Inverse proportion

Question

I am not sure how to approach this question, can I get some help? This is an inverse proportion question.

$y$ is inversely proportional to the square of $x$.

Find the percentage change of $x$ when $y$ is decreased by $36\%$ .

Not sure where to continue after my working:

Old $y$ value = $1 y$

New $y$ value = $0.64 y$

$$y = \frac{k}{x^2}\tag{1}$$

$$0.64y = \frac{k}{x^2}\tag{2}$$

score 1 · Accepted Answer · answered Feb 22 '16 at 03:35

1

Let the initial value of $x$ be $x_1$ and the value after the change in $y$ be $x_2$. Continuing from where you left off, $$y=\frac{k}{{x_1}^2}\tag1$$ $$0.64y=\frac{k}{{x_2}^2}\tag2$$ Dividing $(2)$ by $(1)$, $$0.64=\frac{{x_1}^2}{{x_2}^2}$$ $$0.8=\frac{x_1}{x_2}$$ $$x_1=0.8x_2$$ or $$x_2=1.25x_1$$

Thus, the value of $x$ increases by $25\%$

answered Feb 22 '16 at 03:35

GoodDeeds

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Hey , shouldn't it be X2 ^2 / X1^1 ? Because it's 2 divide by 1 – user307640 Feb 22 '16 at 05:43
@user307640 Since it is $(2)$ divided by $(1)$, the ${x_2}^2$ term which is in the denominator in $(2)$ stays in the denominator. – GoodDeeds Feb 22 '16 at 07:29

Inverse proportion

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