A runs 1 frac{3}{8} times as fast as B. If A | vedclass

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(B) The ratio of the speeds of $A$ and $B$ is $1 \frac{3}{8} : 1 = \frac{11}{8} : 1 = 11 : 8$.Let the speed of $A$ be $11v$ and the speed of $B$ be $8

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$440$

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(B) The ratio of the speeds of $A$ and $B$ is $1 \frac{3}{8} : 1 = \frac{11}{8} : 1 = 11 : 8$.Let the speed of $A$ be $11v$ and the speed of $B$ be $8v$.Let the total distance of the race be $x \, m$.Since $A$ gives $B$ a start of $120 \, m$,$B$ covers a distance of $(x - 120) \, m$ while $A$ covers $x \, m$.Since they reach the goal at the same time,the time taken by both is equal.Time = $\frac{	ext{Distance}}{	ext{Speed}}$.Therefore,$\frac{x}{11v} = \frac{x - 120}{8v}$.Canceling $v$ from both sides,we get $\frac{x}{11} = \frac{x - 120}{8}$.Cross-multiplying gives $8x = 11(x - 120)$.$8x = 11x - 1320$.$11x - 8x = 1320$.$3x = 1320$.$x = 440 \, m$.

$A$ runs $1 \frac{3}{8}$ times as fast as $B$. If $A$ gives $B$ a start of $120 \, m$ and they reach the goal at the same time,the goal is at a distance of (in $m$)

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