INDIRECT VARIATION — MEANING, EXPLANATION AND EXAMPLES

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When two variables, say x and y are related in such a way that whenever one increases, the other decreases, we say that x and y are inversely proportional.

$x$ varies inversely as $y$ can be written as $x\alpha \frac{1}{y}$, which is $x=\frac{k}{y}$ where $k$ is the constant of proportionality.

👉This post extends our series on variation👈

Example 1

$x$ varies indirectly as the square root of $y$. If $y=36$ when $x=5$,  Find

1. The formula connecting $x$ and $y$.

2. Determine the value of $y$ for which $x=\frac{15}{2}$

Solution:

$x\alpha \frac{1}{\sqrt{y}}$

$x=\frac{k}{\sqrt{y}}$

$y=36$ when $x=5$

$5=\frac{k}{\sqrt{36}}$

$5=\frac{k}{6}$

Cross multiply

$k=6\times5$

$k=30$

The formula that connect $x$ and $y$ is

$x=\frac{30}{\sqrt{y}}$

Let's find $y$ when $x=\frac{15}{2}$

$\frac{15}{2}=\frac{30}{\sqrt{y}}$

Cross multiply

$15\sqrt{y}=2\times30$

$15\sqrt{y}=60$

Divide both side by $15$

$\frac{15\sqrt{y}}{15}=\frac{60}{15}$

$\sqrt{y}=4$

To get rid of square root, we square both sides

$(\sqrt{y})^2=4^2$

$y=16$

Example 2

Given that $x$ varies inversely as the cube of $y$ and that $x=240$ when $y=3$, find $x$ when $y=2$

Solution:

$x\alpha \frac{1}{y^3}$

$x=\frac{k}{y^3}$

$x=240$, $ y=3$

$240=\frac{k}{3^3}$

$240=\frac{k}{27}$

Cross multiply

$240\times27=k$

$6480=k$

$k=6480$

$x=\frac{6480}{y^3}$

Let's find $x$ when $y=2$

$x=\frac{6480}{2^3}$

$x=\frac{6480}{8}$

$x=810$

Example 3

Given that $x$ is inversely proportional as $y$ and $x=2$ when $y=60$. Find

1. The value of $x$ when $y=90$

2. Find the value of $y$ when $x=\frac{5}{2}$

Solution:

$x\alpha \frac{1}{y}$

$x=\frac{k}{y}$

$2=\frac{k}{60}$

$2\times60=k$

$k=120$

$x=\frac{120}{y}$

Let's find $x$ when $y=90$

$x=\frac{120}{90}$

$x=\frac{4}{3}$

Now, let's find $y$ when $x=\frac{5}{2}$

$\frac{5}{2}=\frac{120}{y}$

Cross multiply

$2\times120=5\times y$

$240=5y$

Divide both sides by $5$

$\frac{240}{5}=\frac{5y}{5}$

$y=48$

Example 4

$a\alpha \frac{1}{b}$, if $a=\frac{1}{3}$ when $b=4$. Find $b$ when $a=\frac{2}{9}$

Solution:

$a=\frac{k}{b}$

$a=\frac{1}{3}$ when $b=4$

$\frac{1}{3}=\frac{k}{4}$

Cross multiply

$1\times4=k\times3$

$4=3k$

Divide both side by $3$

$\frac{4}{3}=\frac{3k}{3}$

$\frac{4}{3}=k$

$k=\frac{4}{3}$

$a=\frac{4}{3b}$

Let's solve for $b$ when $a=\frac{2}{9}$

$\frac{2}{9}=\frac{4}{3b}$

Cross multiply

$2\times3b=9\times4$

$6b=36$

Divide both side by $6$

$\frac{6b}{6}=\frac{36}{6}$

$b=6$

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Example 5

A variable $x$ varies inversely as another variable $y$ and $x=0.15$ and $y=120$. Find

1. $x$ when $y=4$

2. $y$ when $x=0.12$

Solution:

$x\alpha \frac{1}{y}$

$x=\frac{k}{y}$

$0.15=\frac{k}{120}$

Cross multiply

$0.15\times120=k$

$18=k$

$k=18$

$x=\frac{18}{y}$

Let's find $x$ when $y=45$

$x=\frac{18}{45}$

$x=\frac{2}{5}$

Now, let's solve for $y$ when $x=0.12$

$0.12=\frac{18}{y}$

Cross multiply

$0.12y=18$

Divide both side by $0.12$

$\frac{0.12y}{0.12}=\frac{18}{0.12}$

$y=150$

We just concluded indirect variation. We would our attention to joint variation in  this post.

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