mathewhe/aqua_rat huggingface dataset:
Question: The cross-section of a cannel is a trapezium in shape. If the cannel is 10 m wide at the top and 6 m wide at the bottom and the area of cross-section is 640 sq m, the depth of cannel is?
(A) 28
(B) 77
(C) 99
(D) 80
(E) 21
Rationale: 1/2 * d (10 + 6) = 640
d = 80
Answer: D
Answer: (D) 80
Question: If a man can cover 12 metres in one second, how many kilometres can he cover in 3 hours 45 minutes?
(A) 12 km
(B) 162 km
(C) 6 km
(D) 87 km
(E) 15 km
Rationale: Explanation:
12 m/s = 12 * 18/5 kmph
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Distance = speed * time = 12 * 18/5 * 15/4 km = 162 km.
B)
Answer: (B) 162 km
Question: Two pipes A and B can separately fill a tank in 2 minutes and 15 minutes respectively. Both the pipes are opened together but 4 minutes after the start the pipe A is turned off. How much time will it take to fill the tank?
(A) 33
(B) 10
(C) 99
(D) 73
(E) 23
Rationale: 4/12 + x/15 = 1
x = 10
Answer: B
Answer: (B) 10
Question: Water is poured into a tank so that the tank is being filled at the rate of 4 cubic feet per hour. If the empty rectangular tank is 6 feet long, 4 feet wide, and 3 feet deep, approximately how many hours does it take to fill the tank?
(A) 16
(B) 18
(C) 20
(D) 22
(E) 24
Rationale: The volume the tank is: length*width*depth = 6*4*3= 72 cubic feet.
72 cubic feet / 4 cubic feet per hour = 18 hours.
It will take 18 hours to fill the tank.
The answer is B.
Answer: (B) 18
Question: The salary of a worker is first increased by 40% and afterwards reduced by 40%. What is the net change in the worker's salary?
(A) 8% decrease
(B) 8% increase
(C) 16% decrease
(D) 16% increase
(E) no change
Rationale: Let x be the original salary.
The final salary is 0.6(1.4x) = 0.84x
The answer is C.
Answer: (C) 16% decrease
Question: In September, John sold 2323 more soap than in February. In November, he sold 5 percent more than in September. John's November sales were what percent greater than his February sales?
(A) 10/3%
(B) 65%
(C) 214/3 %
(D) 75%
(E) 514/3%
Rationale: Let february sale= x
september sale= x+2/3x= 5x/3
November sale= 5x+3 + 5*5/100*3x
21/12x= 1.75 x
Hence November sale is 75% > February sale
D is the answer
Answer: (D) 75%
Question: Two sets A and B consists only of positive integers such that mean of set A is 45 . If set R consist of all the elements of both A and B such that mean of R is 80 then the mean of set B is =>
(A) [A] 65
(B) [B] 60
(C) [C] 70
(D) [D] 55
(E) [E] cannot be determined
Rationale: For this question, we have zero information about the size of two sets, sets AB. These two sets may have the same size or may be wildly different. Thus, the answer is (E), the mathematical answer cannot be determined. This is a very weak math question. As someone who writes GMAT practice questions, I would give this a grade of anD-as a question.
Answer E
Answer: (E) [E] cannot be determined
Question: A man can row with a speed of 15 kmph in still water. If the stream flows at 5 kmph, then the speed in downstream is?
(A) 29
(B) 378
(C) 20
(D) 27
(E) 121
Rationale: M = 15
S = 5
DS = 15 + 5 = 20
Answer: C
Answer: (C) 20
Question: At a special sale, 9 tickets can be purchased for the price of 6 tickets. If 9 tickets are purchased at the sale, the amount saved will be What percent of the original price of the 9 tickets?
(A) 20%
(B) 33.3%
(C) 40%
(D) 60%
(E) 66.6%
Rationale: let the price of a ticket be Rs.100, so 6 tickets cost 600 & 9 tickets cost 900
9 tickets purchased at price of 6 tickets ie., for 600, so amount saved s Rs.300,
% of 9 tickets = (300 /900)*100 =33.3 %
ANSWER:B
Answer: (B) 33.3%
Question: In a group containing 6 cows and 4 buffalos, 4 livestock are to be selected in such a way that at least 1 cow should always be present. How many way of doing that are possible?
(A) 209
(B) 277
(C) 297
(D) 161
(E) 108
Rationale: ANSWER:A
Answer: (A) 209
Question: If 47. 2506 = 4*A + 7/B + 2*C + 5/D + 6*E, then the value of 5*A + 3*B + 6*C + D + 3*E is :
(A) 153.6002
(B) 153.6008
(C) 153.6003
(D) 153.6001
(E) 153.6022
Rationale: Explanation:
4*A + 7/B + 2*C + 5/D + 6*E =47.2506
=> 4 * A + 7/B + 2 *C + 5/D + 6*E = 40 + 7 + 0.2 + 0.05 + 0.0006
Comparing the terms on both sides, we get :
4*A =40, 7/B = 7, 2*C = 0.2, 5/D = 0.05, 6*E = 0.0006
A = 10,
B = 1,
C = 0.1,
D = 100,
E = 0.0001.
5*A + 3*B + 6*C + D + 3*E =(5*10) + (3*1) + (6*0.1) +100 +(3*0.0001)
=50 + 3 + 0.6 + 100 + 0.003 = 153.6003.
Answer: C) 153.6003
Answer: (C) 153.6003
Question: Alex takes a loan of $5,000 to buy a used truck at the rate of 2 % simple Interest.Calculate the annual interest to be paid for the loan amount.
(A) 100
(B) 150
(C) 720
(D) 200
(E) 750
Rationale: From the details given in the problem Principle = P = $5,000 and R = 2% or 0.02 expressed as a decimal.
As the annual Interest is to be calculated, the time period T =1.
Plugging these values in the simple Interest formula,
I = P x T x R
= 5,000 x 1 x 0.02
= 100.00
Annual Interest to be paid = $100
answer :A
Answer: (A) 100
Question: A can run a kilometer race in 4 1/2 min while B can run same race in 5 min. How many meters start can A give B in a kilometer race, so that the race mat end in a dead heat?
(A) 887 m
(B) 998 m
(C) 55 m
(D) 100 m
(E) 44 m
Rationale: A can give B (5 min - 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.
Answer:D
Answer: (D) 100 m
Question: A merchant marks goods up by 60% and then offers a discount on the marked price. The profit that the merchant makes after offering the discount is 20%. What % discount did the merchant offer?
(A) 20%
(B) 25%
(C) 30%
(D) 35%
(E) 40%
Rationale: Let P be the original price of the goods and let x be the rate after the markup.
(1.6P)*x = 1.2P
x = 1.2/1.6 = 0.75 which is a discount of 25%.
The answer is B.
Answer: (B) 25%
Question: Find the one which does not belong to that group ?
(A) 41
(B) 43
(C) 53
(D) 47
(E) 57
Rationale: Explanation:
41, 43, 47 and 53 are prime numbers, but not 57.
ANSWER IS E
Answer: (E) 57
Question: The cubic f(x) = x3 +bx2 +cx+d satisfies f(1) = 3, f(2) = 6, and f(4) = 12. Compute f(3).
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Rationale: Note that f(x) - 3x = (x - 1)(x- 2)(x - 4), so f(x) = (x - 1)(x - 2)(x- 4) + 3x
and therefore f(3) = 2 *1 *(-1) + 9 = 7 .
correct answer B
Answer: (B) 7
Question: A sun is divided among X, Y and Z in such a way that for each rupee X gets, Y gets 45 paisa and Z gets 50 paisa. If the share of Y is RS. 54, what is the total amount?
(A) 234
(B) 116
(C) 117
(D) 118
(E) 119
Rationale: x:y:z = 100:45:50
20:9:10
9 --- 54
39 --- ? => 234
ANSWER:A
Answer: (A) 234
Question: The edge of a cube is 6a cm. Find its surface?
(A) 216a2
(B) 24a4
(C) 24a1
(D) 24a2
(E) 24a7
Rationale: 6a2 = 6 * 6a * 6a = 216a2
Answer:A
Answer: (A) 216a2
Question: The difference of the squares of two consecutive even integers is divisible by which of the following integers ?
(A) 5
(B) 18
(C) 7
(D) 45
(E) 4
Rationale: Explanation:
Let the two consecutive even integers be 2n and (2n + 2). Then,
(2n+2)^2 = (2n+2 2n)(2n 2-2n)
= 2(4n + 2)
=4(2n + 1), which is divisible by 4.
ANSWER:E
Answer: (E) 4
Question: 2, 7, 14, 23, ?, 47
(A) 31
(B) 28
(C) 34
(D) 33
(E) 32
Rationale: The given sequence is +5, +7, +9, ——
ie. 2+ 5 = 7, 7 + 7 = 14, 14 + 9 = 23
Missing Number = 23 + 11 = 34. ANSWER:C
Answer: (C) 34
Question: The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is
(A) 45 kg
(B) 50 kg
(C) 53 kg
(D) 55 kg
(E) NOne
Rationale: Sol.
Weight of the teacher
= (35.4 × 25 – 35 × 24)kg
= 45 kg.
Answer A
Answer: (A) 45 kg
Question: If a certain sample of data has a mean of 20.0 and a standard deviation of 2.0, which of the following values is more than 2.5 standard deviations from the mean?
(A)  13.5
(B)  12.0
(C)  17.0
(D)  23.5
(E)  26.5
Rationale: The standard deviation is 2.0. 2.5 standard deviations, which is just 2.5 x 2.0 is 5.0. The only answer more than 6.0 away from the mean is B.
Answer: (B)  12.0
Question: Henry answered 3/5 of the questions he was given for homework in the first hour, 5/19 of the remaining questions in the second hour and the remaining x questions in the third hour where x is an integer. How many questions was Henry given for homework?
(A) x
(B) (95/7)*x
(C) (95/14)*x
(D) (95/21)*x
(E) (95/28)*x
Rationale: Let Q be the number of homework questions.
Q = 3Q/5 + (5/19)(2Q/5) + x
Q = 57Q/95 + 10Q/95 + x
28Q/95 = x
Q = (95/28)*x
The answer is E.
Answer: (E) (95/28)*x
Question: a grazing land have length 200m and breadth 50m.there is grass 0.5kg/m2.
a cow graze 20kg grass daily.if cow graze for 20days.so after 20 days calculate the percentage of grass filled area.
(A) 90
(B) 92
(C) 95
(D) 85
(E) 80
Rationale: area of land=200*50=10000m2.
total grass grazed=400kg.
area grazed=400/0.5=800m2.
%of grass filled area=9200/10000*100=92%
answer B
Answer: (B) 92
Question: The product of three consecutive numbers is 120. Then the sum of the smallest two numbers is ?
(A) 9
(B) 15
(C) 20
(D) 38
(E) 56
Rationale: Product of three numbers = 120
120 = 2*2*5*6 = 4*5*6
So, the three numbers are 4,5 and 6.
And Sum of smallest of these two = 4 + 5 = 9.
ANSWER : OPTION A
Answer: (A) 9
Question: If a*b denotes the greatest common divisor of a and b, then ((16*20)*(18*24)) = ?
(A)  24
(B)  12
(C)  6
(D)  4
(E)  2
Rationale: The greatest common divisor of 16 and 20 is 4. Hence 16*20 = 4 (note that * here denotes the function not multiplication).
The greatest common divisor of 18 and 24 is 6. Hence 18*24 = 6.
Hence ((16*20)*(18*24)) = 4*6.
The greatest common divisor of 4 and 6 is 2.
Answer; E.
Answer: (E)  2
Question: A money lender lent Rs. 800 at 3% per year and Rs. 1400 at 5% per year. The amount should be returned to him when the total interest comes to Rs. 350. Find the number of years.
(A) 3.5
(B) 3.72
(C) 4
(D) 4.25
(E) 4.5
Rationale: (800xtx3/100) + (1400xtx5/100) = 350 â†’ t =3.72 answer B
Answer: (B) 3.72
Question: If the sum of the 4th term and the 12th term of an arithmetic progression is 8, what is the sum of the first 15 terms of the progression?
(A) 60
(B) 120
(C) 160
(D) 240
(E) 840
Rationale: Let a = first term, d = common difference.
4th term = a+3d
12th term = a+11d
a+3d + a+11d = 8
2a + 14d = 8
a + 7d = 4.
Sum of the first 15 terms = 15/2 (2a + 14d) = 15 (a+7d) = 15 (4) = 60. Ans (A)
Answer: (A) 60
Question: If k is the sum of the digits of integer m, and m=12n, where n is an integer, which of the following must be true?
(A) k is not divisible by 9
(B) The sum of the digits of k is divisible by 3
(C) m is a multiple of 2k
(D) k is a multiple of 9
(E) k is a multiple of 6
Rationale: m= 12n i.e n =1,2,3 ,4 etc or -1,-2.-3 ,-4 then m =12,24,36,48 or -12 ,-24 ,-36 ,-48 then k 3,6 9,12 or -(1+2) , -(4+2) , -(3+6) ,-(4+8) .
ans B
Answer: (B) The sum of the digits of k is divisible by 3
Question: A certain roller coaster has 2 cars, and a passenger is equally likely to ride in any 1 of the 2 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 2 times, what is the probability that the passenger will ride in each of the 2 cars?
(A) 0
(B) 1/2
(C) 2/9
(D) 1/3
(E) 1
Rationale: If he is to ride 2 times and since he can choose any of the 2 cars each time, total number of ways is
= 2 * 2
= 4
Now the number of ways if he is to choose a different car each time is
= 2 * 1
= 2
So the probability is
= 2/4
= 1/2
Answer : B
Answer: (B) 1/2
Question: A train passes a bridge in 40 sec and a man standing on the platform in 20 sec. If the speed of the train is 54 km/hr. What is the length of the platform?
(A) 250
(B) 400
(C) 150
(D) 300
(E) 450
Rationale: Speed = 54 * 5/18 = 15 m/sec.
Length of the train = 15 * 20 = 300 m.
Let the length of the platform be x m . Then,
(x + 300)/40 = 15 => x = 300 m.
Answer: D
Answer: (D) 300
Question: Six women can do a work in 10 days. Ten men can complete the same work in 4 days. What is the ratio between the capacity of a man and a woman?
(A) 1:2
(B) 2:1
(C) 2:3
(D) 3:2
(E) None of these
Rationale: Explanation:
(6 Ã— 10) women can complete the work in 1 day.
âˆ´ 1 woman's 1 day's work = 1/60
(10 Ã— 4) men can complete the work in 1 day.
âˆ´ 1 man's 1 day's work = 1/40
so, required ratio =1/60 : 1/40= 3:2
Answer: D
Answer: (D) 3:2
Question: How many pounds of salt at 70 cents/lb must be mixed with 45 lbs of salt that costs 40 cents/lb so that a merchant will get 20% profit by selling the mixture at 48 cents/lb?
(A) 5
(B) 9
(C) 40
(D) 50
(E) 25
Rationale: Selling price is 48 cents/lb
For a 20% profit, cost price should be 40 cents/lb (CP*6/5 = 48)
Basically, you need to mix 40 cents/lb (Salt 1) with 70 cents/lb (Salt 2) to get a mixture costing 45 cents/lb (Salt Avg)
weight of Salt1/weight of Salt2 = (Salt2 - SaltAvg)/(SaltAvg - Salt1) = (70 - 45)/(45 - 40) = 5/1
We know that weight of salt 1 is 45 lbs. Weight of salt 2 must be 9 lbs.
Answer (B)
Answer: (B) 9
Question: A car covers a certain distance at a speed of 40 kmph in 5 hours. To cover the same distance in 2 hr, it must travel at a speed of?
(A) 100 km/hr
(B) 678 km/hr
(C) 782 km/hr
(D) 789 km/hr
(E) 720 km/hr
Rationale: Distance = (40 x 5) = 200 km
Speed = Distance/Time
Speed = 200/2 = 100 km/hr
Answer : A
Answer: (A) 100 km/hr
Question: If n ≠ 0, which of the following must be greater than n?
I. 2n
II. 2 - n
III. n^2
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) None
Rationale: n ≠ 0 => n = negative or n = positive
if n = negative:
2n < n (n = -1 then 2n = -2 => -2 < -1)
n^2 > n (n =-1 then n^2 = 1 => n^2 > n)
2 - n > n (n = -1 then 2--1 = 2+1=3 => 2-n > n)
but if n = positive:
n^2 > n (n = 2 then n^2 = 4 => n^2 > n) => II
2 - n < n(n =2 then 2-2 = 0)
=> only III
=>answer is E
Answer: (E) None
Question: The product of two numbers is 266 and their difference is 5. What is the bigger number ?
(A) 67
(B) 26
(C) 26
(D) 19
(E) 72
Rationale: Explanation:
Let the two numbers be A and B, here A > B
AB = 266
B = 266/A -----------------(I)
Given,
A – B = 5 ----------- (II)
Substitute from (I) in (II), we get
A – 266/A = 5
A2 – 5A + 266 = 0
(A – 19)(A – 14) = 0
Therefore , A = 19 or A = 14
Hence, bigger number = A = 19
ANSWER: D
Answer: (D) 19
Question: If a trader sold two cars each at Rs. 325475 and gains 14% on the first and loses 14% on the second, then his profit or loss percent on the whole is?
(A) 1.44%
(B) 1.74%
(C) 1.84%
(D) 1.96%
(E) 1.24%
Rationale: SP of each car is Rs. 325475, he gains 14% on first car and losses 14% on second car.
In this case, there will be loss and percentage of loss is given by
= [(profit%)(loss%)]/100
= (14)(14)/100 % = 1.96%
Answer:D
Answer: (D) 1.96%
Question: The average (arithmetic mean) weight of 5 friends is 150 pounds. If Joanne leaves the group and the new average of the group is 162 pounds. What is Joanne's weight?
(A) 60
(B) 78
(C) 102
(D) 105
(E) 110
Rationale: The average (arithmetic mean) weight of 5 friends is 150 pounds.
Let W be sum of weights of the four others. and J be Joanne's weight
(W + J)/5 = 150
When Joanne exits, the average is 162
W/4 = 162
W = 648
W + J = 750
J = 102
ANSWER:C
Answer: (C) 102
Question: Rob also compared the Empire State Building and the Petronas
Towers. What is the height difference between the two if the Empire
State Building is 443m tall and the Petronas Towers is 452m tall?
(A) 9
(B) 17
(C) 23
(D) 45
(E) 12
Rationale: 452-443=9.Answer is A.
Answer: (A) 9
Question: If $10 is invested at x percent simple annual interest for n years, which of the following represents the total amount of interest, in dollars, that will be earned by this investment in the n years?
(A) 10n(x/100)
(B) 10,000(x/100)^n
(C) 10,000n(x/100)
(D) 10,000(1+x/100)^n
(E) 10,000n(1+x/100)
Rationale: I think answer D is for compound interest (which i assumed)
and answer A is for simple interest because it is not compounding.
Answer: (A) 10n(x/100)
Question: 40 people are sitting around a circular table. Starting from 1 every 2nd person is killed. This process continues till only one person remains. How is the survivor? (Josephus riddle)
(A) 22
(B) 17
(C) 88
(D) 237
(E) 121
Rationale: Write the given number in Binary form: 101000
Shift the left digit to the right hand side: 010001
Convert this number into decimal and this could be the number of the survivor: 17
Sol 2: Write the given number as a sum of 2K + M where K should be maximum power less than the given number.
Survivor number is given by 2M + 1
this this case 40 < 25 + 8
So M =8, Survivor number = 2(8) + 1 = 17
Answer:B
Answer: (B) 17
Question: There are X girls in a class, and their average height is E inches. In the same class, there are Y boys with an average height of F inches. What is the average height of all the students in the class?
(A) (X + Y)/(M + N)
(B) (M + N)/(X + Y)
(C) (XM + YN)/(M + N)
(D) (XE + YF)/(X + Y)
(E) MN/XY
Rationale: Total height = XE+YF
No of students = X+Y
Avg = (XE+YF)/(X+Y)
Answer is D
Answer: (D) (XE + YF)/(X + Y)
Question: Find the sum of first 35 odd numbers
(A) 4900
(B) 3000
(C) 1250
(D) 1200
(E) 1225
Rationale: Explanation:
n2 = 352 =1225
Answer: Option E
Answer: (E) 1225
Question: In a class of 25 students, 2 students did not borrow any books from the library, 10 students each borrowed 1 book, 8 students each borrowed 2 books, and the rest of the students each borrowed at least 3 books. If the average (arithmetic mean) number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed ?
(A) 8
(B) 10
(C) 12
(D) 13
(E) 15
Rationale: The average number of books per student was 2means that total of 2*25=50 books were borrowed;
2+10+8=20 students borrowed total of 2*0+10*1+8*2=26 books;
So 50-26=24 books are left to distribute among 25-20=5 students, these 5 arethe rest who borrowed at least 3 books;
Tomaximizethe number of books one student from above 5 could have borrowed we shouldminimizethe number of books other 4 students from 5 could have borrowed. Minimum these 4 students could have borrowed is 4 books per student, so total number of books they could have borrowed is 4*3=12 books. So the 5th student could have borrowed is 24-12=12 books.
Answer: C.
Answer: (C) 12
Question: AB + CD = JJJ, where AB and CD are two-digit numbers and JJJ is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?
(A)  1
(B)  3
(C)  7
(D)  9
(E)  Cannot be determined
Rationale: AB and CD are two digit integers, their sum can give us only one three digit integer of a kind of JJJ it's 111.
So, A=1. 1B+CD=111
Now, C can not be less than 9, because no to digit integer with first digit 1 (mean that it's<20) can be added to two digit integer less than 90 to have the sum 111 (if CD<90 meaning C<9 CD+1B<111) --> C=9
Answer: D.
Answer: (D)  9
Question: A single discount equivalent to the discount series of 20%, 15% and 5% is?
(A) 31.6
(B) 35.4
(C) 31.6
(D) 31.1
(E) 31.5
Rationale: 100*(80/100)*(85/100)*(95/100)
= 68.4
100 - 64.6 = 35.4
Answer: B
Answer: (B) 35.4
Question: 325.124 x 12.98 Ã· 3.001 + 21.21 = ?
(A) 1430.08
(B) 1420.06
(C) 781.189
(D) 656.112
(E) 456.512
Rationale: Explanation:
? = 325.124 x 12.98 Ã· 3.001 + 21.21 = ?
â‰ˆ (325.124 x 13/3) + 21.21 â‰ˆ 1408.87 + 21.21 â‰ˆ 1430.080
Answer: Option A
Answer: (A) 1430.08
Question: There is food for 760 men for 22 days. How many more men should join after two days so that the same food may last for 19 days more?
(A) 16
(B) 40
(C) 18
(D) 19
(E) 11
Rationale: 760 ---- 22
760 ---- 20
x ----- 19
x*19 = 760*20
x = 800
760
-------
40
Answer:B
Answer: (B) 40
Question: Suresh and Ramesh started a business investing Rs. 24,000 and Rs. 40,000 respectively. Out of total profit of Rs. 19,000, what is Suresh's share?
(A) s. 7125
(B) s. 8125
(C) s. 9125
(D) s. 9372
(E) s. 9875
Rationale: Explanation:
Ratio of Suresh and Ramesh's share = 24,000 : 40,000 = 3 : 5
Suresh's share = Rs. (19000 x 3 / 8) = Rs. 7125
ANSWER: A
Answer: (A) s. 7125
Question: The difference between simple interest and C.I. at the same rate for Rs.5000 for 2 years in Rs.72. The rate of interest is?
(A) 11
(B) 12
(C) 77
(D) 85
(E) 65
Rationale: 5000 = 72(100/R)2
5 R2 = 720 => R = 12.Answer:B
Answer: (B) 12
Question: 10 x 0.5 = ?
(A) 0.0001
(B) 0.001
(C) 0.01
(D) 0.1
(E) 5
Rationale: Explanation:
10 x 5 = 50.
Sum of decimal places = 1
10 x 0.5 = 5
Answer- E
Answer: (E) 5
Question: If (t-8) is a factor of t^2-kt-43, then k=
(A) 16
(B) 12
(C) 2
(D) 3
(E) 14
Rationale: t^2-kt-48 = (t-8)(t+m) where m is any positive integer.
if 48/8 = 6, then we know as a matter of fact that:
m = +6
and thus k = 8-6 = 3
t^2-kt-m = (t-a)(t+m) where a>m
t^2+kt-m = (t-a)(t+m) where a<m
t^2-kt+m = (t-a)(t-m)
t^2+kt+m = (t+a)(t+m)
D
Answer: (D) 3
Question: If there are 6 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?
(A) 32
(B) 24
(C) 56
(D) 144
(E) 160
Rationale: Firstly, there are 6 ways to choose 3 groups from the 4 groups. Hence 6C3.
Next, with the 3 groups chosen, there are 2^3 choices of choosing either one of the siblings.
Hence, the total number of ways are 6C3 * (2^3) = 160
E
Answer: (E) 160
Question: Find the compound interest on Rs. 10000 at 12% rate of interest for 1 year, compounded half-yearly
(A) Rs. 1036
(B) Rs. 1236
(C) Rs. 1186
(D) Rs. 1206
(E) Rs. 1226
Rationale: Amount with CI = 10000 [1+ (12/2 * 100)]2 = Rs. 11236
Therefore, CI = 11236 – 10000 = Rs. 1236 ANSWER :B
Answer: (B) Rs. 1236
Question: In how many no. between 100 and 1000 exactly one of the digits is 3?
(A)  648
(B)  512
(C)  252
(D)  225
(E)  26
Rationale: Another approach can be
total nos. between 100 and 1000 are 899
nos. in which all digits are 3 = 1
nos. in which digit 3 is in two places = 8 + 9 + 9 = 26 (for thousand , tens and once as different digit)
nos. in which no digit is 3 = 8*9*9 - 1 = 648 - 1 (for 100 not to be included)
required nos = 899 - 1 - 26 - 647 = 225
answer is D
Answer: (D)  225
Question: a manufacturing company has 15% cobalt ,25% led and 60% of copper. if 5kg of led is used in a mixture how much copper we need to use
(A) 12 kg
(B) 13 kg
(C) 22 kg
(D) 14 kg
(E) 15 kg
Rationale: 25%=5kg
100%=20kg
60% of 20 kg= 12kg
12 kg
ANSWER:A
Answer: (A) 12 kg
Question: A boy has nine trousers and 15 shirts. In how many different ways can he select a trouser and a shirt?
(A) 88
(B) 98
(C) 106
(D) 135
(E) 111
Rationale: Explanation:
The boy can select one trouser in nine ways.
The boy can select one shirt in 15 ways.
The number of ways in which he can select one trouser and one shirt is 9 * 15 = 135 ways
D)
Answer: (D) 135
Question: A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
(A) 5.2 km/hr.
(B) 8.2 km/hr.
(C) 7.2 km/hr.
(D) 3.2 km/hr.
(E) 4.2 km/hr.
Rationale: C
7.2 km/hr.
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: (C) 7.2 km/hr.
Question: The expression (3x-2)/(x+3) is equivalent to which of the following?
(A) (5-2)/3
(B) 5 – (2/3)
(C) 3 – (11)/(x+3)
(D) 5 – (17)/(x+3)
(E) 5 + (17)/(x+3)
Rationale: Easiest way is to pick any number and put in (3x-2)/(x+3). Then check the options with the same number.
I took x=1 and got C as the answer.
Answer: (C) 3 – (11)/(x+3)
Question: A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
(A) 35 1/2
(B) 36 1/2
(C) 37 1/2
(D) 38 1/2
(E) 39 1/2
Rationale: Explanation:
Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1)
Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 ---(2)
Work done by B in 1 day = 1/15 â€“ 1/25 = 2/75
=> B can complete the work in 75/2 days = 37 (1/2) days
ANSWER IS C
Answer: (C) 37 1/2