Which of the numbers lies between $2 / 5$ and $3 / 4$?
(a) $1 / 6$
(b) $4 / 3$
(c) $5 / 2$
(d) $4 / 7$
(e) $1 / 4$
The correct option is (d).

Remember that the order between fractions made up of positive integers is determined by the cross product of the integers, that is, $\frac{a}{b}<\frac{c}{d}$ is equivalent to the statement $a \times d<b \times c$. Thus, we have

$$
\frac{1}{6}<\frac{1}{4}<\frac{2}{5}<\frac{4}{7}<\frac{3}{4}<1<\frac{4}{3}<\frac{5}{2}
$$

A number $x$ that “lies between" $2 / 5$ and $3 / 4$ is a number greater than $2 / 5$ and less than $3 / 4$, that is

$$
\frac{2}{5}<x<\frac{3}{4}
$$

since $4<6.5<8, 14<20, 16<21, 3<4$ (twice) and $8<15$, respectively. Thus, $1 / 6$ and $1 / 4$ are to the left of $2 / 5, 4 / 3$ and $5 / 2$ are to the right of $3 / 4$ and only $4 / 7$ lies between $2 / 5$ and $3 / 4$.