25. Tony had a large number of 1 p, 5 p, 10p and 20 p coins in a bag. He removed some of the coins. The mean value of the coins he removed was 13p. He noticed that a 1 p piece in his group of removed coins was damaged so he threw it away. The mean value of the rest of his removed coins was then 14 p.
How many 10 p coins did he remove from the bag?
A 0
B 1
C 2
D 3
E 4
Solution
Let the number of coins Tony removed be $n$ and let the total value of these coins, in pence, be $X$. The initial information in the question tells us that $\frac{X}{n}=13$. Also, since when he threw away a $1 \mathrm{p}$ coin, the mean value of his coins increased to 14 , we have $\frac{X-1}{n-1}=14$. Therefore $X=13 n$ and $X-1=14(n-1)$ and hence $13 n-1=14 n-14$, which has solution $n=13$. Therefore Tony removed 13 coins with a total value of $13 \times 13 \mathrm{p}$, or 169 p.

We have found that the total of the 13 coins Tony removed is 169 p. It is impossible to have a total of 169 p with only 5 p, 10p and 20 p coins as that would give a total that is a multiple of 5. Hence Tony must have removed either four 1 p coins or nine 1 p coins. However, the latter is impossible as it would mean that the remaining four coins would need to have a total value of 160 p and the maximum value from four 5 p, 10p and 20 p coins is only $4 \times 20$ p $=80$ p. Therefore Tony removed four 1 p coins. This means that the total value of the other nine coins is 165 p. Now, since $9 \times 20 p=180$ p, which is too large and $7 \times 20 p+2 \times 10 p=160 p$, which is too small, Tony must have removed eight 20 p coins plus an additional one 5 p coin to have the required total. Therefore Tony did not remove any 10p coins.