24. Viola has been practising the long jump. At one point, the average distance she had jumped was $3.80 \mathrm{~m}$. Her next jump was $3.99 \mathrm{~m}$ and that increased her average to $3.81 \mathrm{~m}$. After the following jump, her average had become $3.82 \mathrm{~m}$. How long was her final jump?
A $3.97 \mathrm{~m}$
B $4.00 \mathrm{~m}$
C $4.01 \mathrm{~m}$
D $4.03 \mathrm{~m}$
E $4.04 \mathrm{~m}$
24. C Let the number of jumps Viola has already made be $n$ and let the total distance she has jumped in these $n$ jumps be $T \mathrm{~m}$. Since the total distance jumped is equal to the average distance jumped multiplied by the number of jumps, the information in the question tells us that $T=3.8 n$ and that $T+3.99=3.81(n+1)$. Therefore $3.8 n+3.99=3.81 n+3.81$ and hence $0.18=0.01 n$, which has solution $n=18$. In order to increase her average distance to $3.82 \mathrm{~m}$ with her final jump, she must jump $x \mathrm{~m}$, where $T+3.99+x=3.82(n+2)$. Therefore $3.8 n+3.99+x=3.82 n+7.64$ and hence $x=0.02 n+3.65$. Since we have already shown $n=18$, we have $x=0.36+3.65=4.01$. Hence the distance she needs to jump with her final jump is $4.01 \mathrm{~m}$.