Exercise 10.2 — Inverse Proportion
Introduction of inverse proportion.
Exercise 10.2 – Solving Problems on Inverse Proportion
Exercise 10.2 of Class 8 Mathematics, Chapter 10 "Direct and Inverse Proportions" (CBSE, Telangana & Andhra Pradesh syllabus), focuses entirely on inverse proportion — situations where one quantity increases while the other decreases, but their product always stays constant. This exercise builds the skill of identifying inverse proportion from a table of values and solving real-life word problems using the relation x₁y₁ = x₂y₂.
By the end of this exercise, you should be confident in checking whether two quantities vary inversely, filling in missing values in a table, and applying the inverse proportion formula to practical situations like budgets, books, and arranging objects in rows and columns.
What is Inverse Proportion?
Two quantities x and y are said to be in inverse proportion if an increase in one causes a proportional decrease in the other, and vice versa — such that their product xy always remains a constant value, usually written as k. This relationship is written as:
x ∝ 1/y or xy = k (constant)
We read "x ∝ 1/y" as "x is inversely proportional to y." If y₁ and y₂ are the values of y corresponding to x₁ and x₂, then the inverse proportion relation can also be written in two equivalent forms:
x₁y₁ = x₂y₂ or equivalently x₁ / x₂ = y₂ / y₁
This problem gives three different tables of x and y values. For each table, the method is the same: multiply each pair of x and y values to get xy, and check whether this product stays constant across all columns.
Table (i)
| x | 50 | 40 | 30 | 20 |
|---|---|---|---|---|
| y | 5 | 6 | 7 | 8 |
| xy | 250 | 240 | 210 | 160 |
Table (ii)
| x | 100 | 200 | 300 | 400 |
|---|---|---|---|---|
| y | 60 | 30 | 20 | 15 |
| xy | 6000 | 6000 | 6000 | 6000 |
Table (iii)
| x | 90 | 60 | 45 | 30 | 20 | 5 |
|---|---|---|---|---|---|---|
| y | 10 | 15 | 20 | 25 | 30 | 25 |
| xy | 900 | 900 | 900 | 750 | 600 | 125 |
Here, the total amount to be spent is fixed at ₹6000. As the price of each book (x) goes up, the number of books the school can buy (y) goes down — and vice versa. This is a textbook example of inverse proportion, because:
Price per book (x) × Number of books (y) = Total amount = 6000
| Price of each book, x (₹) | Number of books, y | Check: x × y |
|---|---|---|
| 40 | 150 | 40 × 150 = 6000 |
| 50 | 120 | 50 × 120 = 6000 |
| 60 | 100 | 60 × 100 = 6000 |
| 75 | 80 | 75 × 80 = 6000 |
| 80 | 75 | 80 × 75 = 6000 |
How Each Missing Value Was Found
Since x₁y₁ = x₂y₂ = 6000 for every pair, each missing value is found by dividing 6000 by the known value in that column:
- Column 2 (price = 50): y = 6000 ÷ 50 = 120 books
- Column 3 (number of books = 100): x = 6000 ÷ 100 = ₹60
- Column 4 (price = 75): y = 6000 ÷ 75 = 80 books
- Column 5 (number of books = 75): x = 6000 ÷ 75 = ₹80
This activity uses a squared paper grid of 48 squares to show inverse proportion visually. If the total number of squares stays fixed at 48, then as the number of rows (R) increases, the number of columns (C) must decrease, because:
R × C = 48 ⟹ C = 48 / R
Completing the Table for 48 Squares
| Number of Rows (R) | 2 | 3 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| Number of Columns (C) | 24 | 16 | 12 | 8 | 6 |
- When R = 2: C = 48 ÷ 2 = 24
- When R = 3: C = 48 ÷ 3 = 16
- When R = 8: C = 48 ÷ 8 = 6
(i) Is R₁ : R₂ = C₂ : C₁?
Taking R₁ = 2 and R₂ = 3, we get R₁ : R₂ = 2 : 3. The corresponding columns are C₁ = 24 and C₂ = 16, so C₂ : C₁ = 16 : 24 = 2 : 3.
(ii) Is R₃ : R₄ = C₄ : C₃?
Taking R₃ = 4 and R₄ = 6, we get R₃ : R₄ = 4 : 6 = 2 : 3. The corresponding columns are C₃ = 12 and C₄ = 8, so C₄ : C₃ = 8 : 12 = 2 : 3.
(iii) Are R and C inversely proportional to each other?
(iv) Repeat the activity with 36 squares
Using the same logic, R × C = 36, so C = 36 / R. Filling in the table for different row counts:
| Number of Rows (R) | 2 | 3 | 4 | 6 | 9 |
|---|---|---|---|---|---|
| Number of Columns (C) | 18 | 12 | 9 | 6 | 4 |
- R = 2 → C = 36 ÷ 2 = 18
- R = 3 → C = 36 ÷ 3 = 12
- R = 4 → C = 36 ÷ 4 = 9
- R = 6 → C = 36 ÷ 6 = 6
- R = 9 → C = 36 ÷ 9 = 4
Exercise 10.2 at a Glance — Key Takeaways
| Problem | Concept Tested | Method Used |
|---|---|---|
| Problem 1 | Identifying inverse proportion from a table | Check if xy is constant for all values |
| Problem 2 | Fixed budget — price vs. quantity | x₁y₁ = x₂y₂ = 6000; solve for missing x or y |
| Problem 3 | Rows and columns of a fixed-area grid | R × C = constant; ratio relationships R₁:R₂ = C₂:C₁ |
Common Mistakes to Avoid in Exercise 10.2
- Confusing direct and inverse proportion: In direct proportion, x/y stays constant; in inverse proportion, x × y stays constant. Mixing these up is the most common error.
- Checking only the first two columns: A table may show a constant product for the first two pairs but break the pattern later — always verify xy for every column, as in Problem 1 Table (iii).
- Wrong cross-multiplication: Remember x₁y₁ = x₂y₂ can be rearranged as x₁/x₂ = y₂/y₁ — note that the y-terms swap order. Writing x₁/x₂ = y₁/y₂ is a frequent mistake.
- Forgetting the real-world direction of change: Before solving, pause and think — does it make sense for one quantity to decrease as the other increases? This helps catch arithmetic errors.
What Exercise 10.2 Prepares You For
Having mastered how to identify and apply inverse proportion, you are now ready to tackle mixed problems that combine both direct and inverse variation in Exercise 10.3, which covers time-and-work and time-and-distance type questions — common favourites in board exams.
If you'd like to revisit the basics of direct proportion before moving on, check out the introduction to direct and inverse proportions, which explains the difference between the two using simple, relatable examples.
The ratio-based reasoning practised here — comparing x₁:x₂ with y₂:y₁ — also strengthens your foundation for Comparing Quantities Using Proportion, where similar ratio techniques are applied to percentages, profit and loss, and simple interest.