Question

The square of X varies inversely as the square root of Y and directly as the variable M. When M = 27 and Y = 16, then X = 9. Find the exact value of Y when M = 7 and X = 2

Answer 1

Answer:

$Y = 441$

Step-by-step explanation:

Given

M = 27 when Y = 16 and X = 9

Required

Find Y when M = 7 and X = 2

We start by getting the algebraic representation of the given statement

$X^2 \alpha \frac{1}{\sqrt Y} \alpha M$

Convert the variation to an equation; we have

$X^2 = \frac{KM}{\sqrt Y}$

Where K is the constant of variation;

When M = 27; Y = 16; X = 9, the expression becomes

$9^2 = \frac{K * 27}{\sqrt{16}}$

This gives

$81 = \frac{k * 27}{4}$

Make K the subject of formula

$K = \frac{81* 4}{27}$

$K = \frac{324}{27}$

$K = 12$

Solving for Y when M = 7 and X = 2

Recall that $X^2 = \frac{KM}{\sqrt Y}$

Substitute values for K, M and X

$2^2 = \frac{12 * 7}{\sqrt{Y}}$

$4 = \frac{84}{\sqrt{Y}}$

Take square of both sides

$4^2 = (\frac{84}{\sqrt{Y}})^2$

$16 = \frac{7056}{Y}$

Make  Y the subject of formula

$Y = \frac{7056}{16}$

$Y = 441$