Study S2 Mathematics Maths - Direct & Inverse proportion - Geniebook

# Direct & Inverse Proportion

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

• Problems involving Direct Proportion
• Problems involving Inverse Proportion

## Direct Proportion & Inverse Proportion

• 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.

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

Is $$P$$ directly proportional to $$d$$?

Solution:

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

Hence,

 Number Of Litres Of Petrol $$(\;P\;)$$ $$1$$ $$2$$ $$3$$ $$4$$ Distance Travelled $$(\;d \;\text{km}\;)$$ $$9$$ $$18$$ $$27$$ $$36$$ \begin{align*} k &= \frac{d}{p} \end{align*} \begin{align*} &= \frac91\\ &=9 \end{align*} \begin{align*} &= \frac{18}{2}\\ &=9 \end{align*} \begin{align*} &= \frac{27}3\\ &=9 \end{align*} \begin{align*} &= \frac{36}{4}\\ &=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 $$(\;s \;\text{km/h}\;)$$ $$20$$ $$30$$ $$50$$ $$90$$ Time Taken $$(\;t \;\text{hours}\;)$$ $$4.5$$ $$3$$ $$1.8$$ $$1$$

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

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

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