RRB Group-D Model Paper 15 with answers for online practice

Let the width of floor be x m
⇒ Length of the floor = x × 125/100 = 1.25x
As we know,
Area of the floor = length × width
According to the question
x × 1.25x = 20
⇒ ${\mathit{x}}^2$ = 20/1.25
⇒ ${\mathit{x}}^2$ = 16
⇒ x = $\sqrt{16}$
⇒ x = 4
∴ Sum of length and width of the floor = x + 1.25x = 2.25x = 2.25 × 4 = 9 m