mathewhe/aqua_rat huggingface dataset:
Question: The average weight of 8 people increases by 2.5 kg when a new person comes in place of one of them weighing 55 kg. What is the weight of the new person?
(A) 75 kg
(B) 85 kg
(C) 95 kg
(D) 65 kg
(E) 55 kg
Rationale: The total weight increase = (8 x 2.5) kg = 20 kg
weight of new person = (55 + 20) kg = 75 kg
The answer is A.
Answer: (A) 75 kg
Question: If a - b = 3 and a^2 + b^2 = 35, find the value of ab.
(A) 12
(B) 13
(C) 10
(D) 18
(E) 13
Rationale: 2ab = (a^2 + b^2) - (a - b)^2
= 35 - 9 = 26
ab = 13.
Answer is B.
Answer: (B) 13
Question: Jack spent 1/6 of his lifetime in adolescence. His facial hair started growing after 1/12 more of his life. He successfully married his girlfriend Diana after 1/7 more of his life. Their son was born after 5 years from then. The son lived just 1/2 of what Jack lived. Jack dies four years after his son died.
How long did Jack live ?
(A) 75
(B) 90
(C) 45
(D) 80
(E) 84
Rationale: E
84
Explanation:
Let us assume that jack lived for p years.
p/6 + p/12 + p/7 + 5 + p/2 + 4 = x
=> p = 84.
Therefore, Jack lived for 84 years.
Answer: (E) 84
Question: A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
(A) 227
(B) 277
(C) 272
(D) 150
(E) 221
Rationale: Speed=(60 * 5/18) m/sec = (50/3) m/sec Length of the train = (Speed x Time) = (50/3 * 9) m = 150 m.
Answer: D
Answer: (D) 150
Question: In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
(A) 1/10
(B) 2/5
(C) 2/7
(D) 5/7
(E) None of these
Rationale: Explanation:
P (getting a prize) =	10/(10 + 25)	=	10/35	= 2/7.
ANSWER : C
Answer: (C) 2/7
Question: Maths, Physics and chemistry books are stored on a library shelf that can accommodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books. Among all the books, 12 books are soft cover and the remaining are hard-cover. If there are a total of 7 hard-cover books among the maths and physics books. What is the probability Q, that a book selected at random is either a hard cover book or a chemistry book?
(A) 1/10
(B) 3/20
(C) 1/5
(D) 1/4
(E) 9/20
Rationale: First phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf. To do so, you have the equations:
m + p + c = 20 (since 4/5 of the 25 spots are full of books)
m = 2p
p = 4 + c
From that, you can use Substitution to get everything down to one variable.
c = p - 4
m = 2p
p = p
Then (p - 4) + 2p + p = 20, so 4p = 24 and p = 6. That means that there are 12 math, 6 physics, and 2 chemistry books on the shelf.
With those numbers, you also know that there are 8 total hardcovers, 1 of which is chemistry. So if your goal is to get either a hardcover or a chemistry, there are 9 ways towin- either one of the 7 hardcovers that aren't chemistry or the two chemistry books. So out of the 20 total, Q=9 provide the desired outcome, making the answer E.
Answer: (E) 9/20
Question: Two trains A and B are 125 m and 150 m long and are moving at one another at 54 Km/hr and 36 Km/hr respectively. Arun is sitting on coach B1 of train A. Calculate the time taken by Arun to completely cross Train B.
(A) 10 s
(B) 11 s
(C) 4 s
(D) 8 s
(E) 12 s
Rationale: DETAILED SOLUTION
Speed of A = 54∗1000/60∗60 = 15 m/s
Speed of B = 36∗1000/60∗60 = 10 m/s
Relative speed = S1 + S2 = 15 + 10 m/s = 25 m/s
The length that needs to be crossed = length of train B = 150 m. Therefore time taken = 150/25 = 6s.
What is the time taken for trains to completely cross each other? The length that needs to be crossed = 125 + 150 = 275 m. Time taken = 275/25 = 11 s.
CORRECT ANSWER B.
Answer: (B) 11 s
Question: The length of the bridge, which a train 170 metres long and travelling at 45 km/hr can cross in 30 seconds, is?
(A) 245
(B) 777
(C) 282
(D) 266
(E) 205
Rationale: Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (170 + x)/30 = 25/2 => 2(170 + x) = 750 => x = 205 m.
Answer: E
Answer: (E) 205
Question: A 270 m long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 sec. What is the length of the other train?
(A) 230
(B) 277
(C) 299
(D) 267
(E) 221
Rationale: Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 270)/9 = 500/9 => x = 230.Answer:A
Answer: (A) 230
Question: If the side length of Square B is five times that of Square A, the area of Square B is how many times the area of Square A?
(A) 5
(B) 15
(C) 25
(D) 35
(E) 45
Rationale: Let x be the side length of Square A.
Then the area of Square A is x^2.
The area of Square B is (5x)^2=25x^2.
The answer is C.
Answer: (C) 25
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If four marbles are picked at random, what is the probability that none is blue?
(A) 33/95
(B) 33/91
(C) 33/99
(D) 33/96
(E) 33/92
Rationale: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. When four marbles are picked at random, then the probability that none is blue is
= 12C4/15C4
= (12 * 11 * 10 * 9)/(15 * 14 * 13 * 12) = 33/91
Answer: B
Answer: (B) 33/91
Question: The ratio of the number of ladies to gents at a party was 1:2 but when 2 ladies and 2 gents left, the ratio became 1:3. How many people were at the party originally?
(A) 44
(B) 11
(C) 12
(D) 98
(E) 58
Rationale: x, 2x
(x-2):(2x-2) = 1:3
3x-6 = 2x-2
x = 4
x+2x = 3x
=> 3*4 = 12
Answer: C
Answer: (C) 12
Question: Sum of 3 consecutive even no.'s is 30 more than the 1st no. of the series. Find the middle no.?
(A) 8
(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+30
3x + 6 = x + 30
2x = 24
\inline \therefore x = 12
\inline \therefore Middle number = x + 2 = 12+2 = 14
C
Answer: (C) 14
Question: How many words can be formed by using all letters of the word “BIHAR”
(A) 452
(B) 320
(C) 365
(D) 456
(E) 120
Rationale: The word BIHAR contains 5 different letters.
Required number of words = 5p5 = 5! = (5x4x3x2x1) = 120
Ans: E
Answer: (E) 120
Question: An investment yields an interest payment of $228 each month. If the simple annual interest rate is 9%, what is the amount of the investment?
(A) $28,300
(B) $30,400
(C) $31,300
(D) $32,500
(E) $35,100
Rationale: Let the principal amount = P
Simple annual interest = 9%
Simple monthly interest = (9/12) = (3/4)%
(3/4)*(P/100) = 228
=>P = (228 *4 * 10^2 )/3
= 76*4*10^2
= 304 * 10^2
Answer B
Answer: (B) $30,400
Question: The average age of the boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is
(A) 18
(B) 67
(C) 02
(D) Data inadequate
(E) 23
Rationale: Explanation:
Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given.
So, the data provided is inadequate.
Answer: D
Answer: (D) Data inadequate
Question: 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 W apply to what fraction of a week?
(A)  1/2
(B)  5/8
(C)  9/14
(D)  16/21
(E)  9/10
Rationale: Number of hours between 8 PM to 8 AM = 12
Number of hours with reduced rates = (12*5)+(24*2)
Hours with reduced rates W/Total number of hours in a week = (12*5)+(24*2)/(24*7) = 108/(24*7) = 9/14
Answer: C
Answer: (C)  9/14
Question: If 10 toys cost Rs, 234, what do 35 toys cost?
(A) 645
(B) 564
(C) 546
(D) 819
(E) None of them
Rationale: Let the required cost be Rs. x. Then,
More toys, More cost	(Direct Proportion)
x=(35 X 234)/10 =819
Answer is D
Answer: (D) 819
Question: What is the greatest positive integer n such that 3^n is a factor of 54^100?
(A) 100
(B) 200
(C) 300
(D) 600
(E) 900
Rationale: 54 = 3^3 * 2.
54^100 = 3^300 * 2^100
The answer is C.
Answer: (C) 300
Question: How many odd integers F between 200 and 600 are there such that their tens digit is even?
(A) 20
(B) 25
(C) 100
(D) 150
(E) 200
Rationale: The hundreds digit can take 4 values: 2, 3, 4 and 5;
The tens digit can take 5 values: 0, 2, 4, 6 and 8;
The units digit can take 5 values: 1, 3, 5, 7 and 9;
Total F: 4*5*5=100.
Answer: C.
Answer: (C) 100
Question: In the presidential election, 55% of the people who voted, voted for the democratic candidate and 35% voted for the republican candidate. If 8% of the number of people eligible to vote, voted for neither the republican nor the democrat, what percent of people eligible to vote actually voted in the elections?
(A) 40
(B) 60
(C) 80
(D) 88
(E) 90
Rationale: let people eligible to vote = y
people actually voted = x
Democratic = 0.55x
Republican = 0.35 x
total voters = 0.90 x
hence 0.10x voters who voted didnt voted for D and R.
now, 0.10 x = 0.08 y
=> x/y = 8/10
hence 80%
Hence C
Answer: (C) 80
Question: If Sharon's weekly salary increased by 15 percent, she would earn $460 per week. If instead, her weekly salary were to increase by 10 percent, how much would she earn per week?
(A) Rs.400
(B) Rs.440
(C) Rs.150
(D) Rs.460
(E) Rs.520
Rationale: Soln:-
460/115)110 =385
In this case long division does not take much time.
(4/1)110=Rs.440
ANSWER:B
Answer: (B) Rs.440
Question: In a certain village, n litres of water are required per household per month. At this rate, if there are m households in the village, how long (in months) will p litres of water last?
(A) p /m
(B) P/mn
(C) mp / n
(D) np / m
(E) npm
Rationale: I find it much easier to understand with real numbers, so choose (almost) any numbers to replace m,n and p:
(p/m*n)
Answer (B)
Answer: (B) P/mn
Question: Rs.1500 is divided into two parts such that if one part is invested at 6% and the other at 5% the whole annual interest from both the sum is Rs.85. How much was lent at 5%?
(A) 399
(B) 700
(C) 288
(D) 500
(E) 211
Rationale: (x*5*1)/100 + [(1500 - x)*6*1]/100 = 85
5x/100 + 90 – 6x/100 = 85
x/100 = 5
=> x = 500
Answer: D
Answer: (D) 500
Question: A city has a population of 3,00,000 out of which 1,80,000 are males. 50% of the population is literate. If 70% of the males are literate, the number of literate females is :
(A) 20,000
(B) 24,000
(C) 30,000
(D) 34,000
(E) 36,000
Rationale: population - 3,00,000,
males -1,80,000,
females-1,20,000
50% of 3,00,000 literate - so 1,50,000 literate
70% of 1,80,000 literate -so 1,26,000 male literates , so rest s
1,50,000 - 1,26,000 = 24,000
ANSWER:B
Answer: (B) 24,000
Question: If you write down all the numbers from 1 to 100, then how many times do you write 3 ?
(A) 11
(B) 18
(C) 20
(D) 21
(E) 22
Rationale: Explanation:
Explanation:
Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
So, required number = 10 + 10 = 20.
Answer: C
Answer: (C) 20
Question: The arithmetic mean of the scores of a group of students in a test was 42. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is?
(A) A)61.4.
(B) B)51.7
(C) C)33.18
(D) D)51.1
(E) E)51.2
Rationale: Let the required mean score be x. Then,
20 * 80 + 25 * 31 + 55 * x = 42 * 100
= 1600 + 775 + 55x = 4200
= 55x = 1825
= x = 33.18
Answer:C
Answer: (C) C)33.18
Question: Murali travelled from city A to city B at a speed of 40 kmph and from city B to city C at 60 kmph. What is the average speed of Murali from A to C given that the ratio of distances between A to B and B to C is 1 : 2?
A. 48 kmph
(A) 33
(B) 360/7
(C) 28
(D) 27/8
(E) 16
Rationale: Let the distances between city A to B and B to C be x km and 2x km respectively.
Total time taken to cover from A to C
= (x)/40 + (2x)/60 = (3x + 4x)/120 = 7x/120 Average speed = (x + 2x)/(7x/120) = 360/7 kmph.
Answer: B
Answer: (B) 360/7
Question: If 22 men do a work in 55 days, in how many days will 10 men do it?
(A) 121
(B) 123
(C) 119
(D) 125
(E) 127
Rationale: 22 * 55 = 10 * x
x = 121 days
Answer: A
Answer: (A) 121
Question: 2 years ago, the average age of a family of 5 members was 16 years. After a baby is born, the average age of family is the same today. Find the present age of the baby.
(A) 4 years
(B) 6 years
(C) 8 years
(D) 8 ½ years
(E) 9 years
Rationale: Hint: (1) First find total age of 5 members 2 years ago (2) Present age of 5 members (3) Total age of 6 members (4) Age of baby = Total age of 6 members - Present age of 5 members
We know that,
Average =Sum of Quantities / Number of Quantities
1) First calculate total age of 5 members 2 years ago = (Average age of 5 members x number of members)
First calculate total age of 5 members 2 years ago = (16 x 5) = 80 years
2) Calculate the present age of 5 members
2 years ago, their total age was 80 years. Present age can be calculated as follows:
Present age of 5 members = [80 + (2 x 5)] = 90 years
3) Calculate total age of 6 members considering baby = (16 x 6 ) = 96 years
4) Age of baby = (96 – 90) = 6 years
ANSWER IS B
Answer: (B) 6 years
Question: A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum?
(A) 287
(B) 276
(C) 279
(D) 646
(E) 370
Rationale: 840 = P [1 + (10*3)/100]
P = 646
Answer: D
Answer: (D) 646
Question: The cost of 10 kg of mangos is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is $25. Find the total cost of 4 kg of mangos, 3 kg of rice and 5 kg of flour?
(A) 347
(B) 987
(C) 1070
(D) 1371
(E) 1667
Rationale: Let the costs of each kg of mangos and each kg of rice be $a and $r respectively.
10a = 24r and 6 * 25 = 2r
a = 12/5 r and r = 75
a = 180
Required total cost = 4 * 180 + 3 * 75 + 5 * 25
= 720 + 225 + 125 = $1070
C
Answer: (C) 1070
Question: Seven years back, a father was 48 years older than his son. At present the father is 4 times as old as the son. How old will the son be three years from now?
(A) 16 years
(B) 17 years
(C) 18 years
(D) 19 years
(E) 20 years
Rationale: Using linear equations to convert the word problem:
** F = fathers current ageS = sons current age
#1 Seven years back, a father was 48 years older than his son: F-7 = 48 + (S - 7)
#2 At present the father is 4 times as old as the son: F = 4*S
How old will the son be three years from now?: S + 3 = ?
With two variables and two linear equations we are able to solve the problem:
(S*4) - 7 = 48 +S -7
3S=48
S = 16
S + 3 = 16 + 3 = 19
Answer: D
Answer: (D) 19 years
Question: Peeta asks Katniss, 'It is 10 9m as per my watch right now.Do you know what will be the time 23, 999, 996 hours later?'
Katniss knows the answer. Do you ?
(A) 5pm
(B) 6pm
(C) 4am
(D) 7pm
(E) 8pm
Rationale: D
7pm
The time after 24, 000, 000 hours will be the same as it is now. We want the time 3 hours before that and thus you can deduct three hours from 9 pm. So the time will be 6 pm.
Answer: (D) 7pm
Question: How many multiples of 7 are there between 700 and 1000, inclusive?
(A) 43
(B) 42
(C) 22
(D) 21
(E) 20
Rationale: There are exactly 301 numbers between 700 to 1000..
701 to 800 = 100
801 to 1000 = 200...Total 300 numbers , count 700 too..
301/7 = 43...
Ans option A.
Answer: (A) 43
Question: The difference between the compound interest and simple interest accrued on an amount at the end of three years at the rate of 12% is Rs.381.888. What is the amount?
(A) Rs. 9500
(B) Rs. 8400
(C) Rs.8000
(D) Rs.8200
(E) Rs.8500
Rationale: C.I = P[(1+R/100)T - 1]
= P[(1+12/100)3 - 1] = P[(28/25)3 - 1]
= P[(21952/15625) - 1]
= 6327P/15625
S.I = (P x 3 x 12)/100 = Rs. 9P/25
6327P/15625 - 9P/25 = 381.888
=> 702P/15625 = 381.888
=> P = (381.888 x 15625)/702
= Rs. 8500
ANSWER IS E
Answer: (E) Rs.8500
Question: The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
(A) 75 kg
(B) 85 kg
(C) 95 kg
(D) 65 kg
(E) None of these
Rationale: Explanation:
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
Answer B
Answer: (B) 85 kg
Question: A man has 1044 candles. After burning, he can make a new candle from 9 stubs left behind. Find the maximum number of candles that can be made.
(A) 116
(B) 120
(C) 130
(D) 140
(E) None
Rationale: Explanation :
Since, 1 candle is made from 9 stubs.
Therefore, from 1044 stubs, 1044/9 i.e 116 candles will be made.
Now, from 116 stubs, 116/9 i.e 12 candles can be made and 8 stubs will be left.
Thus, total stubs left is (12+8) = 20 out of which 20/9 i.e 2 candles can be made and 2 stubs will be left.
Thus, maximum number of candles that can be made is 130 (116+12+2).
Answer : C
Answer: (C) 130
Question: Mary, Peter, and Lucy were picking chestnuts. Mary picked twice as much chestnuts than Peter. Lucy picked 5 kg more than Peter. Together the three of them picked 29 kg of chestnuts. How many kilograms of chestnuts did Mary, Peter, and Lucy pick respectively?
(A) 6, 11 and 12
(B) 12, 11 and 6
(C) 11, 6 and 12
(D) 12, 6 and 11
(E) 6, 12 and 11
Rationale: M = 2P
L = P+5
M + P + L = 29
2P + P + (P+5) = 29
P = 6, M = 12, L = 11
Therefore, Mary, Peter, and Lucy picked 12, 6, and 11 kg, respectively.
The answer is D.
Answer: (D) 12, 6 and 11
Question: A train covers a distance in 50 min, if it runs at a speed of 48kmph on an average. The speed at which the train must run to reduce the time of journey to 10min will be.
(A) 60 km/h
(B) 55 km/h
(C) 240 km/h
(D) 70 km/h
(E) 65 km/h
Rationale: Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=10/60hr=1/6hr
New speed = 40* 6/1 kmph= 240kmph
Answer : C
Answer: (C) 240 km/h
Question: Company P had 15 percent more employees in December than it had in January. If Company P had 450 employees in December, how many employees did it have in January?
(A)  391
(B)  392
(C)  410
(D)  423
(E)  445
Rationale: D = number of employees in December
J = number of employees in January
J x 1.15 = D
J x 1.15 = 450
J = 450/1.15
J = 45,000/115 = 392
Thus B is the correct answer.
Answer: (B)  392
Question: Sum of 3 consecutive even no.'s is 26 more than the 1st no. of the series. Find the middle no.?
(A) 10
(B) 12
(C) 14
(D) 18
(E) 21
Rationale: Let the numbers be x, x+2 and x+4 then
x + x+2 + x+4 = x+26
3x + 6 = x + 26
2x = 20
\inline \therefore x = 10
\inline \therefore Middle number = x + 2 = 10+2 = 12
B
Answer: (B) 12
Question: Anand starts from a point P towards point Q, where PQ = 80 km. After 1 hour, Ram starts from P and catches up with Anand after 2 more hours. After meeting they continue to travel towards Q. On reaching Q, Ram reverses his direction and meets Anand 6 hours after the first meeting. Find Anand's speed.
(A) (45/7) kmph
(B) (160/21) kmph
(C) (40/7) kmph
(D) (30/7) kmph
(E) (65/7) kmph
Rationale: Let the first meeting point be at a distance of x km from P, the remaining distance until Q would be (80-x) km
Anand traveled this x kms in 3 hours, making his speed x/3 kmph
Ram traveled the same x kms in 2 hours, making his speed x/2 kmph
So, in 6 hours:
Anand will cover=6x/3=2x km
Ram will cover=6x/2=3x km
And between their first meeting point and second, they both together covered a distance of 2(90-x) km.
2x+3x=2(80-x)
5x=160-2x
7x=160
x=160/7 km
Anand's speed=x/3=160/(3*7)=160/21 kmph
Ans:B
Answer: (B) (160/21) kmph
Question: The sum of 5 different positive 2-digit integers is 130. What is the highest possible value of the largest of these integers?
(A) 88
(B) 84
(C) 78
(D) 74
(E) 68
Rationale: sum of 5 different 2 digit integers=130
to find out the largest number, we need to make other 4 numbers to the smallest as possible.
smallest 2 digit number is 10
=>10+11+12+13+n=130
=>n=130-46=84
Answer is B
Answer: (B) 84
Question: For any triangle T in the xy–coordinate plan, the center of T is defined to be the point whose x–coordinate is the average (arithmetic mean) of the x–coordinates of the vertices of T and whose y–coordinate is the average of the y–coordinates of the vertices of T. If a certain triangle has vertices at the points (0,0) and (9,3) and center at the point (4,3), what are the coordinates of the remaining vertex?
(A) (3,4)
(B) (3,9)
(C) (4,9)
(D) (6,4)
(E) (3,6)
Rationale: Let (a,b) be the remaining vertex.
x: 0 + 9 + a = 3*4. Then a = 3.
y: 0 + 3 + b = 3*3. Then b = 6.
The answer is E.
Answer: (E) (3,6)
Question: What could be the maximum value of Q in the following equation?
5P9 + 3R7 + 2Q8 = 1114
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Rationale: 5 P 9
3 R 7
2 Q 8 +
———-
11 1 4
———-
=>2+P+Q+R=11
so at max Q, P=R=0,
therefore Q=11-2=9.
ANSWER:E
Answer: (E) 9
Question: In a local school district, the high school and middle school each received r dollars toward funding for the student arts program. The high school enrolled 150 students and the middle school enrolled 75 students. Later, the middle school transferred s dollars to the high school so that they would have received the same funding per student. Which of the following is equivalent to s?
(A) r/8
(B) r/5
(C) r/2
(D) r/3
(E) r/9
Rationale: Total per head = 2r/225
After s transfer both schools have total for head.
So at High School it will be:
r+s = (150)* (2r/225) = 12r/9=4r/3
i.e. s = 4r/3 - r/1 =4r-3r/3=r/3
Answer :D
Answer: (D) r/3
Question: A chemical supply company has 60 liters of a 25% HNO3 solution. How many liters of pure undiluted HNO3 must the chemists add so that the resultant solution is a 50% solution?
(A) 12
(B) 15
(C) 20
(D) 24
(E) 30
Rationale: 60 liters of a 25% HNO3 solution means HNO3 = 15 liters in 60 liters of the solution.
Now, let x be the pure HNO3 added.
As per question,
15 + x = 50% of (60 + x)
or x =30. Hence, E
Answer: (E) 30
Question: The speed at which a man can row a boat in still water is 15 kmph. If he rows downstream, where the speed of current is 3 kmph, what time will he take to cover 60 metres?
(A) 65 seconds
(B) 14 seconds
(C) 14 seconds
(D) 12 seconds
(E) 12 seconds
Rationale: Speed of the boat downstream
= 15 + 3 = 18 kmph
= 18 * 5/18 = 5 m/s
Hence time taken to cover 60 m
= 60/5 =12 seconds.
Answer: D
Answer: (D) 12 seconds
Question: 50, 47, 45, 42, 40, 37, ?
(A) 36
(B) 35
(C) 34
(D) 33
(E) 32
Rationale: This is an alternating number subtraction series. The pattern is -3, -2, -3, -2, ....
The answer is B.
Answer: (B) 35
Question: A 4-letter code word consists of letters X, Y, and Z. If the code includes all the three letters, how many such codes are possible?
(A) 36
(B) 72
(C) 48
(D) 24
(E) 18
Rationale: Note that the correct answer to this question is 36, not 72.
X-XYZ can be arranged in 4!/2!=12 ways;
Y-XYZ can be arranged in 4!/2!=12 ways;
Z-XYZ can be arranged in 4!/2!=12 ways;
Total: 12+12+12=36.
Answer: A.
Answer: (A) 36
Question: If y exceeds x by 22%, then x is less than y by?
(A) 18 2/8%
(B) 18 8/3%
(C) 18 2/61%
(D) 76 2/3%
(E) 17 2/3%
Rationale: X=100	y=122
122------22
100-------? => 18 2/61%
Answer:C
Answer: (C) 18 2/61%
Question: Two persons starting from the same place walk at a rate of 5kmph and 5.5kmph respectively. What time will they take to be 12.5km apart, if they walk in the same direction?
(A) 17 hrs
(B) 22 hrs
(C) 25 hrs
(D) 12 hrs
(E) 14 hrs
Rationale: The relative speed of the boys = 5.5kmph â€“ 5kmph
= 0.5 kmph
Distance between them is 12.5 km
Time= 12.5km / 0.5 kmph = 25 hrs
Answer : C.
Answer: (C) 25 hrs
Question: If n is a positive integer, then n(n-1)(n-2) is +ve even
(A) odd whenever n is odd
(B) divisible by 3 only when n is odd
(C) divisible by 4 whenever n is even
(D) even only when n is odd
(E) even only when n is even
Rationale: n*(n-1)*(n-2)= +ve even integer divisible by 4 ( e.g 6,7&8)
C
Answer: (C) divisible by 4 whenever n is even
Question: In a friendship gang Andrew has 8 gang, in how many ways can he invite one or more of the gang to his house?
(A) 245
(B) 255
(C) 260
(D) 267
(E) 459
Rationale: Andrew can select one or more than one of his 8 gang.
=> Required number of ways = 28 – 1= 255.
B
Answer: (B) 255
Question: What sum of money put at C.I amounts in 2 years to Rs.8820 and in 3 years to Rs.9261?
(A) Rs.8000
(B) Rs.8400
(C) Rs.7500
(D) Rs.7800
(E) None
Rationale: 8820 ---- 441
100 ---- ? => 5%
x *105/100 * 105/100 = 8820
x*1.1025=8820
x=8820/1.1025 => 8000
ANSWER:A
Answer: (A) Rs.8000