If $X$ varies directly as $Y^{2}$ and $X = 4$ when $Y = 5$ , then find $X$ when $Y$ is $15.$

4

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9

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12

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36

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Solution

The correct option is D $36$
$X = k Y^{2} \Rightarrow 4 = k \times 5^{2} \Rightarrow k = \frac{4}{25} X = \frac{4}{25} Y^{2} \Rightarrow X = \frac{4}{25} 15^{2} = \frac{4 \times 225}{25} = 36$

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Question 10
Let y varies directly as x. if y = 12 when x = 4, then write a linear equation. What is the value of y when x = 5?

Question 10
Let y varies directly as x. if y = 12 when x = 4, then write a linear equation. What is the value of y when x = 5?

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