This MCQ module is based on: Direct and Inverse Proportion
TOPIC 9 OF 23
Direct and Inverse Proportion
🎓 Class 8
Mathematics
CBSE
Theory
Ch 3 — Direct and Inverse Proportions
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Direct and Inverse Proportion
Targeting Class 8 level in General Mathematics, with Basic difficulty.
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3.5 Direct Proportion
In many real-life situations, two quantities change together such that when one increases, the other also increases in the same ratio. Similarly, when one decreases, the other decreases in the same ratio. Such quantities are said to be in direct proportion?.
Definition — Direct Proportion
Two quantities \(x\) and \(y\) are in direct proportion if the ratio \(\dfrac{x}{y}\) stays constant. Equivalently, \(x = k \cdot y\) where \(k\) is a constant, called the constant of proportionality. We write: \(x \propto y\).
Example — Bus Fare
The fare for a bus journey is directly proportional to the distance travelled. If a 15 km trip costs ₹45, what is the fare for a 25 km trip?
Ratio \(\dfrac{\text{fare}}{\text{distance}} = \dfrac{45}{15} = 3\) (constant). So fare for 25 km = \(3 \times 25 = \mathbf{\textsf{₹}75}\).
The straight line through origin indicates direct proportion.
3.6 Inverse Proportion
Sometimes, when one quantity increases, the other decreases such that their product stays constant. Think of travel between Lucknow and Kanpur (roughly 90 km). If you cycle at 15 km/h, the trip takes 6 hours. Using a car at 60 km/h, it takes 1.5 hours.
Does it decrease by the same rate (or factor)? Going by bicycle is 3 times faster than walking (15 ÷ 5). The speed has increased 3 times. The travel time has decreased 3 times too (18 ÷ 6). The speed has increased by the same factor by which the travel time has decreased.
Check if this is the case for the other modes of transport:
| Mode | Speed (km/h) | Time (hours) | Speed × Time |
|---|---|---|---|
| Walking | 5 | 18 | 90 |
| Bicycle | 15 | 6 | 90 |
| Two-wheeler | 30 | 3 | 90 |
| Car | 60 | 1.5 | 90 |
Since both quantities, speed and time, change by the same factor, but in opposite directions, or inversely. Such proportions are called inverse proportions?. From the table, we can see that the product of speed and time are the same for all modes of transport, namely 90 km.
Definition — Inverse Proportion
Two quantities \(x\) and \(y\) are in inverse proportion if there exists a relation of the type \(xy = k\), where \(k\) is a constant. We write \(x \propto \dfrac{1}{y}\).If \((x_1, y_1), (x_2, y_2), \ldots\) are values of the pair, then \(x_1 y_1 = x_2 y_2 = \ldots = k\).
A curve (hyperbola) — typical of inverse proportion. Speed × Time = 90 km (constant).
Example 4 — Filling a Tank
2 pumps can fill a tank in 18 hours. How much time will it take to fill the tank if we add 2 more pumps of the same kind?
If we add 2 more pumps, we will have 4 pumps. Let us denote the time taken to fill the tank with 4 pumps by \(x\). If we increase the number of pumps by the same factor \(n\), the time taken will decrease by the same factor \(n\). Since the quantities are inversely proportional, the product remains constant. So,
\(2 \times 18 = 4 \times x \;\Rightarrow\; x = \dfrac{2 \times 18}{4} = 9\) hours.
Example 5 — School Provisions
A school has food provisions to feed 80 students for 15 days. If provisions are to last for 20 days, how many fewer students should the provisions cater to?
More students → fewer days the provisions will last. If \(x\) is the number of students:
\(80 \times 15 = 100 \times x \;\Rightarrow\; x = \dfrac{80 \times 15}{100} = 12\). Wait — \(x \times 20 = 80 \times 15\) gives \(x = 60\).
Correction: The provisions will last for only 12 days if we have 100 students, but we need 20 days, so number of students = \((80 \times 15)/20 = \textbf{60}\) students. The provisions should cater to 60 students (20 fewer).
Example 6 — Shyam & Ram (Math Talk)
If Ram takes 1 hour to cut a green quantity of vegetables and Shyam takes 1.5 hours to cut the same quantity of vegetables, how much time will they take to cut the vegetables if they do it together?
Consider the work done to cut the given quantity of vegetables as 1 unit of work. Let us figure out the work done by each person in 1 hour.
- Ram finishes the work in 1 hour, so in 1 hour he does 1 unit of work.
- Shyam finishes the work in 1.5 hours, so in 1 hour Shyam does \(\frac{1}{1.5} = \frac{2}{3}\) units of work.
So, the work done by both in 1 hour = \(1 + \frac{2}{3} = \frac{5}{3}\) units. Therefore, to complete 1 unit of work, it takes \(\frac{3}{5}\) hours = 36 minutes.
Activity: Pie Chart from Survey
A group of 360 people were asked for their favourite season (rainy, winter, summer). 90 liked summer, 160 liked rainy, 110 liked winter. Make a pie chart.
- Calculate each angle: Rainy = \(\dfrac{160}{360} \times 360° = 160°\); Winter = \(\dfrac{110}{360} \times 360° = 110°\); Summer = \(\dfrac{90}{360} \times 360° = 90°\).
- Draw a circle; mark centre; measure each angle from a fixed radius using a protractor.
- Shade and label the three slices.
Sum = 160 + 110 + 90 = 360° ✓ (a full circle).
The angles are found by computing the fraction of the circle that each portion occupies. This reflects direct proportion: angle is directly proportional to the number of people.
Practice — Check Yourself
Problem: A TV channel has its viewership distributed as: Entertainment = 50%, Sports = 25%, News = 15%, Information = 10%. Convert to a pie chart.
Angles: Entertainment = 180°, Sports = 90°, News = 54°, Information = 36° (sum = 360° ✓).
Angles: Entertainment = 180°, Sports = 90°, News = 54°, Information = 36° (sum = 360° ✓).
Competency-Based Questions
Scenario: A textile factory hires workers to stitch uniforms. 15 workers stitch 60 uniforms in 8 hours. Meanwhile, delivery trucks moving at 40 km/h take 3 hours for a route. Analyse the proportionalities.
Q1. If the factory adds 15 more workers (total 30), how many hours will it take to stitch the same 60 uniforms?
L3 ApplyWorkers vs time = inverse proportion. 15 × 8 = 30 × t → t = 4 hours.
Q2. The truck takes 3 hours at 40 km/h. If a new truck runs at 60 km/h, analyse how long the same route takes.
L4 AnalyseDistance = 40 × 3 = 120 km. Time = 120 / 60 = 2 hours. Speed × time = constant = 120 (inverse proportion).
Q3. Evaluate: a manager says "doubling workers always halves the time". Is this claim always true? Justify.
L5 EvaluateTrue mathematically under ideal inverse proportion. But real-world constraints (shared tools, coordination overhead, space) break the assumption. The claim oversimplifies — inverse proportion is an idealisation.
Q4. Design a situation with THREE quantities in combined proportion (2 direct, 1 inverse) and state the equation.
L6 CreateExample: Cost of fabric \(C\) is directly proportional to length \(L\) and width \(W\), and inversely proportional to discount \(d\): \(C = k \cdot \dfrac{LW}{d}\).
Assertion–Reason Questions
A: If 5 cats eat 5 rats in 5 minutes, then 10 cats eat 10 rats in 5 minutes.
R: The number of rats eaten is directly proportional to the number of cats (time fixed).
R: The number of rats eaten is directly proportional to the number of cats (time fixed).
(a) Each cat eats 1 rat / 5 min. 10 cats eat 10 rats in 5 min. Direct proportion.
A: Speed and time of travel over a fixed distance are in direct proportion.
R: If speed increases, time increases proportionally.
R: If speed increases, time increases proportionally.
(d) Both are FALSE — speed and time (over fixed distance) are inversely proportional. As speed rises, time falls.
A: If \(x \propto y\) and \(y \propto z\), then \(x \propto z\).
R: Direct proportion is transitive because each ratio equals a constant.
R: Direct proportion is transitive because each ratio equals a constant.
(a) Yes — if \(x = k_1 y\) and \(y = k_2 z\), then \(x = k_1 k_2 z\), i.e., \(x \propto z\).
Frequently Asked Questions
What is direct proportion?
Two quantities are in direct proportion if one increases (or decreases) as the other does, keeping their ratio constant. y = kx where k is the constant of proportionality. Example: distance = speed x time, at fixed speed. NCERT Class 8 Chapter 3 defines this.
What is inverse proportion?
Two quantities are in inverse proportion if one increases as the other decreases, keeping their product constant. xy = k. Example: at fixed distance, speed x time = distance; doubling speed halves time. NCERT Class 8 Ganita Prakash Part 2 Chapter 3 explains this.
How to identify direct vs inverse proportion?
Check what happens when one quantity doubles. If the other doubles too, it's direct. If the other halves, it's inverse. If neither, no simple proportion. NCERT Class 8 Chapter 3 uses this test.
Example of inverse proportion in daily life?
Number of workers and time to complete a job: more workers finish faster (inverse). If 4 workers finish in 10 days, 8 workers finish in 5 days (4 x 10 = 8 x 5 = 40). NCERT Class 8 Ganita Prakash Part 2 Chapter 3 gives such examples.
How do you solve direct proportion problems?
Set up a ratio: if 5 pens cost Rs.40, the ratio pens:cost = 5:40 = 1:8. For 8 pens, cost = 8 x 8 = Rs.64. Or use cross-multiplication. NCERT Class 8 Chapter 3 practises direct proportion word problems.
Is price vs quantity always direct proportion?
Usually yes - buy more, pay more at the same per-unit rate. But bulk discounts break strict proportionality. NCERT Class 8 Ganita Prakash Part 2 Chapter 3 acknowledges real-world deviations while teaching the ideal case.
Frequently Asked Questions — Chapter 3
What is Direct and Inverse Proportion in NCERT Class 8 Mathematics?
Direct and Inverse Proportion is a key concept covered in NCERT Class 8 Mathematics, Chapter 3: Chapter 3. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Direct and Inverse Proportion step by step?
To solve problems on Direct and Inverse Proportion, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 3: Chapter 3?
The essential formulas of Chapter 3 (Chapter 3) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Direct and Inverse Proportion important for the Class 8 board exam?
Direct and Inverse Proportion is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Direct and Inverse Proportion?
Common mistakes in Direct and Inverse Proportion include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Direct and Inverse Proportion?
End-of-chapter NCERT exercises for Direct and Inverse Proportion cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 3, and solve at least one previous-year board paper to consolidate your understanding.
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