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A cyclist leaves $A$ at $10$ am and reaches $B$ at $11$ am. Starting from $10:01$ am, every minute a motor cycle leaves $A$ and moves towards $B$. Forty-five such motor cycles reach $B$ by $11$ am. All motor cycles have the same speed. If the cyclist had doubled his speed, how many motor cycles would have reached $B$ by the time the cyclist reached $B$?

- $23$
- $20$
- $15$
- $22$

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Ans is option **(C)**

Let the distance between point A and point B be $d$ km. Time taken by the cyclist to cover $d$ km is $1$hr.

$\therefore$ Initial Speed of the cyclist $v_{c}=\frac{d}{1}=d$ kmph.

Now, every minute from $10:01$ A.M., a bike leaves point A, and $45$ such bikes reach point B after $1$ hr.

It is given that speed of each bike is same. So, last bike will leave at $10:45$ A.M. and reach B at $11:00$ A.M.

$\therefore$ time taken by a bike to get from point A to point B $=15$ minutes.

Now, if the cyclist doubles his speed, he will reach point B in $\frac{d}{2d}=0.5$ hr $=$ $30$ minutes. He will reach point B at $10:30$ A.M.

Now, starting from $10:01$ A.M., first bike reaches point B at $10:16$ A.M.

Second bike reaches point B at $10:17$ A.M. and so on...till $15^{th}$ bike which starts at $10:15$ A.M. reaches point B at $10:30$ A.M.

Hence a total of $15$ bikes will reach B by the time cyclist reaches B.