JAMB 2017 · UTME (set 2) · Q19

Commercial arithmetic

A man covered a distance of 50 miles on his first trip. On a later trip he travelled 300 miles while going 3 times as fast. His new time compared with the old time was

Worked solution (try it first)

Let the old speed be $v$ .
Time is distance over speed, so the old time is $\dfrac{50}{v}$ .
The new speed is $3v$ , so the new time is $\dfrac{300}{3v} = \dfrac{100}{v}$ .
$\dfrac{100}{v}$ is twice $\dfrac{50}{v}$ , so the new time is twice as much, option C.

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