mathewhe/aqua_rat huggingface dataset:
Question: A is twice as fast as B. If B alone can do a piece of work in 10 days, in what time can A and B together complete the work?
(A) 10/3
(B) 8
(C) 9
(D) 7/3
(E) 12
Rationale: A can do the work in10/2 i.e., 5 days.
A and B's one day's work = 1/5+ 1/10 = (2 + 1)/10 = 10/3
So A and B together can do the work in 10/3 days.
Answer: A
Answer: (A) 10/3
Question: If a carpenter completes 3/8th of his work in 6 days, then how many more days will he require completing his remaining work?
(A) 10
(B) 99
(C) 77
(D) 66
(E) 72
Rationale: Explanation:
As 3/8th of the work is completed, the work remaining is 5/8th of the total.
(3/8):6::(5/8):x
x = 6*(5/8)/(3/8)
x = 10
ANSWER: A
Answer: (A) 10
Question: A shop sells two variants of chocolates - one that costs $3 and the other that costs $5. If the shop sold $141 chocolates on a given day, how many different combinations of (number of $3 sold, number of $5 sold) exist?
(A) 5
(B) 7
(C) 9
(D) 11
(E) 13
Rationale: $141 = 27*$5 + 2*$3
We can find other combinations by reducing the $5 chocolates by 3 (which is -$15)and increasing the $3 chocolates by 5 (which is +$15).
The number of $5 chocolates can be 27, 24, 21, 18, 15,...,3 for a total of 9 combinations.
The answer is C.
Answer: (C) 9
Question: Evaluate:59- 12*3*2 =?
(A) 42
(B) 51
(C) 62
(D) 72
(E) 82
Rationale: According to order of operations, 12?3?2 (division and multiplication) is done first from left to right
12**2 = 4* 2 = 8
Hence
59 - 12*3*2 = 59- 8 = 51
correct answer B
Answer: (B) 51
Question: Meg and Bob are among the 3 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?
(A) 24
(B) 30
(C) 3
(D) 90
(E) 120
Rationale: Total # of ways the race can be finished is 3!. In half of the cases Meg finishes ahead of Bob and in other half Bob finishes ahead of Meg. So, ways Meg to finish ahead of Bob is 3!/2=3.
Answer: C.
Answer: (C) 3
Question: The 8 spokes of a custom circular bicycle wheel radiate from the central axle of the wheel and are arranged such that the sectors formed by adjacent spokes all have different central angles, which constitute an arithmetic series of numbers (that is, the difference between any angle and the next largest angle is constant). If the largest sector so formed has a central angle of 80°, what fraction of the wheel’s area is represented by the smallest sector?
(A) 1/72
(B) 1/36
(C) 1/18
(D) 1/12
(E) 1/9
Rationale: Largest angle = 80
Let, Difference between any two angles in A.P. = d
i.e. Sum of all angles will be
80 + (80-d) + (80-2d) + (80-3d) + (80-4d) + (80-5d) + (80-6d) + (80-7d) = 640 - 28d
But sum of all central angles in a circle = 360
i.e. 640 - 28d = 360
i.e. d = 280/28 = 10
Smallest Sector = (80-7d) = 80-7*10 = 10
Smallest sector as Fraction of entire circle = 10/360 = 1/36
Answer: option B
Answer: (B) 1/36
Question: If the equation x^2−6x=-k has at least one solution, which of the following must be true?
(A) k > 9
(B) k < −9
(C) k = 9
(D) k ≤ 9
(E) k ≥ −9
Rationale: x^2−6x=-k
=> x^2−6x + k = 0
For a quadratic equation to have at least one solution, its discriminant (b^2-4ac) should be non-negative.
b^2-4ac ≥0
=> 36 - 4*1*(-k) ≥0
=> 36 - 4k ≥0
=> k ≤9
Answer D
Answer: (D) k ≤ 9
Question: Look at this series: 544, 509, 474, 439, 404 ... What number should come next?
(A) 206
(B) 408
(C) 369
(D) 507
(E) 404
Rationale: 369
This is a simple subtraction series. Each number is 35 less than the previous number.
C
Answer: (C) 369
Question: If 20 men can build a wall 112 metres long in 6 days, what length of a similar wall can be built by 40 men in 3 days?
(A) 65mtr.
(B) 52mtr
(C) 70mtr.
(D) 112mtr.
(E) 17mtr.
Rationale: 20 men is 6 days can build 112 metres
25 men in 3 days can build = 112*(40/20)x(3/6)
= 112 meters
Answer: D.
Answer: (D) 112mtr.
Question: Sherlock Homes and Dr. Watson have to travel from Rajeev Gandhi chowk to Indira
Gandhi International Airport via the metro. They have enough coins of 1, 5, 10 and 25
paise. Sherlock Homes agree to pay for Dr. Watson only if he tells all the possible
combinations of coins that can be used to pay for the ticket.
How many combinations are possible if the fare is 50 paise ?
(A) 52
(B) 49
(C) 45
(D) 44
(E) 42
Rationale: Let the number of 1, 5, 10 and 50 paise coins required are x, y, z and w respectively.
So when paid total amount is: x + 5y + 10z + 25w, which is equal to the fare paid.
So, x + 5y + 10z + 25w = 50
The number of non negative integral solution of this equation will be equal to the
number of ways the fare can be paid.
x + 5y + 10z + 25w = 50
In this equation w can be either 0, 1 or 2
Case 1: When w is 2
No other coins are required to pay
So there is only one way to pay: (0, 0, 0, 2)
Total no of ways : 1
Case 2: When w is 1
x + 5y + 10z = 25
In this case z can have only three values possible: 0, 1 or 2
Case I: When z is 0
x + 5y = 25
possible values are: (25, 0, 0, 1), (20, 1, 0, 1), (15, 2, 0, 1), (10, 3, 0, 1),
(5, 4, 0, 1), (0, 5, 0, 1)
Case II: When z is 1
x + 5y = 15
possible values are: (15, 0, 1, 1), (10, 1, 1, 1), (5, 2, 1, 1), (0, 3, 1, 1)
Case III: When z is 2
x + 5y = 5
possible values are: (5, 0, 2, 1), (0, 1, 2, 1)
Total no of ways : 12
Case 3: When w is 0
x + 5y + 10z = 50
In this case z can have only three values possible: 0, 1, 2, 3, 4 or 5
Case I: When z is 0
x + 5y = 50
possible values are: (50, 0, 0, 0), (45, 1, 0, 0), (40, 2, 0, 0), (35, 3, 0, 0),
(30, 4, 0, 0), (25, 5, 0, 0), (20, 6, 0, 0), (15, 7, 0, 0), (10, 8, 0, 0), (5, 9, 0, 0),
(0, 10, 0, 0)
Case II: When z is 1
x + 5y = 40
possible values are: (40, 0, 1, 0), (35, 1, 1, 0), (30, 2, 1, 0), (25, 3, 1, 0),
(20, 4, 1, 0), (15, 5, 1, 0), (10, 6, 1, 0), (5, 7, 1, 0), (0, 8, 1, 0)
Case III: When z is 2
x + 5y = 30
possible values are: (30, 0, 2, 0), (25, 1, 2, 0), (20, 2, 2, 0), (15, 3, 2, 0),
(10, 4, 2, 0), (5, 5, 2, 0), (0, 6, 2, 0)
Case IV: When z is 3
x + 5y = 20
possible values are: (20, 0, 3, 0), (15, 1, 3, 0), (10, 2, 3, 0), (5, 3, 3, 0), (0,
4, 3, 0)
Case V: When z is 4
x + 5y = 10
possible values are: (10, 0, 4, 0), (5, 1, 4, 0), (0, 2, 4, 0)
Case VI: When z is 5
1 2 3 4 5 6 7 8 9
x + 5y = 0
possible values are: (0, 0, 5, 0)
Total no of ways : 36
So total no of ways = 1 + 12 + 36 = 49
ANSWER:B
Answer: (B) 49
Question: Nirmal and Kapil started a business investing Rs. 9000 and Rs. 12000 respectively. After 6 months, Kapil withdrew half of his investment. If after a year, the total profit was Rs. 4600, what was Kapil’s share initially ?
(A) Rs 2300
(B) Rs 2400
(C) Rs 2500
(D) None of above
(E) Rs 3500
Rationale: Explanation:
Nirmal:Kapil = 9000*12:(12000*6+6000*6) = 1:1
Kapils share = Rs. [4600 *(1/2)) = Rs. 2300
Option A
Answer: (A) Rs 2300
Question: There are t members in a certain department, including Michael. Two representatives are to be selected to attend a company conference. If there are 55 possible combinations in which Michael is not selected, what is the value of t?
(A) 11
(B) 12
(C) 15
(D) 18
(E) 22
Rationale: Combinations of two persons, in which Michael was not selected = 55
number of ways two persons can be selected from m people = m*(m-1)/2
Let m be the number of people excluding Michael, then m*(m-1) = 110 => m=11
Thus, t = m + 1(Michael) = 12
option B
Answer: (B) 12
Question: There are 7 cups placed on a table; 4 are juice cups and the other 3 are tea cups. If 2 cups are to be selected at random from the 7 cups, what is the probability that at least one of the juice cup will be selected?
(A) 6/7
(B) 2/7
(C) 3/7
(D) 4/7
(E) 5/7
Rationale: Total probability=7C2=21
4C2 +4C1*3C1=6+12=18
therefore the probability that at least one of the fashion magazines will be selected= 18/21=6/7
answer is A
Answer: (A) 6/7
Question: The positive integers a and b leave remainders of 4 and 7, respectively, when divided by 9. What is the remainder when a − 2b is divided by 9?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Rationale: a=9x+4
so a could be 4, 13,22, 31,40,49,58,67,76,85
b=9y + 7
so b could be 7,16,25,34,43
Selecting any random number for a and b such that a-2b is positive number,
say a=22, b=7 => 2b=14
Hence a-2b = 8 Hence remainder when divided by 9 =8
say a=67, b=25 => 2b=50
Hence a-2b = 17 Hence remainder when divided by 9 is again 8
say a=49, b=7 => 2b=14
Hence a-2b = 35 Hence remainder when divided by 9 is again 8
Answer is 8.
ANSWER:D
Answer: (D) 8
Question: In a group of 800 people, 2/5 play at least one instrument, 96 play two or more. What is the probability that one student play exactly one instrument?
(A) 2/125
(B) 3/125
(C) C)2/25
(D) 3/25
(E) 7/25
Rationale: P(playing 2 or more instruments) = 96/800 = 3/25.
Then, the probability of playing exactly one instrument is given by:
P(playing 1 or more instruments) - P(playing 2 or more instruments) = 2/5 - 3/25 = 7/25.
Answer E.
Answer: (E) 7/25
Question: A certain sum becomes four times itself at simple interest in nine years. In how many years does it become ten times itself?
(A) 27 years.
(B) 25 years.
(C) 29 years.
(D) 17 years.
(E) 37 years.
Rationale: Let the sum be Rs. x, then it becomes Rs. 4x in eight years Rs. 3x is the interest on x for eight years.
R = (100 * 3x)/(x * 9) = 100/3 %
If the sum becomes ten times itself, then interest is 9x.
The required time period = (100 * 9x)/(x * 100/3) = (100 * 9x * 3)/(x * 300) = 27 years.
ANSWER:A
Answer: (A) 27 years.
Question: Three business people who wish to invest in a new company. Each person is willing to pay one third of the total investment. . After careful calculation, they realize that each of them would pay $ 7400 less if they could find two more equal investors. How much is the total investment in the new business.
(A) a)	$ 64,000
(B) b)	$ 54,000
(C) c)	$ 55,500
(D) d)	$ 5,400
(E) e)	$3,600
Rationale: Initially each invest in x. Hence total investment is 3x.
Total investment is also 5(x-7400).
3x = 5(x-7400)
x= 5*7400/2 = 18500
3x = 55500 and the answer is C.
Answer: (C) c)	$ 55,500
Question: A box contain the number of balls which is as much greater than 20 is less than 30. The number of balls is?
(A) 40
(B) 25
(C) 30
(D) 50
(E) 35
Rationale: Answer
Let the number be x.
Then, x - 20 = 30- x
â‡’ 2x = 30+ 20= 50
â‡’ 2x = 50
âˆ´ x = 25
Correct Option: B
Answer: (B) 25
Question: If x is even and y is odd, then x-y
(A) Even whenever xy is odd
(B) Odd
(C) Odd whenever xy is even
(D) Even
(E) It cannot be determined
Rationale: x is even and y is odd, then x + y is
Let's take x=4 and y=7.
x+y=4+7=11.
x+y=11=odd.
Option: B
Answer: (B) Odd
Question: The price of 2 saris and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
(A) Rs. 2200
(B) Rs. 2300
(C) Rs. 2400
(D) Rs. 2480
(E) Rs. 2490
Rationale: Let the price of a saree and a shirt be Rs. x and Rs. y respectively.
Then, 2x + 4y = 1600 .... (i)
and x + 6y = 1600 .... (ii)
Divide equation (i) by 2, we get the below equation.
=> x + 2y = 800. --- (iii)
Now subtract (iii) from (ii)
x + 6y = 1600 (-)
x + 2y = 800
----------------
4y = 800
----------------
Therefore, y = 200.
Now apply value of y in (iii)
=> x + 2 x 200 = 800
=> x + 400 = 800
Therefore x = 400
Solving (i) and (ii) we get x = 400, y = 200.
Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.
C
Answer: (C) Rs. 2400
Question: The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C,D and E becomes 79 kg. The weight of A is :
(A) 70 kg
(B) 72 kg
(C) 75 kg
(D) 80 kg
(E) none
Rationale: Explanation:
A+B+C = 3×84=252
A+B+C+D= (4×80)=320
D = (320-252)=68 and E = (68+3)=71
Now, B+C+D+E = (4×79)=316
B+C+D=(316-71)=245 kg
So, A = (320-245)=75 kg
Correct Option: C
Answer: (C) 75 kg
Question: The floor of a rectangular room is 15 m long and 12 m wide. The room is surrounded by a verandah of width 2 m on all its sides. The area of the verandah is :
(A) 124 m2
(B) 120 m2
(C) 108 m2
(D) 58 m2
(E) None of these
Rationale: Area of the outer rectangle = 19 × 16 = 304 m2
Area of the inner rectangle = 15 × 12 = 180 m2
Required area = (304 – 180) = 124 m2
Answer A
Answer: (A) 124 m2
Question: If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Rationale: Let 1 men's 1 day work = x and 1 boy's 1 day work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2
Solving these two equations, we get:
x = 1/100 and y = 1/200
(15 men + 20 boys)'s 1 day work = (15/100 + 20/200) = 1/4
15 men and 20 boys can do the work in 4 days.
Answer:A
Answer: (A) 4
Question: How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits S equal to 28?
(A) 14
(B) 24
(C) 28
(D) 48
(E) 96
Rationale: First, look for all 4 digits without repeat that add up to 28. To avoid repetition, start with the highest numbers first.
Start from the largest number possible 9874.
Then the next largest number possible is 9865.
After this, you'll realize no other solution. Clearly the solution needs to start with a 9 (cuz otherwise 8765 is the largest possible, but only equals 26). With a 9, you also need an 8 (cuz otherwise 9765 is the largest possible, but only equals 27). With 98__ only 74 and 65 work.
So you have two solutions. Each can be rearranged in 4!=24 ways. So S=24+24=48.D
Answer: (D) 48
Question: Exactly three sides of a certain 10-sided die are red. What is the probability that Kumar rolls the die 3 times and the die lands with a red side up for the first time on the third roll?
(A) 0.184
(B) 0.16
(C) 0.18
(D) 0.147
(E) 0.148
Rationale: Total no. of sides = 10
Sides that are red =3
Probability that the die lands with red side up = 3/10
Therefore, Probability that the die does not land with red side up = 1 - 3/10 = 7/10
Probability that Kumar rolls the die 3 times and the die lands with a red side up for the first time on the third roll
= (1st roll - non red face) X (2nd roll - non red face) X (3rd roll - red face)
= (7/10) X (7/10) X (3/10)
= 128/1000
= 0.147
D
Answer: (D) 0.147
Question: The value of Underrot 32pi/4 + Underrot 25 is most nearly equal to which of the following integers? (Please refer to the picture below for the exact version of the question)
(A) 6
(B) 35
(C) 41
(D) 30
(E) 23
Rationale: Under root (32pi/4) ~ 25
Under root (25) = 5
Hence, 25+5 = 30 is the approx answer. D is the correct answer.
Answer: (D) 30
Question: A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
(A) 3.6
(B) 7.2
(C) 8.4
(D) 10
(E) 11
Rationale: Speed =(600/5*60)m/sec.
= 2 m/sec.
Converting m/sec to km/hr
=(2*18)/5	km/hr
=7.2 km/hr.
Answer:B
Answer: (B) 7.2
Question: At what rate percent on simple interest will a sum of money double itself in 30 years?
(A) 3 1/3%
(B) 8 1/3%
(C) 3 5/3%
(D) 5 1/3%
(E) 3 1/7%
Rationale: P = (P*30*R)/100
R = 3 1/3%
Answer: A
Answer: (A) 3 1/3%
Question: A small company employs 3 men and 5 women. If a team of 3 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
(A)  1/14
(B)  1/7
(C)  2/7
(D)  5/7
(E)  1/2
Rationale: Total ways to choose 4 employees from 8 employees = 8C3
Two women = 5C2
Two Men = 3C1
P = (5C2 * 3C1)/8C2 = 5/7 hence D.
Answer: (D)  5/7
Question: A train speeds past a pole in 15 sec and a platform 120 m long in 25 sec, its length is?
(A) 238
(B) 180
(C) 988
(D) 177
(E) 171
Rationale: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 120)/25 = x/15 => x = 180 m.
Answer: B
Answer: (B) 180
Question: If 6x^2 + x - 12 = (ax + b)(cx + d), then |a| + |b| + |c| + |d|
(A) 10
(B) 12
(C) 15
(D) 18
(E) 20
Rationale: This is the hard one, definitely a 700+ level question. We need numbers a, b, c, and d such that
6x^2 + x - 12 = (ax + b)(cx + d)
This means that ac = 6, bd = –12, and ad + bc = 1. The a & c pair could be (1, 6) or (2, 3), in some order. The absolute values of the b & d pair could be (1, 12) or (2, 6) or (3, 4), and of course, in each case, one of the two would have to be negative. After some trial and error, we find:
6x^2 + x - 12 = (2x + 3)(3x - 4)
Thus, we see:
|a| + |b| + |c| + |d| = 2 + 3 + 3 + 4 = 12
Answer = B
Answer: (B) 12
Question: The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes 3/4. Find the original fraction?
(A) 3/15
(B) 3/10
(C) 3/11
(D) 3/12
(E) 3/16
Rationale: The denominator be P, the numerator will be (P - 8).
The fraction will be (P - 8)/P.
Adding 3 to the numerator and subtracting 3 from the denominator, (P - 8 + 3)/(P - 3) = 3/4.
(P - 5)/(P - 3) = 3/4
P = 20 - 9 => P = 11.
The fraction is: 3/11.Answer: C
Answer: (C) 3/11
Question: The events A and B are independent, the probability that event A occurs is greater than 0, and the probability that event A occurs is twice the probability that event B occurs. The probability that at least one of events A and B occurs is 3 times the probability that both events A and B occur. What is the probability that event A occurs?
(A) 44/7
(B) 9/8
(C) 25/8
(D) 3/4
(E) 7/8
Rationale: Let us say probability of A occuring is a.
Let us say probability of B occuring is b.
a = 2b
Probability (either A or B or both) = 3 times Probability (A and B)
a*(1-b) + b*(1-a) + ab = 3*ab
Substituting a=2b in the second equation:
2b*(1-b) + b*(1-2b) + 2b*b = 3*2b*b
3b-2b^2 = 6b^2
3b = 8b^2
b = 3/8 = 3/8
So, a = 2b = 3/4
THE CORRECT ANSWER IS D.
Answer: (D) 3/4
Question: Find the invalid no.from the following series 15, 16, 34, 105, 424, 2124, 12756
(A) 1200
(B) 1465
(C) 1765
(D) 1900
(E) 2124
Rationale: 2nd term = (1st term) * 1 + 1 = 15 *1 + 1 = 16.
3rd term = (2nd term) * 2 + 2 = 16 * 2 + 2 = 34
4th term = (3th term) *3 + 3 = 34 * 3 + 3 = 105
5th term = (4th term) * 4 + 4 = 105 * 4 + 4 = 424
6th term = (5th term) * 5 + 5 = 424 * 5 + 5 = 2125
6th term should 2125 instead of 2124.
E
Answer: (E) 2124
Question: When the integer x is divided by the integer y, the remainder is 60. Which of the following is a possible value of the quotient x/y?
I. 15.19
II.18.16
III. 17.17
(A)  I only
(B)  II only
(C)  III only
(D)  I and II only
(E)  I and III only
Rationale: Problem statement was difficult to parse. but some how I managed to understand what we can do.
1) x/y = 15 + 0.19
2) x/y = 18 + 0.16
3) x/y = 17 + 0.17
w.k.t x = yn (i.e Q) + yk=60 (i.e remainder)
therefore :
1) y*0.19=60 -- we can't get y an integer.
2) y*0.16=60 -- we can get y an integer.
3) y*0.17=60 -- we can't get y an integer.
Therefore 2 are possible answers.
Ans B.
Answer: (B)  II only
Question: Two trains 111 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 40 km and the other at the rate of 50 kmph. In what time will they be completely clear of each other from the moment they meet?
(A) 11.04
(B) 10.0
(C) 6.85
(D) 5.85
(E) 6.15
Rationale: T = (111 + 165)/ (40 + 50) * 18/5
T = 11.04
ANSWER:A
Answer: (A) 11.04
Question: An automobile manufacturer offers a station wagon with either a 14-cylinder engine or a 12-cylinder engine and with either a manual transmission or an automatic transmission. A trailer hitch is also offered, but only on a station wagon with a 14-cylinder engine. How many combinations of the five options listed does the manufacturer offer for its station wagon?
(A)  93
(B)  94
(C)  95
(D)  96
(E)  98
Rationale: 14-cylinder engine wagons = 2*2*2*2*2*2= 64 (manual or automatic, with or without trailer);
12-cylinder engine wagons = 2*2*2*2*2 = 32 (manual or automatic).
Total = 64 + 32 = 96.
Answer: D.
Answer: (D)  96
Question: On increasing the price of T.V. sets by 50%, their sale decreases by 20%. What is the effect on the revenue receipts of the shop ?
(A) 4
(B) 5
(C) 56
(D) 20
(E) 7
Rationale: Explanation:
Let the price be = Rs.100, and number of units sold = 100
Then, sale value = Rs.(100 × 100) = Rs.10000
New sale value = Rs.(150 × 80) = Rs.12000
Increase% = 2000/10000 × 100 = 20%
Answer: D
Answer: (D) 20
Question: A trader has 200 kg of sugar, out of which he sells some with 10% profit and the rest with 25% profit. He gains 15% on the whole. Find the quantity of sugar sold at 25% profit?
(A) 120 kg
(B) 96.5 kg
(C) 84 kg
(D) 67 kg
(E) None of these
Rationale: Explanation:
Let CP of each kg. of sugar = Rs.1.
Cost Price of 200 kg of sugar = Rs.200.
Let quantity of sugar sold with 25% profit be X.
Therefore, (200 – X) kg of sugar sold at a profit of 10%
Then,
(125/100 * X) + 110/100 (200 – X) = 115/100 * 200
125X + 110(200 – X) = 115 * 200
125X + 22000 – 110X = 23000
125X – 110X = 23000 – 22000
15X = 1000
X = 1000/15
X = 66.66 kg
Quantity of sugar that was sold at 25% gain is 66.66 kg
Therefore, option nearby for 66.66 is 67 (d)
ANSWER: D
Answer: (D) 67 kg
Question: By selling 15 pencils for a rupee a man loses 15%. How many for a rupee should he sell in order to gain 15%?
(A) 8
(B) 9
(C) 11
(D) 89
(E) 81
Rationale: 85% --- 15
115% --- ?
15/115 * 15 = 11
Answer: C
Answer: (C) 11
Question: How many terminating zeroes T does 200! have?
(A)  40
(B)  48
(C)  49
(D)  55
(E)  64
Rationale: You have 40 multiples of 5, 8 of 25 and 1 of 125. This will give 49 zeros.C
Answer: (C)  49
Question: Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?
(A) 44, 36
(B) 42, 33
(C) 43, 36
(D) 42, 36
(E) None
Rationale: Let the marks secured by them be x and (x + 9)
Then sum of their marks = x + (x + 9) = 2x + 9
Given that (x + 9) was 56% of the sum of their marks
=>(x+9) = 56/100(2x+9)
=>(x+9) = 14/25(2x+9)
=> 25x + 225 = 28x + 126
=> 3x = 99
=> x = 33
Then (x + 9) = 33 + 9 = 42
Hence their marks are 33 and 42
B
Answer: (B) 42, 33
Question: A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:
(A) Rs. 2000
(B) Rs. 10,000
(C) Rs. 15,000
(D) Rs. 20,000
(E) Rs. 26,000
Rationale: Principal = Rs.(100 x 5400)/(12x3)	= Rs. 15000.
Answer: Option C
Answer: (C) Rs. 15,000
Question: Pipes A and B can fill a tank in 4 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in?
(A) 3 hrs
(B) 3 9/77 hrs
(C) 3 9/17 hrs
(D) 3 9/27 hrs
(E) 3 9/19 hrs
Rationale: Net part filled in 1 hour = 1/4 + 1/6 - 1/12
= 4/12 = 1/3
The tank will be full in 3 hrs
Answer:A
Answer: (A) 3 hrs
Question: P is three times as fast as Q and working together, they can complete a work in 20 days. In how many days can Q alone complete the work?
(A) 26 1/2 days
(B) 27 1/2days
(C) 29 3/4days
(D) 14 1/4 days
(E) 10 1/2days
Rationale: P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 20 days, which means Q can do the work in 80 days.
Hence, P can do the work in 26 1/2 days.
Answer:A
Answer: (A) 26 1/2 days
Question: How many 3's are there in the following sequence which are neither preceded by 6 nor immediately followed by 9 ?
9 3 6 6 3 9 5 9 3 7 8 9 1 6 3 9 6 3 9
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: D
9 3 6 6 3 95 9 3 78 9 16 3 9 6 3 9
Answer: (D) 4
Question: Which of the following is closest to (-9/10)^199？
(A) -1
(B) -1/2
(C) 0
(D) 1
(E) 2
Rationale: (-9/10)^4 = 6561/10,000 which is already less than 2/3.
For larger exponents, the expression will get closer and closer to zero.
The answer is C.
Answer: (C) 0
Question: "How old are you, Alchemerion?" asked one of the wizards appearances the wizards answer with a riddle, "I am still Very young as wizards go.I am only three times my son's age. My father is 40 year more than twice of my age. Together the three of us are a 1240 year old". How old is Alchemerion
(A) 300
(B) 320
(C) 340
(D) 360
(E) 380
Rationale: let Alchemerion be A
let son be S
let father be F
A= 3* S
F= 40 +(2*A)
therefore 1240=(S+F+A)
1240=(A/3)+ (40+(2*A) )+ A
A=360
ANSWER:D
Answer: (D) 360