mathewhe/aqua_rat huggingface dataset:
Question: The Royal Hawaiian Hotel decorates its Rainbow Christmas Tree with non-flashing white lights and a series of colored flashing lights - red, blue, green, orange, and yellow. The red lights turn red every 20 seconds, the blue lights turn blue every 30 seconds, the green lights turn green every 45 seconds, the orange lights turn orange every 60 seconds, and yellow lights turn yellow every 1 minute and 20 seconds. The manager plugs the tree in for the first time on December 1st precisely at midnight and all lights begin their cycle at exactly the same time. If the five colored lights flash simultaneously at midnight, what is the next time W all five colored lights will all flash together at the exact same time?
(A) 0:03 AM
(B) 0:04 AM
(C) 0:06 AM
(D) 0:12 AM
(E) 0:24 AM
Rationale: Time taken by each light to on
Red = 20 secs
Blue = 30 secs
Green = 45 secs
Orange = 60 secs
Yellow = 80 secs
We need to take LCM to find the time all five colored lights will flash together.
LCM = 720 secs
=12 mins
=0.12 AM
Ans = D
Answer: (D) 0:12 AM
Question: Rs.100 doubled in 5 years when compounded annually. How many more years will it take to get another Rs.200 compound interest?
(A) 10 years
(B) 5 years
(C) 7.5 years
(D) 15 years
(E) 8 years
Rationale: Explanatory Answer
Rs.100 invested in compound interest becomes Rs.200 in 5 years.
The amount will double again in another 5 years.
i.e., the amount will become Rs.400 in another 5 years.
So, to earn another Rs.200 interest, it will take another 5 years.
Answer B
Answer: (B) 5 years
Question: In a certain province in France there are 9 cities. If a single road segment connects only two cities, how many road segments are required in order to connect the cities so that each city is connected to all other cities with a single road segment?
(A) 36
(B) 105
(C) 330
(D) 14
(E) 15
Rationale: 1 city needs 8 roads to connect directly to the rest => 9 city needs 9*8 roads.
One road is counted twice => the number of road needed is 9*8/2 = 36
Answer A
Answer: (A) 36
Question: If X gets 25% more than Y and Y gets 20% more than Z, the share of Z out of Rs.925 will be:
(A) Rs. 300
(B) Rs. 200
(C) Rs. 250
(D) Rs. 350
(E) None of these
Rationale: Z share = Z,
Y = 1.2Z
X = 1.25Ã—1.2Z,
X+Y+Z =925
(1.25Ã— 1.2+1.2+1)Z=925
3.7Z=925, Z=250
Answer: . C
Answer: (C) Rs. 250
Question: In a survey of 330 employees, 104 of them are uninsured, 54 work part time, and 12.5 percent of employees who are uninsured work part time. If a person is to be randomly selected from those surveyed, what is the probability that the person will neither work part time nor be uninsured?
(A) 37/66
(B) 8/41
(C) 9/348
(D) 1/8
(E) 41/91
Rationale: ---------UI----------------NUI-------Total
PT----(12.5/100)*104 = 13----------- --54
NPT---104-13-------------- x--------276
Total--104----------------------------330
we have to find not part time and not uninsured . in other words not part time and insured = x/330 = (276-104+13)/330
= 37/66
Answer is A.
Answer: (A) 37/66
Question: On 8th Dec, 2007 is Saturday falls. What day of the week was it on 8th Dec. 2006 ?
(A) Sunday
(B) Thursday
(C) Tuesday
(D) Friday
(E) None of these
Rationale: Solution
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.
But, 8th Dec, 2007 is Saturday.
∴ 8th Dec, 2006 is Friday Answer D
Answer: (D) Friday
Question: A basket has 6 apples and 3 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is
(A) 25/84
(B) 35/48
(C) 45/84
(D) 55/47
(E) 65/84
Rationale: Total fruits = 9Since there must be at least two apples,(6C2 * 3C1)/9C3 + 6C3/9C3 = 65/84.Answer:E
Answer: (E) 65/84
Question: Daniel went to a shop and bought things worth Rs. 25, out of which 30 Paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
(A) 19.7
(B) 20
(C) 21.3
(D) 21.5
(E) 22
Rationale: Total cost of the items he purchased = Rs.25
Given that out of this Rs.25, 30 Paise is given as tax
=> Total tax incurred = 30 Paise = Rs.30/100
Let the cost of the tax free items = x
Given that tax rate = 6%
∴ (25−30/100−x)6/100 = 30/100
⇒ 6(25 −0.3 −x) = 30
⇒ (25 − 0.3 − x) = 5
⇒ x = 25 − 0.3 − 5 = 19.7
A
Answer: (A) 19.7
Question: Set A consists of the integers from 5 to 10, inclusive, while set B consists of the integers from 1 to 8, inclusive. How many distinct integers do belong to the both sets at the same time?
(A) 5
(B) 10
(C) 8
(D) 2
(E) 4
Rationale: A = {5,6, 7, 8, 9, 10}
B = {1,2,3,4,5,6, 7,8}
Common elements = {5,6, 7, 8} = 4 elements
Answer: Option E.
Answer: (E) 4
Question: Two trains of equal are running on parallel lines in the same direction at 47 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is?
(A) 50
(B) 88
(C) 66
(D) 55
(E) 22
Rationale: Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 47 - 36 = 11 km/hr.
= 11 * 5/18 = 55/18 m/sec.
2x/36 = 55/18 => x = 55.
Answer:D
Answer: (D) 55
Question: J can dig a well in 16 days. P can dig a well in 24 days. J, P, H dig in 8 days. H alone can dig the well in How many days?
(A) 229
(B) 29
(C) 27
(D) 48
(E) 20
Rationale: Assume the total work = 48 units.
Capacity fo J = 48 / 16 = 3 units / day
Capacity of P = 48 / 24 = 2 units / day
Capacity of J, P, H = 48 / 8 = 6 units / day
From the above capacity of H = 6 - 2 - 3 = 1
So H takes 48 / 1 days = 48 days to dig the well
Answer:D
Answer: (D) 48
Question: A jogger running at 9 km/hr along side a railway track is 180 m ahead of the engine of a 120 m long train running at 45 km/hr in the same direction. In how much time will the train pass the jogger?
(A) 89
(B) 20
(C) 36
(D) 88
(E) 30
Rationale: Speed of train relative to jogger = 45 - 9 = 36 km/hr.
= 36 * 5/18 = 10 m/sec.
Distance to be covered = 180 + 120 = 300 m.
Time taken = 300/10 = 30 sec.
Answer: Option E
Answer: (E) 30
Question: 0,2, 6, 12, 20, 30, 42, ___
(A) 55
(B) 85
(C) 59
(D) 63
(E) 56
Rationale: 0=1*1-1
2=2*2-2
6=3*3-3
12=4*4-4
20=5*5-5
30=6*6-6
42=7*7-7
Similarly 8*8-8=56
ANSWER:E
Answer: (E) 56
Question: A hall 36m long and 15m broad is to be paved with stones, each measuring 8dm by 5 dm. The number of stones required is:
(A) 180
(B) 1800
(C) 18
(D) 18000
(E) 1350
Rationale: Area of the hall= 3600*1500
Area of each stone= (80*50)
Therefore, number of stones= (3600*1500/ 80*50)= 1350
ANSWER:E
Answer: (E) 1350
Question: 6 workers should finish a job in 8 days. After 3 days came 4 workers join them. How many days D do they need to finish the same job?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: Let Rate of one worker be R
=> (6*R)*8 = 1 (Rate * Time = work)
=> R = 1/48
=> Work remaining after 3 days
1 - (3*6)/48 = 30/48
After 4 ppl joined in
((6+4)*Time)/48 = 30/48
Time D = 3 days to finish the task
IMO A
Answer: (A) 3
Question: To fill a tank, 100 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to four-fifth of its present ?
(A) 100
(B) 125
(C) 150
(D) 175
(E) 200
Rationale: Let the capacity of 1 bucket = x.
Then, the capacity of tank = 100x.
New capacity of bucket = 4/5 x
Therefore, Required number of buckets = (100x)/(4x/5)
= (100x) x 5/4x
= 500/4
= 125
Answer is B.
Answer: (B) 125
Question: If population of certain city increases at the rate of 5%. If population in 1981 was 185220, then population in 1978 was?
(A) 160000
(B) 228788
(C) 266891
(D) 267689
(E) 127791
Rationale: X * (105/100) * (105/100) * (105/100) = 185220
X = 185220/1.157625
X = 160000
Answer:A
Answer: (A) 160000
Question: A and B can do a work in 2 days, B and C in 4 days and C and A in 6 days. In how many days will the work be completed, if all three of them work together?
(A) 10/12
(B) 16/28
(C) 24/11
(D) 42/16
(E) 18/12
Rationale: One day work of A and B = 1/2
One day work of B and C = 1/4
One day work of C and A = 1/6
2(A + B + C) = 1/2 + 1/4 + 1/6.
2(A + B + C) = 11/12
(A + B + C) = 11/24
Number of days required = 24/11 days.
Answer: C
Answer: (C) 24/11
Question: A train crosses a platform of 120 m in 15 sec, same train crosses another platform of length 180 m in 18 sec. then find the length of the train?
(A) 227m
(B) 180m
(C) 176m
(D) 126m
(E) 167m
Rationale: Length of the train be ‘X’
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180m
Answer:B
Answer: (B) 180m
Question: A grocer purchased 80 kg of sugar at 13.50 per kg and mixed it with 120 kg sugar at 16 per kg. At what rate should he sell the mixture to gain 20%?
(A) 17 per kg
(B) 18 per kg
(C) 16.5 per kg
(D) 16 per kg
(E) None of these
Rationale: C.P. of 200 kg of mixture = (80 × 13.50 + 120 × 16)
= 3000.
S.P. = 120% of 3000 = (120⁄100 × 3000) = 3600
∴ Rate of S.P. of the mixture = (3600⁄200) per kg
= 18 per kg.
Answer B
Answer: (B) 18 per kg
Question: On Thursday Mabel handled 90 transactions. Anthony handled 10% more transactions than Mabel, Cal handled 2/3rds of the transactions that Anthony handled, and Jade handled 16 more transactions than Cal. How much transactions did Jade handled?
(A) 80
(B) 81
(C) 82
(D) 83
(E) 84
Rationale: Solution:
Mabel handled 90 transactions
Anthony handled 10% more transactions than Mabel
Anthony = 90 + 90 × 10%
= 90 + 90 × 0.10
= 90 + 9
= 99
Cal handled 2/3rds of the transactions than Anthony handled
Cal = 2/3 × 99
= 66
Jade handled 16 more transactions than Cal.
Jade = 66 + 16
= 82
Jade handled = 82 transactions.
Answer: C
Answer: (C) 82
Question: 10 men and 15 women together can complete a work in 3 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) 1 2/7 days
(B) 64 2/7 days
(C) 1 7/7 days
(D) 1 8/7 days
(E) 62 2/5 days
Rationale: 1 man's 1 day work = 1/100
(10 men + 15 women)'s 1 day work = 1/3
15 women's 1 day work = (1/3- 10/100) = 7/30
1 woman's 1 day work = 7/450
1 woman alone can complete the work in 64 2/7 days.
Answer:B
Answer: (B) 64 2/7 days
Question: If 40% of a number exceeds 20% of it by 12, then find the number?
(A) 60
(B) 50
(C) 47
(D) 55
(E) 80
Rationale: Use the elimination method to find the correct option.
Of all the options only 60 fits
40% of 60 = 24
20% of 60 = 12
24 - 12 = 12
Required number is 60.
ANSWER:A
Answer: (A) 60
Question: On six consecutive flips of a coin, what is the probability that all six produce the same result?
(A) 1/16
(B) 1/8
(C) 1/4
(D) 3/8
(E) 1/32
Rationale: TTTTTT case --> P = (1/2)^6 = 1/64
HHHHHH case --> P = (1/2)^6 = 1/64
P(TTTTTT or HHHHHH) = 1/64 + 1/64 = 1/32
Answer: E
Answer: (E) 1/32
Question: Tough and Tricky questions: Work/Rate Problems.
If it takes 90 workers 3 hours to disassemble the exhibition rides at a small amusement park, how many hours would it take 30 workers to do this same job?
(A) 40/3
(B) 11
(C) 7
(D) 7/3
(E) 9
Rationale: 90 Workers = 3 hours then,
1 Worker = 3*90 Hours
30 Workers = (3*90) /(30) = 9 Answer.E
Answer: (E) 9
Question: Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
(A) 4
(B) 5
(C) 6
(D) 8
(E) 10
Rationale: Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
ANSWER IS A
Answer: (A) 4
Question: If it is assumed that 50 percent of those who receive a questionnaire by mail will respond and 300 responses are needed, what is the minimum number of questionnaires that should be mailed?
(A) 400
(B) 420
(C) 480
(D) 500
(E) 600
Rationale: Minimum no of mail to be sent for getting 300 responses at 50%
= 300/0.5 = 600
Option E
Answer: (E) 600
Question: How many leading zeros (zeros after the decimal point but before the first nonzero digit of a number) are in 1/3!?
(A) 1
(B) 3
(C) 5
(D) 6
(E) 0
Rationale: 3! =6
1/10 < 1/6 < 1
0.01 < 1/6 < 1
So 1/6 has 0 leading zeros
Answer: E
Answer: (E) 0
Question: If a job is complete in X days by A, what will be A's one day work?
(A) 1/x
(B) x
(C) 10
(D) 5
(E) x/2
Rationale: A can complete a work in x days.
so one day work of a is 1/x.
Answer is option A
Answer: (A) 1/x
Question: Weights of two friends Ram and Shyam are in the ratio 7:5. If Ram's weight is increased by 10% and total weight of Ram and Shyam become 82.8 kg, with an increases of 15%. By what percent did the weight of Shyam has to be increased?
(A) 19%
(B) 10%
(C) 22%
(D) 16%
(E) None
Rationale: Solution:
given ratio of Ram and Shayam's weight = 7:5
Hence, (x-15)/(15-10) = 7/5
Or, x = 22%.
Answer: Option C
Answer: (C) 22%
Question: The average age of Lion,Tiger & Elephant were 4 years ago was 30 years and the age of lion and elephant were 6 years ago was 20 years. The present age of the Tiger is?
(A) 52 years
(B) 45 years
(C) 30 Years
(D) 50 Years
(E) 25 Years
Rationale: Sum of the present ages of Lion,Tiger & elephant = (30*3+4*3) = 102 years.
Sum of the present ages of X,Y = (20*2+6*2) = 52 years
Present age tiger = (102 - 52) = 50 years.
Answer : D
Answer: (D) 50 Years
Question: In how many R ways can the wordGMATbe arranged so that no such arrangement has A as the first letter?
(A) 6
(B) 12
(C) 18
(D) 24
(E) 30
Rationale: Number of ways of arranging the word GMAT = 4! = 24
Number of ways of arranging the word GMAT so that A is always in the first = 3! = 6
Number of ways of arranging GMAT so that A is not in the beginning R= 24 - 6 = 18
Answer: C
Answer: (C) 18
Question: The average of 25 results is 19. The average of first 12 of those is 14 and the average of last 12 is 17. What is the 13th result?
(A) 74
(B) 75
(C) 103
(D) 78
(E) 45
Rationale: Solution: Sum of 1st 12 results = 12*14
Sum of last 12 results = 12*17
13th result = x (let)
Now,
12*14+12*17+x = 25*19
Or, x = 103.
Answer: Option C
Answer: (C) 103
Question: A certain ball team has an equal number of right- and left-handed players. On a certain day, two-thirds of the players were absent from practice. Of the players at practice that day, one-third were right handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of left handed players who were not at practice?
(A) 1/3
(B) 2/3
(C) 7/5
(D) 5/7
(E) 3/2
Rationale: Say the total number of players is 18, 9 right-handed and 9 left-handed.
On a certain day, two-thirds of the players were absent from practice --> 12 absent and 6 present.
Of the players at practice that day, one-third were right-handed --> 6*1/3=2 were right-handed and 4 left-handed.
The number of right-handed players who were not at practice that day is 9-2=7.
The number of left-handed players who were not at practice that days is 9-4=5.
The ratio = 7/5.
Answer: C.
Answer: (C) 7/5
Question: When n is divided by 27, the remainder is 4. What is the remainder when n + 16 is divided by 7?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Rationale: Assume n = 23
Remainder (n/27) = 4
n+16 = 39
Remainder (39/7) = 4
Option C
Answer: (C) 4
Question: Efrida and Frazer who live 15 miles apart, meet at a restaurant that is directly north of Efrida's home and directly east of Frazer's home. If the restaurant is two miles closer to Efrida's home, than to Frazer's home, how many miles is the restaurant from Frazer's home?
(A) 6
(B) 7
(C) 8
(D) 10
(E) 13
Rationale: It's a simple geometry problem. Look at the diagram below:
Answer: E.
Answer: (E) 13
Question: A car takes 6 hours to cover a distance of 720 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time?
(A) 60
(B) 50
(C) 80
(D) 70
(E) 65
Rationale: Time = 6
Distance = 720
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 720/9 = 80 Kmph
Answer C.
Answer: (C) 80
Question: Anthony and Michael sit on the six member board od directors for compnay X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
(A) 20%
(B) 30%
(C) 40%
(D) 50%
(E) 60%
Rationale: First approach:
Let's take the group with Michael: there is a place for two other members and one of them should be taken by Anthony, as there are total of 5 people left, hence there is probability of 2/5=40%.
Second approach:
Again in Michael's group 2 places are left, # of selections of 2 out of 5 5C2=10 - total # of outcomes.
Select Anthony - 1C1=1, select any third member out of 4 - 4C1=4, total # =1C1*4C1=4 - total # of winning outcomes.
P=# of winning outcomes/# of outcomes=4/10=40%
Third approach:
Michael's group:
Select Anthony as a second member out of 5 - 1/5 and any other as a third one out of 4 left 4/4, total=1/5*4/4=1/5;
Select any member but Anthony as second member out of 5 - 4/5 and Anthony as a third out of 4 left 1/4, total=4/5*1/4=1/5;
Sum=1/5+1/5=2/5=40%
Fourth approach:
Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10
# of groups with Michael and Anthony: 1C1*1C1*4C1=4
P=4/10=40%
Answer: C.
Answer: (C) 40%
Question: A man decides to buy a nice horse. He pays $60 for it, and he is very content with the strong animal. After a year, the value of the horse has increased to $70 and he decides to sell the horse. But already a few days later he regrets his decision to sell the beautiful horse, and he buys it again. Unfortunately he has to pay $80 to get it back, so he loses $10. After another year of owning the horse, he finally decides to sell the horse for $90. What is the overall profit the man makes?
(A) 20 profit
(B) 10 profit
(C) 25 profit
(D) 30 profit
(E) 28 profit
Rationale: 70-60=10 profit
90-80=10 profit
---------------------
Total=20 profit
ANSWER:A
Answer: (A) 20 profit
Question: When throwing 2 dices, and looking at the sum of numbers on the dices - what is the probability to have a sum which is smaller than 5?
(A) 1/9
(B) 1/6
(C) 1/2
(D) 1/3
(E) 32/36
Rationale: a dice is composed of 6 numbers - 1,2,3,4,5,6.
When trowing 2 dices - there are 36 options (6X6=36).
If we want the sum of two dices to be< 5 there are those options: (1,1), (1,2), (2,1), (1,3), (3,1), (2,2) - total of 6 options.
Therefore - the probability to have a sum which is smaller than 5 is 6/36=1/6.
Answer:B
Answer: (B) 1/6
Question: Anup start a bike at 6am. and rides it at 60kmph. Raghu started at 7pm with his car at 80kmph. When will Raghu overtake Anup?
(A) 8:30am
(B) 9:40am
(C) 10:00am
(D) 12:00pm
(E) 11:10am
Rationale: Suppose bike is overtaken by car xhrs after 6am
Distance covered by the bike in x hrs = distance covered by the car in x-1 hr
60x = 80(x-1)
20x = 80
x = 4 hrs
Required time = 6+4 = 10:00am
Answer is C
Answer: (C) 10:00am
Question: Sandy invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years, Sandy got back 2 times the original investment. What is the value of n?
(A) 12.5
(B) 18.5
(C) 24.5
(D) 30.5
(E) 36.5
Rationale: n* 0.08 * principal + principal = 2 * principal
n * 0.08 = 1
n = 12.5 years
The answer is A.
Answer: (A) 12.5
Question: When two dice are rolled, what is the probability that the sum of the two numbers is 8?
(A)  1/9
(B)  2/9
(C)  1/3
(D)  2/3
(E)  None of the above
Rationale: We have 5 outcomes that satisfy the question:
2-6
3-5
4-4
5-3
6-2.
the probability is (1/6*1/6)*5= 5/36.
I think the answer is E
Answer: (E)  None of the above
Question: If a taxi driver charges x cents for the first quarter-mile of a trip and (x/5) cents for each additional quarter-mile, what is the charge, in cents, for a trip whose distance in miles is the whole number y?
(A) (x+xy)/125
(B) (4x + 4xy)/5
(C) (4x+xy)/500
(D) (4x+xy)/5
(E) xy/25
Rationale: let Y be 1 mile
x= 20 cents
as per given info
in 1 mile we have 4 quarter mile
so for the first mile x=20 cents and x/5 cents =20/5=4 cents for every additional quarter mile
so total =20+4+4+4=32 cents charge for complete ride.
we just substitute our values in the given options and we get our ans B.
Answer: (B) (4x + 4xy)/5
Question: A man sells an article at a gain 15%. If he had bought it at 10% less and sold it for 8 less, he would have gained 25%. Find the cost price of the article.
(A) 150
(B) 160
(C) 170
(D) 180
(E) 320
Rationale: Let the C.P. be 100
First S.P. = 115
Second C.P. = 90
Second S.P = 125% of 90 = 112.50
Difference of two selling prices is 115 – 112.50 = 2.50 and C.P. of the article is 100
But actual difference is 8.
∴ C.P. = 100⁄2.50 × 8 = 320.
Answer E
Answer: (E) 320
Question: It is known that no more than 4 children will be attending a party. What is the smallest number of cookies that must be brought to the party so that each child receives the same number of cookies?
(A) 35
(B) 105
(C) 12
(D) 210
(E) 420
Rationale: The question states that there are no more than 4 children in the party. Thus, the possibility is that there could be 1, 2, 3 or 4 children.
If you assume answer as 35 and there are 3 children, you may not distribute be able to distribute cookies equally.
similarly if there were 105 cookies, and 2 children, cookies cannot be distributed equally.
or if there were 210 cookies, and 4 children, cookies cannot be distributed equally.
Thus, the question asks for a number of cookies which can be distributed to any number of children (from 1 to 4).
And therefore the smallest number of cookies would be lcm of (1, 2, 3, 4), i.e.,12
Answer (C)
Answer: (C) 12
Question: The average marks obtained by 120 cndidates in a certainexaminationis 35.if the average marks of passed candidate is 39 and that of the failedcandidates is 15,what is the number of candidates who passed the examination
(A) 14 years
(B) 14 years 5 month
(C) 14 years 8 month
(D) 15 years 1 month
(E) 15 years 8 month
Rationale: (d) Let No. of passed candidates are x
So, x * 39 + (120 – x) 15= 120 * 35
x = 100
2.(a) Total increase = 8 * 2 = 16 years
So, total age of two women = 35 + 45 +16= 96
Average age of two women=96/2=48 years
3(d) Total decrease = 20 * 2 = 40 month
= 3 years 4 month
So, age of new boy = 18 years – 3 years 4 month
= 14 years 8 month
C
Answer: (C) 14 years 8 month
Question: What is the smallest integer X such that 2X+2 is divisible by 33, 44, 55, and 666?
(A) X = 53186760
(B) X = 26593379
(C) X = 73260
(D) X = 36629
(E) X = 6666
Rationale: 33 = 3 * 11
44 = 2 * 2 * 11
55 = 5 * 11
666 = 2 * 3 * 3 * 37
So 2X+2 = 2 * 2 * 3 * 3 * 5 * 11 * 37 = 73260, and X = (73260 - 2)/2 = 36629, which is option D.
Answer: (D) X = 36629
Question: The area of a rectangular plot is 15 times its breadth. If the difference between the length and the breadth is 10 metres, what is its breadth?
(A) 10 metres
(B) 5 metres
(C) 7.5 metres
(D) Data inadequate
(E) None of these
Rationale: L × B = 15 × B
∴ L = 15 m
and L – B = 10
∴ B = 15 – 10 = 5 m
Answer B
Answer: (B) 5 metres
Question: Given that x = 2^b – (8^35 + 16^5), which of the following values for b yields the lowest value for |x|?
(A) 104
(B) 105
(C) 106
(D) 107
(E) 125
Rationale: 8^35 + 16^5 = 2^105 + 2^20
Compared to 2^105, the value of 2^20 is negligible.
2^105 - (2^105 +2^20) will minimize the value of |x|.
The answer is B.
Answer: (B) 105
Question: For any real number x, the operatoris defined as:
(x) = x(2 − x)
If p + 1 =(p + 1), then p =
(A) -2
(B) 0
(C) 1
(D) 2
(E) 3
Rationale: (x) = x(2 − x)
(p + 1) = (p + 1)(2 - p - 1) = (1-p)(p+1)
We are given that p + 1 =(p + 1)
Therefore (1-p)(p+1) = (p + 1)
Or (p + 1) + (p-1)(p+1) = 0
(p + 1)(p - 1) = 0
p = -1,p=1
Option C
Answer: (C) 1
Question: There are 6 working days in a regular week and for each day, the working hours are 10. A man earns Rs. 2.10 per hour for regular work and Rs. 4.20 per hour for overtime. If he earns Rs.525 in 4 weeks, how many hours did he work?
(A) 225
(B) 235
(C) 215
(D) 255
(E) 245
Rationale: egular working hours in 4 weeks = (4 × 6 × 10) = 240 hours
Amount earned by working in these regular working hours
= 240 × 2.10 = Rs.504
Additional amount he earned = 525 - 504 = Rs.21
Hours he worked overtime = 21/4.2 = 210/42 = 5 hours
Total hours he worked = 240 + 5 = 245 hours
Answer is E.
Answer: (E) 245