A train used to take 10 hours to travel from station A to station B, but now takes only 8 hours. By what percentage has its speed increased?

Step-by-step solution

1. 1

   Note that the distance is fixed. With the same route, speed and time are inversely proportional: $v=\dfrac{S}{t}$. That is why the answer is not simply the $20\%$ by which the time fell — the percentages are taken on different bases.

2. 2

   Write both speeds in terms of the common distance. Let the distance be $S$:

   $v_{\text{old}}=\frac{S}{10},\qquad v_{\text{new}}=\frac{S}{8}$

3. 3

   Form the ratio, so S cancels.

   $\frac{v_{\text{new}}}{v_{\text{old}}}=\frac{S/8}{S/10}=\frac{10}{8}=1.25$

4. 4

   Convert the ratio into a percentage increase. A ratio of $1.25$ means the new speed is $125\%$ of the old, i.e. an increase of

   $1.25-1=0.25=25\%$

   $\boxed{25\%}$

5. 5

   Check with concrete numbers. Take $S=80$ km. Then $v_{\text{old}}=8$ km/h and $v_{\text{new}}=10$ km/h, an increase of $2$ km/h on a base of $8$, which is $\tfrac28=25\%$ ✓.

6. 6

   Note why 20% is the tempting wrong answer. The time fell by $\tfrac{10-8}{10}=20\%$, but that percentage is measured against the old time, whereas the speed increase is measured against the old speed. The general relation is: if time is multiplied by $k$, speed is multiplied by $1/k$; here $k=0.8$ and $1/0.8=1.25$.