NCERT Solutions for Class 8 Maths Chapter 11: Direct and Inverse Proportions — Free PDF

Chapter 11 explores two fundamental relationships between quantities: direct proportion (when one increases, the other increases at the same rate) and inverse proportion (when one increases, the other decreases proportionally).

These concepts appear everywhere in daily life — from calculating the cost of buying multiple items (direct) to figuring out how many workers are needed to finish a job in less time (inverse). Understanding proportions is also foundational for percentage calculations, speed-distance-time problems, and work-rate problems that appear in competitive exams.

This chapter has two exercises. Exercise 11.1 focuses entirely on direct proportion problems, while Exercise 11.2 deals with inverse proportion. The key skill is not the arithmetic (which is straightforward) but correctly identifying which type of proportion applies in a given situation.

Direct Proportion: Two quantities $x$ and $y$ are in direct proportion if their ratio remains constant as they change. Mathematically, $\dfrac{x}{y} = k$ (a constant), or equivalently, $\dfrac{x_1}{y_1} = \dfrac{x_2}{y_2}$.

In plain language: when one quantity doubles, the other also doubles. When one triples, the other triples. They increase and decrease together.

Examples of direct proportion: cost and quantity of items (at fixed price per item), distance travelled and petrol consumed (at constant mileage), wages earned and hours worked (at fixed hourly rate).

Inverse Proportion: Two quantities $x$ and $y$ are in inverse proportion if their product remains constant. Mathematically, $x \times y = k$ (a constant), or equivalently, $x_1 \times y_1 = x_2 \times y_2$.

In plain language: when one quantity doubles, the other halves. They change in opposite directions.

Examples of inverse proportion: speed and time (for a fixed distance), number of workers and days to complete a job (for fixed total work), number of pipes and time to fill a tank.

Unitary Method: This is an alternative approach where you first find the value for one unit, then scale up or down. For example, if $5$ pens cost Rs $60$, then $1$ pen costs Rs $12$, so $8$ pens cost Rs $96$. The unitary method works for both direct and inverse proportion.

How to identify the type: Ask yourself: "If I increase one quantity, does the other increase (direct) or decrease (inverse)?" This single question resolves the type every time.

**Q1. If $5$ notebooks cost Rs $120$, how much do $8$ notebooks cost?**

Solution (Direct Proportion):

More notebooks means more cost — this is direct proportion.

Cross-multiplying:

Alternative (Unitary Method): Cost of $1$ notebook $= \dfrac{120}{5} = \text{Rs } 24$. Cost of $8$ notebooks $= 24 \times 8 = \text{Rs } 192$.

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**Q2. A car travels $180$ km in $3$ hours at constant speed. How far will it travel in $5$ hours?**

Solution:

At constant speed, distance and time are directly proportional.

We can verify: speed $= 180 \div 3 = 60$ km/h. In $5$ hours: $60 \times 5 = 300$ km. Correct.

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**Q3. A recipe needs $250$ g of flour for $4$ servings. How much flour is needed for $10$ servings?**

Solution:

More servings require more flour — direct proportion.

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**Q4. If $12$ metres of cloth cost Rs $1,500$, what will $8$ metres cost?**

Solution:

Cost of $1$ metre $= \dfrac{1500}{12} = \text{Rs } 125$.

Cost of $8$ metres $= 125 \times 8 = \text{Rs } 1000$.

Alternatively: $\dfrac{12}{1500} = \dfrac{8}{x} \Rightarrow x = \dfrac{1500 \times 8}{12} = 1000$.

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**Q5. A map uses a scale of $1$ cm $= 50$ km. If two cities are $3.5$ cm apart on the map, what is the actual distance?**

Solution:

Map distance and actual distance are directly proportional.

**Q1. If $12$ workers can build a wall in $10$ days, how many days will $15$ workers take?**

Solution (Inverse Proportion):

More workers means fewer days — inverse proportion.

Verification: Total work $= 12 \times 10 = 120$ worker-days. With $15$ workers: $120 \div 15 = 8$ days. Correct.

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**Q2. A car travelling at $60$ km/h takes $4$ hours for a journey. How long will it take at $80$ km/h?**

Solution:

For a fixed distance, speed and time are inversely proportional.

Verification: Distance $= 60 \times 4 = 240$ km. At $80$ km/h: $240 \div 80 = 3$ hours. Correct.

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**Q3. A pipe fills a tank in $6$ hours. Two identical pipes together will fill it in how many hours?**

Solution:

More pipes means less time — inverse proportion.

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**Q4. A garrison of $500$ soldiers has provisions for $30$ days. If $200$ more soldiers join, how long will the provisions last?**

Solution:

More soldiers means provisions last fewer days — inverse proportion.

Total soldiers now $= 500 + 200 = 700$.

So the provisions will last approximately $21.4$ days.

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**Q5. If $8$ taps can fill a swimming pool in $27$ hours, how many taps are needed to fill it in $18$ hours?**

Solution:

Fewer hours means more taps — inverse proportion.

Sometimes a problem involves more than two quantities. These require careful analysis of each relationship.

**Example 1. If $6$ workers working $8$ hours a day can build a wall in $12$ days, how many days will $8$ workers working $6$ hours a day take?**

Solution:

Total work $= 6 \times 8 \times 12 = 576$ worker-hours.

With $8$ workers at $6$ hours/day: $8 \times 6 \times d = 576$, so $48d = 576$, giving $d = 12$ days.

Note: More workers (inverse with days) but fewer hours per day (direct with days). The effects partially cancel here.

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**Example 2. A factory produces $240$ toys in $4$ days with $10$ machines. How many toys can $15$ machines produce in $6$ days?**

Solution:

Rate per machine per day $= \dfrac{240}{4 \times 10} = 6$ toys.

With $15$ machines in $6$ days: $6 \times 15 \times 6 = 540$ toys.

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**Example 3. A school trip needs $4$ buses (each seating $50$) to transport all students. If they switch to minibuses seating $20$ each, how many minibuses are needed?**

Solution:

Total students $= 4 \times 50 = 200$. Number of minibuses $= \dfrac{200}{20} = 10$.

This is inverse proportion: smaller seating capacity means more vehicles needed.

Mistake 1: Choosing the wrong type of proportion.
This is the most common error. Students sometimes set up a direct proportion when it should be inverse, or vice versa. Always ask: "If one quantity goes up, does the other go up or down?"

Example of the error: "If $5$ workers take $20$ days, how many days will $10$ workers take?" Setting up $\dfrac{5}{20} = \dfrac{10}{x}$ gives $x = 40$ days — which is wrong because more workers should mean fewer days, not more! The correct setup is $5 \times 20 = 10 \times x$, giving $x = 10$ days.

Mistake 2: Forgetting that proportion requires a constant relationship.
Not all quantities that increase together are in direct proportion. For example, the side of a square and its area are NOT directly proportional (area $= s^2$, not $ks$). Direct proportion specifically means a constant ratio.

Mistake 3: Not checking the answer for reasonableness.
Always ask: "Does my answer make sense?" If more workers are deployed, the number of days should decrease. If a car goes faster, it should take less time. A quick sanity check catches many errors.

Mistake 4: Mixing up the cross-multiplication formula.
For direct proportion, $\dfrac{x_1}{y_1} = \dfrac{x_2}{y_2}$. For inverse proportion, $x_1 \times y_1 = x_2 \times y_2$. Students sometimes cross-multiply an inverse proportion equation, leading to an incorrect answer.

Mistake 5: Ignoring units in word problems.
Always check that both sides of your equation use the same units. If one quantity is in hours and another in minutes, convert before setting up the proportion.

Identifying the correct type of proportion is the single most important skill in this chapter. Here is a quick reference table:

The golden test: Ask yourself — "If one quantity goes UP, does the other go UP (direct) or DOWN (inverse)?" This one question resolves every proportion problem.

Mixed proportion (compound proportion): When a problem involves three or more quantities (e.g., workers, days, AND hours per day), handle each pair separately. First calculate the total work (worker-days or machine-hours), then find the unknown.

Q1. If $15$ books cost Rs $675$, how much do $23$ books cost?

Answer: Cost of $1$ book $= 675 \div 15 = \text{Rs } 45$. Cost of $23$ books $= 45 \times 23 = \text{Rs } 1035$.

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Q2. A train travelling at $90$ km/h covers a distance in $5$ hours. At what speed should it travel to cover the same distance in $3$ hours?

Answer: Inverse proportion. $90 \times 5 = x \times 3 \Rightarrow x = 150$ km/h.

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Q3. A photograph is $4$ cm wide and $6$ cm tall. If it is enlarged so the width becomes $10$ cm, what will be the new height (keeping the same proportions)?

Answer: Direct proportion. $\dfrac{4}{6} = \dfrac{10}{h} \Rightarrow h = \dfrac{6 \times 10}{4} = 15$ cm.

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Q4. $18$ men can dig a trench in $10$ days. How many men are needed to dig it in $6$ days?

Answer: Inverse proportion. $18 \times 10 = x \times 6 \Rightarrow x = 30$ men.

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Q5. A car uses $10$ litres of petrol for $120$ km. How much petrol is needed for $300$ km?

Answer: Direct proportion. $\dfrac{10}{120} = \dfrac{x}{300} \Rightarrow x = \dfrac{10 \times 300}{120} = 25$ litres.

• Direct proportion: $\dfrac{x_1}{y_1} = \dfrac{x_2}{y_2}$ (ratio is constant).
  - Inverse proportion: $x_1 \times y_1 = x_2 \times y_2$ (product is constant).
  - The key skill is identifying whether a relationship is direct or inverse.
  - The unitary method (finding value per unit) is a powerful alternative approach.
  - Real-life applications include cost-quantity, speed-time, workers-days, and many more.
  - Always verify that your answer makes sense in context — this catches most errors.

Practise proportion problems on SparkEd for adaptive, exam-style questions!