Question

If $\frac{15a^2+4b^2}{15a^2-4b^2}=\frac{47}{7}$ then find the values of the following ratios.

• A. $\frac{a}{b}$
• B. $\frac{7a-3b}{7a+3b}$
• C. $\frac{b^2-2a^2}{b^2+2a^2}$
• D. $\frac{b^3-2a^3}{b^3+2a^3}$

Answer 1

The correct option is A $\frac{a}{b}$

Consider given the equation,

$\frac{15a^2+4b^2}{15a^2-4b^2}=\frac{47}{7}$

Apply the componendo-dividendo rule

$\frac{15a^2+4b^2+15a^2-4b^2}{(15a^2+4b^2)-(15a^2-4b^2)}=\frac{47+7}{47-7}$

$\frac{30a^2}{8b^2}=\frac{54}{40}$

$\frac{15a^2}{4b^2}=\frac{27}{20}$

$\frac{a^2}{b^2}=\frac{27\times 4}{20\times 15}=\frac{9}{75}$

$\frac{a}{b}=\frac{3}{5\sqrt{3}}$

Hence, this is the answer.