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A man covered 1 3 of the distance to his destination at 20 miles per h
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26 Jan 2024, 16:29
There's no mention of actual distance, only of actual rates. That means we can't determine actual total time, which rules out (B).
Let's say actual total distance = 360.
Distance traveled at 20mph \(= \frac{1}{3} * 360 = 120\)
1/2 the remaining distance \(\frac{1}{2} * (360 - 120) = 120\)
The remaining distance \(= 360 - (120+120) = 120\)
Let a = Time taken to travel the part of the distance at 20 mph, b = Time taken to travel the part of the distance at 12 mph, and c = Time taken to travel the part of the distance at 40 mph.
Using the distance-rate formula, t = d/r :
\(a = 120 / 20 = 6\)
\(b = 120 / 36 = \frac{10}{3}\)
\(c = 120 / 40 = 3\)
Since c = 1/2a, (A) is correct.
The average speed for the entire journey = \(\frac{360}{6 + \frac{10}{3} + 3} = 18.95\), so (C) is incorrect.