$A$ can run around a circular track in $40$ seconds. $B$ , running in the opposite direction, meets $A$ every $15$ seconds. What is $B^{'} s$ time to run around the track, expressed in seconds?

12 \frac{1}{2}

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24

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25

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Solution

The correct option is B $24$
Let $x = B^{'} s$ time in seconds.
Let $l$ represent the length of the track.
We have $\frac{l}{40} (15) + \frac{l}{x} (15) = 1$
$. ∴ x = 24$ .

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A can run around a circular track in 40 seconds. B, running in the opposite direction, meets A every 15 seconds. What is B′s time to run around the track, expressed in seconds?

The correct option is B 24Let x=B′s time in seconds. Let l represent the length of the track.We have l40(15)+lx(15)=1.∴x=24.

$A$ can run around a circular track in $40$ seconds. $B$ , running in the opposite direction, meets $A$ every $15$ seconds. What is $B^{'} s$ time to run around the track, expressed in seconds?

The correct option is B $24$
Let $x = B^{'} s$ time in seconds.
Let $l$ represent the length of the track.
We have $\frac{l}{40} (15) + \frac{l}{x} (15) = 1$
$. ∴ x = 24$ .