• # question_answer 3) If x varies directly as ${{y}^{2}}$ and $x=4$ when$y=5$, then find x when y is 15. A)  4                    B)         9        C)  12                  D)         36

$\frac{x}{{{y}^{2}}}=\frac{4}{{{(5)}^{2}}}=\frac{{{x}_{1}}}{{{(15)}^{2}}}(\because x\,\,\text{varies}\,\,\text{directly)}$ Or ${{x}_{1}}=\frac{15\times 15\times 4}{5\times 5}\Rightarrow {{x}_{1}}=36$