Ratio & Proportion.
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Lesson Objectives
- Solve advanced ratio problems involving three or more quantities
- Share quantities in given ratios and find unknown parts
- Understand and identify direct proportion relationships
- Understand and identify inverse proportion relationships
- Solve direct proportion problems using the constant of proportionality
- Solve inverse proportion problems using the product constant
- Solve mixed proportion problems (combining direct and inverse relationships)
Introduction to Ratio and Proportion
Ratios compare quantities, while proportions state that two ratios are equal. Understanding ratio and proportion is essential for scaling recipes, mixing paints, dividing profits, calculating speed, working with maps and scales, and many other real-world situations. This section covers advanced ratio problems, direct proportion, inverse proportion, and mixed proportion problems.
Key Formulas
Ratio $a : b$ means $\frac{a}{b}$
Direct proportion: $y = kx$ or $\frac{y_1}{x_1} = \frac{y_2}{x_2}$
Inverse proportion: $y = \frac{k}{x}$ or $x_1 y_1 = x_2 y_2$
Ratio $a : b$ means $\frac{a}{b}$
Direct proportion: $y = kx$ or $\frac{y_1}{x_1} = \frac{y_2}{x_2}$
Inverse proportion: $y = \frac{k}{x}$ or $x_1 y_1 = x_2 y_2$
Key Definitions:
• Ratio: A comparison of two or more quantities using division. Written as $a:b$ or $a:b:c$.
• Proportion: An equation stating that two ratios are equal.
• Direct Proportion: Two quantities increase or decrease together at the same rate. $y \propto x$, so $y = kx$.
• Inverse Proportion: One quantity increases as the other decreases. $y \propto \frac{1}{x}$, so $y = \frac{k}{x}$.
• Constant of Proportionality (k): The fixed value relating two proportional quantities.
• Ratio: A comparison of two or more quantities using division. Written as $a:b$ or $a:b:c$.
• Proportion: An equation stating that two ratios are equal.
• Direct Proportion: Two quantities increase or decrease together at the same rate. $y \propto x$, so $y = kx$.
• Inverse Proportion: One quantity increases as the other decreases. $y \propto \frac{1}{x}$, so $y = \frac{k}{x}$.
• Constant of Proportionality (k): The fixed value relating two proportional quantities.
Advanced Ratio Problems
Advanced ratio problems involve three or more quantities, finding unknown parts when given the total or a difference, and working with ratios in different contexts.
Step-by-Step Method for Advanced Ratio Problems:
1. Write the ratio in its simplest form.
2. Find the total number of parts by adding all ratio numbers.
3. Divide the total quantity by the total number of parts to find the value of one part.
4. Multiply each ratio number by the value of one part to find each share.
5. For difference problems, find the difference in parts and use it to find the value of one part.
1. Write the ratio in its simplest form.
2. Find the total number of parts by adding all ratio numbers.
3. Divide the total quantity by the total number of parts to find the value of one part.
4. Multiply each ratio number by the value of one part to find each share.
5. For difference problems, find the difference in parts and use it to find the value of one part.
Example 1: Three-Quantity Ratio
Problem: Three friends share ₦120,000 in the ratio $3:4:5$. How much does each receive?
Solution:
Total parts = $3 + 4 + 5 = 12$ parts
Value of one part = ₦120,000 ÷ 12 = ₦10,000
First friend: $3 \times 10,000 = ₦30,000$
Second friend: $4 \times 10,000 = ₦40,000$
Third friend: $5 \times 10,000 = ₦50,000$
Answer: ₦30,000, ₦40,000, ₦50,000
Problem: Three friends share ₦120,000 in the ratio $3:4:5$. How much does each receive?
Solution:
Total parts = $3 + 4 + 5 = 12$ parts
Value of one part = ₦120,000 ÷ 12 = ₦10,000
First friend: $3 \times 10,000 = ₦30,000$
Second friend: $4 \times 10,000 = ₦40,000$
Third friend: $5 \times 10,000 = ₦50,000$
Answer: ₦30,000, ₦40,000, ₦50,000
Example 2: Ratio with Difference
Problem: Two numbers are in the ratio $7:3$. The difference between them is 24. Find the numbers.
Solution:
Difference in parts = $7 - 3 = 4$ parts
Value of one part = $24 ÷ 4 = 6$
First number = $7 \times 6 = 42$
Second number = $3 \times 6 = 18$
Answer: 42 and 18
Problem: Two numbers are in the ratio $7:3$. The difference between them is 24. Find the numbers.
Solution:
Difference in parts = $7 - 3 = 4$ parts
Value of one part = $24 ÷ 4 = 6$
First number = $7 \times 6 = 42$
Second number = $3 \times 6 = 18$
Answer: 42 and 18
Example 3: Ratio with Given One Quantity
Problem: In a school, the ratio of boys to girls is $5:3$. If there are 300 boys, how many girls are there?
Solution:
Boys ratio = 5 parts = 300 → one part = $300 ÷ 5 = 60$
Girls = $3 \times 60 = 180$
Answer: 180 girls
Problem: In a school, the ratio of boys to girls is $5:3$. If there are 300 boys, how many girls are there?
Solution:
Boys ratio = 5 parts = 300 → one part = $300 ÷ 5 = 60$
Girls = $3 \times 60 = 180$
Answer: 180 girls
Example 4: Changing Ratios
Problem: Two numbers are in the ratio $5:7$. If 10 is added to each, the new ratio becomes $3:4$. Find the original numbers.
Solution:
Let numbers be $5x$ and $7x$. After adding 10: $\frac{5x+10}{7x+10} = \frac{3}{4}$
Cross-multiply: $4(5x+10) = 3(7x+10)$ → $20x + 40 = 21x + 30$ → $40 - 30 = 21x - 20x$ → $10 = x$
Original numbers: $5x = 50$, $7x = 70$
Answer: 50 and 70
Problem: Two numbers are in the ratio $5:7$. If 10 is added to each, the new ratio becomes $3:4$. Find the original numbers.
Solution:
Let numbers be $5x$ and $7x$. After adding 10: $\frac{5x+10}{7x+10} = \frac{3}{4}$
Cross-multiply: $4(5x+10) = 3(7x+10)$ → $20x + 40 = 21x + 30$ → $40 - 30 = 21x - 20x$ → $10 = x$
Original numbers: $5x = 50$, $7x = 70$
Answer: 50 and 70
Watch Out!
When working with ratios, ensure you are comparing the same units. Also, ratios can be simplified like fractions by dividing all terms by their greatest common factor.
When working with ratios, ensure you are comparing the same units. Also, ratios can be simplified like fractions by dividing all terms by their greatest common factor.
Practice for Advanced Ratio Problems
- Share ₦90,000 in the ratio $2:3:4$.
- Two numbers are in the ratio $9:5$. The difference is 28. Find the numbers.
- In a bag, the ratio of red to blue marbles is $4:7$. If there are 28 blue marbles, how many red marbles are there?
- Three angles of a triangle are in the ratio $2:3:4$. Find each angle.
- Two numbers are in the ratio $3:5$. If 8 is added to each, the new ratio is $2:3$. Find the original numbers.
Direct Proportion
Two quantities are in direct proportion if they increase or decrease together at the same rate. The ratio between them remains constant. In symbols, $y \propto x$ means $y = kx$, where $k$ is the constant of proportionality.
Step-by-Step Method for Direct Proportion:
1. Identify the two quantities that are directly proportional.
2. Write the relationship as $y = kx$ (or $\frac{y}{x} = k$).
3. Use given values to find $k$.
4. Use the value of $k$ to find unknown quantities.
5. Alternatively, use $\frac{y_1}{x_1} = \frac{y_2}{x_2}$.
1. Identify the two quantities that are directly proportional.
2. Write the relationship as $y = kx$ (or $\frac{y}{x} = k$).
3. Use given values to find $k$.
4. Use the value of $k$ to find unknown quantities.
5. Alternatively, use $\frac{y_1}{x_1} = \frac{y_2}{x_2}$.
Example 1: Basic Direct Proportion
Problem: If 5 pens cost ₦200, how much do 8 pens cost?
Solution:
Cost is directly proportional to number of pens.
$\frac{200}{5} = \frac{x}{8}$ → $x = \frac{200 \times 8}{5} = 40 \times 8 = 320$
Answer: ₦320
Problem: If 5 pens cost ₦200, how much do 8 pens cost?
Solution:
Cost is directly proportional to number of pens.
$\frac{200}{5} = \frac{x}{8}$ → $x = \frac{200 \times 8}{5} = 40 \times 8 = 320$
Answer: ₦320
Example 2: Finding Constant of Proportionality
Problem: $y$ is directly proportional to $x$. When $x = 4$, $y = 20$. Find $y$ when $x = 7$.
Solution:
$y = kx$ → $20 = k \times 4$ → $k = 5$
$y = 5 \times 7 = 35$
Answer: 35
Problem: $y$ is directly proportional to $x$. When $x = 4$, $y = 20$. Find $y$ when $x = 7$.
Solution:
$y = kx$ → $20 = k \times 4$ → $k = 5$
$y = 5 \times 7 = 35$
Answer: 35
Example 3: Distance and Time
Problem: A car travels at constant speed. It covers 240 km in 3 hours. How far does it travel in 5 hours?
Solution:
Distance $\propto$ time (constant speed)
$\frac{240}{3} = \frac{d}{5}$ → $d = \frac{240 \times 5}{3} = 80 \times 5 = 400$ km
Answer: 400 km
Problem: A car travels at constant speed. It covers 240 km in 3 hours. How far does it travel in 5 hours?
Solution:
Distance $\propto$ time (constant speed)
$\frac{240}{3} = \frac{d}{5}$ → $d = \frac{240 \times 5}{3} = 80 \times 5 = 400$ km
Answer: 400 km
Example 4: Real-World Application
Problem: The cost of petrol is directly proportional to the number of litres. 15 litres cost ₦9,000. How much do 22 litres cost?
Solution:
$\frac{9000}{15} = \frac{x}{22}$ → $x = \frac{9000 \times 22}{15} = 600 \times 22 = 13,200$
Answer: ₦13,200
Problem: The cost of petrol is directly proportional to the number of litres. 15 litres cost ₦9,000. How much do 22 litres cost?
Solution:
$\frac{9000}{15} = \frac{x}{22}$ → $x = \frac{9000 \times 22}{15} = 600 \times 22 = 13,200$
Answer: ₦13,200
Watch Out!
Direct proportion requires that when one quantity doubles, the other doubles. Not all linear relationships are direct proportion (e.g., $y = 2x + 3$ is not direct proportion because it doesn't pass through the origin).
Direct proportion requires that when one quantity doubles, the other doubles. Not all linear relationships are direct proportion (e.g., $y = 2x + 3$ is not direct proportion because it doesn't pass through the origin).
Practice for Direct Proportion
- If 7 books cost ₦2,100, find the cost of 12 books.
- $y$ is directly proportional to $x$. If $y = 30$ when $x = 5$, find $y$ when $x = 12$.
- A car uses 12 litres of fuel to travel 180 km. How many litres are needed for 300 km?
- If 8 workers produce 240 units in a day, how many units will 15 workers produce (at the same rate)?
- The weight of a metal rod is directly proportional to its length. A 3 m rod weighs 12 kg. Find the weight of a 5 m rod.
Inverse Proportion
Two quantities are in inverse proportion if one increases while the other decreases proportionally. The product of the two quantities is constant. In symbols, $y \propto \frac{1}{x}$ means $y = \frac{k}{x}$ or $xy = k$.
Step-by-Step Method for Inverse Proportion:
1. Identify the two quantities that are inversely proportional.
2. Write the relationship as $xy = k$ (or $y = \frac{k}{x}$).
3. Use given values to find $k$.
4. Use the value of $k$ to find unknown quantities.
5. Alternatively, use $x_1 y_1 = x_2 y_2$.
1. Identify the two quantities that are inversely proportional.
2. Write the relationship as $xy = k$ (or $y = \frac{k}{x}$).
3. Use given values to find $k$.
4. Use the value of $k$ to find unknown quantities.
5. Alternatively, use $x_1 y_1 = x_2 y_2$.
Example 1: Basic Inverse Proportion
Problem: 6 workers can complete a job in 10 days. How long will 15 workers take?
Solution:
More workers → less time (inverse proportion)
$x_1 y_1 = x_2 y_2$ → $6 \times 10 = 15 \times d$ → $60 = 15d$ → $d = 4$ days
Answer: 4 days
Problem: 6 workers can complete a job in 10 days. How long will 15 workers take?
Solution:
More workers → less time (inverse proportion)
$x_1 y_1 = x_2 y_2$ → $6 \times 10 = 15 \times d$ → $60 = 15d$ → $d = 4$ days
Answer: 4 days
Example 2: Finding Constant of Proportionality
Problem: $y$ is inversely proportional to $x$. When $x = 3$, $y = 12$. Find $y$ when $x = 9$.
Solution:
$xy = k$ → $3 \times 12 = 36 = k$
$9 \times y = 36$ → $y = 4$
Answer: 4
Problem: $y$ is inversely proportional to $x$. When $x = 3$, $y = 12$. Find $y$ when $x = 9$.
Solution:
$xy = k$ → $3 \times 12 = 36 = k$
$9 \times y = 36$ → $y = 4$
Answer: 4
Example 3: Speed and Time
Problem: A car travelling at 60 km/h takes 2 hours to cover a distance. How long will it take at 80 km/h?
Solution:
Speed × Time = constant (distance)
$60 \times 2 = 80 \times t$ → $120 = 80t$ → $t = 1.5$ hours
Answer: 1.5 hours
Problem: A car travelling at 60 km/h takes 2 hours to cover a distance. How long will it take at 80 km/h?
Solution:
Speed × Time = constant (distance)
$60 \times 2 = 80 \times t$ → $120 = 80t$ → $t = 1.5$ hours
Answer: 1.5 hours
Example 4: Real-World Application
Problem: A school meal programme feeds 240 students for 30 days with a fixed food supply. How many days will the same food last if 320 students are fed?
Solution:
Number of students × Number of days = constant
$240 \times 30 = 320 \times d$ → $7200 = 320d$ → $d = 22.5$ days
Answer: 22.5 days
Problem: A school meal programme feeds 240 students for 30 days with a fixed food supply. How many days will the same food last if 320 students are fed?
Solution:
Number of students × Number of days = constant
$240 \times 30 = 320 \times d$ → $7200 = 320d$ → $d = 22.5$ days
Answer: 22.5 days
Watch Out!
Inverse proportion means the product is constant, not the sum. If one quantity doubles, the other halves. Be careful not to confuse direct and inverse proportion.
Inverse proportion means the product is constant, not the sum. If one quantity doubles, the other halves. Be careful not to confuse direct and inverse proportion.
Direct vs Inverse Proportion Comparison
| Type | Relationship | Equation | Graph Shape | Example |
|---|---|---|---|---|
| Direct | Increase together | $y = kx$ | Straight line through origin | Cost ∝ number of items |
| Inverse | One increases, other decreases | $xy = k$ | Hyperbola | Time ∝ 1/speed |
Practice for Inverse Proportion
- 8 workers can paint a house in 12 days. How long will 6 workers take?
- $y$ is inversely proportional to $x$. If $y = 15$ when $x = 4$, find $y$ when $x = 10$.
- A car travels at 50 km/h and takes 4 hours to reach its destination. How long will it take at 80 km/h?
- A farmer has enough feed for 60 cows for 45 days. How long will the feed last for 75 cows?
- If 12 pipes fill a tank in 8 hours, how many pipes are needed to fill it in 6 hours?
Mixed and Compound Proportion
Compound proportion involves more than two quantities. For example, the number of workers, time, and output are often related. Some quantities may be directly proportional while others are inversely proportional.
Step-by-Step Method for Compound Proportion:
1. Identify all quantities and how they relate to the unknown.
2. Determine which relationships are direct and which are inverse.
3. Write the combined proportion: $\frac{\text{unknown}}{\text{known}} = \text{product of direct ratios} \times \text{product of inverse reciprocals}$.
4. Solve for the unknown.
1. Identify all quantities and how they relate to the unknown.
2. Determine which relationships are direct and which are inverse.
3. Write the combined proportion: $\frac{\text{unknown}}{\text{known}} = \text{product of direct ratios} \times \text{product of inverse reciprocals}$.
4. Solve for the unknown.
Example 1: Work and Time with Workers
Problem: 5 workers take 8 days to build a wall. How many workers are needed to build the same wall in 4 days?
Solution:
Workers ∝ $\frac{1}{\text{days}}$ (inverse proportion)
$\frac{W_1}{W_2} = \frac{D_2}{D_1}$ → $\frac{5}{W_2} = \frac{4}{8}$ → $\frac{5}{W_2} = \frac{1}{2}$ → $W_2 = 10$
Answer: 10 workers
Problem: 5 workers take 8 days to build a wall. How many workers are needed to build the same wall in 4 days?
Solution:
Workers ∝ $\frac{1}{\text{days}}$ (inverse proportion)
$\frac{W_1}{W_2} = \frac{D_2}{D_1}$ → $\frac{5}{W_2} = \frac{4}{8}$ → $\frac{5}{W_2} = \frac{1}{2}$ → $W_2 = 10$
Answer: 10 workers
Example 2: Three Quantities (Workers, Days, Output)
Problem: 6 workers produce 240 units in 5 days. How many units will 10 workers produce in 8 days?
Solution:
Units ∝ Workers × Days (both direct)
$\frac{U_1}{U_2} = \frac{W_1 \times D_1}{W_2 \times D_2}$ → $\frac{240}{U_2} = \frac{6 \times 5}{10 \times 8} = \frac{30}{80} = \frac{3}{8}$
$U_2 = 240 \times \frac{8}{3} = 80 \times 8 = 640$ units
Answer: 640 units
Problem: 6 workers produce 240 units in 5 days. How many units will 10 workers produce in 8 days?
Solution:
Units ∝ Workers × Days (both direct)
$\frac{U_1}{U_2} = \frac{W_1 \times D_1}{W_2 \times D_2}$ → $\frac{240}{U_2} = \frac{6 \times 5}{10 \times 8} = \frac{30}{80} = \frac{3}{8}$
$U_2 = 240 \times \frac{8}{3} = 80 \times 8 = 640$ units
Answer: 640 units
Example 3: Food Consumption (Inverse with Days, Direct with People)
Problem: 8 people consume 20 kg of rice in 15 days. How much rice will 12 people consume in 10 days?
Solution:
Rice ∝ People × Days (both direct)
$\frac{R_1}{R_2} = \frac{P_1 \times D_1}{P_2 \times D_2}$ → $\frac{20}{R_2} = \frac{8 \times 15}{12 \times 10} = \frac{120}{120} = 1$
$R_2 = 20$ kg
Answer: 20 kg
Problem: 8 people consume 20 kg of rice in 15 days. How much rice will 12 people consume in 10 days?
Solution:
Rice ∝ People × Days (both direct)
$\frac{R_1}{R_2} = \frac{P_1 \times D_1}{P_2 \times D_2}$ → $\frac{20}{R_2} = \frac{8 \times 15}{12 \times 10} = \frac{120}{120} = 1$
$R_2 = 20$ kg
Answer: 20 kg
Example 4: Speed, Time, and Distance with Variation
Problem: A train travels at 80 km/h and takes 6 hours. How long will it take at 100 km/h?
Solution:
Speed × Time = constant (inverse proportion)
$80 \times 6 = 100 \times t$ → $480 = 100t$ → $t = 4.8$ hours
Answer: 4.8 hours
Problem: A train travels at 80 km/h and takes 6 hours. How long will it take at 100 km/h?
Solution:
Speed × Time = constant (inverse proportion)
$80 \times 6 = 100 \times t$ → $480 = 100t$ → $t = 4.8$ hours
Answer: 4.8 hours
Watch Out!
In compound proportion problems, carefully identify which quantities are directly related and which are inversely related. Writing the proportion as a fraction helps avoid errors.
In compound proportion problems, carefully identify which quantities are directly related and which are inversely related. Writing the proportion as a fraction helps avoid errors.
Practice for Mixed and Compound Proportion
- 4 workers can complete a job in 9 days. How many workers are needed to complete it in 6 days?
- 5 machines produce 200 items in 4 hours. How many items will 8 machines produce in 6 hours?
- 10 people consume 30 kg of flour in 12 days. How many people are needed to consume 45 kg in 9 days?
- If 6 painters paint a house in 8 hours, how long will 9 painters take?
- A car travels 300 km in 5 hours at 60 km/h. How long will it take to travel 400 km at 80 km/h?
Methods & Techniques
Mastering ratio and proportion requires systematic approaches and verification strategies.
Verification / Checking Strategy:
1. For ratios: Check that the sum of shares equals the total.
2. For direct proportion: Check that $\frac{y}{x}$ is constant.
3. For inverse proportion: Check that $xy$ is constant.
4. For compound proportion: Check that the relationship makes sense (e.g., more workers should reduce time).
5. Use the unitary method: Find the value of one unit first as an alternative check.
1. For ratios: Check that the sum of shares equals the total.
2. For direct proportion: Check that $\frac{y}{x}$ is constant.
3. For inverse proportion: Check that $xy$ is constant.
4. For compound proportion: Check that the relationship makes sense (e.g., more workers should reduce time).
5. Use the unitary method: Find the value of one unit first as an alternative check.
Example: Checking Direct Proportion
Original: 5 pens cost ₦200, 8 pens cost ₦320.
Check:
$\frac{200}{5} = 40$, $\frac{320}{8} = 40$ ✓ constant ratio
Original: 5 pens cost ₦200, 8 pens cost ₦320.
Check:
$\frac{200}{5} = 40$, $\frac{320}{8} = 40$ ✓ constant ratio
Common Pitfalls & How to Avoid Them:
• Pitfall 1: Adding ratio terms incorrectly → Solution: Double-check the sum of parts.
• Pitfall 2: Confusing direct and inverse proportion → Solution: Think: if one doubles, what happens to the other? Same = direct, opposite = inverse.
• Pitfall 3: Forgetting to simplify ratios → Solution: Always divide by the greatest common factor.
• Pitfall 4: In compound proportion, mixing up direct and inverse relationships → Solution: Write each relationship separately before combining.
• Pitfall 5: Using the wrong formula for inverse proportion → Solution: Remember: $x_1 y_1 = x_2 y_2$.
• Pitfall 1: Adding ratio terms incorrectly → Solution: Double-check the sum of parts.
• Pitfall 2: Confusing direct and inverse proportion → Solution: Think: if one doubles, what happens to the other? Same = direct, opposite = inverse.
• Pitfall 3: Forgetting to simplify ratios → Solution: Always divide by the greatest common factor.
• Pitfall 4: In compound proportion, mixing up direct and inverse relationships → Solution: Write each relationship separately before combining.
• Pitfall 5: Using the wrong formula for inverse proportion → Solution: Remember: $x_1 y_1 = x_2 y_2$.
Quick Reference: Proportion Types
| Scenario | Type | Equation | Example |
|---|---|---|---|
| Cost and quantity | Direct | $\frac{C_1}{Q_1} = \frac{C_2}{Q_2}$ | More items cost more |
| Workers and time | Inverse | $W_1 \times T_1 = W_2 \times T_2$ | More workers take less time |
| Workers, time, output | Compound | $\frac{O_1}{O_2} = \frac{W_1 \times T_1}{W_2 \times T_2}$ | Output depends on both |
Technique Practice
- Verify that 20 pens cost ₦1,000 and 35 pens cost ₦1,750 is a direct proportion.
- Check: If 8 workers take 6 days, and 12 workers take 4 days, is this inverse proportion?
- Identify the error: A student said 6 workers in 5 days produce 30 units, so 10 workers in 3 days produce 50 units. Is this correct? Explain.
- For the ratio $4:5:7$ with total 96, check that the shares sum to 96.
Real-World Applications
Ratio and proportion are used extensively in finance, cooking, construction, map reading, and many other areas.
Application 1: Map Scales
Scenario: A map has a scale of 1:50,000. Two towns are 8 cm apart on the map. Find the actual distance in km.
Problem: Direct proportion.
Solution:
1 cm : 50,000 cm = 1 cm : 0.5 km
8 cm : $8 \times 0.5 = 4$ km
Answer: 4 km
Scenario: A map has a scale of 1:50,000. Two towns are 8 cm apart on the map. Find the actual distance in km.
Problem: Direct proportion.
Solution:
1 cm : 50,000 cm = 1 cm : 0.5 km
8 cm : $8 \times 0.5 = 4$ km
Answer: 4 km
Application 2: Mixing Ratios (Concrete)
Scenario: Concrete is made with cement, sand, and gravel in the ratio $1:2:4$. How much cement is needed to make 350 kg of concrete?
Problem: Share in a given ratio.
Solution:
Total parts = $1 + 2 + 4 = 7$ parts
One part = $350 \div 7 = 50$ kg
Cement = $1 \times 50 = 50$ kg
Answer: 50 kg
Scenario: Concrete is made with cement, sand, and gravel in the ratio $1:2:4$. How much cement is needed to make 350 kg of concrete?
Problem: Share in a given ratio.
Solution:
Total parts = $1 + 2 + 4 = 7$ parts
One part = $350 \div 7 = 50$ kg
Cement = $1 \times 50 = 50$ kg
Answer: 50 kg
Application 3: Recipe Scaling (Direct Proportion)
Scenario: A recipe for 6 people requires 400 g of flour. How much flour is needed for 15 people?
Problem: Direct proportion.
Solution:
$\frac{400}{6} = \frac{x}{15}$ → $x = \frac{400 \times 15}{6} = \frac{6000}{6} = 1000$ g
Answer: 1000 g (1 kg)
Scenario: A recipe for 6 people requires 400 g of flour. How much flour is needed for 15 people?
Problem: Direct proportion.
Solution:
$\frac{400}{6} = \frac{x}{15}$ → $x = \frac{400 \times 15}{6} = \frac{6000}{6} = 1000$ g
Answer: 1000 g (1 kg)
Application 4: Work Rate (Inverse Proportion)
Scenario: 4 machines can produce 600 units in 10 hours. How long will 6 machines take to produce the same number of units?
Problem: Inverse proportion (more machines = less time).
Solution:
$4 \times 10 = 6 \times t$ → $40 = 6t$ → $t = \frac{40}{6} = 6.67$ hours
Answer: 6 hours 40 minutes
Scenario: 4 machines can produce 600 units in 10 hours. How long will 6 machines take to produce the same number of units?
Problem: Inverse proportion (more machines = less time).
Solution:
$4 \times 10 = 6 \times t$ → $40 = 6t$ → $t = \frac{40}{6} = 6.67$ hours
Answer: 6 hours 40 minutes
Application 5: Currency Exchange
Scenario: £1 = ₦1,500. How many naira will £250 exchange to?
Problem: Direct proportion.
Solution:
$250 \times 1,500 = ₦375,000$
Answer: ₦375,000
Scenario: £1 = ₦1,500. How many naira will £250 exchange to?
Problem: Direct proportion.
Solution:
$250 \times 1,500 = ₦375,000$
Answer: ₦375,000
Cross-Curricular Connections
- Geography: Map scales, population densities
- Cooking: Recipe scaling, ingredient ratios
- Construction: Concrete mixing, scale drawings
- Business: Profit sharing, cost allocation
- Science: Concentration ratios, dilution problems
Cumulative Practice Exercises
Try these problems on your own. Show all working steps. Use the verification strategies to check your answers.
- Share ₦240,000 in the ratio $3:5:8$.
- Two numbers are in the ratio $11:4$. The difference is 56. Find the numbers.
- If 9 books cost ₦4,500, find the cost of 15 books.
- $y$ is directly proportional to $x$. If $y = 48$ when $x = 6$, find $y$ when $x = 15$.
- 10 workers can build a wall in 24 days. How many workers are needed to build it in 15 days?
- $y$ is inversely proportional to $x$. If $y = 20$ when $x = 9$, find $y$ when $x = 15$.
- 7 machines produce 350 items in 5 hours. How many items will 10 machines produce in 8 hours?
- In a triangle, the angles are in the ratio $1:2:3$. Find each angle.
- A car travels 240 km in 3 hours. How far will it travel in 5 hours at the same speed?
- If 12 people eat 24 kg of rice in 8 days, how many people will eat 36 kg in 9 days?
- Three numbers are in the ratio $2:5:7$. The sum is 84. Find the numbers.
- A map has scale 1:200,000. Two cities are 12 cm apart on the map. Find the actual distance in km.
- Error analysis: A student said if 6 workers take 8 days, then 12 workers take 4 days. Is this correct? Verify using inverse proportion.
- If 4 painters paint a house in 6 hours, how long will 3 painters take?
- A school has a boy-to-girl ratio of $7:5$. If there are 280 boys, how many girls are there?
Answers to Cumulative Exercises
- Problem: Share ₦240,000 in ratio $3:5:8$.
Answer: Total parts = 16, one part = ₦15,000 → ₦45,000, ₦75,000, ₦120,000 - Problem: Ratio $11:4$, difference 56.
Answer: Difference in parts = 7, one part = 8 → numbers: 88 and 32 - Problem: 9 books cost ₦4,500. Cost of 15 books.
Answer: ₦4,500 ÷ 9 = ₦500 per book, 15 × ₦500 = ₦7,500 - Problem: $y \propto x$, $y=48$ when $x=6$, find $y$ when $x=15$.
Answer: $k = 48/6 = 8$, $y = 8 \times 15 = 120$ - Problem: 10 workers in 24 days, workers for 15 days.
Answer: $10 \times 24 = W \times 15$ → $240 = 15W$ → $W = 16$ workers - Problem: $y \propto 1/x$, $y=20$ when $x=9$, find $y$ when $x=15$.
Answer: $k = 20 \times 9 = 180$, $y = 180/15 = 12$ - Problem: 7 machines, 350 items, 5 hours. Items for 10 machines, 8 hours.
Answer: $\frac{350}{I} = \frac{7 \times 5}{10 \times 8} = \frac{35}{80} = \frac{7}{16}$, $I = 350 \times \frac{16}{7} = 50 \times 16 = 800$ items - Problem: Triangle angles ratio $1:2:3$.
Answer: Total parts = 6, one part = $180^\circ/6 = 30^\circ$ → $30^\circ, 60^\circ, 90^\circ$ - Problem: Car travels 240 km in 3 hours. Distance in 5 hours.
Answer: Speed = 80 km/h, distance = $80 \times 5 = 400$ km - Problem: 12 people, 24 kg, 8 days. People for 36 kg in 9 days.
Answer: $\frac{P_1}{P_2} = \frac{36/9}{24/8} = \frac{4}{3}$, $P_2 = 12 \times \frac{3}{4} = 9$ people - Problem: Numbers ratio $2:5:7$, sum = 84.
Answer: Total parts = 14, one part = 6 → numbers: 12, 30, 42 - Problem: Map scale 1:200,000, 12 cm on map.
Answer: $12 \times 200,000 = 2,400,000$ cm = 24 km - Problem: Error analysis: 6 workers, 8 days → 12 workers, 4 days.
Answer: Inverse proportion: $6 \times 8 = 48$, $12 \times 4 = 48$ ✓ Correct - Problem: 4 painters in 6 hours, time for 3 painters.
Answer: $4 \times 6 = 3 \times t$ → $24 = 3t$ → $t = 8$ hours - Problem: Boys:girls = $7:5$, 280 boys.
Answer: One part = $280/7 = 40$, girls = $5 \times 40 = 200$
Conclusion & Summary
Ratio and proportion are powerful tools for comparing quantities and solving problems involving relationships. Direct proportion means quantities increase together ($y = kx$), while inverse proportion means one increases as the other decreases ($xy = k$). Compound proportion combines multiple relationships. Understanding these concepts is essential for scaling, sharing, and many real-world applications.
Key Takeaways:
1. Ratios: Compare quantities; simplify by dividing by common factors.
2. Direct proportion: $y = kx$ or $\frac{y_1}{x_1} = \frac{y_2}{x_2}$.
3. Inverse proportion: $xy = k$ or $x_1 y_1 = x_2 y_2$.
4. Compound proportion: Combine direct and inverse relationships.
5. Unitary method: Find the value of one unit as a reliable approach.
6. Applications: Scales, recipes, work rates, sharing, finance.
Keep practising with different contexts. Ratio and proportion skills are used daily in many professions!
Video Resource
Watch this video for more examples of direct and inverse proportion problems.
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