Two friends A and B simultaneously start running around a circular track. They run in the same direction. A travels at $6$ $m/s$ and B travels at $b$ $m/s$ . If they cross each other at exactly two points on the circular track and $b$ is a natural number less than 30, how many values can $b$ take?

Hint: Find relative velocity of A with respect to B or B with respect to A. Then calculate time taken to meet for the first time and see how many values satisfy the equation.

Complete step by step answer:
Let the total length of the track be equal to $L$ .
Relative speed of A with respect to B $ = b - 6$
Hence time when they meet for the first time $ = \dfrac{L}{{b - 6}}$
Time taken by A to complete one full lap of the track $ = \dfrac{L}{6}$
Time taken by B to complete one full lap of the track $ = \dfrac{L}{b}$
Thus time when they meet at the starting point for the first time $ = \dfrac{L}{{HCF(b,6)}}$
Number of times they meet on starting point = time taken to meet at the starting point / time taken for meeting the first time
 $ = \dfrac{{b - 6}}{{HCF(b,6)}}$
This is equal to $2$ according to the question
Therefore,
 $\dfrac{{b - 6}}{{HCF(b,6)}} = 2$
Since less then $30$ , only values of $b$ that satisfy the above equation are $2,10,18$ .
Hence there are $3$ values that $b$ can take.

Note: We have to notice that there are multiple values of $b$ that are possible but our answer is limited to these three values since the question says that only values that are natural numbers less than $30$ are to be taken. Hence our answer is this. Also it should be noted that the highest common factor needs to be calculated by putting multiple values for $b$ and checking which ones satisfy the equation. The relative speed can be $b - 6$ or $6 - b$ since we are taking it relative so modulus of it is what we have to take.