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 $\frac{7}{10}&lt;\frac{p}{q}&lt;\frac{11}{15}$ 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:
$$0.70&lt;\frac{p}{q}&lt; \approx0.73 \quad\text{ or }\quad \frac{21}{30}&lt;\frac{p}{q}&lt;\frac{22}{30}$$
Thank you in advance!
The way to find the smallest denominator "from scratch" is with continued fractions.

Begin by rendering the proposed bounds thusly:

$\dfrac{7}{10}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{3}}}}$

$\dfrac{11}{15}=\dfrac{1
}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{1+\dfrac{1}{3}}}}}$

The upper layers of the continued fractions are identical but they eventually become different when we get down to the layers in blue.  We may now replace those entries with the smallest whole number lying between, thus

$1+\dfrac{1}{3}&lt;2&lt;3$

So the smallest denominator fraction meeting the betweenness criterion will be

$\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{2}}}}=\dfrac{5}{\color{blue}{7}}$
We can take the common denominator:
$$
\frac{21q}{30q}&lt;\frac{30p}{30q}&lt;\frac{22q}{30q},\qquad\text{or}\qquad 21q&lt;30p&lt;22q.
$$

Then you go from the smallest $q$:
$$
21\cdot7=147&lt;30p&lt;22\cdot7=154, \qquad \ldots 120,150,180\ldots\\
$$
We see that $30p=150$ lies between $147$ and $154$.