Exercise 10.4 — Compound Proportion
Compound proportion and its applications.
Direct and Inverse Proportions — Exercise 10.4 (Compound Proportion)
This lesson gives a complete, easy-to-follow explanation of every concept and every problem in Exercise 10.4 — with formula derivations, step-by-step solutions, tables and visuals — for CBSE, Telangana and Andhra Pradesh Class 8 students preparing for school tests and board exams.
What You Will Learn in This Lesson
Exercise 10.4 introduces one of the most useful real-life applications of proportion — compound proportion, where a quantity changes because of two other quantities changing at the same time. By the end of this lesson you will be able to:
- Explain what compound proportion means and how it is different from simple direct or inverse proportion.
- Identify the three types of compound proportion — direct & direct, inverse & inverse, and mixed (direct & inverse).
- Derive the working formula for each type starting from the basic rules of proportion.
- Solve all five problems of Exercise 10.4 step by step, with a clear given-data table for each.
- Avoid the common mistakes students make when choosing the wrong type of proportion.
What is Compound Proportion?
In the earlier exercises of this chapter, you studied two basic ideas: direct proportion (when one quantity increases, the other increases too — like more notebooks costing more money) and inverse proportion (when one quantity increases, the other decreases — like more workers finishing a job in fewer days). In both cases, only two quantities were compared.
But many real situations involve three quantities at once. For example, the total cost of rice for a family trip depends on both the number of people being fed and the number of days the rice must last. If either of these changes, the cost changes too. When one quantity (let's call it x) depends on changes in two other quantities (y and z) together, the relationship is called a compound proportion, and it is solved using a method called the compound ratio method.
Three Types of Compound Proportion — Quick Reference
Every compound proportion problem in Exercise 10.4 falls into one of three categories, depending on how the main quantity x behaves when y and z change:
| Type | Relationship | Everyday Example | Working Formula |
|---|---|---|---|
| Type 1 Direct & Direct |
x is directly proportional to both y and z | More members & more days → more rice needed → higher cost | x₁ ⁄ (y₁z₁) = x₂ ⁄ (y₂z₂) |
| Type 2 Inverse & Inverse |
x is inversely proportional to both y and z | More workers & more hours/day → fewer days needed to finish | x₁y₁z₁ = x₂y₂z₂ |
| Type 3 Direct & Inverse |
x is directly proportional to one and inversely proportional to the other | A longer road needs more men, but fewer days needs even more men | x₁z₁ ⁄ y₁ = x₂z₂ ⁄ y₂ |
When x is Directly Proportional to Both y and z
This case happens when increasing either of the other two quantities increases x, and decreasing either of them decreases x. For example, the cost of rice grows if you feed more people, and it also grows if the rice has to last more days — both relationships push x up.
Deriving the Formula
- If z is kept fixed, x ∝ y gives us x₁ : x₂ = y₁ : y₂.
- If y is kept fixed, x ∝ z gives us x₁ : x₂ = z₁ : z₂.
- Since both ratios equal x₁ : x₂, we can multiply (compound) them together: x₁ : x₂ = (y₁ × z₁) : (y₂ × z₂).
- Writing this as a proportion gives the formula we actually use to solve problems.
x₁ ⁄ (y₁z₁) = x₂ ⁄ (y₂z₂)When x is Inversely Proportional to Both y and z
This case happens when increasing either of the other two quantities decreases x. For example, if more workers are added to a job, fewer days are needed — and if each worker also works more hours per day, even fewer days are needed. Both relationships push the number of days down as the other quantities go up.
Deriving the Formula
- Since x is inversely proportional to y, x₁y₁ = x₂y₂, which gives x₁ : x₂ = y₂ : y₁.
- Since x is inversely proportional to z, in the same way x₁ : x₂ = z₂ : z₁.
- Compounding the two ratios: x₁ : x₂ = (y₂ × z₂) : (y₁ × z₁).
- Rearranging this proportion gives the formula used for solving.
x₁y₁z₁ = x₂y₂z₂When x is Directly Proportional to One and Inversely Proportional to the Other
This is the trickiest type because x behaves differently with each quantity. For example, the number of men needed to lay a road increases if the road is longer (direct), but if more days are allowed to finish the same length, fewer men are needed (inverse). Two of the five problems in Exercise 10.4 — the road and the canal — are of this type.
Deriving the Formula
- Since x is directly proportional to y: x₁ : x₂ = y₁ : y₂.
- Since x is inversely proportional to z: x₁ : x₂ = z₂ : z₁.
- Compounding the two ratios: x₁ : x₂ = (y₁ × z₂) : (y₂ × z₁).
- Rearranging this proportion gives the formula used for solving.
x₁z₁ ⁄ y₁ = x₂z₂ ⁄ y₂Rice costing ₹480 is enough for 8 members for 20 days. We need to find the cost of rice required for 12 members for 15 days. Here, the cost of rice (x) depends on the number of people being fed (y) and the number of days the rice must last (z). More people need more rice, so cost rises with members — a direct relationship. More days also need more rice, so cost rises with days too — another direct relationship. Since x is directly proportional to both y and z, this is a Type 1 (Direct & Direct) problem.
| Quantity | Symbol | Situation 1 | Situation 2 |
|---|---|---|---|
| Cost of rice | x | ₹480 (x₁) | x₂ = ? |
| Number of members | y | 8 (y₁) | 12 (y₂) |
| Number of days | z | 20 (z₁) | 15 (z₂) |
x₁ ⁄ (y₁z₁) = x₂ ⁄ (y₂z₂)Step-by-Step Solution
- Substitute the known values into the Type 1 formula: 480 ⁄ (8 × 20) = x₂ ⁄ (12 × 15)
- Simplify the left-hand side: 8 × 20 = 160, so the equation becomes 480 ⁄ 160 = x₂ ⁄ 180, i.e. 3 = x₂ ⁄ 180.
- Multiply both sides by 180 to find x₂: x₂ = 3 × 180 = 540.
10 men can lay a 75 km road in 5 days. How many days will 15 men take to lay a 45 km road? Here, the number of men (x) is connected to the road length (y) and the number of days (z). A bigger team can lay a longer road in the same time, so men and road-length move together — direct proportion. But a bigger team finishes the same length of road in fewer days, so men and days move in opposite directions — inverse proportion. With x directly proportional to y and inversely proportional to z, this is a Type 3 (Direct & Inverse) problem.
| Quantity | Symbol | Situation 1 | Situation 2 |
|---|---|---|---|
| Number of men | x | 10 (x₁) | 15 (x₂) |
| Length of road (km) | y | 75 (y₁) | 45 (y₂) |
| Number of days | z | 5 (z₁) | z₂ = ? |
x₁z₁ ⁄ y₁ = x₂z₂ ⁄ y₂Step-by-Step Solution
- Substitute the known values into the Type 3 formula: (10 × 5) ⁄ 75 = (15 × z₂) ⁄ 45
- Simplify the left-hand side: 50 ⁄ 75 = (15 × z₂) ⁄ 45.
- Cross-multiply: 50 × 45 = 75 × 15 × z₂, which gives 2250 = 1125 × z₂.
- Divide both sides by 1125: z₂ = 2250 ⁄ 1125 = 2.
24 men working 8 hours a day finish a piece of work in 15 days. How many days will 20 men working 9 hours a day take for the same work? Here, the number of men (x) is related to the working hours per day (y) and the number of days (z). If there are more workers, the same job needs fewer days — x and z are inversely related. Similarly, if each person works more hours per day, fewer days are needed — x and y are also inversely related. Since x is inversely proportional to both y and z, this is a Type 2 (Inverse & Inverse) problem.
| Quantity | Symbol | Situation 1 | Situation 2 |
|---|---|---|---|
| Number of men | x | 24 (x₁) | 20 (x₂) |
| Working hours per day | y | 8 (y₁) | 9 (y₂) |
| Number of days | z | 15 (z₁) | z₂ = ? |
x₁y₁z₁ = x₂y₂z₂Step-by-Step Solution
- Substitute the known values into the Type 2 formula: 24 × 8 × 15 = 20 × 9 × z₂
- Simplify the left-hand side: 24 × 8 = 192, and 192 × 15 = 2880.
- Simplify the right-hand side coefficient: 20 × 9 = 180.
- Solve for z₂: z₂ = 2880 ⁄ 180 = 16.
175 men can dig a 3150 m canal in 36 days. How many men are required to dig a 3900 m canal in just 24 days? The number of men (x) depends on the length of the canal (y) and the number of days (z). A longer canal needs more workers for the same time period, so men and canal length move together — direct proportion. But if the work must be completed in fewer days, more men are needed, so men and days move in opposite directions — inverse proportion. This makes it another Type 3 (Direct & Inverse) problem, just like Problem 2.
| Quantity | Symbol | Situation 1 | Situation 2 |
|---|---|---|---|
| Number of men | x | 175 (x₁) | x₂ = ? |
| Length of canal (m) | y | 3150 (y₁) | 3900 (y₂) |
| Number of days | z | 36 (z₁) | 24 (z₂) |
x₁z₁ ⁄ y₁ = x₂z₂ ⁄ y₂Step-by-Step Solution
- Substitute the known values into the Type 3 formula: (175 × 36) ⁄ 3150 = (x₂ × 24) ⁄ 3900
- Simplify the left-hand side: 175 × 36 = 6300, and 6300 ⁄ 3150 = 2. So the equation becomes 2 = (x₂ × 24) ⁄ 3900.
- Multiply both sides by 3900: x₂ × 24 = 2 × 3900 = 7800.
- Divide both sides by 24: x₂ = 7800 ⁄ 24 = 325.
14 typists, each typing 6 hours a day, take 12 days to complete the manuscript of a book. How many days will 4 typists, each typing 7 hours a day, take for the same manuscript? The number of typists (x) is related to the working hours per day (y) and the number of days (z). Fewer typists mean the job naturally takes more days — x and z are inversely related. But if each typist also works more hours per day, fewer days are needed — x and y are inversely related too. So, just like Problem 3, this is a Type 2 (Inverse & Inverse) problem — the total "typist-hours" needed to finish the manuscript stays the same in both situations.
| Quantity | Symbol | Situation 1 | Situation 2 |
|---|---|---|---|
| Number of typists | x | 14 (x₁) | 4 (x₂) |
| Working hours per day | y | 6 (y₁) | 7 (y₂) |
| Number of days | z | 12 (z₁) | z₂ = ? |
x₁y₁z₁ = x₂y₂z₂Step-by-Step Solution
- Substitute the known values into the Type 2 formula: 14 × 6 × 12 = 4 × 7 × z₂
- Simplify the left-hand side: 14 × 6 = 84, and 84 × 12 = 1008.
- Simplify the right-hand side coefficient: 4 × 7 = 28.
- Solve for z₂: z₂ = 1008 ⁄ 28 = 36.
Exercise 10.4 at a Glance — All Five Problems Compared
| # | Real-life Situation | x, y, z | Type | Answer |
|---|---|---|---|---|
| 1 | Cost of rice for members & days | Cost, Members, Days | Direct & Direct | ₹540 |
| 2 | Men laying a road | Men, Road length, Days | Direct & Inverse | 2 days |
| 3 | Men completing a work | Men, Hours/day, Days | Inverse & Inverse | 16 days |
| 4 | Men digging a canal | Men, Canal length, Days | Direct & Inverse | 325 men |
| 5 | Typists completing manuscript | Typists, Hours/day, Days | Inverse & Inverse | 36 days |
Common Mistakes to Avoid in Exercise 10.4
- Confusing direct and inverse relationships: Before writing any formula, ask yourself — "If y increases, does x increase or decrease?" Answer this for both y and z separately.
- Picking the wrong "x": The quantity x is always the one you are comparing against the compound ratio of the other two — usually the unknown quantity you need to find.
- Using simple proportion instead of compound proportion: If two quantities change at the same time, a single direct or inverse proportion is not enough — you must use the compound formula.
- Arithmetic slips in large products: Simplify step by step (cancel common factors first) instead of multiplying everything together at once.
- Forgetting units in the final answer: Always write ₹, days, men, km or m as required — board exams award marks for this.
What Exercise 10.4 Prepares You For
Exercise 10.4 builds directly on the ideas covered in Introduction to Direct and Inverse Proportions, and in Exercise 10.1 and Exercise 10.2, where direct and inverse proportions were studied separately. If the ratio and proportion ideas used here feel a little rusty, it helps to revisit Comparing Quantities Using Proportion before attempting more compound proportion questions.
Once you are comfortable identifying all three types shown here, you are ready for Exercise 10.3, which builds the time-and-work foundation these compound proportion problems rely on — and for the mixed word problems on cost, work and distance that continue to appear right up to Class 10 board exams.