6500+ Quantitative Aptitude Questions and Answers With Explanation

Q:

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

A) CHM	B) GLQ
C) KPV	D) NSX

Answer & Explanation Answer: C) KPV

Explanation:

Report Error

View Answer Report Error Discuss

Filed Under: Odd Man Out
Exam Prep: Bank Exams

0 29316

Q:

Entry fee in an exhibition was Rs. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is :

A) 20 %	B) 40 %
C) 60 %	D) 80 %

Answer & Explanation Answer: C) 60 %

Explanation:

Let the total original sale be Rs. 100. Then, original number of visitors = 100.

New number of visitors = 120/0.75 = 160.

Increase % = 60 %.

Report Error

View Answer Report Error Discuss

Filed Under: Percentage

95 29201

Q:

If the circumference of a circle is 88 cm, then what must be its area (in cm2)?

A) 1232	B) 616
C) 704	D) 1408

Answer & Explanation Answer: B) 616

Explanation:

Report Error

View Answer Report Error Discuss

Filed Under: Area
Exam Prep: Bank Exams

3 29197

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.

Report Error

View Answer Report Error Discuss

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

62 29196

Q:

The average temperature of the town in the first four days of a month was 58 degrees. The average for the second, third, fourth and fifth days was 60 degrees. If the temperatures of the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day ?

A) 62 degrees	B) 64 degrees
C) 65 degrees	D) 66 degrees

Answer & Explanation Answer: B) 64 degrees

Explanation:

Sum of temperatures on 1st, 2nd, 3rd and 4th days = (58 * 4) = 232 degrees ... (1)

Sum of temperatures on 2nd, 3rd, 4th and 5th days - (60 * 4) = 240 degrees ....(2)

Subtracting (1) From (2), we get :

Temp, on 5th day - Temp on 1st day = 8 degrees.

Let the temperatures on 1st and 5th days be 7x and 8x degrees respectively.

Then, 8x - 7x = 8 or x = 8.

Temperature on the 5th day = 8x = 64 degrees.

Report Error

View Answer Report Error Discuss

Filed Under: Average

56 29186

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

Report Error

View answer Workspace Report Error Discuss

Subject: Time and Work
Job Role: Bank PO

72 29170

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.

Report Error

View Answer Report Error Discuss

Filed Under: Alligation or Mixture

40 29106

Q:

A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?

A) 315	B) 385
C) 415	D) 485

Answer & Explanation Answer: A) 315

Explanation:

Let the areas of the two parts be x and (700-x) hectares

therefore, $[x - (700 - x)] = \frac{1}{5} [\frac{x + (700 - x)}{2}]$

$\Rightarrow 2 x - 700 = 70$

$\Rightarrow x = 385$

So, the two parts are 385 and 315.

Hence, Area of the smaller = 315 hectares

Report Error

View Answer Report Error Discuss

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

18 29101

Quantitative Aptitude - Arithmetic Ability Questions

Quick Links