question_answer

A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7:30 am. They will first cross each other at

A)  \[8:30\,\,am\]               

B)  \[8:10\,\,am\]

C)  \[7:48\,\,am\]               

D)  \[7:42\,\,am\]

Correct Answer: D

Solution :

[d] As, time taken for 1 round by A = 1 h

\[\therefore \] Time taken by B for 1 round \[=\frac{1}{6}h=10\,\min \]

The angle made by A on centre in 10 min

\[=\frac{360{}^\circ \times 10}{60}=60{}^\circ \]

Since, A and B both started walking from the point X.

So, after 10 min A will be on point Y and B on point X.

Suppose that, B will catch A in Z min.

\[\therefore \] \[\frac{Z}{10}=Z\left( \frac{1}{60} \right)+\frac{60{}^\circ }{360{}^\circ }\Rightarrow \frac{Z}{10}-\frac{Z}{60}=\frac{1}{6}\]

\[\Rightarrow\frac{6Z-Z}{60}=\frac{1}{6}\Rightarrow 5Z=10\Rightarrow Z=\frac{10}{5}=2\]

\[\therefore \] B will hold A in 10+2= 12 min

Thus, the required time is 7: 42 am.