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Answer 


Using a common denominator
$$\frac15 \qquad \frac14 \qquad \frac13\\
 \\
\frac{12}{60} \qquad \frac{15}{60} \qquad \frac{20}{60}$$ The distance from $\dfrac{12}{60}$ to $\dfrac{20}{60}$ is $\dfrac 8{60}$

There are $16$ intervals on the diagram so two make $\dfrac1{60}$

The distance from $\dfrac{12}{60}$ to $\dfrac{15}{60}$ is $\dfrac 3{60}$ so go along $6$ intervals


Finding the size of the intervals
The difference between $\frac{1}{3}$ and $\frac{1}{5}$ is $\frac{1}{3}-\frac{1}{5}= \frac{2}{15}$.

This section of the number line is divided into $16$ intervals, each of length $\frac{2}{15}\div 16 = \frac{1}{120}$.

The difference between $\frac{1}{4}$ and $\frac{1}{5}$ is $\frac{1}{4}-\frac{1}{5}= \frac{1}{20}= \frac{6}{120}$, and hence $\frac{1}{4}$is six smaller intervals from $\frac{1}{5}$. 
 
 
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.