mathewhe/aqua_rat huggingface dataset:
Question: Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
(A) 11
(B) 10
(C) 76
(D) 16
(E) 12
Rationale: Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
Therefore 20x + 25(x - 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.
Answer: B
Answer: (B) 10
Question: When 2 is added to half of one-third of one-fifth of a number, the result is one-fifteenth of the number. Find the number?
(A) 32
(B) 81
(C) 60
(D) 27
(E) 11
Rationale: Explanation:
Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60
Answer:C
Answer: (C) 60
Question: Two trains leave the same train station at 6:00 am and 7:00 am and they travel at 100 kph and 125 kph respectively. How many kilometers from the train station will the two trains be together?
(A) 450
(B) 475
(C) 500
(D) 525
(E) 550
Rationale: At 7:00, the first train is 100 km ahead.
The second train gains 25 km each hour.
The time it takes the second train to catch the first train is 100 / 25 = 4 hours.
In this time, the second train travels 4*125=500 km.
The answer is C.
Answer: (C) 500
Question: A person, by selling an article at three-fourths of the list price incurs a loss of 20%. Find the profit percentage if he sells at the list price?
(A) 25 %
(B) 6.66 %
(C) 5.45 %
(D) 30 %
(E) None
Rationale: Sol.
0.75 MP = 0.8 CP (since 20% loss) So, MP = 1.0666CP ⇒ 6.66% gain.
Answer B
Answer: (B) 6.66 %
Question: The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
(A) 19
(B) 20
(C) 22
(D) 45
(E) 52
Rationale: A
19
Average of 20 numbers = 0. Sum of 20 numbers (0 x 20) = 0. It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
Answer: (A) 19
Question: A high school has 400 students 1/2 attend the arithmetic club, 5/8 attend the biology club and 3/4 attend the chemistry club. 3/8 attend all 3 clubs. If every student attends at least one club how many students attend exactly 2 clubs.
(A) 35
(B) 45
(C) 55
(D) 60
(E) 50
Rationale: A-Club has 200 members (1/2 of 400)
B-Club has 250 members (5/8 of 400)
C-Club has 300 members (3/4 of 400)
We can create an equation to solve this:
200+250+150 = n + x + 2y
where n is the number of students, x is the number of students in two clubs, and y is the number of students in three clubs.
The question provides y for us (150).
750 = 400 + x + 300
x = 50
E
Answer: (E) 50
Question: The average age of a husband and his wife was 17 years at the time of their marriage. After five years they have a one-year old child. The average age of the family now is :
(A) 55/3
(B) 15
(C) 19
(D) 287
(E) 27
Rationale: Explanation:
Sum of the present ages of husband, wife and child = (17 * 2 + 5 * 2) + 1 = 45 years.
Required average = (45/3)= 15years.
Answer: B
Answer: (B) 15
Question: The sum of three consecutive odd numbers is 20 more than the first of these numbers. What is the middle number?
(A) 5
(B) 7
(C) 9
(D) 11
(E) 10
Rationale: Solution
Let the numbers be x, x+2,x+4
Then x,(x+2)+(x+4)= x+20
‹=›2x = 14
‹=›x= 7.
Therefore Middle number = x+2 = 9.
Answer C
Answer: (C) 9
Question: Two cyclists are running on a circular track of 600m.They start from the same point A at the same time.If they meet for the first time at a point 200 m from A, which of the following could be the possible ratio of their speeds?
(A) 1:2
(B) 2:3
(C) 3:5
(D) 5:7
(E) 7:11
Rationale: well, the key aspect of this question is the distance mentioned, when they met for the first time, which is 200m away from the starting point. This distance between their first meeting can be less than 600 in only two circumstance.
1) if both are travelling in the same direction and one of them has been given a head start say by few meters.
2) or if both are travelling in the opposite direction.
now since they start their journey at the same time. hence option 1 is ruled out. hence they must be running in the opposite direction. hence answer should be 1:2
Why?? here is a solution
As here time is constant, therefore, distance traveled will be directly proportional to the the respective velocities/speed of the two runners.(d1/d2=v1/v2)
now one of them has covered 200m, therefore other must have covered 400m, therefore ratio of their speeds/velocities will be 1:2
A
Answer: (A) 1:2
Question: During the break of a football match, the coach will make 3 substitutions. If the team consists of 13 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?
(A)  21/55
(B)  18/44
(C)  42/143
(D)  28/44
(E)  36/55
Rationale: Required probability = no. of favorable outcome/ total no. of outcomes
out of 13 players any 3 can be selected in 13C3 ways
after removing 2 forwards any 3 players can be selected from the remaining 9 players in 9C3 ways
thus required probability = 9C3/13C3 =42/143
Answer: (C)  42/143
Question: A professional janitor can clean a certain high school in (500-20) minutes, working at a constant rate. A student sentenced to detention can clean that same high school in 20 hours, also working at a constant rate. If the student is paid $7 total per hour and the janitor is paid $21 per hour, how much more would it cost the school to pay the janitor to do the job himself than it would to pay the student and the janitor to do the job together?
(A) -$56
(B) -$6
(C) $0
(D) $6
(E) $8
Rationale: A professional janitor can clean a certain high school in (500-20)/60 or 8 hours
So (applying rule #1), the janitor can clean 1/8 of the school in ONE HOUR
A student sentenced to detention can clean that same high school in 20 hours
So (applying rule #1), the student can clean 1/20 of the school in ONE HOUR
So, COMBINED, the student and janitor can clean (1/8 + 1/20) of the school in ONE HOUR
1/8 + 1/20 = 5/40 + 2/40 = 7/40
So, in ONE HOUR they can clean 7/40 of the school.
Applying rule #2, it will takethem40/7 hoursto clean the ENTIRE school.
The janitor earns $21/hour and the student earns $7/hour, so their combined rate is$28/hour.
Theircombined wages= (pay rate)(time) = ($28/hour)(40/7 hours) =$160
Working ALONE, the janitor takes 8 hours and earns $21/hour
So, working alone, the janitor's earnings = (pay rate)(time) = ($21/hour)(8 hours) =$168
$168-$160= $8, so the answer is E
Answer: (E) $8
Question: By selling 12 pencils for a rupee a man loses 20%. How many for a rupee should he sell in order to gain 20%?
(A) 7
(B) 5
(C) 8
(D) 12
(E) 15
Rationale: 80% --- 12
120% --- ?
80/120 * 12 = 8
ANSWER C
Answer: (C) 8
Question: The length of a train and that of a platform are equal. If with a speed of 108 k/hr, the train crosses the platform in one minute, then the length of the train (in meters) is?
(A) 299
(B) 266
(C) 299
(D) 900
(E) 261
Rationale: Speed = [108 * 5/18] m/sec = 30 m/sec; Time = 1 min. = 60 sec.
Let the length of the train and that of the platform be x meters.
Then, 2x/60 = 30 => x = 30 * 60 / 2 = 900
Answer: D
Answer: (D) 900
Question: What is value of x if 32-y=16x-y?
(A) 1
(B) 3
(C) 4
(D) 2
(E) 6
Rationale: 32-y=16x-y
32=16x-y+y
16x=32
x=2
Answer is option D
Answer: (D) 2
Question: A bag contains 15 red jellybeans and 15 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
(A) 9/100
(B) 2/19
(C) 1/8
(D) 3/20
(E) 3/10
Rationale: Method-1
10 red jellybeans and 10 blue jellybeans
Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140
Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120
Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3
Probability = 120 / 1140 = 2/19
Answer: option B
Method-2
Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans]
Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining]
Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining]
Required Probability = (10/20)*(9/19)*(8/18) = 3/10
Answer: option E
Answer: (E) 3/10
Question: If a coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 6 times, what is the probability that it will land heads up on all 6 flips?
(A) 1/6
(B) 1/8
(C) 1/32
(D) 1/2
(E) 1/64
Rationale: (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/64 Answer: E
Answer: (E) 1/64
Question: 4 dice are thrown simultaneously on the board. Find the probability show the same face.
(A) 1/113
(B) 1/115
(C) 1/216
(D) 2/113
(E) 3/111
Rationale: The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
=6×6×6×6=64=6×6×6×6=64
n(S)=64n(S)=64
Let XX be the event that all dice show the same face.
X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}
n(X)=6n(X)=6
Hence required probability,
=n(X)n(S)=664=n(X)n(S)=664
=1/216
C
Answer: (C) 1/216
Question: Steve traveled the first 2 hours of his journey at 30 mph and the last 3 hours of his journey at 80 mph. What is his average speed of travel for the entire journey?
(A) 68 mph
(B) 56.67 mph
(C) 60 mph
(D) 64 mph
(E) 66.67 mph
Rationale: Answer
Average speed of travel = Total distance travelled/Total time taken
Total distance traveled by Steve = Distance covered in the first 2 hours + distance covered in the next 3 hours.
Distance covered in the first 2 hours = speed * time = 30 * 2 = 60 miles.
Distance covered in the next 3 hours = speed * time = 80 * 3 = 240 miles.
Therefore, total distance covered = 60 + 240 = 300 miles.
Total time taken = 2 + 3 = 5 hours.
Hence, average speed = Total distance travelled/Total time taken=300/5 = 60 miles per hour.
Choice C
Answer: (C) 60 mph
Question: A “palindromic integer” is an integer that remains the same when its digits are reversed. So, for example, 43334 and 516615 are both examples of palindromic integers. How many 6-digit palindromic integers are both even and greater than 400,000?
(A) 200
(B) 216
(C) 300
(D) 400
(E) 2,500
Rationale: first digit=last digit=4 or 6 or 8 so three possibilities..
second =fifth= any of ten digits
third =fourth= any of ten digits..
so total integers=3*10*10=300
ANSWER:C
Answer: (C) 300
Question: Last year the range of the annual fees of the 50 courses in a college was $300. If the annual fees of each of the 50 courses this year is 10 percent greater than it was last year, what is the range of the annual fees of the 50 courses this year?
(A)  $270
(B)  $300
(C)  $330
(D)  $360
(E)  $630
Rationale: let the lowest fees x.
Therefore, highest fees is x +300.
Now fees of each course is increased by 10%.
Therefore the fees will remain arranged in the same order as before.
or Lowest fees = 1.1x and Highest = 1.1 * (x+300)
or Range = highest - lowest = 1.1 * (x+300) - 1.1x
= 330, Hence, C
Answer: (C)  $330
Question: If a and b are different positive integers and a + b = a(a + b), then which of the following must be true?
I. a = 1
II. b = 1
III. a = b = 0
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
Rationale: a has to equal a squared, and that only works if a is 1. both cannot be 0.
Answer : A
Answer: (A) I only
Question: 2^5*3^3*2^4*3 =
(A) 3^5*2^5
(B) 2^9*3^4
(C) 2^4*3^9
(D) 5^8
(E) 6^10
Rationale: 2^5*3^3*2^4*3 =
A) 3^5*2^5
B) 2^9*3^4
C) 2^4*3^9
D) 5^8
E) 6^10
If the base is same, powers are added so that follows B) 2^9*3^4.
Hence the answer is B
Answer: (B) 2^9*3^4
Question: The distance between two towns is 380 km. At the same moment, a passenger car and a truck start moving towards each other from different towns. They meet 4 hours later. If the car drives 5 km/hr faster than the truck, what is the truck's speed?
(A) 35km/hr
(B) 40km/hr
(C) 76km/hr
(D) 60km/hr
(E) 50km/hr
Rationale: The main idea used in this kind of problems is that the distance equals speed multiplied by time S=V⋅tS=V⋅t
V (km/hr)	t (hr)	S (km)
Car	x + 5	4	4(x +5)
Truck	X	4	4x
4(x+5)+4x=380
4x+4x=380−20
8x=360
x=360/8
x=45
Therefore the truck's speed is 45 km/hr, and the car's speed is 50 km/hr.
correct answer E
Answer: (E) 50km/hr
Question: The cost price of 20 articles is the same as the selling price of
x articles. If the profit is ,25%, find the value of x
(A) 16
(B) 17
(C) 19
(D) 20
(E) 25
Rationale: 25=(20-X)100/X
X=4(20-X)
X=16
ANSWER A
Answer: (A) 16
Question: A jogger running at 9 kmph along side a railway track is 260 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
(A) 3.6 sec
(B) 38 sec
(C) 36 sec
(D) 72 sec
(E) None of these
Rationale: Speed of train relative to jogger
= (45 – 9) km/h = 36 km/h
= (36 × 5⁄18) m/sec = 10 m/sec
Distance to be covered = (260 + 120) m = 380 m.
∴ Time taken = (380⁄10) sec = 38 sec.
Answer B
Answer: (B) 38 sec
Question: Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:
(A) 2/109
(B) 7/109
(C) 8/223
(D) 14/263
(E) 13/102
Rationale: n(S) =(52 x 51)/(2 x 1)= 1326.
Let E = event of getting 1 spade and 1 heart.
n(E)	= number of ways of choosing 1 spade out of 13 and 1 heart out of 13
= (13 x 13)
= 169.
P(E) = n(E)/n(S)169/1326=13/102
Answer E
Answer: (E) 13/102
Question: A person lent out some money for 1 year at 6% per annum simple interest and after 18 months, he again lent out the same money at a simple interest of 24% per annum. In both the cases, he got Rs.4704. Which of these could be the amount that was lent out in each case if interest is paid half-yearly?
(A) 2200
(B) 3000
(C) 3200
(D) 4200
(E) 5000
Rationale: If interest is paid half yearly,
in first case , r=3% per 6 months
for 18 months, I=( P*3*3)/100 ; here r=3% and t=3 as interest for 18 months in t=18/6
for next 6 month (i.e 24 months) I= (P*3*4)/100 =12P/100
second case,
After 18 months, same amount is lent out
r=12%
for 6 month i.e, first interest for t=1 , I=(12*1*P)/100=12P/100
For both cases , he got 4704
In above case , both cases have same interest as 12P/100
i.e P+12P/100=4704;
P=4200
ANSWER:D
Answer: (D) 4200
Question: If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area?
(A) 120
(B) 278
(C) 292
(D) 234
(E) 287
Rationale: The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle = 1/2 * 24 * 10
= 120 cm2
Answer:A
Answer: (A) 120
Question: In a partnership between A, B and C. A's capital is Rs.10000. If his share of a profit of Rs.1200 is Rs.400 and C's share is Rs.330, what is B's capital?
(A) Rs.20750
(B) Rs.22750
(C) Rs.21750
(D) Rs.21200
(E) Rs.21320
Rationale: 400 + 330 = 730
1200 - 330 = 870
400 ---- 10000
870 ---- ? =>21750
ANSWER:C
Answer: (C) Rs.21750
Question: A and B together have USD. 1210. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?
(A) USD 460
(B) USD 484
(C) USD 550
(D) USD 664
(E) None of above
Rationale: 4/5 A= 2/5 B
=> A=(2/5 * 15/4) B
=> A=3/2 B
=>A/B= 3/2
=> A:B =3:2
Therefore B's Share =USD (1210 * 2/5) = USD 484
Answer is B
Answer: (B) USD 484
Question: The area of sector of a circle whose radius is 12 metro and whose angle at the center is 42° is?
(A) 26.4 m2
(B) 39.6 m2
(C) 52.8 m2
(D) 79.2 m2
(E) 80.2 m2
Rationale: Explanation:
42/360 * 22/7 * 12 * 12 = 52.8 m2
ANSWER IS C
Answer: (C) 52.8 m2
Question: Which of the following options is an appropriate approximate value in the following equation’? 5 2/3 of 234.92+39% of 1205= ? - 625.7
(A) 2326.85
(B) 2426.85
(C) 2526.85
(D) 2436.85
(E) 2427.85
Rationale: 5 2/3 of 234.92+39% of 1205= ? - 625.7
17/3 of 234.92+39/100 of 1205= ? - 625.7
1331.2+469.95= ? - 625.7
1331.2+469.95 + 625.7= ?
≈ 2426.85
Answer B
Answer: (B) 2426.85
Question: What is the greatest possible (straight line) distance, between any two points on a hemisphere of radius 1?
(A) 0.5
(B) 1
(C) π/2
(D) 2
(E) π
Rationale: Maximum distance straight line is diameter
d = 2r = 2..
ANS option D.
Answer: (D) 2
Question: A 600 meter long train crosses a signal post in 40 seconds. How long will it take to cross a 3 kilometer long bridge, at the same speed?
(A) 4 min
(B) 2 min
(C) 8 min
(D) 9 min
(E) 5 min
Rationale: S = 600/40 = 15 mps
S = 3600/15 = 240 sec = 4 min
Answer: A
Answer: (A) 4 min
Question: Tom runs everyday for exercise and running burns 500 calories/mile. If Tom wants to lose 10 pounds, how many miles for how many days does Tom have to run in total. (Assume it takes 3000 calories to lose 1 pound of weight)
(A) 3 miles for 10 days
(B) 1 mile for 20 days
(C) 10 miles for 6 days
(D) 5 miles for 2 days
(E) 15 miles for 10 days
Rationale: It takes 3000 calories to lose 1 pound, which means it takes 30000 calories to lose 10 pounds. If running 1 mile burns 500 calories, Tom would need to run 10 miles = 5000 calories for 6 days = 5000 * 6 = 30000. None of the other answers would give 30000. Correct Option: 10 miles for 6 days. Answer: C
Answer: (C) 10 miles for 6 days
Question: The greatest possible length which can be used to measure exactly the length 7m, 3m 85cm, 12 m 95 cm is
(A) 22
(B) 27
(C) 35
(D) 87
(E) 98
Rationale: Explanation:
Required Length = H.C.F of 700 cm, 385 cm and 1295 c
= 35 cm.
Answer: C) 35 cm
Answer: (C) 35
Question: The number of words from the letters of the words BHARAT in which B and H will never come together, is
(A) 360
(B) 240
(C) 120
(D) None of these
(E) Cannot be determined
Rationale: Solution
There are 6 letters in the word BHARAT, 2 of them are identical.
Hence total number of words with these letter = 360
Also the number of words in which B and H come together = 120
∴ The required number of words = 360 – 120 = 240. Answer B
Answer: (B) 240
Question: Dravid is 9 years older than his daughter Rosy. In 3 years Dravid will be twice as old as Rosy. What is Rosy current age.
(A) 6
(B) 8
(C) 10
(D) 7
(E) 11
Rationale: Now: Rosy = x, Dravid =x+9
In 3 years Rosy = x+3, Dravid = x+9+3 or 2(x+3)
x+12= 2x+6
x= 6
Rosy is 6 years old
Answer : A
Answer: (A) 6
Question: Letters of alphabets no from 1 to 26 are consecutively with 1 assigned to A and 26 to Z. By 27th letter we mean A, 28th B. In general 26m+n, m and n negative intezers is same as the letters numbered n.
Let P = 6, strange country military general sends this secret message according ot the following codification scheme. In codifying a sentence, the 1st time a letter occurs it is replaced by the pth letter from it. 2nd time if occurred it is replaced by P^2 letter from it. 3rd time it occurred it is replaced by p^3 letter from it. What is the code word for ABBATIAL
(A) DFHASDO
(B) HDFAHLOU
(C) LDOIHYAD
(D) GHLKZOIR
(E) LODIHFAD
Rationale: A should be coded as 1+6 = G (it occurred for first time)
B should be coded as 2+6 = H (it occurred for first time)
B Should be coded as 2 + 36 = 38 - 26 = 12 = L (it occurred for second time)
Option D is correct
Answer: (D) GHLKZOIR
Question: Roses can be purchased individually for $6.30, one dozen for $36, or two dozen for $50. What is the greatest number of roses that can be purchased for $680?
(A)  156
(B)  162
(C)  316
(D)  324
(E)  325
Rationale: Buy as many $50 deals as possible. We can by 650/50=13 two dozen roses, thus total of 13*24 = 312 roses.
We are left with 680 - 650 = $30. We can buy 30/6.3 = ~4 roses for that amount.
Total = 312 + 4 = 316.
Answer: C.
Answer: (C)  316
Question: The class mean score on a test was 40, and the standard deviation was 15. If Jack's score was within 2 standard deviations of the mean, what is the lowest score he could have received?
(A) 20
(B) 31
(C) 45
(D) 90
(E) 89
Rationale: 1 SD from the mean is ADDING and Subtrating the amount if standard deviation from the mean one time. 2 SD from the mean is ADDING and Subtracting TWICE.
1SD from the mean ranges from 25 to 55, where 55 is within SD above the mean and 25 within 1SD below the mean
2SD=15 TWICE=30 from the the mean, which is 55 to 25, where 55 is within 2 SD above the mean AND 30 is within 2 SD below the mean.
Answer = A
Answer: (A) 20
Question: John purchased some shirts and trousers for $1300. He paid $500 less for the shirts than he did for the trousers. If he bought 5 shirts and the cost of a shirt is $20 less than that of a trouser, how many trousers did he buy?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 9
Rationale: Given that the total purchase of two items cost 1300
So the average purchase of one item will cost 1300/2 = 650.
Its given as total shirt cost 500$ less. Hence Total Shirt cost = 650 - 250 and Total trouser cost = 650 + 250
5 shirts = 400$ ==> One shirt = 80$
One trouser = 80 + 20 =100$
Total trousers =900 / 100 = 9.
E
Answer: (E) 9
Question: P and Q started a business investing Rs. 45,000 and Rs. 10,000 respectively. In what ratio the profit earned after 2 years be divided between P and Q respectively?
(A) 9:2
(B) 4:5
(C) 9:3
(D) 17:9
(E) 17:4
Rationale: P:Q = 45000 : 10000
= 9:2
Answer:A
Answer: (A) 9:2
Question: A dishonest dealer professes to sell goods at the cost price but uses a false weight and gains 25%. Find his false weight age?
(A) 7
(B) 9
(C) 8
(D) 4
(E) 5
Rationale: 25 = E/(1000 - E) * 100
1000 - E = 4E
1000 = 5E => E = 200
1000 - 200 = 8
Answer:C
Answer: (C) 8
Question: A sum of money is put out at compound interest for 2 years at 20%. It would fetch Rs.723 more if the interest were payable half-yearly, then it were pay able yearly. Find the sum.
(A) 22200
(B) 21897
(C) 28799
(D) 30000
(E) 32782
Rationale: P(11/10)^4 - P(6/5)^2 = 723
P = 30000
Answer: D
Answer: (D) 30000
Question: The negative and positive charge symbols are assigned (in that order) to the:
(A) 22
(B) 77
(C) 77
(D) 88
(E) 21
Rationale: Answer: Option B
Answer: (B) 77
Question: Albert invested Rs. 8000 in a scheme for 2 years at compound interest rate 5% p.a. How much amount will Albert get on maturity of the fixed deposit?
(A) 8840
(B) 2776
(C) 2998
(D) 2662
(E) 8820
Rationale: Amount = [8000 * (1 + 5/100)2]
= 8000 * 21/20 * 21/20
= Rs. 8820
Answer: E
Answer: (E) 8820
Question: P is three times as fast as Q and working together, they can complete a work in 42 days. In how many days can Q alone complete the work?
(A) 16
(B) 77
(C) 56
(D) 55
(E) 778
Rationale: P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 42 days, which means Q can do the work in 168 days.
Hence, P can do the work in 56 days.
Answer:C
Answer: (C) 56
Question: Tom is 7 years older than Sam, and Jack is thrice as old as Tom. If S is Sam’s age and J is Jack’s age, what is the relationship between S and J?
(A) J = 3S + 10
(B) J = 3S + 21
(C) J = 3S + 5
(D) J = 3S − 11
(E) J = 3S − 4
Rationale: T=S+7 and J=3T --> J=3S+21
Answer B
Answer: (B) J = 3S + 21
Question: If the least common addition of two prime numbers x and y is 36, where x < y, then the value of 4x + y is
(A) 32
(B) 45
(C) 59
(D) 73
(E) 87
Rationale: (x+y) = 36 and both x an y are prime.
The only values of x and y can be 17 and 19 (x = 17 and y = 19)
4x + y = 4*17 + 19 = 87
Correct Option: E
Answer: (E) 87
Question: If 15 men, working 9 hours a day, can reap a field in 16 days, in how many
days will 18 men reap the field, working 8 hours a day?
(A) 12
(B) 16
(C) 8
(D) 15
(E) 10
Rationale: Let the required number of days be x.
More men, Less days (indirect proportion)
Less hours per day, More days (indirect proportion)
Men 18 : 15
Hours per day 8: 9 } : :16 : x
(18 x 8 x x)=(15 x 9 x 16)<=> x=(44 x 15)144 = 15
Hence, required number of days = 15.
Answer is D.
Answer: (D) 15
Question: The measures of the 2 acute angles of a triangle are in the ratio of 2:7. What are the measures of the 2 angles?
(A) 60°
(B) 70°
(C) 80°
(D) 90°
(E) 100°
Rationale: If the ratio of the two angles is 2:7, then the measures of two angles can be written as 2x and 7x. Also the two acute angles of a triangle is equal to 90°. Hence
2x + 7x = 90
9x = 90
x = 10
Measures of the two acute angles are
2x = 2 × 10 = 20°
7x = 7 × 10 = 70°
B
Answer: (B) 70°