<s>[INST] What is the smallest possible value of $q$ such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$? 
If $p$ and $q$ are positive integers such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$ then the smallest possible value of $q$ is:
$(A)\quad 60;\quad (B)\quad 30;\quad (C)\quad 25;\quad (D)\quad 7$.

What is the correct way to solve this kind of problems? I have tried to simplify the given inequality:
$$0.70<\frac{p}{q}< \approx0.73 \quad\text{ or }\quad \frac{21}{30}<\frac{p}{q}<\frac{22}{30}$$
Thank you in advance!
 [/INST] If $7$ worked we would have 
\begin{align} 
&\frac{21}{30}<\frac{p}{7}< \frac{22}{30}\\
\iff&\frac{21*7}{30*7}<\frac{30p}{30*7}< \frac{22*7}{30*7}
\end{align}
So $7$ is a solution if and only if there is a multiple of $30$ between $21*7$ and $22*7$. Since $21*7=147$ and $22*7=154$ we have a multiple of $30,$ namely $150$. So $p=\frac{5}{7}$ is between those numbers.
To prove that this is the smallest just notice that every number smaller than $7$ other than $4$ is a divisor of $30$ so their fractions can be written as $\frac{p}{30}.$ And obviously it is impossible to have 
\begin{align*}
\frac{21}{30}<\frac{p}{30}<\frac{22}{30}
\end{align*}
As for $4$ if $4$ was possible we would have 
\begin{align}
&\frac{21}{30}<\frac{p}{4}<\frac{22}{30}\\
\iff & \frac{21*2}{30*2}<\frac{p*15}{4*15}<\frac{22*2}{30*2}
\end{align}
and there are no multiples of $15$ between $42$ and $44$.
</s>