Question 141
If $$\frac{a}{b}=\frac{25}{6}$$, then the value of $$\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$$ is
Solution
Given $$\frac{a}{b}=\frac{25}{6}$$
Squaring on both sides we get,
$$\frac{a^{2}}{b^{2}} = \frac{625}{36}$$.......(1)
We know that, $$\frac{a}{b}$$ can be written as $$\frac{a-b}{a+b}$$
$$\therefore Equation (1) can be written as,
$$\frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{625 - 36}{625 + 36}$$ = $$\frac{589}{661}$$
Hence, option A is the correct answer.
Need AI Help?
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
