# INDIRECT VARIATION — MEANING, EXPLANATION AND EXAMPLES

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 $6$

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

$b=6$

Related post

#### 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.

Do well to subscribe to our telegram channel here.

Help us grow our readership by sharing this post