6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

A man runs round a circular field of radius 50m at the speed of 12 km/hr. What is the time taken by the man to take twenty rounds of the field?

A) 220/7 min	B) 110/7 min
C) 90/7 min	D) 230/7 min

Answer & Explanation Answer: A) 220/7 min

Explanation:

speed = 12 km/h = $12 \times \frac{5}{18} = \frac{10}{3} m / s$

distance covered = $20 \times 2 \times \frac{22}{7} \times 50 = \frac{44000}{7} m$

time taken = distance /speed = $\frac{44000}{7} \times \frac{3}{10} s e c = \frac{4400 \times 3}{7} \times \frac{1}{60} m i n = \frac{220}{7} m i n$

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31 20310

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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13 20266

Q:

The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m.What is the area of the field?

A) 47500	B) 37500
C) 57500	D) 77500

Answer & Explanation Answer: B) 37500

Explanation:

Let length = x meters, then breadth = 0.6x

Given that perimeter = 800 meters

=> 2[ x + 0.6x] = 800

=> x = 250 m

Length = 250m and breadth = 0.6 x 250 = 150m

Area = 250 x 150 = 37500 sq.m

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19 20112

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 20112

Q:

The no of revolutions a wheel of diameter 40cm makes in traveling a distance of 176m is

A) 240	B) 140
C) 40	D) 340

Answer & Explanation Answer: B) 140

Explanation:

distance covered in 1 revolution = $2 πr$ = 2 x (22/7) x 20 = 880/7 cm
required no of revolutions = 17600 x (7/880) = 140

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21 20104

Q:

A can do a piece of work in 30 days. He works at it for 6 days and then B finishes it in 18 days. In what time can A and B together it ?

A) 14 1/2 days	B) 11 days
C) 13 1/4 days	D) 12 6/7 days

Answer & Explanation Answer: D) 12 6/7 days

Explanation:

Let 'B' alone can do the work in 'x' days

6/30 + 18/x = 1

=> x = 22.5

1/30 + 1/22.5 = 7/90

=> 90/7 = 12 6/7 days

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29 20057

Q:

Vinod have 20 rupees. He bought 1, 2, 5 rupee stamps. They are different in numbers by the reason of no change, the shop keeper gives 3 one rupee stamps. So how many stamps Vinod have ?

A) 10	B) 18
C) 12	D) 15

Answer & Explanation Answer: A) 10

Explanation:

Given total rupees = 20 Rs
No. of one rupee stamps = 3
Now, remaining money = Rs. 17
With that he buys only 2 and 5 rupee stamps
Let number of Rs. 5 stamps = K
Let number of Rs. 2 stamps = L
5K + 2L = 17
K = 3, L = 1 (possible)
L = 6, K = 1 (possible)
=> But given that they are different in number so, K is not equal to 3
one rupee stamps = 3
2 two stamps = 6
5 rupee stamps = 1
Total number of stamps = 10.

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Filed Under: Ratios and Proportions
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45 20007

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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30 20003

Quantitative Aptitude - Arithmetic Ability Questions

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