Question

If $\frac{a}{b}=\frac{4}{5}$ and $\frac{b}{c}=\frac{15}{16}$, then $\frac{c^2-a^2}{c^2+a^2}$ is equal to __________.

• A. $\frac{1}{7}$
• B. $\frac{7}{25}$
• C. $\frac{3}{4}$
• D. $\frac{7}{15}$

Answer 1

The correct option is B $\frac{7}{25}$

$\frac{a}{b}=\frac{4}{5}$ --- ( 1 )

$\frac{b}{c}=\frac{15}{16}$ ---- ( 2 )

$\Rightarrow$ $\frac{a}{b}\times \frac{b}{c}=\frac{4}{5}\times \frac{15}{16}$ [ Multiplying ( 1 ) and ( 2 ) ]

$\Rightarrow$ $\frac{a}{c}=\frac{3}{4}$

$\Rightarrow$ $\frac{c}{a}=\frac{4}{3}$ [ Taking reciprocal on both sides ]

$\Rightarrow$ $\frac{c^2}{a^2}=\frac{16}{9}$ [ Taking square on both sides ]

$\Rightarrow$ $\frac{c^2-a^2}{c^2+a^2}=\frac{16-9}{16+9}$ [ Taking Dividendo in numerator and Componendo in denominator ]

$\therefore$ $\frac{c^2-a^2}{c^2+a^2}=\frac{7}{25}$