# Direct & Inverse Proportion

Direct and inverse proportion are mathematical concepts that describe the relationship between two variables.

1. Direct Proportion: Two quantities are directly proportional if an increase in one quantity leads to a proportional increase in the other, and a decrease in one quantity leads to a proportional decrease in the other. In other words, as one variable increases, the other variable increases as well, and vice versa. Mathematically, if two quantities $x$ and $y$ are directly proportional, it can be expressed as:

$y=kx$

Where $k$ is a constant of proportionality. This means that for every unit increase in $x$, $y$ increases by a constant amount, and vice versa.

2. Inverse Proportion: Two quantities are inversely proportional if an increase in one quantity leads to a proportional decrease in the other, and vice versa. Inverse proportionality implies that as one variable increases, the other variable decreases, and vice versa. Mathematically, if two quantities $x$ and $y$ are inversely proportional, it can be expressed as:

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

Where $k$ is a constant of proportionality. This means that as $x$ increases, $y$ decreases, and as $x$ decreases, $y$ increases, while their product remains constant.

## Direct Proportion & Inverse Proportion Formula

• When $y$ is directly proportional to $x$
• \begin{align*} y &= kx \end{align*}, where $k$ is a constant and \begin{align*} k &\ne 0 \end{align*} ​.
• The graph of $y$ against $x$ is a straight line.
• When $y$ is inversely proportional to $x$,
• \begin{align*} y &= \frac {k}{x} \end{align*}, where $k$ is a constant and \begin{align*} k &\ne 0 \end{align*} ​.
• The graph of $y$ against $x$ is a reciprocal curve.

In this chapter, we will be discussing the below-mentioned topics in detail:

• Problems involving Direct Proportion
• Problems involving Inverse Proportion

## A. Identifying problems involving direct proportion

Let’s understand this with the help of some examples:

Question 1:

The table shows the number of litres of petrol, $P$, consumed by a car to travel a distance of $d \;km$.

 Number Of Litres Of Petrol Distance Travelled $(\;P\;)$ $(\;d \;\text{km}\;)$ $1$ $2$ $3$ $4$ $9$ $18$ $27$ $36$

Is $P$ directly proportional to $d$?

Solution:

\begin{align*} d &= kp \\\\ k &= \frac{d}{p } \end{align*}

Hence,

 Number Of Litres Of Petrol Distance Travelled \begin{align*} k &= \frac{d}{p} \end{align*} $(\;P\;)$ $(\;d \;\text{km}\;)$ $1$ $2$ $3$ $4$ $9$ $18$ $27$ $36$ \begin{align*} &= \frac91\\[2ex] &=9 \end{align*} \begin{align*} &= \frac{18}{2}\\[2ex] &=9 \end{align*} \begin{align*} &= \frac{27}3\\[2ex] &=9 \end{align*} \begin{align*} &= \frac{36}{4}\\[2ex] &=9 \end{align*}

Hence, $k$ is a constant and $k ≠ 0$; hence $d$ and $P$ are in direct proportion.

Question 2:

The extension of a spring, $s \;cm$, is directly proportional to the weight, $w \;kg$, attached to it.

If the extension of the spring is $3\;cm$ when a weight of $8 \;kg$ is attached to it, find an equation connecting $s$ and $w$

Hence, find the extension of the spring when a weight of $50 \;kg$ is attached to it.

Solution:

$s = kw$

Substituting $s = 3$, $w = 8$,

\begin{align*} 3 &= k (8) \\ 8k &= 3 \\ k &= \frac {3}{8 } \end{align*}

The equation connecting $s$ and $w$ is

\begin{align*} s &= \frac{3}{8}w \end{align*}

Substituting $w = 50$

\begin{align*} s &= \frac{3}{8}(50) \\ &= 18 \frac{3}{4} \;cm \end{align*}

## B. Identifying problems involving Inverse Proportion

Let’s understand this with the help of some examples:

Question 3:

The table shows the time taken, $t \;\text{hours}$, by John to travel from Singapore to Johor Bahru at varying speeds, $s \;\text{km/h}$.

 Speed $(\;s \;\text{km/h}\;)$ Time Taken $(\;t \;\text{hours}\;)$ $20$ $30$ $50$ $90$ $4.5$ $3$ $1.8$ $1$

Are $s$ and $t$ in inverse proportion?

Solution:

Let \begin{align*} s &= \frac{k}{t} \end{align*}.

Then, \begin{align*} k &= st \end{align*}.

 Speed $(\;s \;\text{km/h}\;)$ Time Taken $(\;t \;\text{hours}\;)$ $k = st$ $20$ $30$ $50$ $90$ $4.5$ $3$ $1.8$ $1$ $= 20 × 4.5$ $= 90$ $= 30 × 3$ $= 90$ $= 50 × 1.8$ $= 90$ $= 90 × 1$ $= 90$

Hence, $k$ is a constant and $k ≠ 0$; hence $s$ and $t$ are in inverse proportion.

Question 4:

The force, $F \;Newtons$, between two particles is inversely proportional to the square of the distance $d \;cm$

If the force is $1.5 \;Newtons$ when the distance between two particles is $4 \;cm$, find

1. a formula for $F$ in terms of $d$.
2. the force when the distance between the particles is $10 \;cm$.
3. the distance between the particles when the force is $96\; \text{Newtons}$.

Solution:

\begin{align*} A)\quad F &= \frac{k}{d^2} \end{align*}

Substituting $F = 1.5, \;d = 4$,

\begin{align*} 1.5 &= \frac{k}{4^2}\\ \\ 1.5 &= \frac{k}{16}\\ \\ k &= 1.5 × 16\\ \\ &= 24 \end{align*}

Hence, \begin{align*} F = \frac{24}{d^2} \end{align*}

\begin{align*} B) \quad F &= \frac{24}{d^2} \end{align*}

Substituting $d = 10$,

\begin{align*} F &= \frac{24}{10^2} \\ \\ &= \frac{24}{100}\\ \\ &= 625 \;\text{Newtons} \end{align*}

$C)$ Substituting $F = 96$,

\begin{align*} 96 &= \frac{24}{d^2}\\ \\ 96 d^2 &= 24\\ \\ d^2 &= \frac{24}{96}\\ \\ &= \frac {1}{4} \\ \\ d &= \frac {1}{2} && \text{or} && - \frac {1}{2} \quad\text{(reject)}\\ \\ d &= \frac {1}{2} \; cm \end{align*}

Question 5:

$9$ workers can clean a beach in $16$ days.

1. How long will it take $6$ workers to clean the same beach?
2. The beach is to be cleaned in $x$ days. Write down an expression, in terms of $x$, for the number of workers needed to clean the beach.

Solution:

A)  Let $w$ be the number of workers.

Let $d$ be the number of days.

\begin{align*} w &= \frac{k}{d } \end{align*}

Substituting $w = 9, \;d = 16$,

\begin{align*} 9 &= \frac{k}{16}\\ \\ k &= 9 × 16\\ \\ &= 144\\ \\ w &= \frac{144}d \end{align*}

Substituting $w = 6$,

\begin{align*} 6 &= \frac{144}d\\ \\ 6d &= 144\\ \\ d &= 24 \;days \end{align*}

B)  \begin{align*} w &= \frac{144}{d} \end{align*} ​

Substituting \begin{align*} d &= x \end{align*}

\begin{align*} w &= \frac{144}{x} \end{align*}  workers

Continue Learning
Algebraic Fractions Direct & Inverse Proportion
Congruence And Similarity Factorising Quadratic Expressions
Further Expansion And Factorisation Quadratic Equations And Graphs
Simultaneous Equation

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