In a race of 1000 meters, A beats B by 200 meters and A beats C by 500 meters. Assuming that the contestants run at constant speeds, by how many meters does B beat C?

(A) 300 (B) 375 (C) 450 (D) 500 (E) 625

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- Feb 8th 2010, 08:35 AMsri340race problem
In a race of 1000 meters, A beats B by 200 meters and A beats C by 500 meters. Assuming that the contestants run at constant speeds, by how many meters does B beat C?

(A) 300 (B) 375 (C) 450 (D) 500 (E) 625 - Feb 8th 2010, 08:52 AMArchie Meade
A speed = $\displaystyle \frac{1000}{t}$

B speed = $\displaystyle \frac{800}{t}$

C speed = $\displaystyle \frac{500}{t}$

After t, B has 200 metres to travel, an extra quarter of his distance already travelled in t (not a fifth of 1000, it's a quarter of 800).

This will take him an extra $\displaystyle \frac{t}{4}$

In $\displaystyle \frac{t}{4}$, C will have travelled a quarter of 500,

which is 125 metres.

Therefore C still has 1000-(500-125) metres to travel, when B crosses the line. - Feb 8th 2010, 01:14 PMSoroban
Hello, sri340!

Another approach (very primitive) . . .

Quote:

In a 1000-m race, $\displaystyle A$ beats $\displaystyle B$ by 200 meters and $\displaystyle A$ beats $\displaystyle C$ by 500 meters.

Assuming that the contestants run at constant speeds, by how many meters does $\displaystyle B$ beat $\displaystyle C$?

. . $\displaystyle (A)\;300 \qquad(B)\;375 \qquad (C)\;450 \qquad (D)\;500 \qquad (E)\; 625$

The race ended like this:Code:`C B A`

+ - - - - - - - - - o - - - - - o - - - o

0 500 800 1000

$\displaystyle C$ ran 500 m in the time that $\displaystyle B$ ran 800 m.

C's speed is $\displaystyle \tfrac{500}{800} = \tfrac{5}{8}$ of $\displaystyle B$'s speed.

When $\displaystyle B$ ran his last 200 m, $\displaystyle C$ ran only: .$\displaystyle \tfrac{5}{8}(200) \,=\,125$ m.

As $\displaystyle B$ crosses the finish line, $\displaystyle C$ is at the 625-m mark.

Therefore, $\displaystyle B$ beats $\displaystyle C$ be 375 m . . .*answer (B).*