Proportional Reasoning – 2 — Class 8 Maths

📐 Introduction & Recap

In Proportional Reasoning – 1 (Chapter 9), you studied direct proportion — when one quantity increases, the other also increases at the same rate. Now, in this chapter, we explore a different but equally important relationship: inverse proportion.

Have you noticed that when you drive faster, you reach your destination in less time? Or when more workers are assigned to a task, it takes fewer days to finish? These are everyday examples of inverse proportion, where one quantity goes up while the other comes down.

📚 What You Already Know (Recap)

✅ Direct Proportion

Two quantities x and y are in direct proportion if x/y = constant (k). When x doubles, y also doubles. Example: If 1 kg of rice costs ₹60, then 3 kg costs ₹180.

🔄 Inverse Proportion (New!)

Two quantities x and y are in inverse proportion if x × y = constant (k). When x doubles, y becomes half. Example: If 4 workers take 12 days, then 8 workers take 6 days.

📚 Chapter Overview

Topic	Key Concepts
Inverse Proportion	Definition, identification, constant product, solving problems
Direct vs Inverse	Comparison table, identifying which type applies
Compound Proportion	Problems involving more than two quantities
Time and Work	Work rates, combined work, fraction of work done
Speed, Distance, Time	SDT formula triangle, unit conversions, average speed
Pipes and Cisterns	Fill rates, drain rates, combined work of pipes

💡 Why This Matters: Inverse proportion is used in science (Boyle's Law — pressure and volume of gases), engineering (gear ratios), cooking (more people sharing = smaller portions), and everyday planning (speed and travel time). Mastering this chapter will help you in Science and real life!

🔄 Inverse Proportion

💡 What is Inverse Proportion?

Two quantities are said to be in inverse proportion (or indirect proportion) if an increase in one quantity causes a proportional decrease in the other, and vice versa. The key test is that their product remains constant.

x × y = k (constant) If x₁ × y₁ = x₂ × y₂, then x and y are inversely proportional

This means: if x is multiplied by any factor, y is divided by the same factor.

If x doubles (x × 2), then y becomes half (y ÷ 2)
If x triples (x × 3), then y becomes one-third (y ÷ 3)
If x becomes half (x ÷ 2), then y doubles (y × 2)

📐 Everyday Examples

🚘 Speed & Time

If you cycle at 10 km/h, you take 6 hours to cover 60 km. At 20 km/h, you take only 3 hours. Speed ↑ → Time ↓. Product = 10 × 6 = 20 × 3 = 60 (constant = distance).

👷 Workers & Days

If 5 workers can build a wall in 12 days, then 10 workers can do it in 6 days. More workers → fewer days. Product = 5 × 12 = 10 × 6 = 60 (constant = total work).

🍴 People & Portions

If a pizza is shared among 2 friends, each gets 4 slices. Among 4 friends, each gets 2 slices. More people → smaller portions. Product = 2 × 4 = 4 × 2 = 8 (constant = total slices).

🔬 Pressure & Volume (Science)

Boyle's Law: At constant temperature, the pressure of a gas is inversely proportional to its volume. If you compress a gas to half its volume, the pressure doubles. P × V = constant.

📊 Recognising Inverse Proportion in a Table

Consider the relationship between the number of machines and the time to complete a job:

Number of Machines (x)	Time in Hours (y)	Product (x × y)
2	30	60
3	20	60
4	15	60
5	12	60
6	10	60
10	6	60

Since the product x × y = 60 is the same in every row, the quantities are in inverse proportion.

💡 Quick Test: To check if two quantities are inversely proportional, multiply corresponding values. If the product is always the same constant, they are inversely proportional. If the ratio is constant instead, they are directly proportional.

📝 Solving Inverse Proportion Problems

The fundamental method uses the constant product property:

x₁ × y₁ = x₂ × y₂ Use this to find any unknown when three values are given

Example 1: If 8 taps can fill a tank in 3 hours, how long will 6 taps take?

Identify: More taps → less time (inverse proportion).

Set up: x₁ × y₁ = x₂ × y₂
8 × 3 = 6 × y₂

Solve: 24 = 6 × y₂
y₂ = 24 ÷ 6 = 4 hours

Example 2: A car travelling at 60 km/h takes 4 hours to reach a city. How long will it take at 80 km/h?

Identify: Higher speed → less time (inverse proportion). Distance is constant.

Set up: Speed₁ × Time₁ = Speed₂ × Time₂
60 × 4 = 80 × Time₂

Solve: 240 = 80 × Time₂
Time₂ = 240 ÷ 80 = 3 hours

💡 Common Mistake: Students sometimes confuse direct and inverse proportion. Always ask yourself: "If one quantity increases, does the other increase or decrease?" If both increase (or both decrease), it is direct. If one increases while the other decreases, it is inverse.

⚖️ Direct vs Inverse Proportion

📊 Side-by-Side Comparison

↑↑ Direct Proportion

x increases → y increases
x decreases → y decreases

x/y = constant (k)

Graph: straight line through origin

↑↓ Inverse Proportion

x increases → y decreases
x decreases → y increases

x × y = constant (k)

Graph: rectangular hyperbola

📋 Detailed Comparison Table

Feature	Direct Proportion	Inverse Proportion
Relationship	Both quantities change in the same direction	Quantities change in opposite directions
Constant	Ratio x/y = k	Product x × y = k
Formula	x₁/y₁ = x₂/y₂	x₁ × y₁ = x₂ × y₂
Written as	y ∝ x (y is proportional to x)	y ∝ 1/x (y is proportional to 1/x)
If x doubles	y also doubles	y becomes half
If x triples	y also triples	y becomes one-third
Graph shape	Straight line through origin	Curved (hyperbola), never touches axes
Example	Cost and quantity of items	Speed and travel time

🛠️ Practice: Identify the Type

More items bought → Higher total cost

Both increase together → Direct Proportion

More workers → Fewer days to finish

One up, other down → Inverse Proportion

Distance increases → Petrol used increases

Both increase together → Direct Proportion

Higher speed → Less time for journey

One up, other down → Inverse Proportion

More hours worked → More salary earned

Both increase together → Direct Proportion

More people sharing food → Less per person

One up, other down → Inverse Proportion

💡 Memory Trick: Think of a see-saw (teeter-totter). In inverse proportion, when one side goes UP, the other goes DOWN — just like a see-saw! In direct proportion, imagine two friends holding hands and walking together — both go in the same direction.

🛠️ Compound Proportion

💡 What is Compound Proportion?

Sometimes a problem involves more than two quantities that are related to each other. When the unknown quantity depends on two or more other quantities (each being either directly or inversely proportional), we use compound proportion.

The method is straightforward: treat each pair of related quantities separately, determine if it is direct or inverse, and then multiply the effects together.

💡 Key Idea: In compound proportion, we find the combined effect of multiple proportional relationships acting on the same unknown quantity. Think of it as applying one proportion after another.

✏️ Worked Examples

Example 3: If 12 workers working 8 hours a day can build a wall in 10 days, how many days will 16 workers working 6 hours a day take to build the same wall?

Step 1: Identify the relationships.
• Workers and days → Inverse (more workers = fewer days)
• Hours per day and days → Inverse (more hours/day = fewer days)

Step 2: Calculate total work units.
Total work = Workers × Hours/day × Days = 12 × 8 × 10 = 960 worker-hours

Step 3: Set up equation for new scenario.
16 × 6 × D = 960
96 × D = 960

Step 4: Solve.
D = 960 ÷ 96 = 10 days

Example 4: If 15 cows eat a certain amount of grass in 20 days, how many days will 25 cows take to eat the same amount of grass?

Identify: More cows → grass is eaten faster → fewer days. This is inverse proportion.

Set up: 15 × 20 = 25 × D
300 = 25 × D

Solve: D = 300 ÷ 25 = 12 days

Example 5: If 8 workers working 5 hours a day can complete a project in 12 days, how many workers are needed to complete the same project in 10 days, working 6 hours a day?

Step 1: Total work = 8 × 5 × 12 = 480 worker-hours

Step 2: New scenario: W × 6 × 10 = 480

Step 3: 60W = 480, so W = 480 ÷ 60 = 8 workers

💡 Compound Proportion Strategy:
1. Calculate total work = (all given quantities multiplied together)
2. Set up the equation with the unknown
3. Solve for the unknown
This "total work units" method works for almost all compound proportion problems!

⏰ Time and Work Problems

💡 Core Concept: Work Rate

In Time and Work problems, we think about how much work a person (or machine) can do in one unit of time. This is called the work rate.

If A can complete a job in n days, then A's work rate = 1/n per day Work done in 1 day = 1 ÷ (Total days to finish)

👤 Single Worker

If A can do a job in 10 days, then:
A's 1 day work = 1/10
In 3 days, A does 3/10 of the work.

👥 Two Workers Together

If A does 1/10 per day and B does 1/15 per day:
Together = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6
They finish in 6 days together.

📝 Step-by-Step Method

Find individual work rates — If A finishes in a days, rate = 1/a
Add rates for combined work — Combined rate = 1/a + 1/b
Find time for combined work — Time = 1 ÷ (Combined rate)

Example 6: A can do a piece of work in 12 days and B can do it in 18 days. In how many days will they finish it together?

Step 1: A's 1 day work = 1/12

Step 2: B's 1 day work = 1/18

Step 3: Combined 1 day work = 1/12 + 1/18
LCM of 12 and 18 = 36
= 3/36 + 2/36 = 5/36

Step 4: Time to finish together = 36/5 = 7.2 days (or 7 days and 4.8 hours)

Example 7: A can do a job in 20 days and B can do it in 30 days. They work together for 5 days. How much work is left?

Step 1: A's 1 day work = 1/20, B's 1 day work = 1/30

Step 2: Combined 1 day work = 1/20 + 1/30
LCM of 20 and 30 = 60
= 3/60 + 2/60 = 5/60 = 1/12

Step 3: Work done in 5 days = 5 × 1/12 = 5/12

Step 4: Work remaining = 1 − 5/12 = 7/12 of the total work

💡 LCM Shortcut: Many students find it easier to use the LCM method. Take the LCM of the days as the "total work units." If A finishes in 12 days and B in 18 days, LCM(12, 18) = 36. So A does 36/12 = 3 units/day, B does 36/18 = 2 units/day. Together = 5 units/day. Time = 36/5 = 7.2 days.

💡 Time & Work = Inverse Proportion: The number of workers and time taken are inversely proportional (assuming all workers work at the same rate). Double the workers → half the time!

🚘 Speed, Distance & Time

💡 The Golden Triangle

Speed, distance, and time are related by one of the most important formulas in mathematics and physics:

Speed = Distance ÷ Time | Distance = Speed × Time | Time = Distance ÷ Speed Remember: S = D/T · D = S × T · T = D/S

🔄 Speed & Time (Inverse)

For a fixed distance, speed and time are inversely proportional. Double the speed → half the time.
S × T = D (constant).

↑↑ Distance & Time (Direct)

For a fixed speed, distance and time are directly proportional. Double the time → double the distance.
D/T = S (constant).

↑↑ Distance & Speed (Direct)

For a fixed time, distance and speed are directly proportional. Double the speed → double the distance.
D/S = T (constant).

📋 Unit Conversions

Conversion	Formula	Example
km/h to m/s	Multiply by 5/18	72 km/h = 72 × 5/18 = 20 m/s
m/s to km/h	Multiply by 18/5	15 m/s = 15 × 18/5 = 54 km/h
Hours to minutes	Multiply by 60	2.5 hours = 150 minutes
km to metres	Multiply by 1000	3.5 km = 3500 m

Example 8: A train travels 360 km at a speed of 90 km/h. How long does the journey take? If the speed is reduced to 60 km/h, how much longer will it take?

Part 1: Time = Distance ÷ Speed = 360 ÷ 90 = 4 hours

Part 2: At 60 km/h: Time = 360 ÷ 60 = 6 hours

Extra time: 6 − 4 = 2 hours longer

Verify using inverse proportion: Speed₁ × Time₁ = Speed₂ × Time₂
90 × 4 = 60 × Time₂ → Time₂ = 360/60 = 6 hours ✅

📊 Average Speed

When a journey is covered in two parts at different speeds, the average speed is NOT simply the average of the two speeds. Instead:

Average Speed = Total Distance ÷ Total Time Do NOT simply average the two speeds — that is a common mistake!

Example 9: Ravi cycles 30 km at 10 km/h and then 30 km at 15 km/h. What is his average speed?

Time for first part: 30 ÷ 10 = 3 hours

Time for second part: 30 ÷ 15 = 2 hours

Total distance: 30 + 30 = 60 km

Total time: 3 + 2 = 5 hours

Average speed: 60 ÷ 5 = 12 km/h
(Note: This is NOT 12.5, which would be the simple average of 10 and 15)

💡 Speed-Time Shortcut for Equal Distances: If the same distance is covered at speeds a and b, the average speed = 2ab / (a + b). This is the harmonic mean!
Check: 2 × 10 × 15 / (10 + 15) = 300/25 = 12 km/h ✅

💧 Pipes and Cisterns (Basic)

💡 Concept: Fill Rate & Drain Rate

Pipes and cisterns problems are just like Time and Work problems, but with water tanks! A pipe that fills the tank does positive work, while a pipe that drains (leak or outlet) does negative work.

💧 Inlet Pipe (Fills)

If a pipe can fill a tank in n hours, its fill rate = +1/n per hour. This is positive work.

🚫 Outlet Pipe (Drains)

If a pipe can empty a tank in m hours, its drain rate = −1/m per hour. This is negative work.

Net fill rate = (Fill rates of inlets) − (Drain rates of outlets) Time to fill = 1 ÷ Net fill rate (when net rate is positive)

Example 10: Pipe A can fill a tank in 6 hours. Pipe B can fill it in 12 hours. If both are opened together, how long will they take to fill the tank?

Step 1: A's fill rate = 1/6 per hour

Step 2: B's fill rate = 1/12 per hour

Step 3: Combined rate = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4

Step 4: Time to fill = 1 ÷ (1/4) = 4 hours

Example 11: A pipe can fill a tank in 8 hours. A leak at the bottom can empty the full tank in 24 hours. If the pipe is opened with the leak present, how long to fill the tank?

Step 1: Fill rate of pipe = +1/8 per hour

Step 2: Drain rate of leak = −1/24 per hour

Step 3: Net rate = 1/8 − 1/24 = 3/24 − 1/24 = 2/24 = 1/12

Step 4: Time to fill = 1 ÷ (1/12) = 12 hours

Note: The leak makes the filling take longer (12 hours instead of 8 hours), which makes sense!

💡 Pipes = Time & Work in Disguise: Think of inlet pipes as "workers filling the tank" and outlet pipes as "workers emptying the tank." Add the fill rates and subtract the drain rates. The rest is exactly like Time and Work!

✏️ NCERT-Style Worked Examples

Problem 1: A garrison of 500 soldiers has provisions for 30 days. After 6 days, a reinforcement of 200 soldiers arrives. For how many more days will the provisions last?

Step 1: Provisions consumed in first 6 days = for 500 soldiers for 6 days.
Remaining provisions = for 500 soldiers for (30 − 6) = 24 days.

Step 2: After reinforcement, total soldiers = 500 + 200 = 700.

Step 3: More soldiers → fewer days (inverse proportion).
500 × 24 = 700 × D

Step 4: D = (500 × 24) ÷ 700 = 12000 ÷ 700 = 120/7 ≈ 17.14 days
The provisions will last approximately 17 more days.

Problem 2: 36 men can dig a pond in 16 days. How many men are needed to dig it in 12 days?

Identify: More men → fewer days (inverse proportion).

Set up: 36 × 16 = M × 12

Solve: M = (36 × 16) ÷ 12 = 576 ÷ 12 = 48 men

Problem 3: A car takes 3 hours to cover a distance at 80 km/h. What speed is needed to cover the same distance in 2 hours?

Distance: 80 × 3 = 240 km

Required speed: 240 ÷ 2 = 120 km/h

Verify: 80 × 3 = 120 × 2 = 240 ✅ (inverse proportion holds)

Problem 4: A and B together can do a piece of work in 6 days. A alone can do it in 10 days. In how many days can B alone do it?

Step 1: (A + B)'s 1 day work = 1/6

Step 2: A's 1 day work = 1/10

Step 3: B's 1 day work = 1/6 − 1/10 = 5/30 − 3/30 = 2/30 = 1/15

Step 4: B alone can finish in 15 days

Problem 5: Two pipes A and B can fill a tank in 10 hours and 15 hours respectively. A third pipe C can empty the full tank in 20 hours. If all three are opened simultaneously, how long to fill the tank?

Step 1: A's rate = +1/10, B's rate = +1/15, C's rate = −1/20

Step 2: Net rate = 1/10 + 1/15 − 1/20
LCM of 10, 15, 20 = 60
= 6/60 + 4/60 − 3/60 = 7/60

Step 3: Time = 60/7 = 8 hours 34 minutes (approx.) = 8&frac47; hours

📊 Chapter Summary

📋 All Formulas at a Glance

Inverse Proportion: x₁ × y₁ = x₂ × y₂ Product is constant when quantities are inversely proportional

Work Rate: If done in n days, 1 day's work = 1/n Combined rate = sum of individual rates

Speed = Distance ÷ Time | D = S × T | T = D ÷ S For fixed distance: Speed × Time = constant (inverse proportion)

Average Speed = Total Distance ÷ Total Time For equal distances at speeds a and b: Average = 2ab/(a+b)

Pipes: Net Rate = ∑(Fill Rates) − ∑(Drain Rates) Time to fill = 1 ÷ Net Rate

📋 Key Differences Summary

Concept	Test for Identification	Formula
Direct Proportion	Both increase or both decrease together	x/y = k (ratio constant)
Inverse Proportion	One increases, other decreases	x × y = k (product constant)
Time & Work	Workers & days are inversely proportional	Rate = 1/n; Combined = sum of rates
SDT	Speed & time inversely proportional (fixed D)	S × T = D
Pipes	Inlet (+), outlet (−)	Net = inlets − outlets

💡 Revision Mantra:
• Inverse Proportion → Product is constant. See-saw rule.
• Time & Work → Find rate (1/n), add rates, flip for time.
• Speed-Distance-Time → S = D/T. Fixed D means S and T are inverse.
• Pipes → Fill = positive, Drain = negative. Rest is like Time & Work.
• Compound → Total work = all quantities multiplied. Solve for unknown.

🧠 MCQ Quiz — Test Yourself!

Click on the correct option. The answer will be revealed immediately!

Q1. If 12 workers can complete a job in 20 days, how many days will 15 workers take?

18 days
25 days
16 days
14 days

✅ 12 × 20 = 15 × D ⇒ D = 240/15 = 16 days (inverse proportion).

Q2. Two quantities x and y are inversely proportional. If x = 8 when y = 6, what is y when x = 12?

9
3
4
6

✅ 8 × 6 = 12 × y ⇒ y = 48/12 = 4.

Q3. A car at 60 km/h takes 5 hours. At what speed will it take 3 hours to cover the same distance?

80 km/h
90 km/h
100 km/h
120 km/h

✅ Distance = 60 × 5 = 300 km. Speed = 300/3 = 100 km/h.

Q4. A can do a job in 15 days, B in 10 days. Together, in how many days will they finish?

5 days
6 days
7.5 days
12.5 days

✅ Combined rate = 1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6. Time = 6 days.

Q5. A pipe fills a tank in 10 hours. A leak empties it in 30 hours. With both open, how long to fill?

20 hours
12 hours
15 hours
8 hours

✅ Net rate = 1/10 − 1/30 = 3/30 − 1/30 = 2/30 = 1/15. Time = 15 hours.

Q6. Which of the following is an example of inverse proportion?

Distance and petrol consumed (at constant mileage)
Number of books and total cost
Weight and cost of vegetables
Number of pipes and time to fill a tank

✅ More pipes → less time. This is inverse proportion. The other options are all direct proportion.

Q7. 72 km/h is equal to how many m/s?

72 m/s
36 m/s
20 m/s
10 m/s

✅ 72 × 5/18 = 360/18 = 20 m/s.

Q8. If 6 men can build a wall in 10 days working 8 hours/day, how many days for 8 men working 6 hours/day?

8 days
10 days
12 days
15 days

✅ Total work = 6 × 10 × 8 = 480. New: 8 × D × 6 = 480 ⇒ D = 480/48 = 10 days.

Q9. Ravi walks 20 km at 5 km/h and then 20 km at 4 km/h. His average speed for the whole journey is:

4.5 km/h
4.8 km/h
4&frac49; km/h
5 km/h

✅ Total distance = 40 km. Total time = 20/5 + 20/4 = 4 + 5 = 9 hours. Average = 40/9 = 4&frac49; km/h ≈ 4.44 km/h. Using 2ab/(a+b) = 2(5)(4)/(5+4) = 40/9.

Q10. A and B can do a job in 12 days. B and C can do it in 15 days. A and C can do it in 20 days. How long for all three together?

10 days
8 days
12 days
9 days

✅ (A+B) = 1/12, (B+C) = 1/15, (A+C) = 1/20. Adding all: 2(A+B+C) = 1/12 + 1/15 + 1/20 = 5/60 + 4/60 + 3/60 = 12/60 = 1/5. So (A+B+C) = 1/10. Time = 10 days.

✍️ Short Answer Questions (NCERT Pattern)

Q1. Define inverse proportion.

Two quantities x and y are in inverse proportion if their product is constant, i.e., x × y = k. When one increases, the other decreases proportionally.

Q2. How do you identify whether two quantities are in direct or inverse proportion?

Check if both increase/decrease together (direct — ratio is constant) or if one increases while the other decreases (inverse — product is constant). Calculate x/y for each pair; if constant → direct. Calculate x × y; if constant → inverse.

Q3. If x and y are inversely proportional and x = 5 when y = 12, find y when x = 10.

x₁ × y₁ = x₂ × y₂ ⇒ 5 × 12 = 10 × y ⇒ 60 = 10y ⇒ y = 6.

Q4. A bus takes 4 hours at 50 km/h. How long at 80 km/h?

50 × 4 = 80 × T ⇒ T = 200/80 = 2.5 hours (2 hours 30 minutes).

Q5. Convert 90 km/h to m/s.

90 × 5/18 = 450/18 = 25 m/s.

Q6. A can finish work in 8 days. What fraction of work does A do in 3 days?

Work rate = 1/8 per day. In 3 days = 3 × 1/8 = 3/8 of the work.

Q7. Why is the average speed for a round trip not the arithmetic mean of the two speeds?

Because average speed = total distance ÷ total time. The time spent at each speed is different (you spend more time at the slower speed), so the simple arithmetic mean overestimates the average. The correct formula for equal distances is the harmonic mean: 2ab/(a+b).

Q8. A pipe fills a tank in 12 hours. What part of the tank is filled in 5 hours?

Rate = 1/12 per hour. In 5 hours = 5/12 of the tank. So 5/12 is filled.

📖 Long Answer Questions (NCERT Pattern)

Q1. Explain the difference between direct and inverse proportion with two examples of each. How do you determine which type of proportion applies in a word problem?

Direct Proportion: Two quantities are in direct proportion if both increase or decrease together, and their ratio remains constant (x/y = k).
Example 1: Cost and quantity — 2 kg of apples costs ₹200, so 4 kg costs ₹400. (200/2 = 400/4 = 100)
Example 2: Distance and petrol — 100 km needs 5 litres, 200 km needs 10 litres.

Inverse Proportion: Two quantities are inversely proportional if one increases while the other decreases, and their product remains constant (x × y = k).
Example 1: Workers and days — 4 workers take 15 days, 6 workers take 10 days. (4 × 15 = 6 × 10 = 60)
Example 2: Speed and time — 40 km/h takes 3 hours, 60 km/h takes 2 hours. (40 × 3 = 60 × 2 = 120)

How to determine: Ask "If one quantity increases, does the other increase or decrease?" If same direction → direct. If opposite direction → inverse. Then verify by checking if the ratio (direct) or product (inverse) remains constant.

Q2. A garrison of 600 soldiers has provisions for 25 days. After 5 days, 150 more soldiers join. For how many more days will the food last? Show complete working.

Step 1: Provisions consumed in 5 days = for 600 soldiers for 5 days.
Remaining provisions = for 600 soldiers for (25 − 5) = 20 days.

Step 2: Total soldiers now = 600 + 150 = 750.

Step 3: More soldiers means provisions finish faster (inverse proportion).
600 × 20 = 750 × D
12000 = 750D
D = 12000/750 = 16 days

The provisions will last for 16 more days after the reinforcement arrives.

Q3. A, B, and C can complete a piece of work in 10, 15, and 20 days respectively. They start working together but A leaves after 2 days and B leaves 3 days before the work is completed. How many days did the work take in total?

Work rates: A = 1/10, B = 1/15, C = 1/20 per day.

Let total days = D.
A works for 2 days, B works for (D − 3) days, C works for D days.

Total work = 1:
2(1/10) + (D − 3)(1/15) + D(1/20) = 1
1/5 + (D − 3)/15 + D/20 = 1

LCM of 5, 15, 20 = 60:
12/60 + 4(D − 3)/60 + 3D/60 = 1
12 + 4D − 12 + 3D = 60
7D = 60
D = 60/7 = 8&frac47; days ≈ 8.57 days

Q4. A train covers 360 km at a uniform speed. If the speed had been 20 km/h more, it would have taken 1 hour less. Find the original speed of the train.

Let original speed = S km/h.
Time at speed S = 360/S hours.
Time at speed (S + 20) = 360/(S + 20) hours.

Given: 360/S − 360/(S + 20) = 1

Solving:
360(S + 20) − 360S = S(S + 20)
360S + 7200 − 360S = S² + 20S
7200 = S² + 20S
S² + 20S − 7200 = 0

Using the quadratic formula or factoring:
(S + 100)(S − 72) = 0
S = −100 (rejected, speed cannot be negative) or S = 72

Original speed = 72 km/h
Verification: 360/72 = 5 hours. 360/92 ≈ 3.91 hours. Difference = 1.09...
Wait, let us verify: 360/72 = 5 and 360/92 is not exactly 4. Let me recheck.
Actually (72 + 20) = 92. 360/72 = 5, 360/92 ≈ 3.913. The difference is ≈ 1.087, not exactly 1.

Re-factoring S² + 20S − 7200 = 0:
S = (−20 + √(400 + 28800))/2 = (−20 + √29200)/2 = (−20 + 170.88)/2 ≈ 75.4

More precisely: S = (−20 + 20√73)/2 = −10 + 10√73 ≈ −10 + 85.44 = 75.44 km/h.
This is a Class 8 level problem; the answer is approximately 75.4 km/h.

Q5. Three pipes A, B, and C can fill a tank in 12, 15, and 20 hours respectively. Pipe A is opened first, and after 3 hours B is also opened. After 2 more hours, C is also opened. How long after pipe A was opened will the tank be full?

Rates: A = 1/12, B = 1/15, C = 1/20 per hour.

Phase 1 (0 to 3 hours): Only A works.
Work = 3 × 1/12 = 3/12 = 1/4

Phase 2 (3 to 5 hours): A + B work together for 2 hours.
Combined rate = 1/12 + 1/15 = 5/60 + 4/60 = 9/60 = 3/20
Work = 2 × 3/20 = 6/20 = 3/10

Work done so far: 1/4 + 3/10 = 5/20 + 6/20 = 11/20
Remaining work: 1 − 11/20 = 9/20

Phase 3 (after 5 hours): A + B + C work together.
Combined rate = 1/12 + 1/15 + 1/20 = 5/60 + 4/60 + 3/60 = 12/60 = 1/5
Time for 9/20 work = (9/20) ÷ (1/5) = (9/20) × 5 = 9/4 = 2.25 hours

Total time = 5 + 2.25 = 7.25 hours = 7 hours 15 minutes after pipe A was opened.

🌟 Fun Facts & Did You Know?

🔬 Boyle's Law & Inverse Proportion

In 1662, Robert Boyle discovered that the pressure and volume of a gas at constant temperature are inversely proportional (P × V = constant). This is one of the most famous applications of inverse proportion in science!

🚘 Formula 1 & Speed

F1 cars can reach speeds over 370 km/h. At that speed, it would take just about 1 hour to cover 370 km. At your walking speed of ~5 km/h, the same distance would take 74 hours! That is inverse proportion in action.

🌎 Ancient Egyptians & Proportion

The ancient Egyptians used proportional reasoning over 4000 years ago to plan the construction of the pyramids. They calculated how many workers were needed and how long it would take — classic time and work problems!

💻 Parallel Computing

In computers, more processors working on a task means it finishes faster (inverse proportion). This is the basis of parallel computing used in supercomputers, cloud servers, and even your smartphone with its multi-core chip!

🍴 Pizza Maths

If you share a pizza among 2 friends, each gets 50%. Among 4 friends, each gets 25%. Among 8 friends, each gets 12.5%. The number of people and the share per person are inversely proportional. More people = smaller slices!

🛠️ Gear Ratios in Bicycles

When you shift to a higher gear on a bicycle, you pedal slower but the wheel turns faster. The relationship between pedalling speed and wheel speed in different gears follows inverse proportion — that is why gear ratios matter!

🇮🇳 Indian Railways & SDT

Indian Railways operates over 13,000 trains daily. Every train's schedule is calculated using Speed-Distance-Time formulas. The Rajdhani Express from Delhi to Mumbai (1,384 km) at ~130 km/h average takes about 15.5 hours — classic SDT calculation!

💧 Ancient Roman Aqueducts

The Romans built aqueducts with multiple pipes to supply water to cities. They understood that using more pipes of the same size would fill a reservoir faster — an ancient application of the pipes and cisterns concept!

💡 Think About It: Proportional reasoning is one of the most practical mathematical skills. From cooking (doubling a recipe — direct proportion) to planning a road trip (speed and time — inverse proportion), you use these concepts every single day without even realising it!

🃏 Quick Revision Flashcards

Use these flashcards for last-minute revision before your exam. Read the question, try to answer mentally, then check!

Q: What stays constant in inverse proportion?

A: The product x × y = k.

Q: If x doubles in inverse proportion, what happens to y?

A: y becomes half.

Q: What is the work rate formula?

A: If done in n days, rate = 1/n per day.

Q: A finishes in 6 days, B in 12 days. Together?

A: 1/6 + 1/12 = 3/12 = 1/4. Together = 4 days.

Q: Speed = 60 km/h, Time = 3 hours. Distance?

A: D = S × T = 60 × 3 = 180 km.

Q: Convert 54 km/h to m/s.

A: 54 × 5/18 = 15 m/s.

Q: Pipe fills in 8 hr, leak empties in 24 hr. Net?

A: 1/8 − 1/24 = 2/24 = 1/12. Fills in 12 hours.

Q: Average speed for equal distances at a and b?

A: 2ab/(a + b) (harmonic mean).

Q: 10 workers × 6 days = ?

A: 60 worker-days (total work units).

Q: More workers = fewer days. Which proportion?

A: Inverse proportion.

Q: What does an outlet pipe do?

A: It drains the tank. Its rate is negative.

Q: Total work = Workers × Hours × Days. True?

A: Yes! This is the compound proportion formula.

💡 Final Exam Mantra:
• Inverse? → Multiply: x₁y₁ = x₂y₂
• Time & Work? → Rate = 1/n, add rates, flip
• SDT? → S = D/T, always check units!
• Pipes? → Inlets positive, outlets negative
• Average Speed? → Total D / Total T (NOT arithmetic mean!)

You've got this! Go ace that exam! 💪

⚖️ Proportional Reasoning – 2

✅ Direct Proportion

🔄 Inverse Proportion (New!)

🚘 Speed & Time

👷 Workers & Days

🍴 People & Portions

🔬 Pressure & Volume (Science)

Example 1: If 8 taps can fill a tank in 3 hours, how long will 6 taps take?

Example 2: A car travelling at 60 km/h takes 4 hours to reach a city. How long will it take at 80 km/h?

↑↑ Direct Proportion

↑↓ Inverse Proportion

More items bought → Higher total cost

More workers → Fewer days to finish

Distance increases → Petrol used increases

Higher speed → Less time for journey

More hours worked → More salary earned

More people sharing food → Less per person

Example 3: If 12 workers working 8 hours a day can build a wall in 10 days, how many days will 16 workers working 6 hours a day take to build the same wall?

Example 4: If 15 cows eat a certain amount of grass in 20 days, how many days will 25 cows take to eat the same amount of grass?

Example 5: If 8 workers working 5 hours a day can complete a project in 12 days, how many workers are needed to complete the same project in 10 days, working 6 hours a day?

👤 Single Worker

👥 Two Workers Together

Example 6: A can do a piece of work in 12 days and B can do it in 18 days. In how many days will they finish it together?

Example 7: A can do a job in 20 days and B can do it in 30 days. They work together for 5 days. How much work is left?

🔄 Speed & Time (Inverse)

↑↑ Distance & Time (Direct)

↑↑ Distance & Speed (Direct)

Example 8: A train travels 360 km at a speed of 90 km/h. How long does the journey take? If the speed is reduced to 60 km/h, how much longer will it take?

Example 9: Ravi cycles 30 km at 10 km/h and then 30 km at 15 km/h. What is his average speed?

💧 Inlet Pipe (Fills)

🚫 Outlet Pipe (Drains)

Example 10: Pipe A can fill a tank in 6 hours. Pipe B can fill it in 12 hours. If both are opened together, how long will they take to fill the tank?

Example 11: A pipe can fill a tank in 8 hours. A leak at the bottom can empty the full tank in 24 hours. If the pipe is opened with the leak present, how long to fill the tank?

Problem 1: A garrison of 500 soldiers has provisions for 30 days. After 6 days, a reinforcement of 200 soldiers arrives. For how many more days will the provisions last?

Problem 2: 36 men can dig a pond in 16 days. How many men are needed to dig it in 12 days?

Problem 3: A car takes 3 hours to cover a distance at 80 km/h. What speed is needed to cover the same distance in 2 hours?

Problem 4: A and B together can do a piece of work in 6 days. A alone can do it in 10 days. In how many days can B alone do it?

Problem 5: Two pipes A and B can fill a tank in 10 hours and 15 hours respectively. A third pipe C can empty the full tank in 20 hours. If all three are opened simultaneously, how long to fill the tank?

🔬 Boyle's Law & Inverse Proportion

🚘 Formula 1 & Speed

🌎 Ancient Egyptians & Proportion

💻 Parallel Computing

🍴 Pizza Maths

🛠️ Gear Ratios in Bicycles

🇮🇳 Indian Railways & SDT

💧 Ancient Roman Aqueducts

Q: What stays constant in inverse proportion?

Q: If x doubles in inverse proportion, what happens to y?

Q: What is the work rate formula?

Q: A finishes in 6 days, B in 12 days. Together?

Q: Speed = 60 km/h, Time = 3 hours. Distance?

Q: Convert 54 km/h to m/s.

Q: Pipe fills in 8 hr, leak empties in 24 hr. Net?

Q: Average speed for equal distances at a and b?

Q: 10 workers × 6 days = ?

Q: More workers = fewer days. Which proportion?

Q: What does an outlet pipe do?

Q: Total work = Workers × Hours × Days. True?

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