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Two friends $\mathrm{A}$ and $\mathrm{B}$ are $30 \mathrm{~km}$ apart and they start simultaneously on motorcycles to meet each other. The speed of A is 3 times that of B. The distance between them decreases at the rate of $2 \mathrm{~km}$ per minute. Ten minutes after they start, A's vehicle breaks down and A stops and waits for $\mathrm{B}$ to arrive. After how much time (in minutes) A started riding, does B meet $\mathrm{A}$ ?
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Verified Answer
The correct answer is:
30
Speed of $\mathrm{B}=\mathrm{V} \mathrm{km} / \mathrm{hr}$
Speed of $\mathrm{A}=3 \mathrm{~V} \mathrm{~km} / \mathrm{hr}$
Given $4 \mathrm{~V}=2 \times 60 \mathrm{~km} / \mathrm{hr}$
$\Rightarrow \mathrm{V}=30 \mathrm{~km} / \mathrm{hr}$
Distance covered by then after $10 \mathrm{~min}=2 \times 10=20 \mathrm{~km}$
so remaining distance $=(30-20) \mathrm{km}=10 \mathrm{~km}$
Time taken by $\mathrm{B}$ to cover $10 \mathrm{~km}=\frac{10}{30 / 60}=20 \mathrm{~min}$.
Total time $=20+10=30 \mathrm{~min}$
Speed of $\mathrm{A}=3 \mathrm{~V} \mathrm{~km} / \mathrm{hr}$
Given $4 \mathrm{~V}=2 \times 60 \mathrm{~km} / \mathrm{hr}$
$\Rightarrow \mathrm{V}=30 \mathrm{~km} / \mathrm{hr}$
Distance covered by then after $10 \mathrm{~min}=2 \times 10=20 \mathrm{~km}$
so remaining distance $=(30-20) \mathrm{km}=10 \mathrm{~km}$
Time taken by $\mathrm{B}$ to cover $10 \mathrm{~km}=\frac{10}{30 / 60}=20 \mathrm{~min}$.
Total time $=20+10=30 \mathrm{~min}$
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