Problem 243

Expressing $\dfrac15$ and $\dfrac14$ with a common denominator of 40, we get $\dfrac15=\dfrac{8}{40}$ and $\dfrac14=\dfrac{10}{40}$.

We require that $\dfrac{n}{40}>\dfrac{8}{40}$ and $\dfrac{n}{40}<\dfrac{10}{40}$, thus $n>8$ and $n<10$.

The only integer $n$ that satisfies both of these inequalities is $n=9$.
Expressing $\dfrac m8$ and $\dfrac13$ with a common denominator of 24, we require $\dfrac{3m}{24}>\dfrac{8}{24}$ and so $3m>8$ or $m>\dfrac83$.

Since $\dfrac83=2\dfrac23$ and $m$ is an integer, then $m\geq3$.

Expressing $\dfrac {m+1}{8}$ and $\dfrac23$ with a common denominator of 24, we require $\dfrac{3(m+1)}{24}<\dfrac{16}{24}$ or $3m+3<16$ or $3m<13$, and so $m<\dfrac{13}{3}$.

Since $\dfrac{13}{3}=4\dfrac13$ and $m$ is an integer, then $m\leq4$.

The integer values of $m$ which satisfy $m\geq3$ and $m\leq4$ are $m=3$ and $m=4$.
At the start of the weekend, Fiona has played 30 games and has $w$ wins, so her win ratio is $\dfrac{w}{30}$.

Fiona’s win ratio at the start of the weekend is greater than $0.5=\dfrac12$, and so $\dfrac{w}{30}>\dfrac12$.

Since $\dfrac12=\dfrac{15}{30}$, then we get $\dfrac{w}{30}>\dfrac{15}{30}$, and so $w>15$.

During the weekend Fiona plays five games giving her a total of $30+5=35$ games played.

Since she wins three of these games, she now has $w+3$ wins, and so her win ratio is $\dfrac{w+3}{35}$.

Fiona’s win ratio at the end of the weekend is less than $0.7=\dfrac{7}{10}$, and so $\dfrac{w+3}{35}<\dfrac{7}{10}$.

Rewriting this inequality with a common denominator of 70, we get $\dfrac{2(w+3)}{70}<\dfrac{49}{70}$ or $2(w+3)<49$ or $2w+6<49$ or $2w<43$, and so $w<\dfrac{43}{2}$.

Since $\dfrac{43}{2}=21\dfrac12$ and $w$ is an integer, then $w\leq21$.

The integer values of $w$ which satisfy $w>15$ and $w\leq21$ are $w=16,17,18,19,20,21$.