17. Bethany has 11 pound coins and some 20 p coins and some 50 p coins in her purse. The mean value of the coins is 52 pence.
Which could not be the number of coins in the purse?
A 35
B 40
C 50
D 65
E 95
Solution
B

We suppose that numbers of $20 \mathrm{p}$ and 50 p coins that Bethany has are $m$ and $n$, respectively. (Note that $m$ and $n$ are positive integers.)
Then Bethany has $c$ coins, where $c=11+m+n$. The total value of these coins is $11+0.20 m+0.50 n$ pounds, which equals $1100+20 m+50 n$ pence.
Because the mean value of Bethany's coins is 52 pence, $\frac{1100+20 m+50 n}{11+m+n}=52$. It follows that $1100+20 m+50 n=52(11+m+n)$ and hence $1100+20 m+50 n=572+52 m+52 n$. This equation may be rearranged as $2 n=528-32 m$, which simplifies to $n=264-16 m$.
It follows that $c=11+m+n=11+m+(264-16 m)=275-15 m$. Therefore $275-c=15 m$. Thus $275-c$ is an integer multiple of 15 .
We now consider the different values for $c$ given by the options in the question.
We have $275-35=240=15 \times 16,275-50=225=15 \times 15,275-65=210=15 \times 14$ and $275-95=180=15 \times 12$. Hence Bethany could have 35 or 50 or 65 or 95 coins in her purse.
However $275-40=235$ which is not an integer multiple of 15 . Hence Bethany could not have 40 coins in her purse.