, 2017

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

We require that $\text{n 40 gt; 8 40}$ and $\text{n 40 lt; 10 40}$, thus $\text{n gt;8}$ and $\text{n lt;10}$.

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 $\text{3m 24 gt; 8 24}$ and so $\text{3m gt;8}$ or $\text{m gt; 83}$.

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 $\text{3(m+1) 24 lt; 16 24}$ or $\text{3m+3 lt;16}$ or $\text{3m lt;13}$, and so $\text{m lt; 13 3}$.

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 $\text{w 30 gt; 12}$.

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

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 $\text{w+3 35 lt; 7 10}$.

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

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

The integer values of $w$ which satisfy $\text{w gt;15}$ and $w\leq21$ are $w=16,17,18,19,20,21$.