4. If $a, b, c$, and $d$ satisfy $\frac{a}{b}=\frac{2}{3}$ and $\frac{c}{b}=\frac{1}{5}$ and $\frac{c}{d}=\frac{7}{15}$, what is the value of $\frac{a b}{c d}$ ?
4. Solution 1
Since $\frac{c}{d}=\frac{7}{15}$, then $c=\frac{7}{15} d$.
Since $\frac{c}{b}=\frac{1}{5}$, then $b=5 c=5 \times \frac{7}{15} d=\frac{7}{3} d$.
Since $\frac{a}{b}=\frac{2}{3}$, then $a=\frac{2}{3} b=\frac{2}{3} \times \frac{7}{3} d=\frac{14}{9} d$.
Thus,
$$
\frac{a b}{c d}=\frac{(14 / 9) d \times(7 / 3) d}{(7 / 15) d \times d}=\frac{(98 / 27) d^{2}}{(7 / 15) d^{2}}=\frac{98}{27} \times \frac{15}{7}=\frac{14}{9} \times \frac{5}{1}=\frac{70}{9}
$$

Solution 2
Suppose that $d=45$.
Since $\frac{c}{d}=\frac{7}{15}$, then $c=\frac{7}{15} d=\frac{7}{15} \times 45=21$.
Since $\frac{c}{b}=\frac{1}{5}$, then $b=5 c=5 \times 21=105$.
Since $\frac{a}{b}=\frac{2}{3}$, then $a=\frac{2}{3} b=\frac{2}{3} \times 105=70$.
Therefore, $\frac{a b}{c d}=\frac{70 \times 105}{21 \times 45}=\frac{10 \times 7}{3 \times 3}=\frac{70}{9}$.
This solution does not show that this is the value of $\frac{a b}{c d}$ for all such values of $a, b, c$, and $d$, but the text of the problem implies that the value is the same for any choice of $a, b, c$, and $d$ that satisfy the restrictions. Solution 1 justifies that this is the value for all choices of $a, b, c$, and $d$.
ANSWER: $\frac{70}{9}$