mathewhe/aqua_rat huggingface dataset:
Question: Five drainage pipes, each draining water from a pool at the same constant rate, together can drain a certain pool in 12 days. How many additional pipes, each draining water at the same constant rate, will be needed to drain the pool in 6 days?
(A)  5
(B)  9
(C)  10
(D)  12
(E)  15
Rationale: This is an inverse proportional problem......
5 pipes in 12 days; So for 6 days, it will be = 12 x 5 / 6 = 10
So, 10-5 = 5
Answer A
Answer: (A)  5
Question: The salaries of A and B together amount to $2000. A spends 95% of his salary and B, 85% of his. If now, their savings are the same, what is A's salary?
(A) $1000
(B) $1250
(C) $2500
(D) $1500
(E) $1200
Rationale: Let A's salary is x
B's salary = 2000-x
(100-95)% of x = (100-85)% of (2000-x)
x = $1500
Answer is D
Answer: (D) $1500
Question: If 213 × 16 = 3408, then 0.016 × 2.13 is equal to :
(A) 0.3408
(B) 3.408
(C) 34.08
(D) 340.8
(E) 0.03408
Rationale: Solution
0.016 × 2.13 = (16/1000x 213/100)= (16/1000x 213/100) = 3408/100000 = 0.03408. Answer E
Answer: (E) 0.03408
Question: A, B and C work on a task. To complete the task alone, B takes twice the time that A would take to complete the task alone and 1/5rd the time that C would take to complete the task alone. If B actually worked for half the number of days that A worked and 3/2 times the number of days that C worked, what proportion of the total work was completed by B?
(A) 1/3
(B) 2/9
(C) 9/49
(D) 7/19
(E) 25/126
Rationale: The first thing to notice is that A is faster than B and B is faster than C.
Since work is proportional to time, in 1 day lets say if A does 2 works, B does 1 work and C does 2/3rd of a work.
If A works for 2 days, B works for 1 day and C works for only 2/3 of the day.
Therefore total work done = (2*2) + (1*1) + (1/5*1/5) = 126/25
Proportion of work done by B = (1*1) / (126/25) = 25/126
Hence Answer E.
Answer: (E) 25/126
Question: A part-time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee's total income did not change. By what percent T should the number of hours worked be increased?
(A) 12.5%
(B) 20%
(C) 25%
(D) 50%
(E) 100%
Rationale: Correct Answer: C
Solution: C. We can set up equations for income before and after the wage reduction. Initially, the employee earns W wage and works H hours per week. After the reduction, the employee earns .8W wage and works X hours. By setting these equations equal to each other, we can determine the increase in hours worked: WH = .8WX (divide both sides by .8W) 1.25H = X We know that the new number of hours worked T will be 25% greater than the original number. The answer is C.
Answer: (C) 25%
Question: A survey was conducted to find out how many people in a housing colony of 140 residents could swim, dance and drive a car. It was found that the number of people who could not swim was 89, the number of people who could not dance was 100 and that the number of people who could not drive a car was 91. If the number of people who could do at least two of these things, was found to be 37 and the number of people who could do all these things was found to be 6, how many people could not do any of these things?
(A) 43
(B) 44
(C) 45
(D) 46
(E) 48
Rationale: T=n(A)+n(B)+n(C)-n(Exactly two of the events)-2*n(All 3 Events)+n(None of the events)
T=140
n(A)=T-n(A')=140-89=51
n(B)=T-n(B')=140-100=40
n(C)=T-n(C')=140-91=49
n(Exactly two of the events)=n(At least 2 Events)-n(All 3 Events)=37-6=31
n(All 3 Events)=6
140=51+40+49 -31-2*6+n(None of the events)
n(None of the events)=140-51-40-49+31+12=43
Ans:A
Answer: (A) 43
Question: If m > 0, which of the following could be true?
I. m^3 > m^2
II. m^2 = m
III. m^2 > m^3
(A) I only
(B) III
(C) IIIII
(D) All of the above
(E) None of the above
Rationale: 1. m^3 > m^2 Well this is true for all m > +1 (but not for values 0 < m < 1)
2. m^2 = m This is true for only one positive number, 1
3. m^2 > m^3 Again this is true only for values of m such that 0 < m < 1
So the answer is D, since there exists at least one value of m which satisfies the equation/ in equation(s)
Answer: (D) All of the above
Question: Find the odd man out. 187, 264, 336, 473, 682, 781
(A) 187
(B) 336
(C) 781
(D) 682
(E) 473
Rationale: In all numbers except 336, the middle digit is the sum of other two digits.Answer is B
Answer: (B) 336
Question: Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. What is the probability that both of them get selected?
(A) 8/35
(B) 34/35
(C) 27/35
(D) 30/35
(E) None of these
Rationale: EXPLANATION:
P(A) = 2/5
P(B) = 4/7
E = {A and B both get selected}
P(E) = P(A)*P(B)
= 2/5 * 4/7
= 8/35
ANSWER IS A
Answer: (A) 8/35
Question: 4 boys and 2 girls are to be seated in a row in such a way that the two girls are always together. In how many different ways can they be seated?
(A) 120
(B) 720
(C) 148
(D) 240
(E) None of these
Rationale: Assume the 2 given students to be together (i.e one).
Now there are five students.
Possible ways of arranging them are = 5! = 120
Now, they (two girls) can arrange themselves in 2! ways.
Hence total ways = 120 × 2 = 240
Answer D
Answer: (D) 240
Question: If each permutation of the digits 1,2,3,4,5,6 are listed in the increasing order of magnitude, then 289thterm will be
(A) 361452
(B) 321546
(C) 321456
(D) 341256
(E) None of these
Rationale: Explanation :
Let's see how many numbers can be formed with the left most digit as 1
The digit '1' is placed at the 1st position (only 1 way of doing this)
Since one digit is placed at the 1st position,
any of the remaining 5 digits can be placed at 2nd position.
Since one digit is placed at the 1st position and another digit is placed
at the 2nd position, any of the remaining 4 digits can be placed at the
3rd position.
So on ...
1	5	4	3	2	1
i.e., total number of ways = (1)( 5)( 4)( 3)( 2)( 1) = 120
i.e., total count of numbers which can be formed
with the left most digit as 1 = 120
Similarly, total count of numbers which can be formed
with the left most digit as 2 = 120
Similarly, total count of numbers which can be formed
with the left most digit as 3 = 120
i.e., 240 numbers (=120 + 120) can be formed
(with left most digit as 1) or (with left most digit as 2)
Similarly, 360 numbers (=120 + 120 + 120) can be formed
(with left most digit as 1) or (with left most digit as 2)
or (with left most digit as 3)
Hence, the left most digit of the 249th number = 3
Now, let's find out how many numbers can be formed
with the left most digit as 3 and next digit as 1
The digit '3' is placed at the 1st position (only 1 way of doing this)
The digit '1' is placed at the 2nd position (only 1 way of doing this)
Any of the remaining 4 digits can be placed at 3rd position.
Since 3 digits are placed in the first three positions, any of the remaining 3 digits
can be placed at the 4th position.
Since 4 digits are placed in the first four positions, any of the remaining 2 digits
can be placed at the 5th position.
Since 5 digits are placed in the first five positions, the remaining 1 digit
can be placed at the 6th position.
1	1	4	3	2	1
i.e., total number of ways = (1)(1)(4)(3)(2)(1) = 24
i.e., Total count of numbers which can be formed
(with the left most digit as 3) and (next digit as 1) = 24
Similarly, total count of numbers which can be formed
(with the left most digit as 3) and (next digit as 2) = 24
Hence, 120 + 120 + 24 + 24 = 288 numbers can be formed
(with left most digit as 1) or (with left most digit as 2)
or (with left most digit as 3 and next digit as 1)
or (with left most digit as 3 and next digit as 2)
Hence, the 289th number is the minimum value number which is formed
with the left most digit as 3 and next digit as 4.
i.e., the number is 341256
Answer : D
Answer: (D) 341256
Question: Siddharth wants to borrow Rs.5000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction?
(A) s.21.60
(B) s.21.29
(C) s.22.29
(D) s.18.00
(E) s.21.12
Rationale: Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.5,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth's income = C.I - S.I
= p[1 + r/ 100]n - p - pnr/100
= p{ [1 + r/ 100]2 - 1 - nr/100
= 5,000{ [1 + 6/100]2 - 1 - 12/100}
= 5,000 {(1.06)2- 1 - 0.12} = 5,000(1.1236 - 1 - 0.12)
= 5,000 (0.0036) = 5 * 3.6 = Rs.18.00
Answer: D
Answer: (D) s.18.00
Question: If the price of a certain computer increased 30 percent from a dollars to 351 dollars, then 2a =
(A) 540
(B) 570
(C) 619
(D) 649
(E) 700
Rationale: Before Price increase Price = a
After 30% Price increase Price = a+(30/100)*a= 1.3a = 351 (Given)
i.e. a= 351/1.3 = $270
i.e. 2a = 2*270 = 540
Answer: option A
Answer: (A) 540
Question: A man has a certain number of small boxes to pack into parcles. If he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcle, none is left over. What is the number of boxes, he may have to pack?
(A) 106
(B) 301
(C) 309
(D) 400
(E) 450
Rationale: Explanation:
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5, or 6 and no remainder when divided by 7. Thus, the number must be of the form (L.C.M of 3, 4, 5, 6) x + 1 i.e., (60x + 1 ) and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So the number is 301.
Answer: B) 301
Answer: (B) 301
Question: A completes a work in 12 days and B complete the same work in 36 days. If both of them work together, then the number of days required to complete the work will be?
(A) 9 days
(B) 11 days
(C) 21 days
(D) 22 days
(E) 24 days
Rationale: If A can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days
Therefore, here, the required number of days = 12 Ã— 36/ 48 = 9 days.
Option 'A'
Answer: (A) 9 days
Question: Average of five numbers is 26. If one number is excluded the average becomes 28. The excluded number is
(A) 10
(B) 12
(C) 16
(D) 20
(E) 18
Rationale: Explanation:
Number is (5*26) - (4*28) = 130-112 = 18
Answer: Option E
Answer: (E) 18
Question: All the faces of cubes are painted with red colour. It is then cut into 64 equal small cubes.Find How many small cubes are there whose two adjacent faces are coloured red ?
(A) 0
(B) 8
(C) 16
(D) 24
(E) 4
Rationale: There are 64 small cubes,Hence one side side of the big cube is 3√64=4 cm
Number of small cubes having two adjacent faces coloured red = (x - 2) x No. of edges
= (4 - 2) x 12
= 24
Answer :D
Answer: (D) 24
Question: Hitesh is 40 years old and Ronnie is- 60 years old. How many years ago was the ratio of their ages 3:5 ?
(A) 5 years
(B) 10 years
(C) 20 years
(D) 15 years
(E) 25 years
Rationale: Suppose, the ratio was 3 : 5, x years ago.
Then,
(40-x)/(60-x)=3/5
=5 (40 - x) = 3 (60 - x)
=200-5x =180-3x
=2x = 20
= x = 10.
Answer: B
Answer: (B) 10 years
Question: Evaluate the expression
m - (m - n) : (-2) ⋅ (-5) when m = -4, n = -6
(A) 8
(B) 9
(C) 5
(D) 7
(E) 6
Rationale: Solution:
m - (m - n) : (-2) ⋅ (-5) = (-4) - [-4 - (-6)] : (-2) ⋅ (-5) = -4 - 2 : (-2) ⋅ (-5) = -4 - (-1) ⋅ (-5) = -4 - (+5) = -9
Answer B
Answer: (B) 9
Question: The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy?
(A) 8
(B) 9
(C) 7
(D) 5
(E) 4
Rationale: 3x + 5x + 7x = 45
x =3
3x = 9
Answer:B
Answer: (B) 9
Question: A certain board game has a row of squares numbered 1 to 100. If a game piece is placed on a random square and then moved 13 consecutive spaces in a random direction, what is the probability the piece ends no more than 13 spaces from the square numbered 49?
(A) 28%
(B) 27%
(C) 30%
(D) 37%
(E) 17%
Rationale: No more than 13 spaces from 49 means in the rangefrom 49-13=36 to 49+13=62, inclusive. Total numbers in this range 62-36+1=27, the probability favorable/total=27/100
Answer: B
Answer: (B) 27%
Question: Walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.
(A) 12 min
(B) 40 min
(C) 60 min
(D) 30 min
(E) 45 min
Rationale: dist = 4*22/4 = 22km
new speed = 16.5 kmph
=>D/S = 11/16.5 = 40 min
ANSWER B
Answer: (B) 40 min
Question: A garrison of 3300 men had provisions for 32 days, when given at the rate of 860 gms per head. At the end of 7 days, a reinforcement arrives and it was for that the provisions wi1l last 17 days more, when given at the rate of 826 gms per head, What is the strength of the reinforcement?
(A) 1900
(B) 1600
(C) 1300
(D) 1500
(E) 1700
Rationale: The problem becomes:
3300 men taking 850 gms per head have provisions for (32 - 7) or 25 days,
How many men taking 825 gms each have provisions for 17 days?
Less ration per head, more men	(Indirect Proportion)
Less days, More men	(Indirect Proportion)
Ration 825 : 850
Days 17: 25 } : : 3300 : x
(825 x 17 x x) = 850 x 25 x 3300 or x = (850 x 25 x 3300)/(825 x 17)=5000
Strength of reinforcement = (5500 - 3300) = 1700.
Answer is E.
Answer: (E) 1700
Question: In a friendship gang Aravind has 2 gang, in how many ways can he invite one or more of the gang to his house?
(A) 25
(B) 49
(C) 3
(D) 87
(E) 36
Rationale: Aravind can select one or more than one of his 8 gang.
=> Required number of ways = 2^2 – 1= 3.
C
Answer: (C) 3
Question: Shyam visited Ram during his brief vacation. In the mornings they both would go for yoga. In the
evenings they would play tennis. To have more fun, they indulge only in one activity per day, i.e.
either they went for yoga or played tennis each day. There were days when they were lazy and
stayed home all day long. There were 24 mornings when they did nothing, 14 evenings when they stayed at home, and a total of 22 days when they did yoga or played tennis. For how many days Shyam stayed with Ram?
(A) 32
(B) 24
(C) 30
(D) None of these
(E) Cannot be determined
Rationale: Explanation :
Let ,
P days : they play tennis.
Y days : they went to yoga.
T days : total duration for which Ram and Shyam were together.
=> P + Y = 22.
Also, (T - Y ) = 24 & (T - P) = 14.
Adding all of them,
2T = 22 + 24 + 14
=> T = 30 days.
Answer : C
Answer: (C) 30
Question: A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of A?
(A) 240
(B) 882
(C) 727
(D) 199
(E) 212
Rationale: (3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 630 = 240
Answer: A
Answer: (A) 240
Question: If 36 men do a work in 70 days, in how many days will 40 men do it?
(A) 64
(B) 63
(C) 65
(D) 66
(E) 67
Rationale: 36 * 70 = 40 * x
x = 63 days
Answer: B
Answer: (B) 63
Question: During a certain season, a team won 85 percent of its first 100 games and 50 percent of its remaining games. If the team won 70 percent of its games for the entire season, what was the total number of games that the team played?
(A) 180
(B) 175
(C) 156
(D) 150
(E) 105
Rationale: We are first given that a team won 85 percent of its first 100 games. This means the team won 0.85 x 100 = 85 games out of its first 100 games.
We are next given that the team won 50 percent of its remaining games. If we use variable T to represent the total number of games in the season, then we can say T – 100 equals the number of remaining games in the season. Thus we can say:
0.5(T – 100) = number of wins for remaining games
0.5T – 50 = number of wins for remaining games
Lastly, we are given that team won 70 percent of all games played in the season. That is, they won 0.7T games in the entire season. With this we can set up the equation:
Number of first 100 games won + Number of games won for remaining games = Total Number of games won in the entire season
85 + 0.5T – 50 = 0.7T
35 = 0.2T
350 = 2T
175 = T
Answer is B.
Answer: (B) 175
Question: The sum of all two digit numbers divisible by 5 is
(A) 945
(B) 545
(C) 745
(D) 645
(E) 845
Rationale: Required numbers are 10,15,20 . . . . . 95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,
10+5n-5=95
5+5n=95
5n=95-5
n = 90/5
then n=18.
Required sum
=n/2(a+l)
=18/2(10+95)
=945
Answer is A.
Answer: (A) 945
Question: Find the principle on a certain sum of money at 5% per annum for 3 1/5 years if the amount being Rs.1160?
(A) Rs.1000
(B) Rs.1100
(C) Rs.1010
(D) Rs.10000
(E) None of these
Rationale: Explanation:
1160 = P [1 + (5*16/5)/100]
P= 1000
Answer: Option A
Answer: (A) Rs.1000
Question: A pump will fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ?
(A) 20 hrs
(B) 21 hrs
(C) 22 hrs
(D) 23 hrs
(E) 24 hrs
Rationale: work done for 1 hr without leak = 1/3
work done with leak = 31/2 ==>7/2
work and leak in 1 hr = 2/7
work leak in 1 hr = 1/3 - 2/7 = 1/21
so 21 hrs
ANSWER B
Answer: (B) 21 hrs
Question: At present the ratio between the ages of Arun and Deepak is 4:3. After 6years, Arun's age will be 26 years. What is the age of Deepak at present?
(A) 15
(B) 21
(C) 23
(D) 12
(E) 25
Rationale: Let the present ages of Arun and Deepak be 4x years and 3x years respectively
4x+6 = 26
4x = 20
x = 5
Deepak's age = 3x = 15years
Answer is A
Answer: (A) 15
Question: A class consists of 100 students, 20 of them are girls and 80 boys; 20 of them are rich and remaining poor; 40 of them are fair complexioned. The probability of selecting a fair complexioned rich girl is
(A) 2/125
(B) 1/59
(C) 1/56
(D) 1/54
(E) 1/53
Rationale: The probability of selecting girl is: 20/100 = 1/5
The probability of selecting rich is: 20/100 = 1/5
The probability of selecting fair complexioned is: 40/100 = 2/5
Three are independent;probability of rich and fair complexioned girl is:
(1/5) Ã—(1/5) Ã— (2/5) = 2/125
Answer: A
Answer: (A) 2/125
Question: Express 25 mps in kmph?
(A) 22
(B) 88
(C) 90
(D) 21
(E) 24
Rationale: 25 * 18/5 = 90 kmph
Answer: C
Answer: (C) 90
Question: At Company K, 15 percent of the employees are secretaries and 60 percent are salespeople. If there are 25 other employees of Company K, how many employees does Company K have?
(A)  160
(B)  180
(C)  190
(D)  200
(E)  100
Rationale: Let the total number of employees in the company be x
% of secretaries = 15%
% of salespeople = 60%
% of of employees other than secretaries and salespeople = 100 - 75 = 25%
But this number is given as 25
so 25% of x = 25
x = 100
Therefore there a total of 100 employees in the company K
Correct answer - E
Answer: (E)  100
Question: A committee of 6 is chosen from 5 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed?
(A) 150
(B) 700
(C) 1404
(D) 2620
(E) 3510
Rationale: The only cases possible are :
1. 2 men and 4 women : 5C2 * 5C4 = 50
2. 3 men and 3 women: 5C3*5C3 = 100
Rest of the cases will either have 1 or 0 men (not allowed) or will have 1 or 2 or 0 women (not allowed)
Total possible combinations = 50+100 = 150. Thus A is the correct answer.
Answer: (A) 150
Question: How many even multiples of 45 are there between 449 and 901?
(A) 5
(B) 6
(C) 9
(D) 10
(E) 11
Rationale: 450 = 10*45
900 = 20*45
The even multiples are 45 multiplied by 10, 12, 14, 16, 18, and 20 for a total of 6.
The answer is B.
Answer: (B) 6
Question: If a number T between 0 and 1 is selected at random, which of the following will the number most likely be between?
(A) 0 and 3/20
(B) 3/20 and 1/5
(C) 1/5 and 1/4
(D) 1/4 and 3/10
(E) 3/10 and 1/2
Rationale: The number T will most likely be between the largest range.
A. 0 and 3/20 --> range 3/20;
B. 3/20 and 4/20 --> range 1/20;
C. 4/20 and 5/20 --> range 1/20;
D. 5/20 and 6/20 --> range 1/20;
E. 6/20 and 10/20 --> range 4/20.
Answer: E.
Answer: (E) 3/10 and 1/2
Question: How many ways are there to split a group of 8 girls into two groups of 4 girls each? (The order of the groups does not matter)
(A) 30
(B) 40
(C) 35
(D) 45
(E) 50
Rationale: the combination is 8C4 /2
= 8!/4!*4! *2 = 35
C
Answer: (C) 35
Question: 49 members attended the party. In that 22 are males, 27 are females. The shake hands are done between males, females, male and female. Total 12 people given shake hands. How many such kinds of such shake hands are possible?
(A) 66
(B) 291
(C) 26
(D) 29
(E) 11
Rationale: If only 12 people shaked their hands, then total hand shakes are 12C212C2 = 66
Answer:A
Answer: (A) 66
Question: A Laptop was sold at a loss of 10%. If it was sold for Rs.5000 more, there would have been a gain of 10%. What is the cost price?
(A) Rs.18000
(B) Rs.21000
(C) Rs.22000
(D) Rs.20000
(E) Rs.25000
Rationale: 90%
110%
--------
20% ---- 5000
100% ---- ? => Rs.25000
Answer: E
Answer: (E) Rs.25000
Question: What is the remainder when 1271 * 1276 * 1277 *1285 is divided by 12?
(A)  0
(B)  1
(C)  11
(D)  9
(E)  4
Rationale: 1271*1275*1277*1285 = (1272 - 1)(1272 + 4)(1284 - 7)(1284 + 1) = (multiple of 12 - 1)(multiple of 12 + 3)(multiple of 12 - 7)(multiple of 12 + 1)
Now, if we expand this expression all terms but the last one will have a multiple of 12 in them, thus will leave no remainder upon division by 12, the last term will be (-1)(4)(-7)(1) = 28. Thus the question becomes: what is the remainder upon division of 28 by 12? The answer to this question is 4.
Answer: E
Answer: (E)  4
Question: A reading list for a certain course consists of 10 books, of which 4 are poems and the rest are novels. Each student is required to read a selection of 4 books from the list, including at most 1 novels. How many selections of 4 books satisfy the requirements?
(A) 90
(B) 115
(C) 130
(D) 144
(E) 195
Rationale: Ans: B
Solution: Selection requires at most 2 novels, means=
0N4P + 1N3P + 2N2P
6c0*4c4+ 6c1*4c3 + 6c2*4c2 +6c3*4c1
= 1+ 24 + 90+80
= 195
Ans: E
Answer: (E) 195
Question: Martin has to secure 80% marks to clear his exam of class 9th.
He got 200 marks and failed by 200 marks.
What is the maximum marks ?
(A) 500
(B) 505
(C) 510
(D) 515
(E) 525
Rationale: A
500
To pass the exam ravish needs 200 + 200 = 400 marks.
=> (400/80) *100 = 500
Answer: (A) 500
Question: 96 can be expressed as a product of primes as
(A) 2 × 2 × 3 × 3 × 3 × 7
(B) 2 × 3 × 3 × 3 × 7 × 7
(C) 2 × 2 × 2 × 2 × 2 × 3
(D) 2 × 3 × 3 × 3 × 3 × 7
(E) None of these
Rationale: Explanation:
It is clear that 96 = 2 × 2 × 2 × 2 × 2 × 3
Answer: Option C
Answer: (C) 2 × 2 × 2 × 2 × 2 × 3
Question: S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk–1 + 2(10k–1). If p is the sum of the first 30 terms of S, what is the tenth digit of p, counting right to left from the units digit?
(A) 1
(B) 2
(C) 4
(D) 6
(E) 9
Rationale: C
Sum of unit digits of first 30 terms = 60
Sum of tens digits of first 30 terms = 58
Sum of thousands digits of first 30 terms = 56
and so on..
p1 = 0
p2 = (6+58) = 4
p3 = (6+56) = 2
p4 = (6+54) = 0
p5 = (6+52) = 8
p6 = (5+50) = 5
p7 = (5+48) = 3
p8 = (5+46) = 1
p9 = (5+44) = 9
p10= (4+42) = 6
ANSWER : D
Answer: (D) 6
Question: Vijay bought 160 shirts at the rate of Rs. 225 per shirt. The transport expenditure was Rs. 1400. He paid an octroi at the rate of Rs. 1.75 per shirt and labour charges were Rs. 320. What should be the selling price of one shirt, if he wants a profit of 20%?
(A) Rs.282
(B) Rs.229
(C) Rs.208
(D) Rs.285
(E) Rs.229
Rationale: Total CP per shirt = 225 + 1400/160 + 1.75 + 320/160 = Rs. 237.5
SP = CP[(100 + profit%)/100]
= 237.5 * [(100 + 20)/100]
= Rs.285.
Answer:D
Answer: (D) Rs.285
Question: An accurate clock shows 8 o'clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 1 o'clock in the afternoon?
(A) 30
(B) 60
(C) 90
(D) 120
(E) 150
Rationale: The hour hand will move 5/12 of a circle.
(5/12)*360 = 150 degrees.
The answer is E.
Answer: (E) 150
Question: In a throw of dice what is the probability of geÆ«ng number greater than 4
(A) 1/2
(B) 1/3
(C) 1/5
(D) 1/6
(E) None of these
Rationale: Explanation:
Number greater than 4 is 5 & 6, so only 2 number
Total cases of dice = [1,2,3,4,5,6]
So probability = 2/6 = 1/3
Answer: B
Answer: (B) 1/3
Question: A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum?
(A) 338
(B) 277
(C) 229
(D) 646
(E) 261
Rationale: 840 = P [1 + (10*3)/100]
P = 646
Answer:D
Answer: (D) 646