A can beat B by 20 yards in a race of 200 yards. B can beat C by 10 yards in a race of 250 yards. By how many yards can A beat C in a race of 100 yards?

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- Nov 13th 2012, 07:26 PMhisajeshRace problem
A can beat B by 20 yards in a race of 200 yards. B can beat C by 10 yards in a race of 250 yards. By how many yards can A beat C in a race of 100 yards?

- Nov 13th 2012, 08:27 PMSorobanRe: Race problem
Hello, hisajesh!

Quote:

A can beat B by 20 yards in a 200-yard race.

B can beat C by 10 yards in a 250-yard race.

By how many yards can A beat C in a 100-yard race?

Let $\displaystyle \begin{Bmatrix}a &=& \text{A's speed} \\ b &=& \text{B's speed} \\ c &=& \text{C's speed} \end{Bmatrix}$

In the time A has run 200 yards, B has run only 180 yards.

B's speed is $\displaystyle \tfrac{180}{200} = \tfrac{9}{10}$ of A's speed: .$\displaystyle b = \tfrac{9}{10}a$ .[1]

In the time B gas run 250 yards, C has run only 240 yards.

C's speed is $\displaystyle \tfrac{240}{250} = \tfrac{24}{25}$ of B's speed: .$\displaystyle c = \tfrac{24}{25}b$ .[2]

Substitute [1] into [2]: .$\displaystyle c \:=\:\tfrac{24}{25}(\tfrac{9}{10}a) \quad\Rightarrow\quad c \:=\:\tfrac{108}{125}a$

. . That is, C's speed is $\displaystyle \tfrac{108} {125}$ of A's speed.

When A has run 100 yards, C has run only $\displaystyle \tfrac{108} {125}(100) \:=\:86.4$ yards.

Therefore, A wins by $\displaystyle 13.6$ yards.

- Nov 17th 2012, 07:16 PMhisajeshRe: Race problem
@Soroban: Thanks for the detailed explanation!