Answer: Option C
Distance covered by smaller wheel in 1 revolution :
$$\eqalign{
& = \left( {2\pi \times 15} \right)cm \cr
& = \left( {30\pi } \right)cm \cr} $$
Distance covered by larger wheel in 1 revolution :
$$\eqalign{
& = \left( {2\pi \times 25} \right)cm \cr
& = \left( {50\pi } \right)cm \cr} $$
Let,
$$k \times 30\pi = 15 \times 50\pi $$
Then,
$$k = \left( {\frac{{15 \times 50\pi }}{{30\pi }}} \right) = 25$$
∴ Required number of revolution = 25