divelab/Sys2Bench huggingface dataset:
Question: A and B can do a piece of work in 12 days. With the help of C they finish the work in 3 days. C alone can do that piece of work in?
Options:
A)40 days
B)2 days
C)4 days
D)8 days
E)40 days
Rationale: C
30 days
C = 1/3 – 1/12 = 1/4 => 4 days
Answer: C
Question: A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected, if it should have 5 seniors and 5 juniors?
Options:
A)2C7 * 10C6
B)2C7 * 17C5
C)2C7 * 10C5
D)2C9 * 10C5
E)2C7 * 30C5
Rationale: Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected 10C5 ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= 12C5 = 12C7
Answer:C
Answer: C
Question: In a bag, there are 6 brown, 4 blue and 10 orange pens. One pen is picked up randomly. What is the probability that it is neither brown nor blue?
Options:
A)1/4
B)1/5
C)1/3
D)1/2
E)3/4
Rationale: Explanation :
Neither brown nor blue means the pen drawn is orange.
Total number of outcomes = (6+4+10) = 20.
Number of Favourable outcomes = 10 = Number of orange pens.
Hence, Probability of the event = 10/20 = 1/2.
Answer : D
Answer: D
Question: A and B are walking with 25 kmph and 30 kmph. A is walking towards east and B is towards south. They both started at 7:00 A.M at what time they are 130 metres apart.
Options:
A)130m
B)140m
C)150m
D)160m
E)170m
Rationale: let after t sec they are 130m apart then
(25* 5/18 *t)^2 + (30* 5/18 *t)^2 = 130^2
=> t * (5/18) * 5 * √ 61 = 130
=> t = 12.98425 sec ~ 13 sec
so, at 7:00:13 AM they are 130m apart.
ANSWER:A
Answer: A
Question: Thefactor countof an integer n greater than 1 is the number of distinct prime factors of n. For example, the factor count of 36 is 2, since 2 and 3 are the distinct prime factors of 36 = 2 x 2 x 3 x 3. For which of the following integers is the factor count greatest E?
Options:
A) 60
B) 61
C) 62
D) 63
E) 64
Rationale: 60 = 2*2*3*5 factor count =3
61= prime number
62 = 2*31 factor count=2
63=3*3*7 factor count=2
64=2*2*2*2*2*2 factor count=1=E
answer is A.
Answer: A
Question: Rs.1500 is divided into two parts such that if one part is invested at 6% and the other at 5% the whole annual interest from both the sum is Rs.82. How much was lent at 5%?
Options:
A)899
B)866
C)878
D)800
E)877
Rationale: (x*5*1)/100 + [(1500 - x)*6*1]/100 = 82
5x/100 + 90 – 6x/100 = 82
x/100 = 8
=> x = 800
Answer:D
Answer: D
Question: How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
Options:
A)41
B)23
C)56
D)99
E)87
Rationale: Explanation:
30 * 33/2 = 5/2 * 2 * x => x = 99
Answer: D
Answer: D
Question: The sum of all the integers g such that -26 < g < 24 is
Options:
A)0
B)-2
C)-25
D)-49
E)-51
Rationale: Easy one -
-25, -24, -23,-22,...... -1,0, 1, 2...., 22, 23
Cancel everyhitng and we're left with - -25 and -24 g= -49.
D is the answer.
Answer: D
Question: A can do a piece of work in 12 days. When he had worked for 2 days B joins him. If the complete work was finished in 8 days. In how many days B alone can finish the work?
Options:
A)18
B)99
C)77
D)62
E)12
Rationale: 8/12 + 6/x = 1
X = 18 days
Answer:A
Answer: A
Question: In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?
Options:
A)57.5 m
B)127.5 m
C)150.7 m
D)98.6 m
E)95 m
Rationale: When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = (900 * 700)/800 = 6300/8 = 787.5 m.
In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = (600 * 212.5)/1000 = 127.5 m.
ANSWER:B
Answer: B
Question: The areas of the two spheres are in the ratio 1: 4. The ratio of their volume is?
Options:
A)1:4
B)1:16
C)1:8
D)1:64
E)2:3
Rationale: 4πr12 : 4πr22 = 1 : 4
r1 : r2 = 1 : 2
4/3 πr13 : 4/3 πr23
r13 : r23 = 1 : 8
ANSWER:C
Answer: C
Question: Let f(x) = x^2 + bx + c. If f(4) = 0 and f(-3) = 0, then b + c =
Options:
A)18
B)15
C)-13
D)-21
E)-24
Rationale: f(x) = x^2 + bx + c. If f(4) = 0 and f(-3) = 0, then b + c =
f(4) = 0= 16+4b+c--- Taking 16 to the other side
-> 4b+c= -16
f(-3) = 0= 9-3b+c --- taking -3b+c to the other side
-> 3b-c=9
When we add these 2 equations, we get 7b= -7---> b= -1
And while substituting b= -1 we get c= -12.
b+c= -13--- Answer C
Answer: C
Question: An accurate clock show 8 o’ clock in the morning. The hour hand then rotates by 90o. The time in the clock is:
Options:
A)9 o’clock
B)10 o’ clock
C)11 o’clock
D)12 o’clock
E)8 o’clock
Rationale: as from 9 o clock to 12 o clock is 90 degrees so 8 is one less than nine and hence tha answer will be one less than 12 that is 11 o clock
ANSWER:C
Answer: C
Question: If x is negative and y is positive, which of the following cannot be positive?
Options:
A)xy
B)-xy
C)x/-y
D)x^2/y
E)x^4/y
Rationale: - * + = -
xy is not positive.
Answer : A
Answer: A
Question: Find large number from below question The difference of two numbers is 1395. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder
Options:
A)1235
B)1345
C)1678
D)1767
E)1671
Rationale: Let the smaller number be x. Then larger number = (x + 1395).
x + 1395 = 6x + 15
5x = 1380
x = 276
Large number = 276+1395 = 1671
E
Answer: E
Question: The average wages of a worker during a fortnight comprising 15 consecutive working days was $90 per day. During the first 7 days, his average wages was $87 per day and the average wages during the last 7 days was $90 per day. What was his wage on the 8th day?
Options:
A)$83
B)$90
C)$92
D)$97
E)$111
Rationale: Average daily wage of a worker for 15 consecutive working days = 90$
During the first 7 days , the daily average daily wage = 87$
During the last 7 days , the daily average daily wage = 90$
Wage on 8th day = 90*15 -( 87*7 + 90 *7)
= 1350 - (609 + 630)
=1350 - 1239
= 111
Answer E
Answer: E
Question: There are 6561 balls out of them 1 is heavy. Find the min. no. of times the balls have to be weighedfor finding out the heavy ball.
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: 6561 if u notice is 3^8...n can be divided by 3(prime factor) only...we keep on dividing it into sets of 3 and in 1 weighing itself you can determine which set is heavy. Divide that set into further 3 lots...keep doing that..
6561
2187 2187 2187 1st weighing
729 729 729
243 243 243
81 81 81
27 27 27
9 9 9
3 3 3
1 1 1
ANswer is 8 attempts
ANSWER:E
Answer: E
Question: A boy goes to his school from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming. The distance between his house and school is?
Options:
A)288
B)268
C)208
D)860
E)270
Rationale: CP = SP + 1CP = SP - g
800 + x = 980 - 2x
3x = 180 => x = 60
CP = 800 + 60 = 860
Answer: D
Answer: D
Question: A train which has 250 m long, is running 40 kmph. In what time will it cross a person moving at 4 kmph in same direction ?
Options:
A)30 sec
B)28 sec
C)29 sec
D)35 sec
E)25 sec
Rationale: Time taken to cross a moving person = length of train/ relative speed
Time taken = 250/((40-4) (5/18) = 250/ 36*(5/18) = 250/10= 25 sec
Answer: E
Answer: E
Question: If f(f(n)) + f(n) = 2n+3, f(0) = 1 then f(2015) = ?
Options:
A)2016
B)2088
C)270
D)1881
E)1781
Rationale: f (f(0)) + f(0) = 2(0) + 3 ⇒⇒ f(1) = 3-1 = 2, f(1) = 2
f(f(1)) + f(1) = 2(1) + 3 ⇒⇒ f(2) = 5-2 = 3, f(2) = 3
f(f(2)) + f(2) = 2(2) + 3 ⇒⇒ f(3) = 7-3 = 4, f(3) = 4
..............
f(2015) = 2016
Ans: A
Answer: A
Question: The positive integers y and z leave remainders of 2 and 3, respectively, when divided by 6. y > z. What is the remainder when y– z is divided by 6?
Options:
A)1
B)2
C)5
D)3
E)4
Rationale: Let's test out some values of p and q that satisfy the given information.
y leaves are remainder of 2 when divided by 6
So, y COULD equal 8
z leaves are remainder of 3 when divided by 6
So, z COULD equal 3
What is the remainder when y – z is divided by 6?
So, y - z = 8 - 3 = 5, and when we divide 5 by 6, we get 0 with remainder 5
Answer: C
Answer: C
Question: My watch was correct at noon, after which it started to lose 17 minutes per hour until six hours ago it stopped completely. It now shows the time as 2.52 pm. What time is it now?
Options:
A)11 pm
B)10 pm
C)01 am
D)03 am
E)09 pm
Rationale: B
10 pm
12 noon = 12 noon,
1 pm = 12.43,
2 pm = 1.26,
3 pm = 2.09,
4 pm = 2.52 + 6 hours = 10 pm.
Answer: B
Question: How many miles is 320km?
Options:
A)150miles
B)175miles
C)200miles
D)225miles
E)250miles
Rationale: 200 miles
320/1.6=200
Answer:C
Answer: C
Question: Kathleen can paint a room in 3 hours, and Anthony can paint an identical room in 5 hours. How many hours would it take Kathleen and Anthony to paint both rooms if they work together at their respective rates?
Options:
A)8/15
B)4/3
C)15/8
D)9/4
E)15/4
Rationale: kat---3 hrs---1 room ; so rate is work/time= 1/3
Anthony---5 hrs---1 room ; rate= 1/5
rate of kat and anthony together=1/3+1/5 = 8/15
But we are told that they both paint 2 rooms which are identical.
So; Work is 2 rooms, rate is 8/15,
total time required for both of them together to complete the work = 2/(8/15)= 30/8= 15/4
Hence Answer= E
Answer: E
Question: Which of the following will not be a correct option for b given the equation: b = x/2 + y/2^4 + z/4^3, where x,y and z are each equal to 1 or 0.
Options:
A)1/4
B)1/16
C)9/32
D)1/32
E)3/32
Rationale: b = x/2 + y/2^4 + z/4^3
= x/2 + y/16 + z/64
= (16x + 2y + z) /64
Depending on whether a, b and c take 0 or 1
substitute 0 or 1 for the different values of x,y and z.
Answer E
Answer: E
Question: Lindsay can paint a room in 8 hours at a constant rate. If Lindsay paints the room for 4 hours and Joseph paints the rest of the room, who works at 1/2 constant rate of Lindsay. How long will it take for Joseph alone to paint the rest of the room?
Options:
A)4hrs
B)8hrs
C)10hrs
D)12hrs
E)16hrs
Rationale: Lindsay will do 4/8 in 4 hrs, so Lindsay does 1/2 of the work ..
Therefore, Joseph will do the remaining 1/2 work alone..
As the speed of Joseph is 1/2 rate of Lindsay, Joseph will do the 1/2 work in same time that Lindsay takes to complete full job...
ans 8
B
Answer: B
Question: If a man walks to his office at ¾ of his usual rate, he reaches office 1/3 of an hour late than usual. What is his usual time to reach office?
Options:
A)1 hour
B)8 hour
C)9 hour
D)6 hour
E)2 hour
Rationale: Speed Ratio = 1:3/4 = 4:3
Time Ratio = 3:4
1 -------- 1/3
3 --------- ? è 1 hour.Answer:A
Answer: A
Question: Series D(n) is such that i*D(i) = j*D(j) for any pair of positive integers (i, j). If D(1) is a positive integer, which of the following is possible?
I. 2*D(100) = D(99) + D(98)
II. D(1) is the only integer in the series
III. The series does not contain negative numbers
Options:
A)I only
B)II only
C)IIII only
D)IIIII only
E)I, IIIII
Rationale: First thing I want to understand is this relation: i*D(i) = j*D(j) for any pair of positive integers. I will take examples to understand it.
When i = 1 and j = 2, D(1) = 2*D(2)
So D(2) = D(1)/2
When i = 1 and j = 3, D(1) = 3*D(3)
So D(3) = D(1)/3
I see it now. The series is: D(1), D(1)/2, D(1)/3, D(1)/4 and so on...
II and III are easily possible. We can see that without any calculations.
II. D(1) is the only integer in the series
If D(1) = 1, then series becomes 1, 1/2, 1/3, 1/4 ... all fractions except D(1)
III. The series does not contain negative numbers
Again, same series as above applies. In fact, since D(1) is a positive integer, this must be true.
I. 2*D(100) = D(99) + D(98)
2*D(1)/100 = D(1)/99 + D(1)/98 (cancel D(1) from both sides)
2/100 = 1/99 + 1/98
Not true hence this is not possible
Answer (D)
Answer: D
Question: Product of two co-prime numbers is 117. Their L.C.M should be
Options:
A)222
B)117
C)278
D)767
E)298
Rationale: H.C.F of co-prime numbers is 1. So, L.C.M = \inline \fn_jvn \frac{117}{1} =117
Answer: B
Answer: B
Question: A train 400m long takes 10 sec to cross a man walking at 5kmph in a direction opposite to that of the train. Find the speed of the train?
Options:
A)A)175kmph
B)B)150kmph
C)C)162kmph
D)D)149kmph
E)E)100kmph
Rationale: Let the speed of the train be x kmph
Speed of the train relative to man = x+5 = (x+5)*5/18 m/sec
400/[(x+5)*5/18] = 10
10(x+5) = 80*18
x = 149kmph
Answer is D
Answer: D
Question: In the faculty of Reverse-Engineering, 100 second year students study numeric methods, 300 second year students study automatic control of airborne vehicles and 200 second year students study them both. How many students are there in the faculty if the second year students are approximately 40% of the total?
Options:
A)550
B)500
C)600
D)650
E)700
Rationale: Total number of students studying both are 100 + 300 - 200 = 200 (Subtracting the 200 since they were included in the both the other numbers already).
So 40% of total is 200, so 100% is 500
Answer is B
Answer: B
Question: When Rahul was born, his father was 35 years older than his brother and his mother was 25 years older than his sister. If Rahul's brother is 5 years older than him and his mother is 3 years younger than his father, how old was Rahul's sister when he was born ?
Options:
A)11 years
B)10 years
C)12 years
D)9 years
E)8 years
Rationale: When Rahul was born, his brother's age = 5 years; his father's age = (5 + 35) years = 40 years,
his mother's age = (40 - 3) years = 37 years;
his sister's age = (37 - 25) years = 12 years.
Answer: Option C
Answer: C
Question: If (x+3)^2/(3x+65)=2, then the difference between the two possible values of x is:
Options:
A)16
B)18
C)20
D)22
E)24
Rationale: (x+3)^2/(3x+65)=2
(x+3)^2=2(3x+65)
x^2+6x+9=6x+130
x^2-121=0
(x-11)(x+11)=0
x=11 or x=-11
The answer is D.
Answer: D
Question: By selling an umbrella for Rs. 300, a shop keeper gains 20%. During a clearance sale, the shopkeeper allows a discount of 10% on the marked price. His gain percent during the sale is?
Options:
A)7%.
B)9%.
C)8%.
D)2%.
E)3%.
Rationale: Explanation:
Marked price = Rs. 300
C.P. = 100/120 * 300 = Rs. 250
Sale price = 90% of Rs. 300 = Rs. 270
Required gain % = 20/250 * 100 = 8%.
Answer: C
Answer: C
Question: Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 2/3 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?
Options:
A)1/16
B)7/8
C)9/64
D)5/64
E)3/64
Rationale: P(Xavier will solve)=1/4
P(Yvonne will solve)=2/3
P(Zelda will NOT solve) = 1- 5/8 = 3/8.
Now, we need to multiply all this Ps to find an answer:
p= (1/4)*(2/3)*(3/8) = 1/16.
Ans. A.
Answer: A
Question: A man invested Rs. 5050 in 5% stock at 99 and sold it when the price rose to Rs. 101. He invested the sale proceeds in 8% stock at 88. Find the change in man's income if the Brokerage is Rs. 2.
Options:
A)150
B)160
C)170
D)180
E)190
Rationale: Purchase price of the first stock = Rs.99 + Rs.2 = Rs.101
Number of stocks purchased in this case = 5050/101 = 50
Since the face value is not given, it can be taken as Rs.100. So, dividend per share = Rs.5
Income = 50 × 5 = Rs.250
Sale Price of the stock = 101-2 = 99
Amount received by selling the stock = 50 × 99 = 4950
Then he invests this Rs.4950 in 8% stock at 88
Purchase price of this stock = 88+2 = 90
Number of stocks purchased in this case = 4950/90 = 55
Since the face value is not given, it can be taken as Rs.100. So, dividend per share = Rs.8
Income = 55 × 8 = Rs.440
Change in income = 440-250 = Rs.190
Answer is E.
Answer: E
Question: There are 24 students in Chadâ€™s class. He brought 30 cookies to pass out
for his birthday. How many cookies will each student get? Will there be
any cookies left over?
Options:
A)1-9
B)1-7
C)2-8
D)1-6
E)3-10
Rationale: 30/24 = 1 R 6 Chad will give each student 1 cookie and there will be 6
cookies left over
correct answer D
Answer: D
Question: The cost of one photocopy is $0.02. However, a 25% discount is offered on orders of more than 100 photocopies. If Steve and Dinley have to make 80 copies each, how much will each of them save if they submit a single order of 160 copies?
Options:
A)$0.32
B)$0.40
C)$0.45
D)$0.48
E)$0.54
Rationale: If Steve and Dinley submit separate orders, each would be smaller than 100 photocopies, so no discount. Each would pay (80)*($0.02) = $1.60, or together, a cost of $3.20 --- that's the combinedno discount cost.
If they submit things together as one big order, they get a discount off of that $3.20 price ---- 25% or 1/4 of that is $0.80, the discount on the combined sale. They each effective save half that amount, or $0.40. Answer =(B).
Answer: B
Question: 3log3(-5) = ?
Options:
A)1
B)undentified
C)3
D)7
E)5
Rationale: Since -5 is not in the domain of function log3(x),
3log3(-5) is undefined
correct answer B
Answer: B
Question: Of the five numbers in a sequence, the first term is 5,000, and each of the following terms is 20% of the previous term. What is the value range of the five numbers?
Options:
A)4,992
B)9,750
C)9,975
D)9,984
E)10,736
Rationale: 1st number= 5,000
2nd number = 20% of 5,000= 1000
3rd number= 20% of 1000= 200
4th number = 20% of 200= 40
5th number= 20% of 40= 8
Range = 5,000-8= 4992
A is the answer
Answer: A
Question: In the first week of last month, Company X realized an average wholesale profit of $4284 per day from the sale of q units of Product Y. Which of the following CANNOT be the difference between Product Y’s sale price and cost per unit?
Options:
A)$3
B)$4
C)$7
D)$13
E)$51
Rationale: The key operator that everyone seems to have missed is the average. The average profit of $4284 per day is equivalent to a total profit of $29988 for the whole week.
Therefore, the difference between sale price and cost per unit is equal to $29988, which is divisible by all of the answer choices, except for 13.
So, D is the answer.
Answer: D
Question: What profit percent is made by selling an article at a certain price, if by selling at 2/3rd of that price, there would be a loss of 20%?
Options:
A)20%
B)29%
C)10%
D)80%
E)90%
Rationale: SP2 = 2/3 SP1
CP = 100
SP2 = 80
2/3 SP1 = 80
SP1 = 120
100 --- 20 => 20%
Answer: A
Answer: A
Question: Three models (S, T, and P) of cars are distributed among three showrooms. The number of cars in each showrooms must be equal and each model must be represented by at least one car in every showroom. There are 19 cars of Model S, 17 cars of Model T, and 15 cars of Model P. What is the maximum number of cars of model S in any showroom ?
Options:
A)17
B)16
C)15
D)14
E)13
Rationale: The total number of cars is 51. So each showroom has 17 cars(since the number of cars in each showrooms should be equal 51/3=17).
Moreover that the number of model S is maximum means that the numbers of model T and P should be minimum. Since each model must be represented by at least one car in every showroom that minimum number should be 1.
So Maximum number of model S is 17-2= 15.
The answer is (C)
Answer: C
Question: A shopkeeper sells his goods at cost price but uses a faulty meter that weighs 950 grams. Find the profit percent.
Options:
A)100/17%
B)100/23%
C)100/19%
D)100/13%
E)None of these
Rationale: Explanation:
(100 + g) / (100 + x) = True measure / Faulty measure
x = 0
true measure = 1000
faulty measure = 950
100 + g / 100 + 0 = 1000 / 950
100+g = 20/19*100
g = 100/19 %
ANSWER: C
Answer: C
Question: If the speed of X kilometers per second is equivalent to the speed of Y kilometers per hour, what is Y in terms of X? (1 kilometer=1000 meters)
Options:
A)15x/18
B)6x/5
C)18x/5
D)60x
E)3600000x
Rationale: x=y/3600
y = 3600x
Answer : D
Answer: D
Question: A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7.30 a.m. They shall first cross each other at?
Options:
A)12
B)66
C)77
D)81
E)71
Rationale: Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two.
Relative speed of A and B = 6 - 1 = 5 rounds per hour.
Time taken to complete one round at this speed = 1/5 hr = 12 min.
Answer: A
Answer: A
Question: In an examination, 300 students appeared. Out of these students; 27 % got first division, 54 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed.
Options:
A)33
B)57
C)54
D)99
E)01
Rationale: The number of students with first division = 27 % of 300 = 27/100 × 300 = 8100/100 = 81
And, the number of students with second division = 54 % of 300
= 54/100 × 300
=16200/100
= 162
Therefore, the number of students who just passed = 300 – (81 + 162) = 57
Answer:B
Answer: B
Question: If grapes are 93% water and raisins are 16% water, then how many kilograms did a quantity of raisins, which currently weighs 8 kilograms, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)
Options:
A)92
B)96
C)100
D)104
E)108
Rationale: Let x be the original weight.
The weight of the grape pulp was 0.07x.
Since the grape pulp is 84% of the raisins, 0.07x = 0.84(8).
Then x = 12*8 = 96 kg.
The answer is B.
Answer: B
Question: In an examination, there are 100 questions divided into three groups A, B and C such that each group contains at least one question. Each question in group A carries 1 mark, each question in group B carries 2 marks and each question in group C carries 3 marks. It is known that the questions in group A together carry at least 60% of the total marks.
If group B contains 23 questions, then how many questions are there in Group C?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: Explanation :
Since group B contains 23 questions, the marks associated with this group are 46.
Now, check for option (1), if group (C) has one question, the marks associated with this group will be 3. Thus total marks will be 49.
total number of questions = 100, group (A) will have 76 questions, the corresponding weightage being 76 marks.
Answer : A
Answer: A
Question: A person travelled from his house to office at 45km/hr and office to house at 55km/hr. Then find the average speed?
Options:
A)49km/hr
B)49.5km/hr
C)50km/hr
D)50.5km/hr
E)51km/hr
Rationale: Average speed = (2*45*55)/(45+55)= 49.5km/hrAnswer is B
Answer: B
Question: Five years ago the average of the ages of A and B was 40 years and now the average of the ages of B and C is 48 years. What will be the age of the B ten years hence?
Options:
A)22
B)77
C)26
D)96
E)51
Rationale: Let the present ages of A, B and C be a, b and c respectively.
Given, [(a - 5) + (b - 5)] / 2 = 40 => a + b = 90 --- (1)
(b + c)/2 = 48 => b + c = 96 --- (2)
From (1) and (2), we cannot find b.
Answer: D
Answer: D
Question: Alok ordered 16 chapatis, 5 plates of rice, 7 plates of mixed vegetable and 5 ice-cream cups. The cost of each chapati is Rs.6, that of each plate of rice is Rs.45 and that of mixed vegetable is Rs.70. The amount that Alok paid the cashier was Rs.961. Find the cost of each ice-cream cup?
Options:
A)25
B)30
C)29
D)12
E)20
Rationale: Explanation:
Let the cost of each ice-cream cup be Rs.x
16(6) + 5(45) + 7(70) + 6(x) = 961
96 + 225 + 490 + 5x = 961
5x = 150 => x = 30.
Answer:B
Answer: B
Question: Working alone at its constant rate, pump P pumped out ¼ of the water in a tank in 2 hours. Then pumps Q and R started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump Q, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump R, working alone at its constant rate, to pump out all of the water that was pumped out of the tank?
Options:
A)6
B)12
C)15
D)18
E)24
Rationale: Rate of pump P = 1/8
3 hours are required to pump out the remaining (3/4)ths of tank --> 1 hr to pump out 1/4
Rate of P + Rate of Q + Rate of R = 1/4
Rate of Q + Rate of R = 1/4 - 1/8 = 1/8
Q takes 18 hours to pump out the remaining (3/4)ths of tank --> 6 hrs per (1/4)ths --> 24 hrs to pump out fully.
Rate of Q = 1/24
1/24 + Rate of R = 1/8
Rate of R = 1/8 - 1/24 = 1/12
Time required to pump out all the water by R = 12 hrs
Answer: B
Answer: B