6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

How many window coverings are necessary to span 50 windows, if each window covering is 15 windows long?

A) 4	B) 15
C) 3	D) 50

Answer & Explanation Answer: A) 4

Explanation:

Given that,

Number of windows = 50

Each window covering covers 15 windows

=> 50 windows requires 50/15 window coverings

= 50/15 = 3.333

Hence, more than 3 window coverings are required. In the options 4 is more than 3.

Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.

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Filed Under: Average
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Analyst , Bank Clerk , Bank PO , Database Administration , IT Trainer

13 19912

Q:

A man can row 9 km/h in still water. It takes him twice as long as to row down. Find the rate of stream of the river ?

Answer

$\inline \frac{Time\: taken\: in\: upstream}{Time\: taken\: in\: downstream}$ = $\inline \frac{2}{1}$

$\inline \therefore$ $\inline \frac{Downstream\: speed}{upstream\: speed}$ = $\inline \frac{2}{1}$ where $\inline \frac{B+R}{B-R}=\frac{2}{1}$

B ---> speed of boat in still water

R---->speed of current

$\inline \Rightarrow \frac{B}{R}=\frac{3}{1}$ ( By componendo and dividendo)

$\inline \Rightarrow \frac{9}{R}=\frac{3}{1}\: \: \Rightarrow R= 3\: km/h$

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Subject: Boats and Streams

113 19889

Q:

The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

A) 1/4	B) 1/2
C) 1/6	D) 1/3

Answer & Explanation Answer: C) 1/6

Explanation:

P(X) = $\frac{1}{5}$ , P(Y) = $\frac{1}{4}$ , P(Z) = $\frac{1}{3}$

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

= $(\frac{1}{4} * \frac{1}{3} * \frac{4}{5}) + (\frac{3}{4} * \frac{1}{3} * \frac{1}{5}) + (\frac{2}{3} * \frac{1}{4} * \frac{1}{5}) + (\frac{1}{4} * \frac{1}{3} * \frac{1}{5})$

= $\frac{4}{60} + \frac{3}{60} + \frac{2}{60} + \frac{1}{60}$ = $\frac{10}{60}$ = $\frac{1}{6}$

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Filed Under: Probability

19 19839

Q:

Find the remainder in the expression $\frac{557 x 653 x 672}{9}$ ?

A) 0	B) 3
C) 5	D) 6

Answer & Explanation Answer: D) 6

Explanation:

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Filed Under: Problems on Numbers
Exam Prep: Bank Exams

1 19807

Q:

If a number 6573547x45 is divisible by 9, then find the value of x.

A) 7	B) 0
C) 8	D) 6

Answer & Explanation Answer: C) 8

Explanation:

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Filed Under: Problems on Numbers
Exam Prep: Bank Exams

1 19800

Q:

The ratios of areas of two squares, one having its diagonal double than the other is

A) 2:1	B) 2:3
C) 3:1	D) 4:1

Answer & Explanation Answer: D) 4:1

Explanation:

ratio = $\frac{(1 / 2) \times {(2 d)}^{2}}{\frac{1}{2} d^{2}} = 4 : 1$

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

49 19796

Q:

In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case ?

A) 693	B) 741
C) 528	D) 654

Answer & Explanation Answer: C) 528

Explanation:

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k

As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)

x = ([(1200)(30)(3)] / (25)(2.5)) - 1200 => x = 528.

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Filed Under: Ratios and Proportions
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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30 19727

Q:

From a tank of petrol , which contains 200 liters of petrol, the seller replaces each time with kerosene when he sells 40 liters of petrol(or mixture). Everytime he sells out only 40 liters of petrol (pure or impure). After replacing the petrol with kerosene 4th time, the total amount of kerosene in the mixture is

A) 81.92L	B) 96L
C) 118.08L	D) None of these

Answer & Explanation Answer: C) 118.08L

Explanation:

The amount of petrol left after 4 operations = $200 \times {(1 - \frac{40}{200})}^{4} =$ 81.92

Hence the amount of kerosene = 200 - 81.92 = 118.08 liters

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Filed Under: Alligation or Mixture

24 19722

Quantitative Aptitude - Arithmetic Ability Questions

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