mathewhe/aqua_rat huggingface dataset:
Question: Maxwell leaves his home and walks toward Brad's house at the same time that Brad leaves his home and runs toward Maxwell's house. If the distance between their homes is 72 kilometers, Maxwell's walking speed is 6 km/h, and Brad's running speed is 12 km/h. What is the distance traveled by Maxwell when they meet in the middle?
(A) 14
(B) 24
(C) 16
(D) 18
(E) 20
Rationale: Consider Max starts from Point A and Brad Starts from Point B and move towards each other.
Assume they shall meet at point O after Time 'T'. The question asks us to find OA.
From the question stem we can make out :- Distance OA = 50km - Distance OB
=> 6xT = 72 - 12xT (i.e distance =Speed x Time)
=> 18T = 72
Hence T = 4
OA = 6 x 4 = 24 Km
Answer : B
Answer: (B) 24
Question: How many prime numbers between 1 and 100 are factors of 67830?
(A) 4
(B) 5
(C) 6
(D) 3
(E) 2
Rationale: factor of 67830 = 2*3*5*7*17*19--- 6 prime numbers
C
Answer: (C) 6
Question: What is the sum of 20 consecutive integers from -9 inclusive, in a increasing order?
(A) -9
(B) 9
(C) -10
(D) 10
(E) 20
Rationale: From -9 to -1 --> 9 nos.
zero --> 1 number
from +1 to +9 --> 9 nos.
When we add up nos. from -9 to +9 sum will be zero. Total 19 nos will be added.
20th number will be 10.
Sum of these 20 nos. = 10.
D is the answer.
Answer: (D) 10
Question: The area of a triangle is with base 4m and height 5m?
(A) 15
(B) 19
(C) 56
(D) 10
(E) 36
Rationale: Explanation:
Area=1/2*b*h
1/2 * 4 * 5 = 10 m^2
Answer : D
Answer: (D) 10
Question: Car A runs at the speed of 65km/hr & reaches its destination in 8hr. Car B runs at the speed of 70 km/h & reaches its destination in 4h. What is the respective ratio of distances covered by Car A & Car B?
(A) 3 : 7
(B) 4 : 9
(C) 13 : 7
(D) 5 : 7
(E) 6 : 11
Rationale: Sol. Distance travelled by Car A = 65 × 8 = 520 km
Distance travelled by Car B = 70 × 4 = 280 km
Ratio = 520/280 = 13 : 7
C
Answer: (C) 13 : 7
Question: There are 8 periods in each working day of a school. In how many ways can one organize 7 subjects such that each subject is allowed at least one period?
(A) 112456
(B) 145678
(C) 141120
(D) 11456
(E) None of these
Rationale: 7 subjects can be arranged in 8 periods in 8P7 ways.
Remaining 1 period can be arranged in 7P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (8P7 x 7P1)/2! = 141120
ANSWER:C
Answer: (C) 141120
Question: Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.1600?
(A) 3000
(B) 4200
(C) 2025
(D) 1205
(E) 3251
Rationale: 1800* 9/8 * 9/8
= 2025
Answer: C
Answer: (C) 2025
Question: (1000)7÷(10)17=?
(A) 10
(B) 100
(C) 1000
(D) 10000
(E) None of these
Rationale: Explanation:
=(103)7/(10)17
=(10)21/(10)17=10(4)=10000
Option D
Answer: (D) 10000
Question: If u and v are both negative and uv < v^2, which of the following must be true?
(A) u < v < u^2 < v^2
(B) u < v < v^2 < u^2
(C) v < u < u^2 < v^2
(D) u^2 < v^2 < v < u
(E) v^2 < u^2 < v< u
Rationale: Because uv < v^2 and both are negative, I thought u < v. So I crossed off answers c), d), and e).
And because u< v , u^2 < v^2.
ans C
Answer: (C) v < u < u^2 < v^2
Question: The greatest ratio out of 2:3, 5:8, 3:1, 8:9 and 8:5 is?
(A) 2:3
(B) 3:1
(C) 5:8
(D) 8:5
(E) 8:9
Rationale: 2/3=0.666
5/8=0.625
3/1=3
8/9=0.888
8/5=1.6
ANSWER:B
Answer: (B) 3:1
Question: A fruit shake costs R 2,00 more than a chocolate milk shake. If 3 fruit shakes and 5 chocolate milk shakes cost R 78,00, determine the individual prices.
(A) x=6,y=9
(B) x=4,y=4
(C) x=7,y=5
(D) x=5,y=11
(E) x=9,y=11
Rationale: Let the price of a chocolate milkshake be x and let the price of a fruitshake be y.
Set up a system of equations
y=x+2.........(1)
3y+5x=78............(2)
Substitute equation (1) into equation (2)
3(x+2)+5x=78
Rearrange and solve for x
3x+6+5x=78
8x=72
x=9
Substitute the value of x back into equation (1) and solve for y
y=x+2
=9+2
=11
One chocolate milkshake costs R 9,00 and one fruitshake costs R 11,00.
Answer is E.
Answer: (E) x=9,y=11
Question: How many two-digit numbers yield a remainder of 1 when divided by both 4 and 17?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Rationale: Easier to start with numbers that are of the form 17p+1 ---> 18,35,52,69,86.
Out of these, there is only one (69) is also of the form 4q+1. Thus 1 is the answer. B is the correct answer.
Answer: (B) 1
Question: In 2 bags, there are to be put together 5 red and 12 white balls, neither bag being empty. How must the balls be divided so as to give a person who draws 1 ball from either bag- the least chance of drawing a red ball ?
(A) 1/32
(B) 2/32
(C) 3/32
(D) 4/32
(E) 5/32
Rationale: Least Chance.
P= 1/2 * 5/16 + 1/2 * 0 = 5/32
ANSWER:E
Answer: (E) 5/32
Question: A man sells two articles for Rs.3600 each and he gains 35% on the first and loses 35% on the next. Find his total gain or loss?
(A) 9% loss
(B) 400
(C) 12.25%loss
(D) 324
(E) 8% LOSS
Rationale: (35*35)/100 = 12.25%loss
ANSWER:C
Answer: (C) 12.25%loss
Question: Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?
(A) Rs. 8600
(B) Rs. 8620
(C) Rs. 8820
(D) Rs. 8000
(E) None of these
Rationale: Explanation:
Amount
= Rs.[8000x(1+5/100)2]
=Rs.(8000x21/20x21/20)
=Rs.8820.
ANSWER IS C
Answer: (C) Rs. 8820
Question: There are two box,one containing 39 red balls and the other containing 26 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is
(A) .60
(B) .50
(C) .70
(D) .80
(E) .90
Rationale: Here we are allowed to transfer the balls between the two boxes, so we keep only one red ball in first box and move all the remaining balls to the second box, so that we get maximum probability of drawing red ball from first box
So first box--- only one red ball
second box--- 38 red and 26 green
Probability of choosing any box is 1/ 2.
probability of choosing the red ball from first box is 1
probability of choosing the red ball from second box is 38/64
So probability of taking one red ball = [(1/2)×(1)]+[(1/2)*(38/64)]≃0.8
ANSWER:D
Answer: (D) .80
Question: If a car went the first third of the distance at 80 kmh, the second third at 30 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?
(A) 36 kmh
(B) 40 kmh
(C) 42 kmh
(D) 45 kmh
(E) 50 kmh
Rationale: Assume D/3 = 240 (this number is convenient because it is divisible by 80, 30 and 48)
So:
240 = 80*T1 = 3 hrs
240 = 30*T2 = 8 hrs
240 = 48*T3 = 5 hrs
T = T1 + T2 + T3 = 16 hrs
D = RT
(240*3) = R*16
R = 45
ANSWER: D
Answer: (D) 45 kmh
Question: Calculate the speed of a stream if a boat going upstream at
4km/hr and downstream at 8km/hr
(A) 1 kmph
(B) 2 kmph
(C) 3 kmph
(D) 6 kmph
(E) 8 kmph
Rationale: Sol.
Speed of stream = 1/2(8 -4) kmph = 2 kmph.
Answer B
Answer: (B) 2 kmph
Question: A batch of cookies was divided among three tins: 2/3 of all the cookies were placed in either the blue tin or the green tin, and the rest were placed in the red tin. If 1/4 of all the cookies were placed in the blue tin, what fraction R of the cookies that were placed in the other tins were placed in the green tin?
(A) 15/2
(B) 9/4
(C) 5/9
(D) 7/5
(E) 9/7
Rationale: Blue tin or red tin: 2/3 (N)
Red tin: (1/3)N
Blue tin: (1/4)N
What the last statment meant, is it wants this fraction: (# of cookies in green tin)/(# of cookies in red and green tin)
# of cookies in green tin = 2N/3 - N/4 = 8N- 3N/12 = 5N/12
# of cookies in red and green tin = N/3 + 5N/12 = 9N/12
Fraction R= 5N/12 * 12/9N = 5/9 (C)
Answer: (C) 5/9
Question: Which of the following is closest to the difference between sum Q of all proper fractions (fractions less than 1) in the form 1/x , where x is a positive digit, and the product of all proper fractions in the form y/(y+1), where y is a positive digit?
(A) 2.82
(B) 2.72
(C) 1.82
(D) 1.72
(E) 0.82
Rationale: Sum Q of all proper fractions (fractions less than 1) in the form 1/x, where x is a positive digit:
1/1 + 1/2 + 1/3 +.....+ 1/9
This is a harmonic progression. Harmonic progression is inverse of arithmetic progression.
Approximate sum of a harmonic progression with even number of terms = number of terms * (average of middle 2 terms)
Approximate sum of a harmonic progression with odd number of terms = number of terms * (middle term)
The actual sum will be slightly more than the approximation.
Here we have 9 terms (odd).
Sum = 9 * 1/5
= 9/5
Product of all proper fractions in the form y/(y+1), where y is a positive digit:
1/2 * 2/3 * 3/4 *.....*9/10
We will be left with 1/10.
Required = 9/5 - 1/10
= 1.8 - 0.1
= 1.7
Closest is 1.72
Answer (D).
Answer: (D) 1.72
Question: A can run a kilometer race in 4 1/2 min while B can run same race in 5 min. How many meters start can A give B in a kilometer race, so that the race mat end in a dead heat?
(A) 400 m
(B) 500 m
(C) 250 m
(D) 100 m
(E) 150m
Rationale: Explanation:
A can give B (5 min - 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.
ANSWER D
Answer: (D) 100 m
Question: If x > 3000, then the value of (3x)/(2 x^1.21 - 1111^1.3) is closest to?
(A) 1/6
(B) 1/3
(C) 10/21
(D) 4/10
(E) 3/2
Rationale: assume x = 3002
(3x)/(2 x^1.21 - 1111^1.3) = 9006 / (2 * (3002)^1.21 - 1111^1.3)
=9006 / 23154
= = 4 /10
D
Answer: (D) 4/10
Question: Q. How many 9-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6, 6, 7?
(A) 180
(B) 140
(C) 210
(D) 270
(E) 300
Rationale: the number is less than 3000000 so we will have the digit on the left most side to be either 1 or 2
i.e your number can be 1 _ _ _ _ _ _ or 2 _ _ _ _ _ _
case 1 > 1 _ _ _ _ _ _ here the units digit can be either 2 or 6
when the units digit is 2 i.e 1 _ _ _ _ _ 2
number of ways this can be done would be 5! / 2! (as 5 is repeated twice) = 60
when the units digit is 6 i.e. 1 _ _ _ _ _ 6, number of ways would be 5!/(2! * 2!) {both 2 and 5 repeat twice} = 30
case 2 > 2 _ _ _ _ _ _ (here units digit may be 2 or 6)
number of ways this can be done would be 5! / (2!) for 2 =60
and 5! / 2! for 6 ... = 60
adding all of these ... gives the answer 140 ...
the explanation looks difficult but this technique helps if you are familiar with the concept of permutations when the numbers/alphabets repeat
B
Answer: (B) 140
Question: Find the 100th place digit of a number forming from digits 1 to 9 without repeatation and number passes the divisibility test of 11 means difference of sum of alternative digits is 0 or divisible by 11.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: Largest no is 987652413
So digit will be 4
ANSWER:D
Answer: (D) 4
Question: n is a whole number which when divided by 4 gives 3 as remainder. What will be the remainder when 2*n is divided by 4 ?
(A) 8
(B) 2
(C) 6
(D) 7
(E) 1
Rationale: Explanation:
Let n=4*q + 3. Then, 2*n = 8*q + 6 = 4(2*q + 1) + 2.
Thus when 2*n is divided by 4, the reminder is 2.
Answer: B) 2
Answer: (B) 2
Question: A candidate got 30% of the votes polled and he lost to his rival by 4000 votes. How many votes were cast?
(A) 7500
(B) 2028
(C) 2775
(D) 10000
(E) 6851
Rationale: 30%-----------L
70%-----------W
------------------
40%----------4000
100%---------? => 10000
Answer:D
Answer: (D) 10000
Question: The average of first 10 odd numbers is?
(A) 12
(B) 10
(C) 87
(D) 23
(E) 34
Rationale: Sum of 10 odd no.
= 100
Average = 100/10
= 10
Answer:B
Answer: (B) 10
Question: A cricketer scored 138 runs which included 12 boundaries and 2 sixes. What percent of his total score did he make by running between the wickets.
(A) 46.2%
(B) 54.54 %
(C) 63%
(D) 56.52%
(E) None of these
Rationale: Explanation :
Number of runs made by running = 138 - (12 x 4 + 2 x 6)
= 138 - (60)
= 78
Now, we need to calculate 72 is what percent of 138.
=> 78/138 * 100 = 56.52 %
Answer : D
Answer: (D) 56.52%
Question: If 36 men can do a piece of work in 25 hours, in how many hours will 15 men do it ?
(A) 10 hours
(B) 40 hours
(C) 20 hours
(D) 60 hours
(E) None of them
Rationale: Let the required number of hours be x. Then,
Less men, More hours	(Indirect Proportion)
15 : 36 : : 25 : x = (15 x x) = (36 x 25) = (36 x 25)/15 = 60
Hence, 15 men can do it in 60 hours.
Answer is D.
Answer: (D) 60 hours
Question: Calculate the value of q from the below equation: q^3âˆ’25q+5=173
(A) 7
(B) 4
(C) 6
(D) 9
(E) 8
Rationale: Use elimination method to find the correct option. you find that of all the options 7 is the correct value for q
Answer: A
Answer: (A) 7
Question: If 45% of z is 90% of y and y is 75% of x, what percent of x is z?
(A) 200
(B) 160
(C) 100
(D) 65
(E) 150
Rationale: (45/100)z = (90/100)y
and y = (75/100)x i.e. y = (3/4)x
i.e. (45/100)z = (90/100)*(3/4)x
i.e. z = (90*3)x/(45*4)
i.e. z = (1.5)x = (150/100)x
i.e. z is 150% of x
Answer: Option E
Answer: (E) 150
Question: The area of a rectangular field is equal to 200 square meters. Its perimeter is equal to 60 meters. Find the width of this rectangle.
(A) 5
(B) 10
(C) 15
(D) 20
(E) 25
Rationale: L * W = 200 : area , L is the length and W is the width.
2 L + 2 W = 60 : perimeter
L = 30 - w : solve for L
(30 - W) * W = 200 : substitute in the area equation
W = 10 and L = 20 :
correct answer B
Answer: (B) 10
Question: 417 x 417 + 383 x 383=?
(A) 320578
(B) 80578
(C) 80698
(D) 81268
(E) None of them
Rationale: =(417)^2 + (383)^2
=(400+17)^2 + (400-17)^2
=2[(400)^2+(17)^2]
=2[160000+289]
=2 x 160289 = 320578
Answer is A
Answer: (A) 320578
Question: The average weight of a group of boys is 30 kg. After a boy of weight 38 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ?
(A) A)4
(B) B)8
(C) C)6
(D) D)2
(E) E)7
Rationale: Let the number off boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 38 kg joins the group, total weight of boys = 30x + 38
So 30x + 38 = 31(x + 1) = > x = 7.
Answer:E
Answer: (E) E)7
Question: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negative integers. What is the value of (a + b) + 5, when a # b = 100?
(A) 5
(B) 8
(C) 10
(D) 13
(E) 17
Rationale: OFFICIAL SOLUTION:
(B) We know that a # b = 100 and a # b = 4a² + 4b² + 8ab. So
4a² + 4b² + 8ab = 100
We can see that 4a² + 4b² + 8ab is a well-known formula for (2a + 2b)². Therefore
(2a + 2b)² = 100.
(2a + 2b) is non-negative number, since both a and b are non-negative numbers. So we can conclude that 2(a + b) = 10. (a + b) + 5 = 10/2 + 5 = 10.
The correct answer is C.
Answer: (C) 10
Question: If the median of a list of numbers is m, the first quartile of the list is the median of the numbers in the list that are less than m. What is the first quartile of the list of numbers 42, 24, 30, 28, 26, 19, 33 and 35 ?
(A) 33
(B) 28
(C) 27
(D) 25
(E) 23
Rationale: It is given that a quartile is the middle number of all numbers less than median..
so lets arrange the number in ascending order- 42, 24, 30, 28, 26, 19, 33 and 35
19, 24, 26, 28, 30 , 33, 35, 42...
numbers less than median are 19, 24, 26, 28..
the median of these numbers = center of 24 and 26 = 25
D
Answer: (D) 25
Question: A vessel of capacity 90 litres is fully filled with pure oil. Nine litres of oil is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure oil in the final oil solution?
(A) 83.9 litres
(B) 54.9 litres
(C) 37.9 litres
(D) 72.9 litres
(E) 27.9 litres
Rationale: D
72.9 litres
Let the initial quantity of oil in vessel be T litres.
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of oil finally in the vessel is then given by [(T - y)/T]n * T
For the given problem, T = 90, y = 9 and n = 2.
Hence, quantity of oil finally in the vessel
= [(90 - 9)/90]2 (90) = 72.9 litres.
Answer: (D) 72.9 litres
Question: A set of football matches is to be organized in a "round-robin" fashion, i.e., every
participating team plays a match against every other team once and only once. If 21
matches are totally played, how many teams participated?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Rationale: 2 ways to solve this problem..
summation(x)=n(n-1)/2
n(n-1)/2=21;
n^2-n-42=0
factors : 7,-6
Ans : 7
ANSWER:D
Answer: (D) 7
Question: If each letter in the English alphabet is assigned an even number such as A = 2, B = 4 and so on, then what will be the total value of the letters in the word LEAD?
(A) 42
(B) 40
(C) 46
(D) 44
(E) 48
Rationale: GIven the Condition:
L=24,E=10, A=2 and D=8
Hence the addition is 44
ANSWER:D
Answer: (D) 44
Question: What is the unit digit in the product (3^65 x 6^59 x 7^71)?
(A) 4
(B) 8
(C) 6
(D) 2
(E) 1
Rationale: Unit digit in 3^4 = 1 Unit digit in (3^4)16 = 1
Unit digit in 3^65 = Unit digit in [ (3^4)16 x 3 ] = (1 x 3) = 3
Unit digit in 6^59 = 6
Unit digit in 7^4 ==> Unit digit in (7^4)17 is 1.
Unit digit in 7^71 = Unit digit in [(7^4)17 x 7^3] = (1 x 3) = 3.
Required digit = Unit digit in (3 x 6 x 3) = 4.
Answer A)4.
Answer: (A) 4
Question: Divide $300 among A,B in the ratio 1:2. How many $ that A get?
(A) $50
(B) $500
(C) $150
(D) $250
(E) $100
Rationale: Sum of ratio terms = 1+2 = 3
A = 300*1/3 = $100
Answer is E
Answer: (E) $100
Question: A man can row
9 1/3
kmph in still water and finds that it takes him thrice as much time to row up than as to row, down the same distance in the river. The speed of the current is.
(A) 3 2/3kmph
(B) 4 2/3kmph
(C) 5 2/3kmph
(D) 6 2/3kmph
(E) None of these
Rationale: Explanation:
Friends first we should analyse quickly that what we need to calculate and what values we require to get it.
So here we need to get speed of current, for that we will need speed downstream and speed upstream, because we know
Speed of current = 1/2(a-b) [important]
Let the speed upstream = x kmph
Then speed downstream is = 3x kmph [as per question]
speed in still water = 1/2(a+b)
=>1/2(3x+x)
=>2x as per question we know,
2x=9 1/3
=>2x=28/3
=>x=14/3
So,
Speed upstream = 14/3 km/hr, Speed downstream 14 km/hr.
Speed of the current
=1/2[14−14/3]
=14/3=4 2/3kmph
Option B
Answer: (B) 4 2/3kmph
Question: A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 800 sq.ft, how many feet of fencing will be required ?
(A) 100feet
(B) 120feet
(C) 150feet
(D) 180feet
(E) 186feet
Rationale: Explanation:
We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 800
=> Breadth = 40 feet
Area to be fenced = 2B + L = 2*40 + 20
= 100 feet
Answer: Option A
Answer: (A) 100feet
Question: A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?
(A) 79
(B) 83
(C) 85
(D) 87
(E) 88
Rationale: Consider 4 teams.
Max possible games if a team is eliminated after 2 loses = 3 + 2 = 5
Similarly, if 5 teams are there,
Max possible games = 4 + 3 = 7.
So going forth we can play 15,17,19,21 max possible games for 9,10,11,12 teams respectively.
Adding them up gives 72. Now there are two teams left in each group still. So four more games to give a winner in each team.
So 72+ 4 = 76.
Now of the four teams we can have max 3 games if its a knock out round.
So 76 + 3 = 79.
Answer A = 79.
Answer: (A) 79
Question: For the symbol, m”n=n^2−m for all values of m and n. What is the value of 6”3?
(A) 5
(B) 3
(C) 2
(D) 1
(E) 0
Rationale: 6”3= 9 - 6=3
Answer: B
Answer: (B) 3
Question: A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?
(A) 33381
(B) 27991
(C) 92211
(D) 25000
(E) 27821
Rationale: 25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000
Answer: D
Answer: (D) 25000
Question: Find the number of zeroes in the expression 15*32*25*22*40*75*98*112*125
(A) 12
(B) 9
(C) 14
(D) 7
(E) 8
Rationale: (3*5)*(2*2*2*2*2)*(5*5)*(2*11)*(2*2*2*2*5)*(5*5*3)*(2*7*7)*(2*2*2*2*7)*(5*5*5)
there are 9 (5*2) pairs which gives zero
so.no of zeros 9
ANSWER:B
Answer: (B) 9
Question: 1904 x 1904 = ?
(A) 3623216
(B) 3624216
(C) 3624316
(D) 3625116
(E) 3625216
Rationale: 1904 x 1904 = (1904)2
= (1900 + 4)2
= (1900)2 + (4)2 + (2 x 1900 x 4)
= 3610000 + 16 + 15200.
= 3625216.
E)
Answer: (E) 3625216
Question: A fraudulent dairy guy declares to market his dairy at price range but he combined it with water and therefore benefits 25%. The percentage of water in the combination is:
(A) 20%
(B) 30%
(C) 50%
(D) 35%
(E) 60%
Rationale: Explanation :
-> If Capability Price of 1 L of milk = Rs.1
And S.Price of 1 L of combination = Rs.1
Benefits = 20%
Capability Price of 1 L of combination =[ (100 / 100+25)*1 ]
= Rs. 100/125 = Rs. 4/5
-> From the rule of allegation
i. Capability Price of 1 L of water = 0
ii. Capability Price of 1 L of milk = 1
iii. (p) = 4/5
iv. d – m = 1 – 4/5 = 1/5
v. m – c = 4/5 – 0 = 4/5
Ratio of milk to water = 4/5 : 1/5 = 4:1
% of water in the combination = (1/5 x 100)% = 20%
Answer : A
Answer: (A) 20%
Question: On Monday Joshua sold 90 phones, Rob sold 30% more phones than Joshua, Prerna sold 11 more phones than Rob, and Declan sold 3/4 of what Prerna sold. How many phones did Declan sell?
(A) 111
(B) 75
(C) 29
(D) 128
(E) 96
Rationale: E
96
Joshua sold 90 phones.
Rob sold 30 % more phones than Joshua
Rob = 90 + 90 × 30%
= 90 + 90 × 0.30
= 90 + 27
= 117
Prerna sold 11 more phones than Rob
Prerna = 117 + 11
= 128
Declan sold 3/4 of what Prerna sold.
Declan= 128 x 3/4
=96
Declan sold 96 phones.
The correct answer is E.
Answer: (E) 96
Question: Today John is twice as old as Sam and Sam is 3 years younger than Mike. If John, Sam, and Mike are all alive 4 years from today, which of the following must be true on that day. I. John is twice as old as Sam II. Sam is 3 years younger than Mike III. John is older than Mike
(A) I, III
(B) I
(C) II
(D) I,II & III
(E) III
Rationale: Say Mike age is 10
Sam age is 7
John age is 14
After 4 years
Mike age is 14
Sam age is 11
John age is 28
i.e ) after 4 years
Sam is 3 years younger than Mike
John is twice as old as Sam
John is older than Mike
Answer : D
Answer: (D) I,II & III
Question: Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability W that Kate has more than $10 but less than $15?
(A) 5/16
(B) 1/2
(C) 12/30
(D) W=15/32
(E) W=3/8
Rationale: For 21 heads Kate will end up with < $10 and we want her to win. Therefore, only possibilities W are 3 or 4 heads.
I made an educated guess and it worked fine.
Ans- 'D'
MGMAT's anagram helped here as well.
HHHHT = 5!/4!*1! = 5
HHHTT = 5!/3!*2! = 10
Total acceptable cases = 15
Total cases = 32
P = 15/32.D
Answer: (D) W=15/32
Question: Find unit digit of{(6374)^1793 x (625)^317 x (341)^491}.
(A) 5
(B) 15
(C) 0
(D) 20
(E) 69
Rationale: Explanation:
Unit digit of 6374 is 4.
Unit digit of 625 is 5.
Unit digit of 341 is 1.
Just multiply 4*5*1 = 20
So unit digit of 20 is 0.
Answer C.
Answer: (C) 0
Question: The ratio by weight, measured in pounds, of books to clothes to electronics in a suitcase initially stands at 5:4:2. Someone removes 7 pounds of clothing from the suitcase, thereby doubling the ratio of books to clothes. How many pounds do the electronics in the suitcase weigh?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Rationale: The weights of the items in the suitcase are 5k, 4k, and 2k.
If removing 7 pounds of clothes doubles the ratio of books to clothes, then 7 pounds represents half the weight of the clothes.
2k = 7 pounds and then k = 3.5 pounds.
The electronics weigh 2(3.5) = 7 pounds.
The answer is B.
Answer: (B) 7