# 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

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.

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

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

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