KarunMendiratta wrote:
Car A, B, and C participate in a car race. Car A beats car B by 45m, car B beats car C by 50m, and car A beats car C by 90m.
Quantity A |
Quantity B |
Distance over which the race was conducted |
380m |
Let the distance = d.
Car A beats car B by 45m.In the time it takes A to travel d meters, B travels d-45 meters.
Thus:
\(\frac{A}{B }= \frac{d}{d-45}\)
Car A beats car C by 90m.In the time it takes A to travel d meters, C travels d-90 meters.
Thus:
\(\frac{A}{C} = \frac{d}{d-90}\)
Combining the resulting ratios, we get:
\(\frac{B}{C} = \frac{B}{A} * \frac{A}{C} = \frac{d-45}{d} * \frac{d}{d-90} = \frac{d-45}{d-90}\)
Car B beats car C by 50m.In the time it takes B to travel d meters, C travels d-50 meters.
Thus:
\(\frac{B}{C} = \frac{d}{d-50}\)
Since\( \frac{B}{C} = \frac{d-45}{d-90}\) and \(\frac{B}{C} = \frac{d}{d-50}\), we get:
\(\frac{d-45}{d-90} = \frac{d}{d-50}\)
Cross-multiplying, we get:
\(d^2 - 90d = d^2 - 95d + 2250\)
\(5d = 2250\)
\(d = 450\)
Quantity A is greater.
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