RRB NTPC - Problems on Areas : Aptitude Test

1.

A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? 

   A.) 98
   B.) 88
   C.) 99
   D.) 89

Answer: Option 'B'

Given that area of the field = 680 sq. feet
=> lb = 680 sq. feet
Length(l) = 20 feet
=> 20 × b = 680
=> b = 680/20 = 34 feet
Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet

2.

The area of the largest triangle that can be inscribed in a semicircle pf rasius r Cm, is:

   A.) r3 Cm2
   B.) r2 Cm3
   C.) r2 Cm2
   D.) r3 Cm3

Answer: Option 'C'

Area of the largest triangle = (1/2 × 2r × r) cm2 = r2 cm2

3.

A girl walking at the rate of 9 Km per hour crosses a square field diagonally in 12 seconds. The area of the field is: 

   A.) 460 sq.m
   B.) 600 sq.m
   C.) 510 sq.m
   D.) 450 sq.m

Answer: Option 'D'

Distance covered in (9×1000)/(3600) × 12 = 30 m 
Diagonal of squarre field = 30 m. 
area of square field = 302/2
= 900/2 = 450 Sq.m

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 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

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