RRB NTPC - Problems on Areas : Aptitude Test

1.

The area of the largest circle that can be drawn inside a square of side 28 Cm in length is:

   A.) 616 Cm3
   B.) 626 Cm2
   C.) 619 Cm2
   D.) 616 Cm2

Answer: Option 'D'

Radius of required circle = 14 Cm 
its area = 22/7 × 14 × 14 
44 × 14 = 616 Cm2

2.

A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet? a

   A.) 126 sq. ft.
   B.) 128 sq. ft.
   C.) 136 sq. ft.
   D.) 116 sq. ft.

Answer: Option 'A'

Let l = 9 ft.
Then l + 2b = 37
=> 2b = 37 - l = 37 - 9 = 28
=> b = 28/2 = 14 ft
Area = lb = 9 × 14 = 126 sq. ft.

3.

The ratio of the areas of two squares, one having double its diagonal then the other is:

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

Answer: Option 'D'

Lenth of the diagonals be 2x and x units. 
areas are 1/2 × (2x)2 and (1/2 × x2
Required ratio = 1/2 × 4 x2 : 1/2 x2 = 4 : 1

4.

A large field of 900 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.) 395
   B.) 385
   C.) 415
   D.) 405

Answer: Option 'C'

Let the areas of the parts be x hectares and (900 - x) hectares. 
Difference of the areas of the two parts = x - (900 - x) = 2x - 900
one-fifth of the average of the two areas = 1/5[x+(900−x)]/2 
= 1/5 × (900/2) = 450/5 = 90 
Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x - 900 = 90
=> 2x = 990 
⇒ x = 990/2 = 495
Hence, area of smaller part = (900 - x) = (900 – 495) = 405 hectares.

5.

The diagonal of a rhombus are 65 m and 60 m. Its area is:

   A.) 1950 cm^3
   B.) 1950 cm^2
   C.) 1960 cm^2
   D.) 1960 cm^3

Answer: Option ''

Area of the rhombus = 1/2 d1d2 = ( 1/2 × 65 × 60 ) cm2 
= 65 × 30 = 1950 cm2

Areas in Mensuration Download Pdf