JAMB 2018 · UTME · Q39

Commercial arithmetic

Two cars $X$ and $Y$ start at the same point and travel towards a point $P$ which is $150\text{ km}$ away. If the average speed of $Y$ is $60\text{ km/h}$ and $X$ arrives at $P$ 25 minutes earlier than $Y$ , what is the average speed of $X$ ?

Worked solution (try it first)

Time is distance over speed: $Y$ takes $\dfrac{150}{60} = 2\frac12$ hours.
$X$ arrives 25 minutes earlier. 25 minutes is $\frac{25}{60} = \frac{5}{12}$ hour, so $X$ takes $2\frac12 - \frac{5}{12} = \frac{25}{12}$ hours.
Speed of $X$ $= 150 \div \dfrac{25}{12}$
$= 150 \times \dfrac{12}{25}$
$= 72$ km/h, option B.

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