CAT 2019 Quant Question with Solution 19

Correct Answer: Option: 3

The distance travelled by bicycle A in one revolution = $2\pi {{r}_{a}}=2\pi \times 30=60\pi \ cm$

The distance travelled by bicycle B in one revolution = $2\pi {{r}_{b}}=2\pi \times 40=80\pi \ cm$

Let B makes ‘n’ revolutions to cover the distance. Then, A would make (n+5000) to cover the same distance.

$\therefore n\times 80\pi =(n+5000)\times 60\pi $ $\Rightarrow n=15000$

Distance travelled by B = $n\times 80\pi \ cm=\frac{15000\times 80\pi }{{{10}^{5}}}\ km=12\ km$

Time taken by B= 45 min = $\frac{45}{60}=\frac{3}{4}hrs$

Hence the speed of B = $\frac{12\pi }{3/4}=16\pi \ km/h$