mathewhe/aqua_rat huggingface dataset:
Question: X can do a piece of work in 40days. He works at it for 8days and then Y finished it in 36days. How long will Y take to complete the work?
(A) 45
(B) 12
(C) 15
(D) 18
(E) 20
Rationale: Work done by X in 8days = 8*1/40 = 1/5
Remaining work = 1 - 1/5 = 4/5
4/5 work is done by Y in 36days
Whole work will be done by Y in 36*5/4 = 45 days
Answer is A
Answer: (A) 45
Question: Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 12x + 3 = 0?
(A) 6
(B) 7
(C) 23
(D) 17
(E) 42
Rationale: a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab
= [(a + b)2 - 2ab]/ab
a + b = -12/1 = -12
ab = 3/1 = 3
Hence a/b + b/a = [(-12)2 - 2(3)]/3 = 126/3 = 42.
E)
Answer: (E) 42
Question: By selling 150 mangoes, a fruit-seller gains the selling price of 30 mangoes. Find the gain percent?
(A) 26%
(B) 25%
(C) 85%
(D) 24%
(E) 22%
Rationale: SP = CP + g
150 SP = 150 CP + 30 SP
120 SP = 150 CP
120 --- 30 CP
100 --- ? => 25%
Answer: B
Answer: (B) 25%
Question: A cylindrical water tower with radius 2.5 m and height 4 m is 3/4 full at noon. Every minute, .08π m3 is drawn from tank, while .03π m3 is added. Additionally, starting at 1pm and continuing each hour on the hour, there is a periodic drain of 4π m3. From noon, how many hours will it take to drain the entire tank?
(A) 20 2/7
(B) 20 6/7
(C) 21
(D) 21 3/7
(E) 3
Rationale: initial volume = (3/4)×∏×2.5²×4 = 18.75∏
Relative Drain/ min = .08∏ - .03∏ = .05∏ m³/min drain
Relative drain / hour = .05∏×60 = 3∏ m³/ hr
Every one hour starting from 1pm, 4∏ m³ of water is drained. It means that only at the hour the water is drained and NOT “in that 1 hour“
So after 1 hr the relative drain would be 3∏ + 4∏ = 7∏m³ of water drain
What i did initially was formed an equation 18.75∏ = 7∏×n (n is the number of hrs) so ended up with 2.67. This wrong
Look at this way
after 2 hrs the amount of water drain will be 2.67×7∏ = 14∏ m³
Left over water in the tank after 2 hrs = 3∏ m³
From above we know that it take 1 more hour to drain that 3∏ m³.
So ans is 3hrs
E
Answer: (E) 3
Question: Two cars were driven by R and B. They were 580 miles apart when they started driving towards each other.R's car had traveled at a speed of 20 miles per hour, 4 hours per day for 5 days when it had met B's car. If B had driven 3 hours per day for 5 days, what was B's speed in miles per hour?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
Rationale: 1. At 20mph for 4 hours a day, then driver R was traveling 80 miles per day. When R met with B, R had already traveled 400 miles.
2. This means that driver B had traveled 580-400 = 180 miles upon meeting up with R. B traveled a total of 3 x 5 = 15 hours (3hrs/day for 5 days). 180 miles / 15 hours = 12mph
3. Answer E!
Answer: (E) 12
Question: From the given equation find the value of x: x² − 3x + 2
(A) -1
(B) 1
(C) -2
(D) 3
(E) 4
Rationale: (x − 1)(x − 2)
x = 1 or 2.
B
Answer: (B) 1
Question: In a certain company, the ratio of the number of managers to the number of non-managers in any department must always be greater than 7 : 24. In the company, what is the maximum number of non-managers in a department that has 8 managers?
(A) 26
(B) 27
(C) 28
(D) 29
(E) 30
Rationale: 8/7 * 24 = 27.4
The answer is B.
Answer: (B) 27
Question: A cistern is filled by a tap in 3 1/2 hours. Due to leak in the bottom of the cistern, it takes half an hour longer to fill the cistern. If the cistern is full how long will it take the leak to empty it?
(A) 56.27 %
(B) 56.65 %
(C) 56.65 %
(D) 58.25 %
(E) 52.25 %
Rationale: a = 100 a2 = 10000
a = 125 a2 = 15625
----------------
10000 --------- 5625
100 -------? => 56.25 %
Answer: C
Answer: (C) 56.65 %
Question: The ages of Patrick and Michael are in the ratio of 3 : 5 and that of Michael and Monica are in the ratio of 3 : 5. If the sum of their ages is 196, what is the difference between the ages of Patrick and Monica?
(A) 27
(B) 64
(C) 45
(D) 72
(E) 18
Rationale: Ages of P and Mi=3x:5x
Ages of Mi and Mo=3x:5x
Rationalizing their ages. ratio of their ages will be 9x:15x:25x
Sum=49x=196
x=4
Difference if ages of Pa and Mo=25x-9x=16x=16*4=64
Answer B
Answer: (B) 64
Question: A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is
(A) No profit, no loss
(B) 5%
(C) 8%
(D) 10%
(E) None of these
Rationale: C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.
S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.
Gain =(80/1600)x 100	% = 5%.
ANSWER:B
Answer: (B) 5%
Question: The present age of a father is 3 years more than three times the age of his son. Three years hence ,father's age will be 10 years more than twice the age of the son , Find the present age of the father.
(A) 33 years
(B) 35 years
(C) 31 years
(D) 35 years
(E) 40 years
Rationale: Explanation :
Explanation :
Sol. Let the son's present age be xx years.
Then , father's present age = (3x+3)(3x+3) years
âˆ´âˆ´ (3x+3+3)=2(x+3)+10(3x+3+3)=2(x+3)+10
â‡” 3x+6 =2x+16â‡” 3x+6 =2x+16
â‡” x=10â‡” x=10
Hence , father's present age = (3x+3)(3x+3) = (3 x 10 + 3) years = 33 years.
ANSWER IS A
Answer: (A) 33 years
Question: The average of 6 no.'s is 2.80. The average of 2 of them is 2.4, while the average of the other 2 is 2.3. What is the average of the remaining 2 no's?
(A) 2.3
(B) 2.6
(C) 3.6
(D) 3.7
(E) 4.6
Rationale: Sum of the remaining two numbers = (2.80 * 6) - [(2.4 * 2) + (2.3 * 2)]
= 16.80 - (4.8 + 4.6) = 16.80 - 9.40 = 7.40
Required average = (7.4 / 2) = 3.7
Answer : D
Answer: (D) 3.7
Question: Mother,her daughter and her grand child weighs 150 kg. daughter and her daughter(child) weighs 60 kg. child is 1/5th of her grand mother. What is the age of the daughter?
(A) 46
(B) 47
(C) 42
(D) 49
(E) 50
Rationale: mother + daughter + child = 150kg
daughter + child = 60kg
mother = 150 - 60 = 90kg
child = 1/5th of mother = (1/5)*90 = 18kg
So now daughter = 120 -(mother + child) = 150 - (90+18) = 42kg
ANSWER:C
Answer: (C) 42
Question: Divide $900 among A,B in the ratio 1:2. How many $ that A get?
(A) $50
(B) $300
(C) $150
(D) $250
(E) $600
Rationale: Sum of ratio terms = 1+2 = 3
A = 900*1/3 = $300
Answer is B
Answer: (B) $300
Question: The sub-duplicate ratio of 9:25 is
(A) 4:3
(B) 1:2
(C) 3:5
(D) 1:4
(E) 2:3
Rationale: root(9):root(25)=3:5
ANSWER:C
Answer: (C) 3:5
Question: Look at this series: 7, 10, 8, 11, 9, 12, ... What number should come next?
(A) 10
(B) 8
(C) 7
(D) 9
(E) 12
Rationale: Alternating addition and subtraction series. in first added second is subtracted.
=> 10 is the next one
ANSWER A
Answer: (A) 10
Question: Rs. 6000 is lent out in two parts. One part is lent at 6% p.a simple interest and the other is lent at 10% p.a simple interest. The total interest at the end of one year was Rs. 450. Find the ratio of the amounts lent at the lower rate and higher rate of interest?
(A) 5:1
(B) 5:5
(C) 5:8
(D) 5:4
(E) 5:3
Rationale: Let the amount lent at 6% be Rs. x
Amount lent at 10% is Rs. (6000 - x)
Total interest for one year on the two sums lent
= 6/100 x + 10/100 (6000 - x) = 600 - 4x/100
=> 600 - 1/25 x = 450 => x = 3750
Amount lent at 10% = 2250
Required ratio = 3750 : 2250 =75:45=15:9=5:3
Answer:E
Answer: (E) 5:3
Question: Sara bought both German chocolate and Swiss chocolate for some cakes she was baking. The Swiss chocolate cost $3.6 per pound, and German chocolate cost $3 per pound. if the total the she spent on chocolate was $36 and both types of chocolate were purchased in whole number of pounds, how many total pounds of chocolate she purchased?
(A) 7
(B) 8
(C) 10
(D) 15
(E) 36/3
Rationale: If there were all the expensive ones,3.6...
There would be 36/3.6 or 10 of them but since 3 $ ones are also there, answer has to be >10.....
If all were 3$ ones, there will be 36/3
Ans E
Answer: (E) 36/3
Question: What is the tens' digit W of the sum of the first 40 terms of 1, 11, 111, 1111, 11111, 111111, ...?
(A) W=2
(B) W=3
(C) 4
(D) 8
(E) 9
Rationale: All of the first 40 terms have 1 at the units place.
Except the first term, the tens digit of all the remaining 39 terms is 1.
So now if you do the addition
1 1
1 1
. . .
. . .
. . .
1 1
0 1
3 0
Answer is B
Answer: (B) W=3
Question: Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77 but three tickets from city A to B and two tickets from city A to C cost Rs. 73. What are the fares for cities B and C from A ?
(A) Rs. 4, Rs. 23
(B) Rs. 13, Rs. 17
(C) Rs. 15, Rs. 14
(D) Rs. 17, Rs. 13
(E) Rs. 19, Rs. 15
Rationale: Explanation:
Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.
Then, 2x + 3y = 77 ...(i) and
3x + 2y = 73 ...(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.
Putting y = 17 in (i), we get: x = 13.
ANSWER IS B
Answer: (B) Rs. 13, Rs. 17
Question: For each color copy, Print Shop X charges $1.25 and Print Shop Y charges $2.75. How much greater is the charge for 40 color copies at Print Shop Y than at Print Shop X?
(A) $48
(B) $54
(C) $60
(D) $66
(E) $72
Rationale: The difference in the two prices is $2.75 - $1.25 = $1.50 for each color copy.
Each color copy will cost an extra $1.50 at Print Shop Y.
40*$1.50 = $60
The answer is C.
Answer: (C) $60
Question: Two trains 150 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet?
(A) 7.18
(B) 7.12
(C) 7.16
(D) 7.55
(E) 7.82
Rationale: T = (150 + 165)/ (80 + 65) * 18/5
T = 7.82
Answer:E
Answer: (E) 7.82
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
(B) 7
(C) 6
(D) 8
(E) 9
Rationale: S = 600/40 = 15 mps
S = 3600/15 = 240 sec = 4 min
Answer: A
Answer: (A) 4
Question: A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is?
(A) 37
(B) 50
(C) 77
(D) 55
(E) 22
Rationale: Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45 ==> x
= 50 km/hr.
Answer: B
Answer: (B) 50
Question: How many of the factors of 168 are divisible by 2?
(A) 4
(B) 8
(C) 10
(D) 12
(E) 16
Rationale: 168 = 2^3*3*7
The total number of factors is (1+3)(1+1)(1+1) = 16
The odd factors are 1, 3, 7, and 21.
The number of even factors is 16-4 = 12
The answer is D.
Answer: (D) 12
Question: It is the New Year and Mandy has made a resolution to lose weight this year. She plans to exercise and do yoga. For exercise she plans to workout at the gym and ride her bicycle in the ratio of 2:3 everyday. She will also do yoga in the ratio, yoga:exercise = 2:3. If she rides her bike for 12 minutes, how much time will she spend doing yoga? (rounded to minutes)
(A) 10 min.
(B) 41 min.
(C) 17 min.
(D) 23 min.
(E) 20 min.
Rationale: The ratio is 2:3 = gym:ride, so (12)(3/2) = 18 minutes at the gym, and 18 + 12 = 30 minutes exercise, so (2/3)(30) = 20 minutes yoga.
Answer: E
Answer: (E) 20 min.
Question: A set consist of 2b-1 element. What is the number of subsets of this set which contain at most b-1 elements?
(A) 2^(2b-2)
(B) 2^(2b) - 2
(C) 2^(2b) -1
(D) 2^(2b)
(E) 2^(2b-1)
Rationale: I used b=3, so then we have
5!/2!3! + 5!/4!1! + 5!/0!5!
10 + 5 + 1 = 16
So our target is 16
Now replace in answer choices
A gives us 2^4 = 16
Hence A is the correct option
Read carefully it says at most so keep in mind that picking a small number such as 3 will help you save time since you have to list fewer outcomes
Avoid 2 since you will get 1 arrangement (b-1) and may be risky since 1 is a number with certain unique properties
Answer: (A) 2^(2b-2)
Question: A man goes from A to B at a speed of 35 kmph and comes back to A at a speed of 28 kmph. Find his average speed for the entire journey?
(A) 38.9 kmph
(B) 38.8 kmph
(C) 38.1 kmph
(D) 37.1 kmph
(E) 37.8 kmph
Rationale: Distance from A and B be 'd'
Average Speed = total distance/total time
Average Speed = (2d)/[(d/35) + (d/28]
= (2d)/[11d/140) => 38.1 kmph.
Answer:C
Answer: (C) 38.1 kmph
Question: If 76 is divided into four parts proportional to 7, 5, 3, 4, then the smallest part is:
(A) 12
(B) 88
(C) 27
(D) 92
(E) 32
Rationale: Explanation:
Given ratio = 7 : 5 : 3 : 4, Sum of ratio terms = 19.
Smallest part = = 12
Answer: A) 12
Answer: (A) 12
Question: J is 25% less than P and 20% less than T. T is w% less than P. What is the value of w?
(A) 93.5
(B) 90
(C) 6.75
(D) 6.25
(E) 2
Rationale: Usually We can solve every question of this type by choosing appropriate value of the Variable and deriving the value of other related variables.
Let, P = 400
then J = (75/100)*400 = 300
also J = (80/100)*T
i.e. T = 300*100/80 = 375
and T = [1-(w/100)]*P
i.e. 100-w = 100*T/P = 100*375/400 = 93.75
i.e. w = 6.25
Answer: Option D
Answer: (D) 6.25
Question: Bob had x number of sweets and wanted to share it with his 6 friends. However, he could not divide x into 7 parts equally. He also found that he could not divide x equally into 6,5,4,3 or 2 parts either. He decided to throw one sweet away and found that he had enough sweets to divide into 7,6,5,4,3 or 2 equal parts. What is the smallest possible number of x?
(A) 421
(B) 422
(C) 423
(D) 420
(E) 541
Rationale: x-1 is lcm of 7,6,5,4,3,2,1 i.e. 420
x-1=420
x=421
ANSWER:A
Answer: (A) 421
Question: Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution P in the liquid in the tank after 3 minutes?
(A) 3/11
(B) 2/11
(C) 1/11
(D) 4/11
(E) 1/11
Rationale: Part filled by (A + B + C) in 3 minutes = 3(1/30 + 1/20 + 1/10) = 11/20
Part filled by A in 3 minutes = 3/30
Required ratio = 3/30 * 20/11 = 2/11
Answer:B
Answer: (B) 2/11
Question: JOHN and ANDREW can finish John work 30 days if they work together. They worked together for 20 days and then ANDREW left. JOHN finished the remaining work in another 20 days. In how many days JOHN alone can finish the work?
(A) 30 days
(B) 60 days
(C) 65 days
(D) 78 days
(E) 80 days
Rationale: Amount of work done by JOHN and ANDREW in 1 day = 1/30
Amount of work done by JOHN and ANDREW in 20 days = 20 × (1/30) = 20/30 = 2/3
Remaining work – 1 – 2/3 = 1/3
JOHN completes 1/3 work in 20 days
Amount of work JOHN can do in 1 day = (1/3)/20 = 1/60
=> JOHN can complete the work in 60 days
B
Answer: (B) 60 days
Question: Two trains travelling in the same direction at 108 and 90 kmph completely pass off another in 1 minute. If the length of the first train is 100 m, what is the length of the second train?
(A) 228
(B) 266
(C) 175
(D) 200
(E) 171
Rationale: RS = 108 â€“ 90 = 18 * 5/18 = 5 mps
T = 60 sec
D = 5 * 60 = 300 m
100
--------
200 m
Answer: D
Answer: (D) 200
Question: Jonathan’s grandparents lend $10,000 to Jonathan at a simple annual interest rate of 10%. He pays them back 5 years later. How much do they receive?
(A) 10,000∗0.1^5
(B) 10,000∗0.5
(C) 10,000∗1.5
(D) 10,000∗1.1^5
(E) (10,000∗0.1)^5
Rationale: Let the principal be $1
So Interest for the $1 for 5 years at 10% Pa is -
1∗5∗10/100 => 0.50
Now Principal is $1 and Interest is $ 0.50 , so Amount to be returned is - $1 + $0.50 => $1.50
So, for $ 10,000 , amount to be returned is 10,000 * 1.50; Answer is (C)
Answer: (C) 10,000∗1.5
Question: If K persons take y days to complete z similar jobs, how long does it take y persons to complete 1 such job?
(A) z
(B) x
(C) x/y
(D) z/x
(E) y/z
Rationale: the Q doesn't seem to be having the any correct answer as a choice..
since we are looking for days, lets keep days in end and work out a solution..
K person can do z jobs in y days..
so K person can do 1 job in y/z days..
and 1 person can do 1 job in y^2/z days..
so y person can complete 1 job in y^2/zy=y/z..
so the answer should be y/z=E
Answer: (E) y/z
Question: What least number mustbe subtracted from 427398 so that remaining no.is divisible by 15
(A) 354665677
(B) 436557678
(C) 457676788
(D) 725117481
(E) 567768788
Rationale: Explanation:
On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted
Answer: Option D
Answer: (D) 725117481
Question: A car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 120 kilometers per hour. At what speed, in kilometers per hour, is the car traveling?
(A) 121.5
(B) 122
(C) 122.5
(D) 113
(E) 112.5
Rationale: E
120 * T = 1 km => T = 1/120 km/h
V*(T+2/3600) = 1
V (1/120 + 2/3600) = 1 => V = 112.5 km/h
Answer: (E) 112.5
Question: The greatest ratio out of 2:3, 5:8, 3:2, 8:9 and 8:1 is?
(A) 2:3
(B) 3:2
(C) 5:8
(D) 8:1
(E) 8:9
Rationale: 2/3=0.666
5/8=0.625
3/2=1.5
8/9=0.888
8/1=8
ANSWER:D
Answer: (D) 8:1
Question: Machine A and machine B are each used to manufacture 880 sprockets. It takes machine A 10 hours longer to produce 880 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machineAproduce?
(A) 6
(B) 6.6
(C) 8.8
(D) 100
(E) 110
Rationale: time taken by B = t
time taken by A = t+10
qty produced by A = q
qty produced by B = 1.1 q
for B: t(1.1q) =880
qt=800
for A: (t+10)(q) = 880
qt+10q=880
800+10q=880
q=8
so A can produce 8/hour.
then B can produce = 8(1.1)=8.8/hour.
C
Answer: (C) 8.8
Question: A wildlife preserve is being planned for 6,000 rhinoceroses. The preserve is to contain a total of 10,000 acres of watering area, plus 100 acres of grazing area for each rhinoceros. If the number of rhinoceroses is expected to increase by 10 percent, how many thousand acres should the preserve have in order to provide for the increased population?
(A)  340
(B)  330
(C)  670
(D)  310
(E)  300
Rationale: Number of rhinos = 6000
Watering area = 10,000 acres
Number of rhino to increase by 10 percent ,then number of rhino = 6600
Grazing area for a rhino = 100
Total grazing area for 6600 rhinos= 6600 * 100 = 6,60,000
Total area required for the wildlife preserve = 6,60,000 + 10,000 = 6,70,000 = 670 * 1000
Answer C
Answer: (C)  670
Question: Donald duck can can swim his pool downstream(with the pool current helping time) in exact 40 seconds and upstream(against the pool current) in a pool in exact 60 seconds. The length of pool is 2 kilometers.
How long Donald duck can cover distance of one side at a still pool (with no current).
(A) 44 minutes
(B) 45 minutes
(C) 43 minutes
(D) 48 minutes
(E) 50 minutes
Rationale: D
48 minutes.
Donald duck's speed = X km/seconds
pool current speed = Y km/seconds
2/(X+Y) = 40
2/(X-Y) = 60
Solving the simultaneous equations gives X = 1/24
Therefore, to cover 2km will take 2/X = 48 seconds
Answer: (D) 48 minutes
Question: A woman named Ranjani started a business investing Rs. 5,000. Roja joined him after six months with an amount of Rs. 5,000 and Sachin joined them with Rs. 10,000 after another six months. The amount of profit earned should be distributed in what ratio among Ranjani, Roja and Sachin respectively, 3 years after A woman started the business ?
(A) 7:6:10
(B) 12:15:16
(C) 42:45:56
(D) 6 : 5 : 8
(E) 6 : 4 : 3
Rationale: Explanation :
Akash : Roja : Sachin = (5,000 x 36) : (5,000 x 30) : (10,000 x 24)
= 36 : 30 : 48 => 6 : 5 : 8
Answer : D
Answer: (D) 6 : 5 : 8
Question: In the quadrilateral PQRS d=8 cm, h1=4.4 cm and
h2=2.2 cm. Find the area
(A) 21
(B) 22
(C) 23
(D) 24
(E) 26.4
Rationale: area of quad.=1/2*any diagonal*(sum of 2 perpendiculars which is drawn on that diagona)
so 1/2*8*(4.4+2.2)=26.4
ANSWER:E
Answer: (E) 26.4
Question: In printing test papers for Excel, Gayatri found that if she used Arial font size 10 instead of Times Roman font size 10 there was a reduction of 18% in the number of pages required for test papers. Further, if she reduced the font size from 10 to 9, the savings were 14% and 10% in Times Roman and Arial fonts respectively. If an test papers printing in Times Roman font size 9 is converted to Arial font size 9, what is the percentage reduction in the number of pages?
(A) 14.1
(B) 14.2
(C) 14.3
(D) 14.4
(E) 14.5
Rationale: Ans.
From TR-10 to A - 10 there is a reduction of 18% in the number of pages. and From TR-10 to TR-9 14% reduction, A - 10 to A - 9 10% reduction.
Assume Test paper in TR- 10 is 100 pages. So in TR- 9 the number of pages are 100X 86% = 86
Now in A-10 total pages are 100 X 82% = 82
Now in A - 9 total pages are 82 X 90% = 73.8
So total change in the number of pages = (86−73.8)/86×100 = 14.1
Answer A
Answer: (A) 14.1
Question: John went on a diet 6 months ago when he weighed 224 pounds. If he now weighs 200 pounds and continues to lose at the same average monthly rate, in approximately how many months will he weigh 180 pounds?
(A)  3
(B)  3.5
(C)  4
(D)  4.5
(E)  5
Rationale: 224 - 200 = 24 pounds lost in 6 months
24/6 = 4, so Joe is losing weight at a rate of 4 pounds per month.
...in approximately how many months will he weigh 180 pounds?
A simple approach is to just list the weights.
Now: 200 lbs
In 1 month: 196 lbs
In 2 months: 192 lbs
In 3 months: 188 lbs
In 4 months: 184 lbs
In 5 months: 180 lbs
Answer: E
Answer: (E)  5
Question: If 4 spiders make 4 webs in 4 days, then 1 spider will make 1 web in how many days?
(A) 1
(B) 3
(C) 5
(D) 4
(E) 6
Rationale: Explanation :
Let the required number days be x.
Less spiders, More days (Indirect Proportion)
Less webs, Less days (Direct Proportion)
Spiders 1:4|
|::4:1
Webs 4:1|
=> 1 * 4 * x = 4 * 1 * 4
=> x = 4
Answer : D
Answer: (D) 4
Question: The ratio of investments of A, B and C is 5:6:7. If they share the profit in the ratio of 5:12:14, then find the ratio of their periods of investments.
(A) 1:2:2
(B) 3:5:6
(C) 3:2:4
(D) 2:4:3
(E) 2:5:3
Rationale: The ratio of investments of A, B and C is 5:6:7.
The ratio of shares of A, B and C is 5:12:14.
Therefore, Ratio of their periods of investments is
(5/5) : (12/6) : (14/7) = 1:2:2
ANSWER:A
Answer: (A) 1:2:2
Question: Which of the following is closest to (9!-8!)/(9!+8!)?
(A) 0.01
(B) 0.8
(C) 0.2
(D) 1
(E) 10
Rationale: (9!-8!)/(9!+8!)
8! (9-1)/8! (9+1)
8/9 = 0.8
B is the answer
Answer: (B) 0.8
Question: If each term in the sum a1+a2+a3+.....+an is either 7 or 77 and the sum equals 364, which of the following could be equal to n?
(A) 38
(B) 39
(C) 40
(D) 41
(E) 42
Rationale: Since, there is no 52 in the answer choices (364/7 = 52), we know there is at least one 77.
364 - 77 = 287
287/7 = 41
41+1 = 42.
If 42 wasn't there, I would have subtracted 77 from 287 and continued in a similar way.
Ans. E
Answer: (E) 42
Question: How many digits are there in the integer 8^7 × 25^12?
(A) 27
(B) 24
(C) 21
(D) 19
(E) 15
Rationale: 8^7 × 25^12
This can be simplified to
2^21 × 5^24 (since 2^3 = 8 and 5^2 = 25)
Now 2 x 5 =10
So, simplifying further
(2^21 × 5^21) x 5^3
= 10^21 x 5^3
= 125 x 10^21
10^21 has 1 one and 21 zeros which when multiplied with 125 gives 125 followed by 21 zeros
So, in total, 24 digits
Correct Option: B
Answer: (B) 24
Question: Tickets numbered from 1 to 21 are mixed and then a ticket is selected randomly. What is the probability that the selected ticket bears a number which is a multiple of 3?
(A) 1/3
(B) 2/5
(C) 3/10
(D) 3/7
(E) 1/7
Rationale: Here, S = [1, 2, 3, 4, …., 19, 20, 21]
Let E = event of getting a multiple of 3 = [3, 6, 9, 12, 15, 18, 21]
P (E) = n (E) / n (S) = 7 / 21 = 1/3
The answer is A.
Answer: (A) 1/3
Question: The average of 11numbers is 60. Out of 11numbers the average of first 6no. is 78, and last 6numbers is 75 then find 6th number?
(A) A)258
(B) B)65
(C) C)58
(D) D)62
(E) E)48
Rationale: 6th number = sum of 1st 6no.s + sum of last 6no.s - sum of 11no.s
answer = 6*78+6*75-11*60 = 258
Answer is A
Answer: (A) A)258
Question: A, B and C invests Rs.4000, Rs.3000 and Rs.2000 in a business. After one year B removed his money; A and C continued the business for one more year. If the net profit after 2 years be Rs.2400, then B's share in the profit is?
(A) 276
(B) 289
(C) 480
(D) 400
(E) 278
Rationale: 4*24 : 3*12 : 2*24
8: 3: 4
3/15 * 2400 =480
Answer: C
Answer: (C) 480