WAEC 2021 · Paper 2 · Q2

(a)
A man left town $M$ at 10:00 a.m. and travelled by car to town $N$ at an average speed of $72\text{ km/h}$ . He spent 2 hours for a meeting and returned to town $M$ by bus at an average speed of $40\text{ km/h}$ . If the distance covered by the bus was $2\text{ km}$ longer than that of the car and he arrived at town $M$ at 1:55 p.m., calculate the distance from $M$ to $N$ .

Worked solution (try it first)

From 10:00 a.m. to 1:55 p.m. is 3 hours 55 minutes.
Take off the 2-hour meeting: he spent 1 hour 55 minutes travelling, which is $1\frac{55}{60} = \frac{23}{12}$ hours.
Let the car's distance from $M$ to $N$ be $x$ km.
The bus travelled $x + 2$ km.
Time $= \frac{\text{distance}}{\text{speed}}$ , so the travelling time is $\frac{x}{72} + \frac{x + 2}{40} = \frac{23}{12}$ .
Multiply every term by 360: $5x + 9(x + 2) = 690$ .
So $14x + 18 = 690$ , $14x = 672$ and $x = 48$ .
The distance from $M$ to $N$ is 48 km.