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 \(d \;km\).

Is \(P\) directly proportional to \(d\)?

Solution:

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

Hence, 

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, \(s \;\text{km/h}\).

Are \(s\) and \(t\) in inverse proportion?

Solution: 

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

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

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