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Chethan92
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mohitranjan05
In a 100m race, A beats B by 20m. B beats C by 20m. What is the distance by which A beats C ?

A. 20
B. 36
C. 40
D. 50
E. cannot be determined

The prompt should state that the rates are constant

In a 100m race, A beats B by 20m.
Since B is 20 meters behind A when A completes the 100-meter race, B travels 80 meters in the time it takes A to travel 100 meters:
\(\frac{A}{B} = \frac{100}{80} = \frac{5}{4}\)

B beats C by 20m.
Since C is 20 meters behind B when B completes the 100-meter race, C travels 80 meters in the time it takes B to travel 100 meters:
\(\frac{B}{C} = \frac{100}{80} = \frac{5}{4}\)

Thus:
\(\frac{A}{C} = \frac{A}{B }* \frac{B}{C} = \frac{5}{4} * \frac{5}{4} = \frac{25}{16} = \frac{100}{64}\)

Since \(\frac{A}{C} = \frac{100}{64}\), C travels 64 meters in the time it takes A to complete the 100-meter race, with the result that A beats C by 36 meters.

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Hi. I could not see the question right below the timer. It was the solution provided by Chethan92 directly. Kindly help.
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Hi. I could not see the question right below the timer. It was the solution provided by Chethan92 directly. Kindly help.
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Fixed the issue. Thank you!
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