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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

 


 

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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