Question

There are two concentric circular tracks of radii $100\mathrm{m}$ and $102\mathrm{m}$ respectively. A runs on the inner track and goes once round the track in $1$ minute $30$ seconds while B runs on outer track in $1$ minute $32$ seconds. Who runs faster ?

Answer 1

Step 1: Determine the speed of A.

The circumference of the circle $\left(C\right)$ is calculated by $C=2\mathrm{\pi r}$ where $r$ represents the radius of the circle.

Circumference of the inner circle

$\begin{array}{rcl}& =& 2\times \mathrm{\pi }\times \times 100\\ & =& 200\mathrm{\pi }\mathrm{m}\end{array}$

A covers a distance equal to the circumference of the inner track in $1$ minute $30$ seconds i.e. in $\frac{3}{2}$ minutes.
So, the distance travelled by A in $\frac{3}{2}$ minutes $=200\pi \mathrm{m}$

Distance travelled by A in $1$ minute $=200\mathrm{\pi }\times \frac{2}{3}=133.33\mathrm{\pi }\mathrm{m}$

$\mathrm{Speed}\mathrm{of}\mathrm{A}=133.33\mathrm{\pi }\mathrm{m}/\min$

Step 2: Determine the speed of B.

$\begin{array}{rcl}\mathrm{Circumference}\mathrm{of}\mathrm{the}\mathrm{outer}\mathrm{circle}& =& 2\times \mathrm{\pi }\times \times 102\\ & =& 204\mathrm{\pi }\mathrm{m}\end{array}$

B covers a distance equal to the circumference of the outer track in $1$ minute $32$ seconds i.e. in $1+\frac{32}{60}=\frac{23}{15}$ minutes.
So, the distance travelled by B in $\frac{23}{15}$ minutes $=204\pi \mathrm{m}$

Distance travelled by B in $1$ minute $=204\mathrm{\pi }\times \frac{15}{23}=133.04\mathrm{\pi }\mathrm{m}$

$\mathrm{Speed}\mathrm{of}\text{B}=133.04\mathrm{\pi }\mathrm{m}/\min$

Hence, A runs faster because the speed of A is greater than the speed of B.