mathewhe/aqua_rat huggingface dataset:
Question: From January 1, 1991, to January 1, 1993, the number of people enrolled in health maintenance organizations increased by 25 percent. The enrollment on January 1, 1993, was 45 million. How many million people, to the nearest million, were enrolled in health maintenance organizations on January 1, 1991 ?
(A)  36
(B)  39
(C)  40
(D)  41
(E)  42
Rationale: 1.25x=45 --> 5/4*x=45 --> x=45*4/5=180/5=36.
Answer: A.
Answer: (A)  36
Question: James can finish a job in 20 dasy he worked for 10 days alone and completed the remaining part working with David 2 days how many days would both David and James together take to complete the entire job?
(A) 4
(B) 5
(C) 10
(D) 12
(E) 14
Rationale: Explanation:
James and David complete half work in 2 days
=> they can complete whole work in 4 days
Answer: Option A
Answer: (A) 4
Question: The difference between the local value and face value of 7 in the numeral 657903 is:
(A) 0
(B) 7896
(C) 6993
(D) 903
(E) 897
Rationale: (Local value)-(Face value)= (7000-7)=6993.
Answer: C
Answer: (C) 6993
Question: A shopkeeper sold 50articles at the cost price of 60 articles. Then find the profit% or lost%
(A) 20%
(B) 22%
(C) 25%
(D) 30%
(E) 35%
Rationale: here 50 articles selling price = 60 articles cost price
so the difference = 60-50 = 10
% of profit = 10*100/50 = 20%
correct option is A
Answer: (A) 20%
Question: x is a positive integer of S less than 500. When x is divided by 7, the remainder is 1; when x is divided by 3, the remainder is 2. How many x are there?
(A) 21
(B) 22
(C) 23
(D) 24
(E) 25
Rationale: Took me more than 2 mins to solve.
That's how I did it.
x is between 1-499 included.
Smallest number that can be written as 7n+1 (7*0+1) is 1
Largest number that can be written as 7n+1 is (7*71 +1) is 498
So there are total 72 numbers that can be written as 7n+1
Because x can also be written as 3m+2, we have to see how many numbers that can be written as 7n+1 also can be written as 3m+2
7*0+1 can not be written in the form 3m+2
7*1+1 can be written in the form 3m+2
7*2+1 can not be written in the form 3m+2
7*3+1 can not be written in the form 3m+2
7*4+1 can be written in the form 3m+2
Here we see a sequence that one out of every 4 numbers can be written as 3m+1
72/4= 24 numbers can be written as 3m+1
D is the answer
Answer: (D) 24
Question: Rs.200 contained in a box consists of one rupee, 50 paisa and 25 paisa coins in the ratio 4:5:6. What is the number of 25 paisa coins?
(A) 388
(B) 266
(C) 120
(D) 150
(E) 267
Rationale: 4x 5x 6x
100 50 25
400x + 250x + 150x = 20000
x = 25
6x = 150
Answer: D
Answer: (D) 150
Question: If m > 0, y > 0, and x is m percent of 2y, then, in terms of y, m is what percent of x?
(A) y/200
(B) 2y
(C) 50y
(D) 50/y
(E) 5000/y
Rationale: Answer:
A = Y/200,
Done this by taking numbers:
Let 10 (x) be 5% (m) of 2 * 100 (y)
Then m/x*100 = 5/10*100 = 50% or 1/2
50% or 1/2 in terms of Y = 100/200 = Y/200
ANSWER:A
Answer: (A) y/200
Question: A boat running downstream cover a distance of 16km in 2 hr while covering the same distance upstream it takes 4 hr then the speed of the boat still in the water
(A) 43km/hr
(B) 78km/hr
(C) 9km/hr
(D) 7km/hr
(E) 3km/hr
Rationale: rate down stream=(16/2)= 8 kmph
rate upstream=(16/4)= 4 kmph
speed still water =1/2(8+4)kmph=km/hr
Answer(C)
Answer: (C) 9km/hr
Question: If the two-digit integers G and N are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of G and N?
(A) 181
(B) 165
(C) 121
(D) 99
(E) 44
Rationale: Since the question asks for the answer that CANNOT be the sum of G and N, and the answers are numbers, we can use a combination of TESTing VALUES and TESTing THE ANSWERS to eliminate the possible values and find the answer to the question.
We're told that G and N are two-digit positive integers and have the SAME DIGITS but in REVERSE ORDER. We're asked which of the 5 answers CANNOT be the SUM of M and N.
Let's start with the 'easiest' answer first:
44. Can we get to 44 in the manner described?
Yes, if the numbers are 13 and 31.....13+31 = 44. Eliminate Answer E
Now let's work through the rest of the list....
Can we get to 99 in the manner described?
Yes, there are several ways to do it. For example, if the numbers are 18 and 81.....18+81 = 99. Eliminate Answer D
Can we get to 121 in the manner described?
Yes, there are several ways to do it. For example, if the numbers are 38 and 83.....38+83 = 121. Eliminate Answer C
Can we get to 165 in the manner described?
Yes, there are a couple of ways to do it. For example, if the numbers are 78 and 87.....78+87 = 165. Eliminate Answer B
There's only one answer left....
A
Answer: (A) 181
Question: The ratio of daily wages of two workers is 4 : 3 and one gets daily Rs 9 more than the other, what are their daily wages?
(A) Rs 32 and Rs 24
(B) Rs 60 and Rs 45
(C) Rs 80 and Rs 60
(D) Rs 36 and Rs 27
(E) Rs 36 and Rs 24
Rationale: let us take second worker wages= x
so from the given data
first worker wages= x+9
from the qstn we know that wages of first/ wages of second=4/3
so (x+9)/x = 4/3
==> 3x+27=4x
==> x=27
so x+9= 36
so the answer is 36rs and 27rs
ANSWER:D
Answer: (D) Rs 36 and Rs 27
Question: In an election between two candidates, one got 60% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 2000, the number of valid votes that the other candidate got, was:
(A) 1200
(B) 750
(C) 800
(D) 600
(E) 900
Rationale: Number of valid votes = 80% of 2000 = 1600.
Valid votes polled by other candidate = 40% of 2000
=(40/100)X2000 =800
ANSWER = C
Answer: (C) 800
Question: If x = y^2, and x and y are positive integers, which of the following numbers could be the value of x?
(A) 9,827
(B) 10,816
(C) 11,433
(D) 12,908
(E) 13,482
Rationale: The square of an integer can have a units digit of 0, 1, 4, 5, 6, or 9.
The answer is B.
Answer: (B) 10,816
Question: If the selling price of 7 Candle is same as the cost price of 14 candle. Find the gain or loss percentage?
(A) 85%
(B) 65%
(C) 75%
(D) 35%
(E) 100%
Rationale: Let the C.P of each candle be Re 1. Then,
S.P of 7 candle = C.P of 14 candle= Rs. 14
Now, C.P of 7 candle= Rs.7, S.p of 7 candle= Rs 14
Gain = Rs ( 14 -7 ) = Rs 7.
Gain% = ( 7/7× 100 ) % = 100%
Answer : E
Answer: (E) 100%
Question: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
(A) 207
(B) 209
(C) 300
(D) 305
(E) 307
Rationale: We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number
of ways	= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
= (24 + 90 + 80 + 15)
=209
ANSWER B
Answer: (B) 209
Question: By selling 99 pens, a trader gains the cost of 11 pens. Find his gain percentage?
(A) 11 1/3%
(B) 11 7/3%
(C) 11 1/8%
(D) 11 1/3%
(E) 11 1/9%
Rationale: Let the CP of each pen be Rs. 1.
CP of 99 pens = Rs. 99
Profit = Cost of 11 pens
= Rs. 11
Profit% = 11/99 * 100
= 11 1/9%
Answer: E
Answer: (E) 11 1/9%
Question: How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg?
(A) 36 kg
(B) 42 kg
(C) 54 kg
(D) 63 kg
(E) 60 kg
Rationale: S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.
C.P. of 1 kg of mixture = Rs.((100/110) x 9.24) = Rs. 8.40
C.P. of 1 kg sugar of 1st kind Rs.9
Cost of 1 kg sugar of 2nd kind Rs.7
Mean Price Rs. 8.40
Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.
Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.
Then, 7 : 3 = x : 27
x =	(7 x 27)/3	= 63 kg.
Answer:D
Answer: (D) 63 kg
Question: Find the principle on a certain sum of money at 5% per annum for 2 years if the amount being Rs.1120?
(A) 1018
(B) 1067
(C) 1977
(D) 1056
(E) 1097
Rationale: 1120 = P [1 + (5*2)/100]
P= 1018
Answer: A
Answer: (A) 1018
Question: Two trains one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 64 hours and 121 hours respectively. The ratio of their speeds is
(A) 8 : 7
(B) 11 : 8
(C) 6 : 7
(D) 9 : 16
(E) None
Rationale: Sol.
Let us name the trains as A and B. Then,
(A's speed) : (B's speed) = √b : √a
= √121 : √64
= 11 : 8.
Answer B
Answer: (B) 11 : 8
Question: The average age of a family of 5 was 32 . It fell to 18 when the grandmother died.How old was she ?
(A) 86
(B) 87
(C) 88
(D) 89
(E) 90
Rationale: Avg of 5 = 32
total/5=32
Total age of 5 = 160
After grandma died, total ll be
total/4=18
Total age of 4 = 72
Age of grandma ll be => 160-72 = 88
ANSWER:C
Answer: (C) 88
Question: How many times the keys of a writer have to be pressed in order to write first 250 counting no's?
(A) 632
(B) 642
(C) 652
(D) 662
(E) 672
Rationale: 1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 250 = 151 * 3 = 453
-----------
642
B
Answer: (B) 642
Question: Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is?
(A) 3/8
(B) 3/2
(C) 3/5
(D) 3/6
(E) 3/7
Rationale: Let the speeds of the two trains be x m/sec and y m/sec respectively. Then, length of the first train
= 27 x meters, and length of the second train = 17 y meters. (27 x + 17 y) / (x + y) = 23
==> 27 x + 17 y
= 23 x + 23 y
==> 4 x = 6 y
==> x/y
= 3/2.
Answer: B
Answer: (B) 3/2
Question: Raju age after 45 years will be 5 times his age 5 years back, What is the present age of Raju
(A) 28.5
(B) 17.2
(C) 10.2
(D) 17.5
(E) 66.2
Rationale: Clearly,
x+45 = 5(x-5)
<=> 4x = 70 => x = 17.5
Answer:D
Answer: (D) 17.5
Question: The output of a factory is increased by 10% to keep up with rising demand. To handle the holiday rush, this new output is increased by 60%. By approximately what percent would the output of the factory now have to be decreased in order to restore the original output?
(A) 20%
(B) 24%
(C) 30%
(D) 43%
(E) 79%
Rationale: take it as original output =100.
to meet demand increase by 10%, then output=110.
to meet holiday demand , new output increase by 60% then output equals 176
To restore new holidy demand output to original 100.
final -initial/final*100
=76/176*100=8/33*100=43% approxiamately.
option D is correct.
Answer: (D) 43%
Question: The value of 4.5*1.8+4.5*8.2/1.5*4.5+1.5*5.5
(A) 7
(B) 8
(C) 5
(D) 3
(E) 2
Rationale: Exp: 4.5*1.8+4.5*8.2/1.5*4.5+1.5*5.5 = 4.5 (1.8+8.2)/1.5 (4.5+5.5)
= 4.5*10/1.5*10
= 45/15 =3
ANSWER:D
Answer: (D) 3
Question: When positive integer x is divided by 12, the quotient is y and the remainder is 3. When 2x is divided by 11, the quotient is 3y and the remainder is 6. What is the value of 21y – x?
(A) 3
(B) 2
(C) -1
(D) -2
(E) -3
Rationale: (1) x = 12y+3
(2) 2x = 33y+6
(2)-(1): x=21y+3
21y-x=-3
The answer is E.
Answer: (E) -3
Question: What is the remainder when 5632 is divided by 6?
(A) 2
(B) 1
(C) 0
(D) 4
(E) 3
Rationale: f the number is divisible by both 2 and 3, then number will be divisible by 6.
As there is 2 an even number in units place it is divisible by 2.
sum of digits is:5+6+3+2=16. 16 is not divisible by 3 hence number is not divisible by 3.
The nearest multiple of 6 is 5630.
5632-5630=2.
Answer:2
Answer: (A) 2
Question: The cost to park a car in a certain parking garage is $20.00 for up to 2 hours of parking and $1.75 for each hour in excess of 2 hours. What is the average (arithmetic mean) cost per hour to park a car in the parking garage for 9 hours?
(A)  $3.58
(B)  $1.67
(C)  $2.25
(D)  $2.37
(E)  $2.50
Rationale: Total cost of parking for 9 hours = 20$ for the first 2 hours and then 1.75 for (9-2) hours = 20+7*1.75 = 32.25
Thus the AVERAGE parking price = 32.25/9 = 3.58 $
A is the correct answer.
Answer: (A)  $3.58
Question: For dinner at a restaurant, there are x choices of appetizers, y + 1 main courses, and z+1 choices of dessert. How many total possible choices are there if you choose 1 appetizer, 1 main course, and 1 dessert for your meal?
(A) x + y + z + 1
(B) xyz + xz
(C) xy + z + 1
(D) xz+x+xyz+xy
(E) xyz + 1/2
Rationale: x appetizers,
y + 1 main courses,
z+1 dessert
Required: 1 appetizer, 1 main course, and 1 dessert
Number of ways possible = xC1*(y+1)C1*(z+1)C1 {NC1 = N! / (N-1)!*1! = N}
Hence, number of ways = x(y+1)(z+1) = (x + xy)(z+1)=xz+x+xyz+xy
Correct Option: D
Answer: (D) xz+x+xyz+xy
Question: A company has two types of machines, type R and type S. Operating at a constant rate a machine of R does a certain job in 36 hours and a machine of type S does the job in 36 hours. If the company used the same number of each type of machine to do job in 12 hours,how many machine R were used?
(A) 3
(B) 4
(C) 6
(D) 9
(E) 12
Rationale: yes there is a typo in the question, I got the same ques on my GMAT PREP last week, and the questions goes as:
A company has two types of machines,type R and type S.Operating at a constant rate a machine of R does a certain job in 36 hours and a machine of type S does the job in 36 hours .If the company used the same number of each type of machine to do job in2 hours,how many machine R were used?
So for a job to be done in 2 hours R = 1/2
R_a(Rate of Machine R) = 1/36
R_s(Rate of Machine S) = 1/36
Lets say x machines are used to attain the desired rate, thus
x/36+x/36 = 1/2 (Desired R =1/2 i.e. to complete the job in 2 hours)
(x+x)/36 = 1/2
2x/36 = 1/2
x = 9.
QA = 9 (ANSWER D)
Answer: (D) 9
Question: At a certain college 25% of the students join a band, 25% play a sport and 20% both join the band and play a sport. if a randomly selected student is a member of the band, what is the probablility that the student also plays a sport?
(A) 1) 3/4
(B) 2)1/5
(C) 3)2/5
(D) 4)1/2
(E) 5)4/5
Rationale: This is the way to approach the question:
25% of the students join a band => 25% of the students either only join a band or both join a band and play sports
25% of the students play sports => 25% of the students either only play sports or both play sports and join a band
If a randomly selected student is a member of the band, then probability that the student also plays a sport = percentage of cases in which the student can do both/percentage of cases in which the student can just play sports or do both
=20%/25% = 4/5
The answer is therefore (E)
Answer: (E) 5)4/5
Question: A trumpet section includes five musicians. If they hold an audition to determine chair order within the section, and there are to be no ties, how many distinct chair orders can result?
(A) 240
(B) 125
(C) 120
(D) 100
(E) 25
Rationale: We have a word problem using permutations. We must identify the number of total items in the group. There are five possible musicians who could be placed in the first chair, four possible musicians for the second, three for the third, two for the fourth, and one for the fifth. To determine the number of possible orderings, we multiply these numbers, which we can express as 5!. Now 5! = (5)(4)(3)(2)(1) = 120.
ANS:C
Answer: (C) 120
Question: There are four temples, one plant of flower and a pond.
A boy goes there and before entering the first temple he plucks the flower and has to take a bath in the pond before entering into every temple.
The specialty of the pond is that if a person is carrying flowers with him and he takes a dip in the pond the number of the flowers gets doubled.
He takes a dip and offers some flowers in the first temple, then he again takes a dip and offers flowers in second temple , again the same in the third temple, and again the same in the fourth temple
The boy offers EQUAL number of flowers in all the four temples and goes back with EMPTY hand after offering flower in the fourth temple..How many flowers could he have offered in each temples?
(A) 20
(B) 24
(C) 30
(D) 48
(E) 60
Rationale: Note : all temples received equal flowers
lets start with the 4th temple , lets say all x flowers were given, then before dip x/2 flowers must have been left after offering in the 3rd temple. ( only then x/2 flowers could be doubled to become x and all offered in the 4th temple with nothing left over )
so x/2 flowers were left after offering made in the 3rd temple
going backwards .....
(x/2) left after 3rd temple + x given in the 3rd temple =(3x/2) {after dip, total flowers just before offering in the 3rd temple}
so before dip (3x/4)
(3x)/4 left after 2nd temple + x given in the 2nd temple = (7x)/4{after dip, total flowers just before offering in the 2nd temple}
so before dip (7x)/8
(7x)/8 left after 1st temple + x given in the 1st temple = (15x)/8{after dip, total flowers just before offering in the 1st temple}
so before dip (15x)/16
now since number of flowers have to be an integer then x has to be a multiple of 2,4,8 and 16
only 48 satisfies
Hence Answer = D
Answer: (D) 48
Question: How many times are the hands of a clock at right angle in 5 days?
(A) 202
(B) 220
(C) 210
(D) 212
(E) 222
Rationale: In 1 day, they are at right angles 44 times.
In 5 days, they are at right angles 220 times. Answer: Option B
Answer: (B) 220
Question: 10 men and 15 women together can complete a work in 4 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
(A) 100 days
(B) 667 days
(C) 177 days
(D) 187 days
(E) 225 days
Rationale: 1 man's 1 day work = 1/100
(10 men + 15 women)'s 1 day work = 1/4
15 women's 1 day work = (1/4 - 10/100) = 3/20
1 woman's 1 day work = 1/100
1 woman alone can complete the work in 100 days.
Answer:A
Answer: (A) 100 days
Question: A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that exactly three bulbs are good?
(A) 20/66
(B) 20/63
(C) 20/87
(D) 20/98
(E) 20/12
Rationale: Required probability = (5C3 . 4C1)/9C4
= (10 * 4)/126
= 20/63
Answer:B
Answer: (B) 20/63
Question: Two cars are traveling towards each other. If car A is traveling at a speed of 50 mph and car B is traveling 12% slower, how much time will it take the cars to meet if the initial distance between the two is 705 miles?
(A) Six hours and 30 minutes.
(B) Seven hours and 30 minutes.
(C) Eight hours and 20 minutes.
(D) Nine hours and 15 minutes.
(E) Ten hours and 20 minutes.
Rationale: since they are travelling towards each other, the total speed is 50 mph + 50*0.88 = 94 mph (0.88 since, the other car is 12% less than first one)
Distance to be covered is 705, so time taken = distance/speed = 705/94 = 7.5 hours or 7 hours and 30 mins.
Answer is B: Seven hours and 30 minutes.
Answer: (B) Seven hours and 30 minutes.
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) 200 m
(B) 700 m
(C) 800 m
(D) 100 metre
(E) 1600 m
Rationale: Explanation:
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 metre
Question: Of 70 players on a football team, 34 are throwers. The rest of the team is divided so one third are left- handed and the rest are right handed. Assuming that all throwers are right handed, how many right- handed players are there total?
(A) 54
(B) 58
(C) 63
(D) 71
(E) 92
Rationale: Total = 70
Thrower = 34
Rest = 70 - 34 = 36
Left Handed = 36/3 = 12
Right handed = 24
If all Thrower are right handed then total right handed is 34 + 24 = 58
so B. 58 is the right answer
Answer: (B) 58
Question: If
1 = 6
2 = 12
3 = 18
4 = 24
5 = 30
6 = 36
Then 12 = ?
Hint: Its a logic Riddle not a mathematical riddle
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: B
2
As stated
2=12 => 12=2
Answer is B
Answer: (B) 2
Question: The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.3 years, The ratio of the number of boys to the number of girls in the class is
(A) 7 : 3
(B) 2 : 3
(C) 9 : 3
(D) 6 : 3
(E) 5 : 6
Rationale: Explanation:
Let the ratio be k : 1. Then,
k * 16.4 + 1 * 15.3 = (k + 1) * 15.8
<=> (16.4 - 15.8) k = (15.8 - 15.3) <=> k = 0.5/0.6 = 5/6.
Required ratio = 5/6 : 1 = 5 : 6.
Answer: E
Answer: (E) 5 : 6
Question: What will come in place of the x in the following Number series? 18, 22, 30, x ,78, 142
(A) 26
(B) 32
(C) 46
(D) 49
(E) 43
Rationale: (C)
The pattern is +4, +8, +16, +32, +64
So the missing term is = 30 + 16 = 46 .
Answer: (C) 46
Question: The sum of Two Numbers is 15 and the product is 56. Find the sum of the squares of the Numbers.
(A) 15
(B) 125
(C) 89
(D) 65
(E) 113
Rationale: let the numbers be x, y.
x + y = 15
Square of (x + Y) = X Square + Y Square + 2XY
225 = X Square + Y Square + 112
X Square + Y Square = 225 - 112 = 113
so the answer is option E
Answer: (E) 113
Question: The area of a rectangle is 360 Sq.m. If its length is increased by 10m and its width is decreased by 6m, then its area does not change. Find the perimeter of the original rectangle.
(A) 74m
(B) 75m
(C) 76m
(D) 77m
(E) 78m
Rationale: let l and b are the length and width of the original rectangle.
since,l*b=360sq.m-(1)
now,a/c to the problem
(l+10)*(b-6)=360sq.m
therefore,
b=6(1+l/10)-(2)
l*l+10l-600=0 {by using equations 1 and 2}
thereafter,
(l-20)(l+30)=0
hence, l=20m
b=18m {by using equation 2}
now, Perimeter of the original rectangle=2(l+b)=2(20+18)=76m
ANSWER:C
Answer: (C) 76m
Question: A shopkeeper sells 400 metres of cloth for Rs. 18000 at a loss of Rs.5 per metre. Find his cost price for one metre of cloth?
(A) 12
(B) 67
(C) 26
(D) 50
(E) 12
Rationale: SP per metre = 18000/400 = Rs. 45 Loss per metre = Rs. 5 CP per metre = 45 + 5
= Rs. 50
Answer:D
Answer: (D) 50
Question: If 100!/x is not an integer,which of the following could be the value of x?
(A) 5^24
(B) 7^16
(C) 11^9
(D) 13^6
(E) 17^6
Rationale: Checking the options is the best way to go. The answer will be a number which doesn't completely divide 100!
No of 5's in 100! = 24
No of 7's in 100! = 16
No of 11's in 100! = 9
No of 13's in 100! = 7
No of 17's in 100! = 5
There are 5 seventeens in 100! , however we are dividing by 17^6 , so this can not divide 100!
Hence our answer will be E
Answer: (E) 17^6
Question: If x-1 – 1 is divided by x - 1 the quotient is:
(A) – 1/y
(B) – 1/z
(C) – 1/9
(D) – 1/k
(E) – 1/x
Rationale: x-1 – 1 is divided by (x – 1)
= x-1 – 1 × 1/(x-1)
= 1/x – 1 × 1/(x-1)
= (1 – x/x) × 1/(x-1)
= - (1 – x)/x × 1/(x-1)
= – 1/x [cancel x – 1 from the numerator and denominator].
Answer: E
Answer: (E) – 1/x
Question: Rs.8000 become Rs.9261 in a certain interval of time at the rate of 5% per annum of C.I. Find the time?
(A) 6 years
(B) 8 years
(C) 9 years
(D) 3 years
(E) 2 years
Rationale: 9261 = 8000(21/20)N
(21/20)3 = (21/20)N => N = 3
5. At the end of three years what will be the comp.Answer: OD
Answer: (D) 3 years
Question: In how many ways can three boys and three girls sit in a row such that all girls sit together?
(A) A)160
(B) B)144
(C) C)96
(D) D)120
(E) E)154
Rationale: Treat all girls as one unit. Now there are four students and they can be arranged in 4! ways. Again three girls can be arranged among themselves in 3! ways.
Required number of arrangements
= 4! * 3! = 24 * 6
= 144.
Answer:B
Answer: (B) B)144
Question: A shopkeeper sells 20% of his stock at 10% loss ans sells the remaining at a profit of 10%. He incurred an overall profit of Rs. 1500. Find the total worth of the stock?
(A) 27500
(B) 26000
(C) 24000
(D) 25000
(E) 23500
Rationale: Let the total worth of the stock be Rs. x.
The SP of 20% of the stock = 1/5 * x * 9/10 = 9x/50
The SP of 80% of the stock = 4/5 * x * 1.1 = 44x/50
Total SP = 9x/50 + 44x/50 = 53x/50
Overall profit = 53x/50 - x = 3x/50
3x/50 = 1500 => x = 25000
Answer:D
Answer: (D) 25000
Question: Peter’s salary is reduced by 30% and then reduced salary is increased by 30%. Find ,how many percentage his present salary is less as compared to his previous salary?
(A) 6%
(B) 7%
(C) 8%
(D) 9%
(E) 10%
Rationale: 30*30
-------- = 9%
100
Answer: D
Answer: (D) 9%