Exercise 10.3 — Inverse Applications
Applications of inverse proportion.
Exercise 10.3 – Word Problems on Inverse Proportion
Exercise 10.3 of Class 8 Mathematics, Chapter 10 "Direct and Inverse Proportions" (CBSE, Telangana & Andhra Pradesh syllabus), takes the inverse proportion relation you learned in Exercise 10.2 and applies it to real-life word problems — shopping, food stock, work and time, pipes filling tanks, speed and time, school periods, and even algebraic and percentage-based variations.
Every problem in this exercise follows the same three-step approach: identify the two quantities, confirm they vary inversely (one increases while the other decreases), and apply the formula x₁y₁ = x₂y₂ to find the unknown value.
x₁y₁ = x₂y₂ (Product of the first pair = Product of the second pair)
Since Siri's money is fixed, if the price per kg goes up, the quantity she can buy must go down. Price (x) and quantity (y) are therefore in inverse proportion.
| Price of potatoes (x) | Quantity of potatoes (y) |
|---|---|
| ₹8 (x₁) | 5 kg (y₁) |
| ₹10 (x₂) | ? (y₂) |
Applying x₁y₁ = x₂y₂:
8 × 5 = 10 × y₂ ⟹ 40 = 10 × y₂ ⟹ y₂ = 40 ÷ 10 = 4
The total food stock is constant. If more people join the camp, the stock will run out sooner — so the number of people (x) and the number of days the stock lasts (y) are in inverse proportion.
| Number of people (x) | Number of days (y) |
|---|---|
| 500 (x₁) | 70 (y₁) |
| 500 + 200 = 700 (x₂) | ? (y₂) |
500 × 70 = 700 × y₂ ⟹ y₂ = (500 × 70) ÷ 700 = 50
Fewer workers naturally take more time to complete the same amount of work. So the number of men (x) and the number of days (y) are inversely proportional.
| Number of persons (x) | Number of days (y) |
|---|---|
| 36 (x₁) | 12 (y₁) |
| 9 (x₂) | ? (y₂) |
36 × 12 = 9 × y₂ ⟹ y₂ = (36 × 12) ÷ 9 = 48
Adding more pipes of the same size fills the tank faster, so the number of pipes (x) and the time taken (y) are in inverse proportion.
| Number of pipes (x) | Time to fill the tank (y) |
|---|---|
| 5 (x₁) | 80 min (y₁) |
| 8 (x₂) | ? (y₂) |
5 × 80 = 8 × y₂ ⟹ y₂ = (5 × 80) ÷ 8 = 50
A nautical mile is a unit used for sea distances, equal to 1852 metres. For a fixed distance, increasing speed reduces the travel time — so speed (x) and time (y) are inversely proportional.
| Speed of the boat (x) | Time to reach (y) |
|---|---|
| 16 nautical mph (x₁) | 10 hr (y₁) |
| ? (x₂) | 8 hr (y₂) |
16 × 10 = x₂ × 8 ⟹ x₂ = (16 × 10) ÷ 8 = 20
To compare times correctly, convert both durations to minutes: 1½ hours = 90 minutes, and half an hour = 30 minutes. More pumps mean less time, so the number of pumps (x) and time (y) are inversely proportional.
| Number of pumps (x) | Time to fill (y) |
|---|---|
| 5 (x₁) | 1½ hr = 90 min (y₁) |
| ? (x₂) | ½ hr = 30 min (y₂) |
5 × 90 = x₂ × 30 ⟹ x₂ = (5 × 90) ÷ 30 = 15
To finish the same wall in fewer hours, more workers are needed — so the number of workers (x) and the time taken (y) are inversely proportional.
| Number of workers (x) | Time (y) |
|---|---|
| 15 (x₁) | 48 hr (y₁) |
| ? (x₂) | 30 hr (y₂) |
15 × 48 = x₂ × 30 ⟹ x₂ = (15 × 48) ÷ 30 = 24
The total number of school hours stays the same. If the number of periods decreases, each remaining period must become longer — so the number of periods (x) and the duration of each period (y) are inversely proportional.
| Number of periods (x) | Duration of each period (y) |
|---|---|
| 8 (x₁) | 45 min (y₁) |
| 6 (x₂) | ? (y₂) |
8 × 45 = 6 × y₂ ⟹ y₂ = (8 × 45) ÷ 6 = 60
This problem combines both direct proportion (z ∝ x) and inverse proportion (z ∝ 1/y) in a single relation. We handle each effect separately as a ratio, then combine the two ratios.
Step 1: Express the New Values of x and y
After a 12% increase, the new value of x is x₁ = (112/100)x, so:
x / x₁ = 100 / 112
After a 20% decrease, the new value of y is y₁ = (80/100)y, so:
y₁ / y = 80 / 100
Step 2: Apply the Direct Proportion Part (z ∝ x)
Since z ∝ x, the ratio z/z₁ equals x/x₁:
z / z₁ = x / x₁ = 100 / 112 ⟹ z : z₁ = 100 : 112
Step 3: Apply the Inverse Proportion Part (z ∝ 1/y)
Since z ∝ 1/y, we have zy = z₁y₁, which gives z/z₁ = y₁/y:
z / z₁ = y₁ / y = 80 / 100 ⟹ z : z₁ = 80 : 100
Step 4: Combine Both Ratios (Compound Ratio)
The overall change in z is the compound ratio of the two effects above — multiply the corresponding terms of both ratios:
z : z₁ = (100 × 80) : (112 × 100) = 8000 : 11200
Inverting this ratio to express z₁ in terms of z:
z₁ / z = 11200 / 8000 = 140 / 100 ⟹ z₁ = (100 + 40)/100 × z
This is the same "men and days" idea as Problem 3, but with algebraic expressions instead of plain numbers. The number of men (x) and the number of days (y) are still in inverse proportion, so the same formula x₁y₁ = x₂y₂ applies — only now we simplify an algebraic expression instead of a numeric one.
| Number of men (x) | Time (y) |
|---|---|
| x + 1 (x₁) | x + 1 days (y₁) |
| x + 2 (x₂) | ? (y₂) |
(x + 1) × (x + 1) = (x + 2) × y₂ ⟹ (x + 1)² = (x + 2) × y₂
Dividing both sides by (x + 2) gives:
y₂ = (x + 1)² / (x + 2)
Since the perimeter is fixed at 24 m, we know 2(length + breadth) = 24, so length + breadth = 12 m. This means breadth = 12 − length, and the area is simply length × breadth.
breadth = 12 − length | Area = length × breadth
Completed Table of Values
| Length (cm) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Width (cm) | 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 |
| Area (cm²) | 11 | 20 | 27 | 32 | 35 | 36 | 35 | 32 | 27 |
Observation
As the length increases, the width decreases steadily — but unlike the earlier problems, the area does not follow a simple inverse proportion pattern. Instead:
- From length 1 to 6, as length increases, width decreases and area also increases.
- At length = 6, width = 6 (length and breadth become equal), and the area reaches its maximum value of 36 cm² — this is a square.
- From length 7 onward, as length continues to increase, both width and area start decreasing.
Exercise 10.3 at a Glance — All Problems Compared
| Problem | Real-Life Context | Answer |
|---|---|---|
| 1 | Price vs. quantity of potatoes | 4 kg |
| 2 | Camp food stock vs. number of people | 50 days |
| 3 | Workers vs. days to complete work | 48 days |
| 4 | Pipes vs. time to fill a tank | 50 minutes |
| 5 | Ship speed vs. travel time | Increase by 4 nmph |
| 6 | Pumps vs. time to fill a tank | 15 pumps |
| 7 | Workers vs. hours to build a wall | 24 workers |
| 8 | School periods vs. duration | 60 minutes |
| 9 | Combined direct & inverse variation (%) | 40% increase |
| 10 | Algebraic men-and-days problem | (x+1)²/(x+2) days |
| 11 | Fixed-perimeter rectangle — area vs. length | Max area at length = width = 6 |
Common Mistakes to Avoid in Exercise 10.3
- Mismatched units: As in Problem 6, always convert hours and minutes (or other mixed units) to the same unit before applying x₁y₁ = x₂y₂.
- Treating every variation as inverse: Problem 11 shows that even when one quantity (length) and another (width) sum to a constant, a third quantity (area) derived from them may not vary inversely — always check the actual relationship, not just assume.
- Sign errors with percentage increase/decrease: In Problem 9, a 12% increase becomes the multiplier 112/100, and a 20% decrease becomes 80/100 — mixing these up gives a wrong final percentage.
- Forgetting to interpret the final answer: In Problem 5, the answer x₂ = 20 is the new speed, not the increase. The actual increase required is 20 − 16 = 4 nmph.
- Leaving algebraic answers unsimplified: In Problem 10, always present the final answer as a clean fraction, (x+1)²/(x+2), rather than leaving it as an unsimplified equation.
What Exercise 10.3 Prepares You For
Having practised both checking for inverse proportion (Exercise 10.2) and solving real-life inverse proportion word problems here in Exercise 10.3, you now have a complete toolkit for Chapter 10. These skills — especially the work-and-time and speed-and-time problems — form the foundation for more advanced ratio and proportion questions in Comparing Quantities Using Proportion.
The percentage-based reasoning used in Problem 9 also prepares you for percentage applications in profit, loss, and simple interest problems, while the optimisation idea from Problem 11 (maximum area for a fixed perimeter) reappears when you study mensuration and algebraic identities in Algebraic Expressions.