Management Aptitude Test Sep 2013 Paper

(b) Radius of inner circle = x m

Radius of outer circle = (x + 1.22) m
Now, inner wheel taken 1 revolution for every 1$\frac{1}{2}$ revolution of outer wheel
So, distance covered by outer wheel is 1$\frac{1}{2}$ times the inner wheel.
∴ $\frac{3}{2}$ x 2πx = 2π(x + 122)
3πx= 2π(x + 122)
3x = 2x + 2.44 ⇒ x = 2.44
∴ Distance covered by outer wheel
= 2π(1.22 + 2.44) = 2π(3.66)
= 2 x $\frac{22}{7}$ x 3.66 ≈ 23 m