to share equally the cost of a rental van. If one person withdraws from the arrangement and the others share equally the cost of the rental van, then the share of each of the remaining people will increase by...?
(A) 1/12
(B) 1/8
(C) 1/11
(D) 12/11
(E) 10/11
Rationale: Let P = total cost of the rental van.
The original share per person is P/12.
The new share per person is P/11.
P/11 = P/12 * 12/11 = (1 + 1/11)*original share
The answer is C.
Answer: (C) 1/11
Question: In a trapezium ABCD, AB is parallel to DC, AB = 3*DC, and the diagonals of the trapezium intersect at O. The ratio of the area of ∆OCD to the area of ∆OAB is:
(A)  1:9
(B)  1:3
(C)  3:1
(D)  1:10
(E)  Cannot be determined
Rationale: triangles are similar , so ratio will be 1:9
A
Answer: (A)  1:9
Question: There are 5 red shoes, 4 green shoes. If one draw randomly a shoe what is the probability of getting a red shoe
(A) 1/9
(B) 2/9
(C) 5/9
(D) 4/9
(E) 2/3
Rationale: total 9 from take one.
9c1=9
red 5 from one
5c1=5
5/9
ANSWER:C
Answer: (C) 5/9
Question: The average (arithmetic mean) monthly income of four workers is $3,000. After one worker’s income increases by 20 percent the new average income is $3,300. What was the original income of the worker whose monthly income increased?
(A) $6,500
(B) $6,400
(C) $6,000
(D) $6,200
(E) $6,600
Rationale: Increase in total income was 300*4=$1200, we know that this increase was 20% (1/5th) of the workers original income, thus his/her original income was 1200*5=$6,000.
Answer: C
Answer: (C) $6,000
Question: Find the one which does not belong to that group ?
(A) 36
(B) 49
(C) 64
(D) 81
(E) 100
Rationale: Explanation:
36 = 62, 49 = 72, 64 = 82, 81 = 92 and 100 = 102.
36, 64, 81 and 100 are squares of composite numbers, but not 49.
ANSWER IS B
Answer: (B) 49
Question: A can give B 100 meters start and C 150 meters start in a kilometer race. How much start can B give C in a kilometer race?
(A) 10.55 meters
(B) 11.55 meters
(C) 33.55 meters
(D) 55.55 meters
(E) None of these
Rationale: Explanation:
A runs 1000 meters while B runs 900 meters and C runs 850 meters.
Therefore, B runs 900 meters while C runs 850 meters.
So, the number of meters that C runs when B runs 1000 meters = (1000 x 850)/900 = 944.44 meters
Thus, B can give C (1000 - 944.44) = 55.55 meters start
Answer:D
Answer: (D) 55.55 meters
Question: Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a.for the next five years and 13% p.a.for the period beyond eight years If the total interest paid by him at the end of eleven years is 8160, how much money did he borrow?
(A) 8000
(B) 10,000
(C) 12,000
(D) Data inadequate
(E) None of these
Rationale: Let the sum be x. Then,
(x×6×3/100)+(x×9×5/100)+(x×13×/)=8160
⇒ 18x + 45x + 39x = (8160 × 100)
⇒ 102x = 816000
⇒ x = 8000.
Answer A
Answer: (A) 8000
Question: If w is 40 percent less than t, t is 40 percent less than y, and z is 46 percent less than y, then z is greater than w by what percent of w?
(A) 4%
(B) 18%
(C) 36%
(D) 40%
(E) 50%
Rationale: GIven w= 0.6t, t=0.6y, z=0.54y, substituting, w=2/3 Z----> z= 1.5w and thus z is 50% greater than w. E is the correct answer.
Answer: (E) 50%
Question: What percent of 5 letter combinations that can be made from the letters of the word FERMONT and in which each letter can be used only once are the combinations in which the first letter is a vowel and the last letter is a consonant?
(A) 23.8%
(B) 20%
(C) 25%
(D) 30.2%
(E) 35%
Rationale: a) noof combinations when all 5 letters are different = 7*6*5*4*3
b) noof combinations when first letter is vowel and last is consonant = 2c1*5c1*5*4*3
b/a = 5/21= 0.238
A
Answer: (A) 23.8%
Question: A childrens identification code consists of a vowel followed by a 3-digit number greater than 200. Exactly 2 of the 3 digits in the code should be identical. How many different codes is it possible to form?
(A) 211
(B) 216
(C) 1075
(D) 1080
(E) 2160
Rationale: A childrens identification code will be of the type -,-,-,-
first digit can be selected from any of the 5 vowels in 5C1 ways
now for the remaining three digit lets consider the following two cases
case 1: when the number is greater than 200 but less than 300
number will be of the type 2,_,_. now suppose repeating number is same as first digit number i.e. 2. and the third number is one of the remaining 9 numbers (we are rejecting 2 here, because it will result in 222, which is not acceptable as per the given condition). thus these two number can arrange themselves in two blank spaces in 2! ways. hence total number of numbers in which repeating digit is same as the first digit = 1.9.2! =18
now, suppose that repeating number is different than first digit. thus possible case in this case are 8 as listed below:
211
233
244
255
266
277
288
299
here again we have rejected 200( because number must be greater than 200) and 222 ( exactly two repeating digits are allowed)
thus total possible cases are 18 + 8 =26
case 2: number ranging from 300 to 999
here for first digit we have 7 cases (3,4,5,6,7,8,9)
now if the repeating number is same as the first number then we will have 18 cases ( same reasoning as mentioned in the previous case)
if the repeating number is different than first digit number then we will have 9 cases ( because here number ending with two zeros are allowed)
hence total number of ways = 7(18+9) = 189
thus different number of codes = 5(189+26) = 1075
hence C
Answer: (C) 1075
Question: The length of the bridge, which a train 150 m long and traveling at 45 km/hr can cross in 30 sec is?
(A) 377
(B) 225
(C) 237
(D) 245
(E) 267
Rationale: Speed = 45 * 5/18 = 25/2 m/sec.
Time = 30 sec
Let the length of bridge be x meters.
Then, (150 + x)/30 = 25/2
x = 225 m.
Answer: B
Answer: (B) 225
Question: If a = 1, what is the value of –(a^2 + a^3 + a^4 + a^5+a^6)?
(A) -14
(B) -4
(C) 0
(D) 4
(E) -5
Rationale: if a = -1, then putting values in equation
= -[ (1)^2 + (1)^3 + (1^4) + (1^5) + (1^6)]
= -[ 1 + 1 + 1 + 1+1 ]
= -5
Answer = E = -5
Answer: (E) -5
Question: A town's oldest inhabitant is x years older than the sum of the ages of the Lee triplets. If the oldest inhabitants is now J years old, how old will one of the triplets W be in 20 years?
J-X-40/3 is my answers.
J = X + L + L + L is the initial situation
After 20 years
J + 20 = X + L + L + L + 60 ...20 years for each triplet so 60 years totally.
J - X - 40/ 3 = L is my answer.
What wrong am i doing ? Since the age asked is after 20 years i also consider adding 20 years to J .
(A) (J - 50)/3
(B) 3(J + 20)/x
(C) (J + x - 50)/3
(D) (J - x + 60)/3
(E) (J + x - 20)/3
Rationale: here it goes:
Oldest inhabitant = sum of age of triplets + X
J = 3L + X so L = (J - X)/3
After 20 years = L + 20
= (J - X)/3 + 20
W= (J - X + 60)/3=D
Answer: (D) (J - x + 60)/3
Question: If John takes 7 minutes to eat x raisins, how many seconds will it take him to eat z raisins, assuming he continues to eat at the same rate?
(A) 420xz
(B) 7z/x
(C) 420z/x
(D) 7xz
(E) 420x/z
Rationale: It will take (z/x)(7) minutes which is (z/x)(7)(60) seconds = 420z/x seconds.
The answer is C.
Answer: (C) 420z/x
Question: The average of first 25 natural numbers is?
(A) 14
(B) 13
(C) 15
(D) 18
(E) 12
Rationale: Sum of 25 natural no.
= 650/2 = 325
Average = 325/25
= 13
Answer:B
Answer: (B) 13
Question: The distance from Steve's house to work is 30 Km. On the way back Steve drives twice as fast as he did on the way to work. Altogether, Steve is spending 6 hours a day on the roads. What is Steve's speed on the way back from work?
(A) 5.
(B) 10.
(C) 14.
(D) 15
(E) 20.
Rationale: Time is in the ratio 2:1 :: to: fro office
Therefore, 2x+1x=6 hrs
Time take to come back - 2 hrs, distance travelled - 30 km => Speed = 15 kmph
ANSWER:D
Answer: (D) 15
Question: Two men, Vikas and Vishal, working separately can mow a field in 8 and 12 hours respectively. If they work in stretches of one hour alternately, Vikas beginning at 8 a.m, when will the mowing be finished?
(A) 232
(B) 787
(C) 2369
(D) 277
(E) 281
Rationale: Answer:B
Answer: (B) 787
Question: A man complete a journey in 10 hrs. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km?
(A) 334
(B) 224
(C) 667
(D) 2678
(E) 112
Rationale: Let the total distance be x km.
Then, [(1/2 x) / 21 + (1/2 x) / 24] = 10
15x = 168 * 20 => x = 224 km
Answer: B
Answer: (B) 224
Question: The length of a rectangular garden is 2 feet longer than 3 times its width. If the perimeter of the garden is 100 feet, find the width and the length of the garden.
(A) 80
(B) 90
(C) 100
(D) 70
(E) 110
Rationale: Let L and W be the length and width of the garden. The statement "the length of a rectangular garden is 2 feet longer than 3 times its width" may be formulated by
L = 2 + 3 W
The formula for the perimeter is given by
P = 2 L + 2 W
Substitute P and L in the above equation by 100 and 2 + 3 W respectively to obtain
100 = 2(2 + 3 W) + 2 W
Solve for W and L
W = 12 and L = 2 + 3 W = 38.
Check that the perimeter of the rectangular garden is 100
P = 2 L + 2 W = 76 + 24 = 100
Answer C
Answer: (C) 100
Question: A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
(A) 45 km/hr
(B) 55 km/hr
(C) 35 km/hr
(D) 50 km/hr
(E) 56 km/hr
Rationale: Speed of the train relative to man =125/10	m/sec
25/2	m/sec.
25/2	x 18/15	km/hr
= 45 km/hr.
Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.
x - 5 = 45 x = 50 km/hr. Answer : D
Answer: (D) 50 km/hr
Question: What is the probability of getting a number less than 4 when a die is rolled?
(A) 2
(B) 1
(C) 3/2
(D) 2/3
(E) 1/2
Rationale: Total number of outcomes possible when a die is rolled=6
That is any one face out of the 6 faces is n(s)=6
e=getting a number less than 4={1,2,3}.
Hence,n(e)=3
p(e)=n(e)/n(s)
=3/6=1/2
Answer:option E
Answer: (E) 1/2
Question: A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is
(A) 1/20
(B) 1/26
(C) 1/23
(D) 1/50
(E) 1/45
Rationale: Total no of cases = 52
fav cases = 2
prob = 2/56 = 1/26
ANSWER B
Answer: (B) 1/26
Question: The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the 22nd term in the sequence?
(A) 4
(B) 7
(C) 6
(D) 8
(E) 9
Rationale: Following the rule of Infinite sequence we obtain the terms of the sequence as mentioned below
02,07,22,67, 202, 607, 1822, 5467...
Observe the last two digits of the sequence which have the cyclicity of 4 and the last two digits repeat in the order {02, 07, 22, 67}
Also 22th Term =4*5 +2
i.e. 35th Terms will have same Last two digits as 2 nd term of the sequence = 07
i.e. Sum of lat two digits of 2ndth Term = 0+7 = 7
B
Answer: (B) 7
Question: Two trains start from stations A and B spaced 50 kms apart at the same time and speed. As the trains start, a bird flies from one train towards the other and on reaching the second train, it flies back to the first train. This is repeated till the trains collide. If the speed of the trains is 25 km/h and that of the bird is 100 km/h. How much did the bird travel till the collision.
(A) 100
(B) 288
(C) 127
(D) 27
(E) 2691
Rationale: ince the trains is travelling at 25 kmph, at each other, the relative speed is 50 kmph.
Speed = 50 kmph
Distance = 50 km
Time to collision = distance / speed = 1 hr
Speed of bird = 100 kmph
Time flying = 1 hr (the bird is flying till the trains collide)
Distance travelled = speed × time = 100 km
Answer:A
Answer: (A) 100
Question: S1 = 6 S2 = 9 ... Sn = 3n + 4 For the sequence above, in which any term n is defined as 3n + 4, what is the value of n for the first term in the sequence to exceed 50?
(A) 14
(B) 12
(C) 13
(D) 16
(E) 15
Rationale: term 16 = 3*16+4 = 52
Answer : D
Answer: (D) 16
Question: The distance between Delhi and Mathura is 215 kms. A starts from Delhi with a speed of 20 kmph at 7 a.m. for Mathura and B starts from Mathura with a speed of 25 kmph at 8 p.m. from Delhi. When will they meet?
(A) 10.50 a.m.
(B) 00.20 p.m.
(C) 10.30 a.m.
(D) 11.40 a.m.
(E) 1.40 p.m.
Rationale: D = 215 â€“ 20 = 195
RS = 20 + 25 = 45
T = 195/45 = 4 1/3 hours
8 a.m. + 4 hrs 20 min = 00.20 p.m.
ANSWER:B
Answer: (B) 00.20 p.m.
Question: A, B and C can do a work in 9 days, 8 days and 12 days respectively. In how many days can all three of them working together, complete the work?
(A) 3 3/23
(B) 3 3/28
(C) 2 2/0
(D) 2 2/2
(E) 2 2/1
Rationale: Work done by all three of them in one day = 1/9 + 1/8 + 1/12 = 23/72.
The number of days required = 72/23 = 3 3/23 days.
Answer:A
Answer: (A) 3 3/23
Question: If an integer "k" is divisible by 2,5 and 13.what is the next largest number that is divisible by all the three given numbers?
(A) k+13
(B) k+130
(C) 2k
(D) 2k+13
(E) 2k+65
Rationale: LCM of 2,5,13 is 130
so k=130
2k also the correct answer,but we need to find next largest number
therefore ans is k+130
ANSWER:B
Answer: (B) k+130
Question: If the number of boys in a class are 8 times the number of girls. If the number of people in the class are 45, how many girls are in the class?
(A) A - 5
(B) B - 6
(C) C - 7
(D) D - 8
(E) E - 9
Rationale: Let the number of girls = x and the number of boys = 8x Then, total number of students = x + 8x = 9x = 45Hence, x = 45/9 = 5 i.e., the total number of girls in the class are 5. Answer is A.
Answer: (A) A - 5
Question: Eric throws 2 dice,and his score is the sum of the values shown.Sandra throws one dice and her score is the square of the value shown. what is the probabilty that Sandras score will be strictly higher than Erics score??
(A) 117/216
(B) 127/126
(C) 137/126
(D) 147/126
(E) 157/126
Rationale: sandra score can be like 1,4,9,16,25,36
eric score less then 1-->0
eric score less then 4=(1,1),(1,2)(2,1)-->3
eric score less then 9 are (1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(4,1)(4,2)(4,3)(4,4)(5,1)(5,2)(5,3)(6,1)(6,2)-->26
eric score will always be less then 16--->36
eric score will always be less then 25--->36
eric score will always be less then 36--->36
total favorable outcomes=3+26+36+36+36=137
total possible outcomes=216(36*6)
probability=137/216
ANSWER:C
Answer: (C) 137/126
Question: Two spies agreed to meet at a gas station between noon and 1pm, but they have both forgotten
the arranged time. Each arrives at a random time between noon and 1pm and stays for 6
minutes unless the other is there before the 6 minutes are up. Assuming all random times are
equally likely, what is the probability that they will meet within the hour (noon to 1pm)?
(A) 0.12
(B) 0.15
(C) 0.17
(D) 0.19
(E) 0.25
Rationale: The probability that they do not meet is represented by the total of the areas of the
two outer triangles in the figure below, which is 0.81. So the probability of a meeting is 1 -
0.81 =0 .19.
correct answer D
Answer: (D) 0.19
Question: Raja's club with a total membership of 30 has formed 3 committees, M, S and R which have 8,12, and 5 members respectively. If no member of committee M is on either of the other 2 committees, what is the greatest possible number of members in the club who are on none of the committees.
(A) 5
(B) 7
(C) 8
(D) 10
(E) 12
Rationale: Raja's club with a total membership of 30 has formed 3 committees Given no member of committee M is on either of the other 2 committees -- hence (M n R) and (M n S) and (M n R n S) is zero.
Hence only M is 8. Now we need to consider only S, (S n R) and R.
(M U R U S) = M + R + S - (M n R) - (M n S) - (R n S) + (M n R n S) + Neither.
30 = 8 + (12 - x) + (5 - x) + 0 + 0 + x + 0 + Neither.
22 = (12-x) + (5-x) + x + Neither.
Now the max value of x could be 5 and the min value of x could be 0.
When x is 5 (max), Neither is 10.
When x is 0 (min), Neither is 5. We need maximum no of people who do not belong to any group. Hence max value of neither is 10.
Answer 10 (D).
Answer: (D) 10
Question: The sum of the present ages of a father and his son is 50 years. five years ago, father's age was five times the age of the son. After 5 years, son's age will be:
(A) 16
(B) 16.2
(C) 16.1
(D) 16.66
(E) 17
Rationale: Let the present ages of son and father be x and (50 -x) years respectively.
Then, (50 - x) - 5 = 5(x - 5)
45 - x = 5x - 25
6x = 70
x = 70/6= 11.66 years
Son's age after 6 years = (x+ 5) = 16.66 years.
answer :D
Answer: (D) 16.66
Question: A committee of two people is to be chosen from 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
(A) 16
(B) 24
(C) 26
(D) 30
(E) 32
Rationale: One of the approaches:
Each couple can send only onerepresentativeto the committee. Let's see in how many ways we can choose 2 couples (as there should be 2 members) out of 4 to send only onerepresentativesto the committee: 4C2=6.
But each of these 2 couples can send two persons (husband or wife): 2*2=2^2=4.
Total # of ways: 4C2*2^2=24.
Answer: B.
Answer: (B) 24
Question: There are 2 sections A and B in a class, consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, Find the average of the whole class?
(A) 35.29kg
(B) 37.25kg
(C) 42.45kg
(D) 55.12kg
(E) 29.78kg
Rationale: Total weight of 36+44 students = 36*44 + 44*35 = 2980
Average weight of the class is = 2980/80 = 37.25 kg
Answer is B
Answer: (B) 37.25kg
Question: Of 25 applicants for a job, 14 had at least 4 years' experience, 18 had degrees, and 3 had less than 4 years' experience and did not have a degree. How many of the applicants had at least 4 years' experience and a degree?
(A)  14
(B)  10
(C)  9
(D)  7
(E)  5
Rationale: set A : people with more than 4 years exp
set B : people with degree
AUB = total - (less than 4 exp and no degree)
AUB = 25-3=22
AUB = A + B - AiB
AiB = 18+14-22
=10
Answer B
Answer: (B)  10
Question: Working alone at its constant rate, machine A produces x boxes in 10 minutes and working alone at its constant rate, machine B produces 2x boxes in 5 minutes. How many minutes does it take machines A and B, working simultaneously at their respective constant rates, to produce 5x boxes?
(A) 3 minutes
(B) 4 minutes
(C) 5 minutes
(D) 10 minutes
(E) 12 minutes
Rationale: Rate = Work / Time
Given Rate of Machine A = X / 10 min
Machine B Produces 2x boxes in 5 min hence , Machine B produces 4x boxes in 10 min .
Rate of Machine B = 4x / 10
we need tofind the combined time that machines A and B, working simultaneouslytakeat their respective constant rates
let's first find the combined Rate of Machine A and B
Rate of Machine A = X / 10 min + Rate of Machine B = 4x / 10 = 5X/10
Now combine Time = combine work needs to be done / Combine Rate = 5x/5x * 10 = 10 Min
Ans: D
Answer: (D) 10 minutes
Question: In a business school case competition, the top three teams receive cash prizes of $1000, $ 2000, and $ 3000. while remaining teams are not ranked and do not receive any prizes, there are 6 participating teams, named A, B, C, D, E, F. If team A wins one of the prizes, team B wins also one of the prizes. How many outcomes of the competition are possible ?
(A) 18
(B) 28
(C) 36
(D) 84
(E) 120
Rationale: if team A wins one of the prizes, team B wins also one of the prizes". Consider following cases:
A wins one of the prizes, then B must also win one of the prizes, and in this case we can have 4 triplets: {ABC}, {ABD}, {ABE}, {ABF}. Each triplet can be arranged in 3!=6 ways. Hence in the case when A wins one of the prizes 4*6=24 arrangements are possible.
A does NOT win one of the prizes, then three winners must be from other 5 teams. 3 winners out of 5 (B, C, D, E, F) teams can be chosen in C35=10C53=10 ways and each case (for example {CDE}) can be arranged in 3!=6 ways, hence in the case when A does NOT win one of the prizes 10*6=60 arrangements are possible.
Total = 24+60 = 84.
Answer: D.
Answer: (D) 84
Question: A mixture of 20 kg of spirit and water contains 10% water. How much water must be added to this mixture to raise the percentage of water to 25%
(A) 4 kg
(B) 5 kg
(C) 8 kg
(D) 30 kg
(E) 7 kg
Rationale: Water in the given mixture = 10*20/100 =2 kg,
And spirit = (20-2) = 18 kg
Let x kg of water added, Then, x+2/20+x *100 ==25
4x+8 =20+x, or x =4 kg
ANSWER:A
Answer: (A) 4 kg
Question: In a kickball competition of 9 teams, how many possible matches can each team play with each other?
(A) 9
(B) 16
(C) 24
(D) 36
(E) 54
Rationale: Each team can play with each other in 9C2 ways
9C2 = 9*8 / 1*2 = 36
Option D
Answer: (D) 36
Question: Working alone, Mary can pave a driveway in 4 hours and Hillary can pave the same driveway in 3 hours. When they work together, Mary thrives on teamwork so her rate increases by 33.33%, but Hillary becomes distracted and her rate decreases by 50%. If they both work together, how many hours will it take to pave the driveway?
(A) 2 hours
(B) 4 hours
(C) 5 hours
(D) 6 hours
(E) 7 hours
Rationale: Initial working rates:
Mary = 1/4 per hour
Hillary = 1/3 per hour
Rate when working together:
Mary = 1/4 + (1/3*1/4) = 1/3 per hour
Hillary = 1/3 - (1/2*1/3) = 1/6 per hour
Together they work 1/3 + 1/6= 1/2 per hour
So they will need 2 hours to complete the driveway. The correct answer is A.
Answer: (A) 2 hours
Question: A and B can do a piece of work in 12 days and 16 days respectively. Both work for 3 days and then A goes away. Find how long will B take to complete the remaining work?
(A) 6
(B) 7
(C) 5
(D) 9
(E) 3
Rationale: 3/12 + (3 + x)/16 = 1
x = 9 days
Answer: D
Answer: (D) 9
Question: The edge of a cube is 9a cm. Find its surface?
(A) 153a2 cm2
(B) 143a2 cm2
(C) 233a2 cm2
(D) 243a2 cm2
(E) 253a2 cm2
Rationale: 6a2 = 6 * 9a * 9a = 243a2
Answer:D
Answer: (D) 243a2 cm2