6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

A) B=900;H=300	B) B=300;H=900
C) B=600;H=700	D) B=500;H=900

Answer & Explanation Answer: A) B=900;H=300

Explanation:

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
$13.5 * 10000 m^{2} = 135000 m^{2}$
Let altitude = x metres and base = 3x metres.
Then, $\frac{1}{2} * 3 x * x = 135000 \Rightarrow x^{2} = 90000 \Rightarrow x = 300$

Base = 900 m and Altitude = 300 m.

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Filed Under: Area
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62 28672

Q:

In the following question, select the odd letter/letters from the given alternatives.

A) CGJ	B) IMQ
C) OSW	D) JNR

Answer & Explanation Answer: A) CGJ

Explanation:

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Q:

The diameter of the driving wheel of a bus is 140 cm. How many revolution, per minute must the wheel make in order to keep a speed of 66 kmph ?

A) 150	B) 250
C) 350	D) 550

Answer & Explanation Answer: B) 250

Explanation:

Circumference = No.of revolutions * Distance covered

Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.
Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.

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47 28522

Q:

A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.

A) 551/15134	B) 1/2
C) 552/15379	D) 1/9

Answer & Explanation Answer: A) 551/15134

Explanation:

Total number of elementary events = $50 C_{5}$
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = $29 C_{2}$ ways and remaining two in $20 C_{2}$ ways.

Therfore,favourable number of events = $29 C_{2} * 20 C_{2}$

Hence,required probability = $29 C_{2} * 20 C_{2}$ $/ 50 C_{5}$ =551 / 15134

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72 28486

Q:

A and B can do a piece of work in 40 and 50 days. If they work at it an alternate days with A beginning in how many days, the work will be finished ?

Answer

(A+B)'s two days work = $\frac{1}{40} + \frac{1}{50} = \frac{9}{200}$

Evidently, the work done by A and B duing 22 pairs of days

i.e in 44 days = $22 \times \frac{9}{200} = \frac{198}{200}$

Remaining work = $1 - \frac{198}{200}$ = 1/100

Now on 45th day A will have the turn to do 1/100 of the work and this work A will do in $40 \times \frac{1}{100} = \frac{2}{5}$

Therefore, Total time taken = 44 $\frac{2}{5}$ daya

Workspace

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Subject: Time and Work
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72 28457

Q:

Select the odd figure out of the following series.

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

Answer & Explanation Answer: A) 3

Explanation:

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Filed Under: Odd Man Out
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Q:

In the following question, select the odd word from the given alternatives.

A) Ball	B) Bat
C) Wickets	D) Ground

Answer & Explanation Answer: D) Ground

Explanation:

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Q:

From a container of wine, a thief has stolen 15 liters of wine and replaced it with same quantity of water.He again repeated the same process. Thus in three attempts the ratio of wine and water became 343:169. The initial amount of wine in the container was:

A) 75 liters	B) 100 liters
C) 150 liters	D) 120 liters

Answer & Explanation Answer: D) 120 liters

Explanation:

$\frac{W i n e (l e f t)}{W a t e r (a d d e d)} = \frac{343}{169}$

It means $\frac{W i n e (l e f t)}{W i n e (i n i t i a l a m o u n t)} = \frac{343}{512}$ $[∵ 343 + 169 = 512]$

Thus ,

$343 x = 512 x {(1 - \frac{15}{K})}^{3}$

$\Rightarrow \frac{343}{512} = {(\frac{7}{8})}^{3} = {(1 - \frac{15}{k})}^{3}$

=> K = 120

Thus the initial amount of wine was 120 liters.

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Filed Under: Alligation or Mixture

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Quantitative Aptitude - Arithmetic Ability Questions

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