6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

A tin a mixture of two liquids A and B in the proportion 4 : 1. If 45 litres of the mixture is replaced by 45 litres of liquid B, then the ratio of the two liquids becomes 2 : 5. How much of the liquid B was there in the tin? What quantity does the tin hold?

A) 58 l	B) 65 l
C) 50 l	D) 62 l

Answer & Explanation Answer: C) 50 l

Explanation:

Let the tin contain 5x litres of liquids

$\frac{(5 x - 45) x \frac{4}{5}}{(5 x - 45) x \frac{1}{5} + 45} = \frac{2}{5}$

=> 5(4x - 36) = 2(x + 36)

=> 20x - 180 = 2x + 72

=> x = 14 litres

Hence, the initial quantity of mixture = 70l

Quantity of liquid B

= $(70 - 45) x \frac{1}{5} + 45$

= 50 litres.

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Filed Under: Alligation or Mixture
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29 9153

Q:

Sripad has scored average of 65 marks in three objects. In no subjects has he secured less than 58 marks. He has secured more marks in Maths than other two subjects. What could be his maximum score in Maths ?

A) 79	B) 77
C) 76	D) 73

Answer & Explanation Answer: A) 79

Explanation:

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

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Filed Under: Average
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5 9153

Q:

Rs. 69 were divided among 115 family members so that each lady gets 50 paise less than each gentleman, and each gentleman received twice the paise as each lady received. Find the number of gentlemen in the family?

A) 23	B) 92
C) 48	D) 52

Answer & Explanation Answer: A) 23

Explanation:

Total number of female members = 115

Let the number of males = x

Number of females = y

=> x + y = 115 ......(1)

According to the given data,

x + y(0.50) = 69 ......(2)

From (1) & (2), we get

0.5y = 46

=> y = 46 x 2 = 92

=> x = 115 - 92 = 23

Hence, required number of gentlemen in the family = x = 23.

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Filed Under: Percentage
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8 9144

Q:

${(31)}^{31} \times {(31)}^{- 27} = ?$

A) 961 x 861	B) 961 x 961
C) 961 x 2	D) 29791

Answer & Explanation Answer: B) 961 x 961

Explanation:

We know that $a^{m} \times a^{n} = a^{m + n}$

$∴ {(31)}^{31 - 27}$ = $31^{4}$ = 961 x 961.

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Filed Under: Numbers
Exam Prep: CAT
Job Role: Bank PO

11 9141

Q:

The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?

A) 3/5	B) 1/3
C) 2/5	D) 3/4

Answer & Explanation Answer: A) 3/5

Explanation:

Let A be the event of A will tell truth. B be the event of B tell truth

$P (A) = \frac{2}{3}, P (A^{c}) = 1 - P (A) = \frac{1}{3}$

$P (B) = \frac{4}{5}, P (B^{c}) = 1 - P (B) = \frac{1}{5}$

When both agree then they say true or they say false together, that is

$A \cap B o r A^{c} \cap B^{c}$

Also these events will be mutually exclusive :

$P (A \cap B) + P (A^{c} \cap B^{c}) = P (A) P (B) P (C) = \frac{2}{3} * \frac{4}{5} + \frac{1}{3} * \frac{1}{5} = \frac{3}{5}$

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Filed Under: Probability

6 9140

Q:

Two boys are playing on a ground. Both the boys are less than 10 years old. Age of the younger boy is equal to the cube root of the product of the age of the two boys. If we place the digit representing the age of the younger boy to the left of the digit representing the age of the elder boy, we get the age of the father of the younger boy. Similarly, we place the digit representing the age of the elder boy to the left of the digit representing the age of the younger boy and divide the figure by 2, we get the age of the mother of the younger boy. The mother of the younger boy is younger than his father by 3 years. Then, what are the ages of elder and younger boys ?

A) E = 15 & Y = 3	B) E = 14 & Y = 12
C) E = 40 & Y = 22	D) E = 4 & Y = 2

Answer & Explanation Answer: D) E = 4 & Y = 2

Explanation:

Let the the age of the elder boy = E

Let the the age of the younger boy = Y

Given that Y = cube root of EY

=> $Y^{3}$ = EY => E = $Y^{2}$ .....(1)
By the condition of number replacement the age of the father is YE

The Mother's age = EY/2

But she is 3 years less than father => EY/2 + 3 = YE
2YE = EY + 6 ......(2)

Then now from the given options we can identify which satisfies the all the conditions.

Here Y =2 and E = 4 satisfies all the conditions.

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Filed Under: Problems on Ages
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8 9132

Q:

A laborer is engaged for 30 days on the condition that he receives Rs.25 for each day he works and is fined Rs.7.50 for each day is absent. He gets Rs.425 in all. For how many days was he absent ?

A) 10 days	B) 14 days
C) 11 days	D) 9 days

Answer & Explanation Answer: A) 10 days

Explanation:

For 30 days he receives

30 x 25 = 750

Actually he gets 425

Therefore, Fine is 750 - 425 = 325

=> No of days absent = 325/(25+7.50)= 10

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Filed Under: Time and Work
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5 9121

Q:

The True discount on Rs1760 due after a certain time at 12% per annum is Rs160. The time after which it is due is?

A) 15months	B) 5months
C) 12months	D) 10months

Answer & Explanation Answer: D) 10months

Explanation:

P.W = Rs. (1760 - 160) = Rs.1600

S.I on Rs.1600 at 12% is Rs. 160

Time = (100 x 160)/(1600 x 12) = 5/6 years = (5/6) x 12 months = 10 months

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Filed Under: True Discount

2 9117

Quantitative Aptitude - Arithmetic Ability Questions

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