Question

If $\frac{a}{a+b}=\frac{17}{23},$ what is $\frac{a+b}{a-b}$ equal to?

• A. $\frac{11}{23}$
• B. $\frac{17}{32}$
• C. $\frac{23}{11}$
• D. $\frac{23}{17}$

Answer 1

The correct option is C $\frac{23}{11}$

Given, $\frac{a}{a+b}$$=\frac{17}{23}$, then $23a=17a+17b$

$\Rightarrow 6a=17b$

$\Rightarrow \frac{a}{b}$$=\frac{17}{6}$

Using componendo-dividendo, we get

$\frac{a+b}{a-b}$$=\frac{17+6}{17-6}=\frac{23}{11}$