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. |
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