mathewhe/aqua_rat huggingface dataset:
Question: Fox jeans regularly sell for $15 a pair and Pony jeans regularly sell for $18 a pair. During a sale these regular unit prices are discounted at different rates so that a total of $8.55 is saved by purchasing 5 pairs of jeans: 3 pairs of Fox jeans and 2 pairs of Pony jeans. If the sum of the two discount rates is 22 percent, what is the discount rate on Pony jeans?
(A) 7%
(B) 10%
(C) 11%
(D) 12%
(E) 15%
Rationale: Let x be the discount on Pony jeans. Then 0.22-x is the discount on Fox jeans.
3(0.22-x)(15) + 2x(18) = 8.55
9.9 - 45x + 36x = 8.55
9x = 1.35
x = 0.15
The answer is E.
Answer: (E) 15%
Question: (786×74) ÷ ? = 4847
(A) A)4.8
(B) B)12
(C) C)58
(D) D)68
(E) E)48
Rationale: Explanation:
58164/x = 4847 => x = 58164/4847 = 12
Answer: Option B
Answer: (B) B)12
Question: How many odd numbers between 10 and 2000 are the squares of integers?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20
Rationale: The numbers are the squares of 5, 7, 9,...,43 which includes 20 numbers.
The answer is E.
Answer: (E) 20
Question: If x^2−2ax+a^2=0 get value of x/a
(A) 1
(B) 2
(C) 3
(D) 4
(E) 7
Rationale: a is obviously non-zero. We divide the equation by
a^2
we get (x/a-1)^2=0
x/a=1
ANSWER A 1
Answer: (A) 1
Question: If 75 percent of a class answered the first question on a certain test correctly, 70 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percent answered both correctly?
(A)  10%
(B) (8) 20%
(C)  30%
(D)  50%
(E)  65%
Rationale: {Total} = {First} + {Second} - {Both} + {Neither}
100% = 75% + 70% - {Both} + 20% --> {Both} = 65%.
Answer: E.
Answer: (E)  65%
Question: If (a-5)(b-2) =0, which of the following must be true ?
I. a=5 and b=2
II. if a is not 5, then b=2
III. if a=5, then b is not=2
(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
Rationale: (a-5)(b-2)=0
So either a=5 or b=2 or both. It is 'or' not 'and', so it could be any of the three possibilities. Thus I and III are not necessarily true.
The answer is B.
Answer: (B) II only
Question: If the compound interest on a certain sum of money for 5 years at 10% per annum be Rs. 993, what would be the simple interest?
(A) Rs. 1500
(B) Rs. 890
(C) Rs. 895
(D) Rs. 900
(E) None
Rationale: Let P = Principal
A - Amount
We have a = P(1 + R/100)3 and CI = A - P
ATQ 993 = P(1 + R/100)3 - P
? P = 3000/-
Now SI @ 10% on 3000/- for 5 yrs = (3000 x 10 x 5)/100
= 1500/-
Answer: A.
Answer: (A) Rs. 1500
Question: 10 camels cost as much as 24 horses, 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants. If the cost of 10 elephants is Rs.150000, find the cost of a camel?
(A) s.9800
(B) s.3800
(C) s.9800
(D) s.6000
(E) s.6880
Rationale: Cost of the camel = P
10 camels = 24 horses
16 horses = 4 oxen
6 oxen = 4 elephants
10 elephants = Rs.150000
P = Rs.[(24 * 4 * 4 * 150000)/(10 * 16 * 6 * 10)]
P = Rs.(57600000/9600) => P = Rs.6000
Answer:D
Answer: (D) s.6000
Question: √{(6.3)(10^9)} is closest to which of the following?
(A) 1,500
(B) 6,000
(C) 25,000
(D) 80,000
(E) 150,000
Rationale: √{(6.3)(10^9)} = √{(63)(10^8)} = √(63)√(10^8) = (√63) * 10,000 which is almost 80,000.
The answer is D.
Answer: (D) 80,000
Question: In an examination, there were 2,000 candidates, out of which 900 candidates were girls and rest were Boys. If 30% of the boys and 32% of the girls passed, then the total percentage of failed candidates is?
(A) 35.67%
(B) 64.75%
(C) 68.57%
(D) 69.10%
(E) none of these
Rationale: Girls = 900, Boys = 1100
Passed = ( 30% of 1100 ) + ( 32% of 900)
= 330 + 288 = 618
Failed = 2000 - 618 = 1382
Failed % = [(1382/2000) x 100]%
= 69.1%.
ANSWER:D
Answer: (D) 69.10%
Question: If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum Q of the 3 greatest of these temperatures, in degrees Fahrenheit, could be
(A)  6x
(B)  4x
(C)  5x/3
(D)  3x/2
(E)  3x/5
Rationale: Note that we have 5positivetemperatures.
Next, as the average is x then the sum of the temperatures is 5x and as all the temperatures are positive then the sum of the 3 greatest must be more than (or equal to) 3x (as the average of the 3 greatest must be at least x) and less than 5x: 3x<Q<5x --> only option B fits.
Answer: B.
Answer: (B)  4x
Question: There are 79 people that own pets. 15 people own only dogs, 10 people own only cats, 5 people own only cats and dogs, 3 people own cats, dogs and snakes. How many total snakes are there?
(A) 2
(B) 4
(C) 8
(D) 49
(E) 32
Rationale: lets assign variables to all the areas in Venn Diagram of Three.
three different Units are DOG, Cat, Snake= Total = 79
Only Dog= d=15
Only Cat = c=10
Only Snake = s
Exactly Dog and Cat= 5
Exactly Dog and Snake = x
Exactly Cat and Snake = y
all three= 3
so 79=15+10+5+3+x+y+s
we need to know total snakes= x+y+s+3 = 49
answer:D
Answer: (D) 49
Question: A is 20% more efficient than B .how much time they will working together take to complete a job which A alone could have done in 15 days?
(A) 5 6/11 days
(B) 8 2/11 days
(C) 6 4/8 days
(D) 4 days
(E) 17 days
Rationale: The ratio of times taken by A and B=100:120=5:6
suppose B can do work in x days
then 5:6::15:x
x=6*15/5
x=18
A's 1 day's work=1/15
B's 1 day's work=1/18
(A+B)'s 1 day's work =1/15+1/18
A and B together can do work in 8 2/11 days
Answer(B)
Answer: (B) 8 2/11 days
Question: If the average (arithmetic mean) of a and b is 45 and the average of b and c is 60, what is the value of c âˆ’ a?
(A) 25
(B) 30
(C) 90
(D) 140
(E) It cannot be determined from the information given
Rationale: the arithmetic mean of a and b =(a+b)/2=45 -- a+b=90 --1
similarly for b+c=120 --2
subtracting 1 from 2 we have
c-a=30;
ANSWER:B
Answer: (B) 30
Question: Two members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of the 2 members, how many members does the club have?
(A) 20
(B) 27
(C) 40
(D) 57
(E) 95
Rationale: Below is an answer to your question.
C2n=190Cn2=190 --> n!/2!∗(n−2)!=190. Now, notice that n!=(n−2)!∗(n−1)∗n so (n−2)!∗(n−1)∗n2!∗(n−2)!=190 --> (n−1)∗n/2=190 (n−1)n=380 --> n=20
Answer: A.
Answer: (A) 20
Question: In how many ways 3 boys and 3 girls can be seated in a row so that they are alternate.
(A) 144
(B) 288
(C) 36
(D) 256
(E) None
Rationale: Solution: Let the Arrangement be,
B G B G B G B
34 boys can be seated in 3! Ways.
Girl can be seated in 3! Ways.
Required number of ways,
= 3!*3! = 36.
Answer: Option C
Answer: (C) 36
Question: A car driver travels from the plains to the hill station, which are 200 km apart at an average speed of 55km/hr. in return trip, he covers the same distance at an average speed of 20km/hr. the average speed of the car over the entire distance of 400 km is
(A) 8.95km/hr
(B) 26.7km/hr
(C) 27.8km/hr
(D) 29.3km/hr
(E) 8.45km/hr
Rationale: average speed = (2*55*20)/55+20 km/hr
=29.3 km/hr
Answer (D)
Answer: (D) 29.3km/hr
Question: Which of the following expressions CANNOT have a negative value
(A) |a + b| – |a – b|
(B) |a + b| – |a|
(C) a^2 + b^2
(D) |2a + b| – |a + b|
(E) |a^3 + b^3| – a – b
Rationale: Whenever you are required to find if the equation is a perfect square or not, try to bring everything in the form of a perfect square as they are always positive.
Of the given options, only option D can be written in from of a perfect square.
a^2+b^2
This will always be positive.
Correct Option: C
Answer: (C) a^2 + b^2
Question: (2^2-1)(2^2+1)(2^4+1)(2^10+1)=
(A) 2^18 - 1
(B) 2^16 + 1
(C) 2^32 - 1
(D) 2^128 - 1
(E) 2^16(2^16-1)
Rationale: A: is correct
(2^2-1)* (2^2+1)=(2^4-1)
(2^4-1)*(2^4+1)=2^8-1
(2^8-1)*(2^10 +1)=2^18-1
You can apply to this formula: a^2- b^2=(a-b)*(a+b)
Answer: (A) 2^18 - 1
Question: A committee of three people is to be chosen from 6 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
(A) 16
(B) 24
(C) 26
(D) 30
(E) 160
Rationale: One of the approaches:
Each couple can send only onerepresentativeto the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) out of 6 to send only onerepresentativesto the committee: 6C3=20.
But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.
Total # of ways: 6C3*2^3=160.
Answer: E.
Answer: (E) 160
Question: Ramesh can finish a piece of work in 12 days. Rohan is twice as efficient as Ramesh.If they work together how many days will they need to finish the same amount of work?
(A) 8/3 days
(B) 4 days
(C) 5 days
(D) 11/3 days
(E) 7/3 days
Rationale: Ramesh can finish a piece of work in 12 days. Rohan is twice as efficient as Ramesh.
So Rohan can finish a piece of work in 6 days.
days needed to finish the same amount of work = 1/(1/6+1/12)=4 days
ANSWER:B
Answer: (B) 4 days
Question: An article is bought for Rs.600 and sold for Rs.500, find the loss percent?
(A) 16 2/6%
(B) 16 8/3%
(C) 16 7/3%
(D) 16 2/3%
(E) 12 2/3%
Rationale: 600 ---- 100
100 ---- ? => 16 2/3%
Answer:D
Answer: (D) 16 2/3%
Question: he population of a city is 130,000. If it increases at the rate of 6% per annum, then what will be its population 2 years hence?
(A) 146069
(B) 146067
(C) 156068
(D) 146068
(E) 179718
Rationale: Population after n years = P [1 + (R/100)]2
Population after 2 years = 130000 * [1 + 6/100]2 = (130000 x 53/50 x 53/50) = 146068
ANSWER: D
Answer: (D) 146068
Question: In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
(A) 1/10
(B) 2/5
(C) 2/7
(D) 5/7
(E) None
Rationale: Solution
P(getting a prize)	= 10 / (10+25)
‹=› 10 / 35
‹=› 2 / 7.
Answer C
Answer: (C) 2/7
Question: The ages of two persons differ by 20 years. If 5 years ago, the elder one be 5 times as old as the younger one, their present ages (in years) are respectively
(A) 11 years
(B) 77 years
(C) 88 years
(D) 10 years
(E) 66 years
Rationale: Let their ages be x and (x + 20) years.
5 (x - 5) = (x + 20 - 5) or 4x = 40 or x = 10.
Their present ages are 30 years and 10 years
Answer: D
Answer: (D) 10 years
Question: A profit of Rs. 600 is divided between X and Y in the ratio of 1/2:1/3 . What is the difference between their profit shares ?
(A) s. 220
(B) s. 120
(C) s. 320
(D) s. 50
(E) s. 90
Rationale: A profit of Rs. 600 is divided between X and Y in the ratio of 1/2 : 1/3 or 3:2.
so profits are 360 and 240.
difference in profit share = 360-240=120
ANSWER:B
Answer: (B) s. 120
Question: The average age of 15 students of a class is 15 years. Out of these, the average age of 4 students is 14 years and that of the other 10 students is 16 years. The age of the 15th student is
(A) 9 years
(B) 11 years
(C) 14 years
(D) 21 years
(E) 9 years
Rationale: Solution
Age of the 15th student	=[15x15-(14x4+16x10)]
= (225 - 216)
= 9 years.
Answer E
Answer: (E) 9 years
Question: If x and y are positive real numbers such that x + y = 1, which of the following could be the value of 100x + 700y?
I. 90
II. 250
III. 699
(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III
Rationale: We know that x and y are both between 0 and 1.
The minimum limit of 100x + 700y is 100, when x is close to 1 and y is close to 0.
The maximum limit of 100x + 700y is 700, when x is close to 0 and y is close to 1.
The expression can take on any value between 100 and 700.
The answer is E.
Answer: (E) II and III
Question: In the first 8 overs of a cricket game, the run rate was only 2.3 What should be the rate in the remaining 20 overs to reach the target of 260 runs?
(A) 12.08
(B) 12.18
(C) 12.0
(D) 6.19
(E) 6.21
Rationale: Required run rate
= [260- (2.3 * 8)]/20
= 241.60/20
= 12.08
Answer:A
Answer: (A) 12.08
Question: After an ice began to melt out from the freezer, in the first hour lost 3/4, in the second hour lost 3/4 of its remaining. If after two hours, the volume is 0.2 cubic inches, what is the original volume of the cubic ice, in cubic inches?
(A) 2.5
(B) 3.2
(C) 4.0
(D) 6.5
(E) 8.0
Rationale: Let initial volume of ice be = x
Ice Remaining after 1 hour = x - 0.75x = 0.25x
Ice Remaining after 2 hour = (1/4)x - (3/4 * 1/4 * x) = (1/16)x
(1/16)x = 0.2
x = 3.2
Alternate solution: Try to backsolve.
Initial volume = 3.2
After one hour --> (1/4)3.2 = 0.8
After two hours --> (1/4)0.8 = 0.2
Answer: B
Answer: (B) 3.2
Question: 5, 25, 61, 113, ?
(A) 182
(B) 337
(C) 278
(D) 266
(E) 181
Rationale: ANSWER: A
Answer: (A) 182
Question: In a box, there are 10 red, 6 blue and 5 green balls. One ball is picked up randomly. What is the probability that it is neither red nor blue?
(A) 4/51
(B) 10/21
(C) 7/12
(D) 6/21
(E) 13/17
Rationale: Total number of balls = 10+6+5 = 21
E = event that the ball drawn is neither red nor blue
= event that the ball drawn is blue
n(E) = 6
P(E) = 6/21
Answer is D
Answer: (D) 6/21
Question: A man has Rs. 192 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
(A) 37
(B) 36
(C) 22
(D) 90
(E) 28
Rationale: Let number of notes of each denomination be x.
Then, x + 5x + 10x = 192 16x = 192 x = 12.
Hence, total number of notes = 3x = 36.
Answer:B
Answer: (B) 36
Question: If a, b, q are integers, 33 = abq, and a < b < q, what is the smallest possible value of a?
(A) -33
(B) -3
(C) 1
(D) 3
(E) 11
Rationale: Given : a * b * q = 33.
33 is a Positive Odd number.Side Note:Thus, all 3 numbers will be odd.
Prime factorization will also give you numbers. 1, 3, 11. The 3 numbers will be combination of these 3 numbers with possible variation in sign(+/-).
We could have any 2 digits as negative. --> negative * negative * positive. We have negative values in answer option, so eliminate C, D and E from answer choices.
-33 * -1 * 1 = 33.
Correct answer is A
Answer: (A) -33
Question: Six points lie on the circumference of a ellipse. What is the positive difference between the number of triangles and the number of quadrilaterals that can be created by connecting these points?
(A) 4
(B) 6
(C) 5
(D) 15
(E) 20
Rationale: Number of triangles can be formed out of 6 points= 6C3=20
Number of quadrilaterals can be formed out of 6 points= 6C4=15
20-15=5.
Answer=C
Answer: (C) 5
Question: What is the value of: 7^9 - 7^11 ?
(A)  -7^20
(B)  7^-3
(C)  -49^20
(D)  -48(7^9)
(E)  50(7^9)
Rationale: 7^9 - 7^11 = 7^9 (1 - 7^2) = 7^9 (1 - 49) = -48 (7^9), Answer D.
Answer: (D)  -48(7^9)
Question: Two letters from the word FARTHING are selected at random. What is the probability that at least one of the selected letters is a consonant?
Two letters from the word FARTHING are selected at random. What is the probability that at least one of the selected letters is a consonant?
(A) A)1/500
(B) B)1/28
(C) C)27/28
(D) D)124/125
(E) E)499/500
Rationale: There are 6 consonants and 2 vowels in FARTHING.
probability that at least one of the selected letters is a consonant = 1 - (Probability of both vowels)
Probability of both vowels = 2C2/8C2 = 1/28
So, Answer is 1 - 1/28 = 27/28 ------> C.
Answer: (C) C)27/28
Question: A train running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train.
(A) 150 m
(B) 155 m
(C) 180 m
(D) 190 m
(E) 205 m
Rationale: Answer: Option A
Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (distance) = speed * Time
(50/3) * 9 = 150 meter
Answer: (A) 150 m
Question: To reach her destination, Jeanette must drive 90 miles. If she drives 6 miles every 7 minutes, how much time will it take her to reach her destination?
(A) 2 hours and 2 minutes
(B) hours and 45 minutes
(C) 2 hours and 10 minutes
(D) 2 hours and 12 minutes
(E) 2 hours and 15 minutes
Rationale: 7 minute for 6 miles how much for 90? minutes for 1(7/5) mile multiplied by 90 miles.
7*90/6= 105> 1 hours 45 min answer is B
Answer: (B) hours and 45 minutes
Question: When the number 1y92355 is exactly divisible by 11, then the smallest whole number in place of y is?
(A) 1
(B) 3
(C) 7
(D) 9
(E) 0
Rationale: The given number =1y92355
Sum of the odd places =5+3+9+1=18
Sum of the even places = 5+2+y
(Sum of the odd places)- (Sum of even places) = Number (exactly divisible by 11)
18-(7+y) = divisible by 11
11 � y = divisible by 11.
Y must be 0, to make given number divisible by 11.
E
Answer: (E) 0
Question: Sonika deposited Rs.8000 which amounted to Rs.9200 after 3 years at simple interest. Had the interest been 2% more. She would get how much?
(A) 9680
(B) 9687
(C) 96809
(D) 9683
(E) 9682
Rationale: (8000*3*2)/100 = 480
9200
--------
9680
Answer: A
Answer: (A) 9680
Question: In the faculty of Reverse-Engineering, 226 second year students study numeric methods, 423 second year students study automatic control of airborne vehicles and 134 second year students study them both. How many students are there in the faculty if the second year students are approximately 70% of the total?
(A) 515.
(B) 545.
(C) 618.
(D) 644.
(E) 736.
Rationale: Answer is E :736
Solution : Total number of students studying both are 423 + 226 - 134 = 515 (Subtracting the 134 since they were included in the both the other numbers already).
So 70% of total is 515, so 100% is approx. 736.
Answer: (E) 736.
Question: Eighty percent of the lights at Hotel California are on at 8 p.m. a certain evening. However, forty percent of the lights that are supposed to be off are actually on and ten percent of the lights that are supposed to be on are actually off. What percent of the lights that are on are supposed to be off?
(A) 22(2/9)%
(B) 16(2/3)%
(C) 11(1/9)%
(D) 10%
(E) 5%
Rationale: a= number of lights supposed to be on
b= number of lights supposed to be off
total = a+b
no of lights that are on = 0.8(a+yb
Light that are supposed to be on but are actually off = 10% of a
Light that are supposed to be on are on = 90% of a
Light that aare supposed to be but are on = 40% of b
Light that aare supposed to be off are off = 60% of b
0.9a + 0.4b = 0.8 (a+b)
a = 4b
Suppose total Lights = 100
so, a = 80 and b = 20
No of lights that are on are on but are supposed to be off = 40% of b = 0.4 (20) = 5
So, the % of lights that are on are supposed to be off = 5/100 = 5%
ANSWER:E
Answer: (E) 5%
Question: If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
Rationale: Thus proceeding with the "300 + k" assumption, we get:
P = 30!
Option A: 310
=> 310 = 2 * 5 * 31
=> 31 is a prime and is not present in 30!
Thus 310 is not a factor of P
=> A is Eliminated
Option B: 312
=> 312 = 2^3 * 3 * 13
Thus 312 is a factor of P
=> B is the correct answer !
Option C: 314
=> 314 = 2 * 157
=> 157 is a prime and is not present in 30!
Thus 314 is not a factor of P
=> C is Eliminated
Option D: 316
=> 316 = 2^2 * 79
=> 79 is a prime and is not present in 30!
Thus 316 is not a factor of P
=> D is Eliminated
Option E: 318
=> 310 = 2 * 3 * 53
=> 53 is a prime and is not present in 30!
Thus 318 is not a factor of P
=> E is Eliminated
ANS: B
Answer: (B) 12
Question: A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?
(A) 111.12
(B) 888.55
(C) 200.05
(D) 400.04
(E) 250.5
Rationale: A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 800)/900 = 8000/9 = 888.88 m.
B can give C = 1000 - 888.88 = 111.12 m.
ANSWER A
Answer: (A) 111.12
Question: A train 2500 m long can cross an electric pole in 50 sec and then find the speed of the train?
(A) 178
(B) 180
(C) 182
(D) 184
(E) 186
Rationale: Length = Speed * time
Speed = L/T
S = 2500/50
S = 50 M/Sec
Speed= 50*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 180 Kmph
Answer:B
Answer: (B) 180
Question: How many kilograms 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) 60
(B) 50
(C) 55
(D) 63
(E) 70
Rationale: Selling price of 1kg = 924
profit = 10%
cost = 100/(100+profit%)*SP
100/(100+10)*9.24= 8.4
Mean price = (8.4-7=1.4) (9-8.4=0.6)
=>1.4:0.6=14:6 = 7:3
3x=27*7=63
ANSWER D
Answer: (D) 63
Question: A train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 40 seconds. What is the length of the train in meters?
(A) 1777 meters
(B) 1822 meters
(C) 400 meters
(D) 1111 meters
(E) None of these
Rationale: Explanation :
When a train overtakes another object such as a motorbike, whose length is insignificant compared to the length of the train, the distance traveled by the train is equal to the length of the train.
Because the motorbike is also moving, we have to take the relative speed between the train and the motorbike and not just the speed of the train.
The length of the train = distance traveled by the train while overtaking the motorbike = relative speed between the train and the motorbike * time taken
In this case, as both the objects i.e., the train and the motorbike are moving in the same direction, the relative speed between them = difference between their respective speeds = 100 - 64 = 36 kmph.
Distance traveled by the train while overtaking the motorbike = 36 kmph * 40 seconds.
The final answer is in meters and the speed is given in kmph and the time in seconds.
So let us convert the given speed from kmph to m/sec.
1 kmph = 518518 m/sec
Therefore, 36 kmph = 36 * 518518 = 10 m/sec.
Relative speed = 10 m/sec. Time taken = 40 seconds.
Therefore, distance traveled = 10 * 40 = 400 meters.
Answer: (B) 1822 meters
Question: A man can row 9 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and back. What is the total distance traveled by the man?
(A) 6.24 km
(B) 6 km
(C) 8.84 km
(D) 5.66 km
(E) 10 km
Rationale: M = 9
S = 1.2
DS = 10.2
US = 7.8
x/10.2 + x/7.8 = 1
x = 4.42
D = 4.42 * 2 = 8.84
ANSWER:C
Answer: (C) 8.84 km
Question: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 3125 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
(A) 1631
(B) 1531
(C) 1533
(D) 1535
(E) 1536
Rationale: Trees increase by 1/4 the number of trees in preceding year. Hence, correct answer must be divisible by 4. Based on divisibility rules, if last 2 digits are divisible by 4 then the number is divisible by 4. Thus, we can eliminate A, B, C,D The answer to be E
Again, trees increase by 1/4 the number of trees in preceding year. Hence, the number of trees increase by 5/4 times the number of trees the preceding year.
If x = initial number of trees = 1536
Year 1 = 5/4x
Year 2 = (5/4)(5/4)x
Year 3 = (5/4)(5/4)(5/4)x
Year 4 = (5/4)(5/4)(5/4)(5/4)x
Only for Answer D: (5/4)(5/4)(5/4)(5/4)1536 = 3750
Hence, correct answer = E
Answer: (E) 1536
Question: 32% of 5/8 =
(A) 0.2
(B) 0.5
(C) 0.6
(D) 0.75
(E) 1.0
Rationale: Should be simple.
0.32*5/8 = 1.6/8 = 0.2
Correct Option : A
Answer: (A) 0.2
Question: The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
(A) 17 kg
(B) 20 kg
(C) 26 kg
(D) 31 kg
(E) 32 kg
Rationale: Explanation:
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
B's weight = 31 kg.
ANSWER IS D
Answer: (D) 31 kg
Question: A piece of work can finish by a certain number of men in 100 days. If however, there were 10 men less,
it would take 10 days more for the work to be finished. How many men were there originally?
(A) 75
(B) 82
(C) 100
(D) 110
(E) 120
Rationale: Expl: Originally let there be x men.
Less men, more days (Indirect)
:. (x-10): x : : 100 :110 or x-10/x = 100/110
or 11x-110 = 10x or x =110
So, originally there were 110 men.
Answer: D
Answer: (D) 110
Question: There are 24 students in Chad’s class. He brought 29 cookies to pass out
for his birthday. How many cookies will each student get? Will there be
any cookies left over?
(A) 1-9
(B) 1-7
(C) 2-8
(D) 1-5
(E) 3-10
Rationale: 29/24 = 1 R 5 Chad will give each student 1 cookie and there will be 5
cookies left over
correct answer D
Answer: (D) 1-5