Question

$P={116}^2-{16}^2,Q={124}^2-{24}^2$ and $R=120$ then the value of $\sqrt{\frac{P}{100}+\frac{Q}{100}+R}=$

• A. 20
• B. 30
• C. 40
• D. 10

Answer 1

The correct option is A 20

$P=(116+16)(116-16)\Rightarrow \left(100\right)\left(132\right)$

$Q=(124+24)(124-24)\Rightarrow \left(100\right)\left(148\right)$

$\because (a^2-b^2)=(a+b)(a-b)$

From the Question

$\Rightarrow \sqrt{\frac{P}{100}+\frac{Q}{100}+R}=x$

$\Rightarrow \sqrt{\frac{100\times 132}{100}+\frac{100\times 148}{100}+R}=x$

$\Rightarrow \sqrt{132+148+120}=x$

$\Rightarrow \sqrt{400}=x$

$x=20$

$\text{correct option is A.}$