SSC CHSL 2023 PYQ — Question 51

The correct answer is option 4 i.e. $(\frac{x}{6}+\frac{y}{4})^2$.

$4x^2+9y^2+12xy\over144$

$={(2x)^2+(3y)^2+2\times 2x\times 3y\over12^2}$

$={(2x+3y)^2\over12^2}$

$=\left({2x+3y\over12}\right)^2$

$=\left(\frac{x}{6}+\frac{y}{4}\right)^2$

Hence, the value of the expression is $\left(\frac{x}{6}+\frac{y}{4}\right)^2$.