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
- a formula for \(F\) in terms of \(d\).
- the force when the distance between the particles is \(10 \;cm\).
- 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.
- How long will it take \(6\) workers to clean the same beach?
- 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 | |
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Algebraic Fractions | Direct & Inverse Proportion |
Congruence And Similarity | Factorising Quadratic Expressions |
Further Expansion And Factorisation | Quadratic Equations And Graphs |
Simultaneous Equation |