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Answer: 20%


Using inverse operations
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


Using a speed-time graph

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

 
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.