Review question

# How long was the walk to Glastonbury Tor? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6121

## Solution

A school party walked from their camp to Glastonbury Tor at an average speed of $\quantity{4}{km/h}$, and returned by bus at an average speed of $\quantity{30}{km/h}$. The return journey was $\quantity{5}{km}$ longer than the outward journey, but took two hours less. How far did they walk?

Let us say that the distance that the school party walked is $\quantity{x}{km}$. We will express the information given to us by the question algebraically.

Firstly, we know that the school party took $\frac{\quantity{x}{km}}{\quantity{4}{km/h}} = \quantity{\frac{x}{4}}{hours}$ to walk to Glastonbury Tor.

Secondly, we know that the return journey by bus was $\quantity{(x + 5)}{km}$ in length.

Thirdly, we know that the time on the bus journey, therefore: $\frac{\quantity{(x + 5)}{km}}{\quantity{30}{km/h}} = \quantity{\frac{x+5}{30}}{hours},$ was $2$ hours less than the outward journey on foot. That is, we know that $\frac{x+5}{30} = \frac{x}{4} - 2.$

We can rearrange this final equation to find the value of $x$: \begin{align*} \frac{x+5}{30} = \frac{x}{4} - 2 &\implies \frac{x}{4} - \frac{x}{30} = \frac{5}{30} + 2 \\ &\implies \frac{15x - 2x}{60} = \frac{1 + 12}{6} \\ &\implies \frac{13x}{60} = \frac{13}{6} \\ &\implies 13x = 130 \\ &\implies x = 10. \end{align*}

That is, the school party walked $\quantity{10}{km}$.