100% + 25% = 1 + $\frac14$ = $\frac54$

speed $\times$ time = distance

$\times \frac54$ $\times$ ? $\times$ 1 because distance does not change

(new speed) $\times$ (new time) = distance

$\frac 54 \times$ ? = 1, so ? = $\frac45$ = 80%

80% of original time is a 20% reduction

The area on a speed-time graph represents distance.

Yellow rectangle - before 1 year training

Green rectangle - after 1 year training

The rectangles have the same area since the distance is still the same

Vertical scale factor $\times$ 125% = 1.25

So horizontal scale factor is the inverse

$\div$ 1.25 = $\div \frac54$ = $\times \frac45$ = $\times$ 80%

80% of original time is a 20% reduction

**Using algebra**

Before: speed $v$ time $\dfrac{26}{v}$

After: speed $\frac54V$ time $\dfrac{26}{\frac54v}$

$=\dfrac{4\times26}{5v}$

$=\dfrac45\times\dfrac{26}v$

80% of original time is a 20% reduction

Vertical scale factor $\times$ 125% = 1.25

So horizontal scale factor is the inverse

$\div$ 1.25 = $\div \frac54$ = $\times \frac45$ = $\times$ 80%

80% of original time is a 20% reduction

Before: speed $v$ time $\dfrac{26}{v}$

After: speed $\frac54V$ time $\dfrac{26}{\frac54v}$

$=\dfrac{4\times26}{5v}$

$=\dfrac45\times\dfrac{26}v$

80% of original time is a 20% reduction

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.