Chapter Overview
Direct
Proportion Type
Inverse
Proportion Type
x = ky
Direct Formula
xy = k
Inverse Formula
What You'll Learn
Direct Proportion
Understanding how quantities increase or decrease together, with formulas like \( \frac{x}{y} = k \).
Inverse Proportion
Exploring when one quantity increases as the other decreases, using \( xy = k \).
Real-Life Applications
Applying proportions to scenarios like cost, speed, and work.
Problem Solving
Solving examples involving scales, shadows, and mixtures.
Historical Context
This chapter introduces variations in quantities, starting with everyday examples like preparing tea or arranging chairs. It explains direct proportion where ratios remain constant, and inverse proportion where products are constant, building on basic math concepts.
Key Highlights
Key formulas include direct: \( x = ky \), inverse: \( xy = k \). Applications range from cost calculations to time-speed-distance problems, with activities to reinforce understanding.
Comprehensive Chapter Summary
1. Introduction to Proportions
The chapter begins with examples like Mohan preparing tea for varying numbers of people, illustrating how quantities change together. It lists situations where one quantity affects another, such as cost increasing with articles or time decreasing with speed. More content: Five additional situations include fuel consumption with distance, wages with hours worked, crop yield with land area, electricity bill with usage, and pages printed with ink used.
2. Direct Proportion
Definition and Examples
Direct proportion occurs when \( \frac{x}{y} = k \) (constant). Examples: Cost of sugar (\( \frac{36}{1} = 36 \)), petrol and distance (\( \frac{4}{60} = \frac{1}{15} \)). Formula: \( \frac{x_1}{y_1} = \frac{x_2}{y_2} \).
Activities and Observations
Clock hand activity shows angle proportional to time (\( \frac{T}{A} \) constant). Age table shows not always proportional. More formulas: Simple interest \( I = P \times \frac{r}{100} \times t \), compound interest \( A = P(1 + \frac{r}{100})^t \).
Solved Examples
Example 1: Cloth cost. Example 2: Pole shadow (\( x = 21 \) m). Example 3: Paper weight (\( x = 750 \)). Example 4: Train speed (\( x = 25 \) km, \( y = 200 \) min). Example 5: Map scale (\( y = 1200 \) km).
3. Inverse Proportion
Definition and Examples
Inverse proportion: \( xy = k \). Examples: Workers and time, speed and time. Zaheeda's travel: Time decreases as speed increases. Books and price: \( 40 \times 150 = 6000 \).
Activities
Counters arrangement shows rows and columns inverse. Tables check inverse pairs.
Solved Examples
Example 7: Pipes (\( x = 96 \) min). Example 8: Food provisions (\( y = 16 \) days). Example 9: Workers (\( y = 24 \)).
4. Additional Formulas and Content
Key Formulas Expanded
Direct: \( x \propto y \), \( x = ky \). Inverse: \( x \propto \frac{1}{y} \), \( xy = k \). Unitary method for solving. Scale in maps: \( \frac{\text{map distance}}{\text{actual distance}} = \text{scale} \).
5. Applications and Discussions
Real-World Use
Proportions in interest, maps, shadows, work rates. Discussion: Variables not always proportional (e.g., human growth).
Challenges
Identifying proportion type, solving with constants.
6. Summary of Variations
Direct: Ratios constant. Inverse: Products constant. More content: Examples like pressure-volume (inverse), force-acceleration (direct).
Questions and Answers from Chapter
Short Questions
Q1. Check if the parking charges are in direct proportion to the parking time. (Ex 11.1 Q1)
Answer: No.
Q2. In a mixture of paint, find parts of base for 20 parts red pigment. (Ex 11.1 Q2)
Answer: 160.
Q3. How much red pigment with 1800 mL base? (Ex 11.1 Q3)
Answer: 24.
Q4. Bottles filled in five hours? (Ex 11.1 Q4)
Answer: 700.
Q5. Actual length of bacteria? (Ex 11.1 Q5)
Answer: \( 10^{-4} \) cm.
Q6. Length of model ship? (Ex 11.1 Q6)
Answer: 21 cm.
Q7. Crystals in 5 kg sugar? (Ex 11.1 Q7 i)
Answer: \( 2.25 \times 10^7 \).
Q8. Distance covered in map? (Ex 11.1 Q8)
Answer: 4 cm.
Q9. Shadow length for 10 m 50 cm pole? (Ex 11.1 Q9 i)
Answer: 5.6 m.
Q10. Distance in 5 hours? (Ex 11.1 Q10)
Answer: 42 km.
Q11. Which are in inverse proportion? (Ex 11.2 Q1 i)
Answer: Yes.
Q12. Prize for 20 winners? (Ex 11.2 Q2)
Answer: 5000.
Q13. Angle for 12 spokes? (Ex 11.2 Q3)
Answer: 30°.
Q14. Sweets each if 20 children? (Ex 11.2 Q4)
Answer: 6.
Q15. Food last for 30 animals? (Ex 11.2 Q5)
Answer: 4 days.
Medium Questions
Q1. Find enlarged length if 20,000 times. (Ex 11.1 Q5)
Answer: 2 cm. Calculation: Actual length \( 10^{-4} \) cm, enlarged 20,000 times: \( 10^{-4} \times 20000 = 2 \) cm. (3 marks)
Q2. Crystals in 1.2 kg sugar? (Ex 11.1 Q7 ii)
Answer: \( 5.4 \times 10^6 \). Proportion: \( \frac{9 \times 10^6}{2} \times 1.2 = 5.4 \times 10^6 \). (3 marks)
Q3. Height of pole for 5m shadow. (Ex 11.1 Q9 ii)
Answer: 9.375 m. Ratio: \( \frac{5.6}{3.2} = \frac{x}{5} \), \( x = 9.375 \). (3 marks)
Q4. Which are in inverse proportion? (Ex 11.2 Q1 ii)
Answer: No, time and distance at uniform speed are direct. (3 marks)
Q5. Prize for 8 winners. (Ex 11.2 Q2)
Answer: 12500. Inverse: \( 100000 \div 8 = 12500 \). (3 marks)
Q6. Spokes for 40° angle. (Ex 11.2 Q3 iii)
Answer: 9. Inverse: \( 360 \div 40 = 9 \). (3 marks)
Q7. Time for 4 persons to rewire. (Ex 11.2 Q6)
Answer: 3 days. Inverse: \( 4 \div 3 \times 3 = 3 \). (3 marks)
Q8. Boxes with 20 bottles. (Ex 11.2 Q7)
Answer: 15. Inverse: \( 25 \times 12 \div 20 = 15 \). (3 marks)
Q9. Machines for 54 days. (Ex 11.2 Q8)
Answer: 36. Inverse: \( 42 \times 63 \div 54 = 36 \). (3 marks)
Q10. Time at 80 km/h. (Ex 11.2 Q9)
Answer: 1.5 hours. Inverse: \( 2 \times 60 \div 80 = 1.5 \). (3 marks)
Q11. Time for one person. (Ex 11.2 Q10 i)
Answer: 6 days. Inverse: \( 3 \times 2 = 6 \). (3 marks)
Q12. Persons for one day. (Ex 11.2 Q10 ii)
Answer: 6. Inverse: \( 3 \times 2 = 6 \). (3 marks)
Q13. Period length for 9 periods. (Ex 11.2 Q11)
Answer: 40 min. Inverse: \( 45 \times 8 \div 9 = 40 \). (3 marks)
Q14. Check inverse for area and crop. (Ex 11.2 Q1 iii)
Answer: No, direct. (3 marks)
Q15. Angle for 15 spokes. (Ex 11.2 Q3 ii)
Answer: 24°. Inverse: \( 360 \div 15 = 24 \). (3 marks)
Long Questions
Q1. Following are the car parking charges near a railway station up to 4 hours ₹60, 8 hours ₹100, 12 hours ₹140, 24 hours ₹180. Check if in direct proportion. (Ex 11.1 Q1)
Answer: Ratios: \( \frac{60}{4} = 15 \), \( \frac{100}{8} = 12.5 \), \( \frac{140}{12} \approx 11.67 \), \( \frac{180}{24} = 7.5 \). Not constant, so not direct proportion. Explanation: In direct proportion, ratio should be constant, but here it decreases, indicating not proportional.
Q2. Mixture of paint: 1 red with 8 base. Find base for 4,7,12,20 red. If 1 red needs 75 mL base, how much red for 1800 mL base? (Ex 11.1 Q2, Q3)
Answer: Base: 32,56,96,160. Red for 1800 mL: 24. Explanation: Ratio 1:8, so base = 8 × red. For Q3: \( \frac{1}{75} = \frac{x}{1800} \), x=24.
Q3. Machine fills 840 bottles in 6 hours. How many in 5 hours? (Ex 11.1 Q4)
Answer: 700. Explanation: Rate = 840/6 = 140 per hour. In 5 hours: 140 × 5 = 700. Direct proportion: \( \frac{840}{6} = \frac{x}{5} \), x=700.
Q4. Bacteria enlarged 50,000 times is 5 cm. Actual length? Enlarged 20,000 times length? (Ex 11.1 Q5)
Answer: Actual: \( 10^{-4} \) cm. Enlarged: 2 cm. Explanation: Actual = 5 / 50000 = 0.0001 cm. For 20000: 0.0001 × 20000 = 2 cm.
Q5. Model mast 9 cm, actual 12 m. Ship length 28 m, model length? (Ex 11.1 Q6)
Answer: 21 cm. Explanation: Ratio 9/1200 = x/2800, wait no: Mast ratio 9 cm : 12 m = 9:1200 = 3:400. Ship: 28 m = 2800 cm. Model = 2800 × (3/400) / (wait, correct: \( \frac{9}{12} = \frac{x}{28} \), but in m/cm mix. Actual: Ratio 9 cm / 12 m = 9/1200 = 0.0075. Model ship = 28 × 0.0075 × 100 = 21 cm.
Q6. 2 kg sugar has 9 × 10^6 crystals. In 5 kg? In 1.2 kg? (Ex 11.1 Q7)
Answer: 5 kg: 2.25 × 10^7. 1.2 kg: 5.4 × 10^6. Explanation: Per kg: 4.5 × 10^6. 5 × 4.5 × 10^6 = 2.25 × 10^7. 1.2 × 4.5 × 10^6 = 5.4 × 10^6.
Q7. Map scale 1 cm = 18 km. Distance on map for 72 km? (Ex 11.1 Q8)
Answer: 4 cm. Explanation: \( \frac{1}{18} = \frac{x}{72} \), x=4.
Q8. 5.6 m pole shadow 3.2 m. Shadow for 10.5 m pole? Height for 5 m shadow? (Ex 11.1 Q9)
Answer: i) 6 m. ii) 8.75 m. Explanation: Ratio 5.6/3.2. For i: (10.5 × 3.2)/5.6 = 6. For ii: (5 × 5.6)/3.2 = 8.75.
Q9. Truck 14 km in 25 min. Distance in 5 hours? (Ex 11.1 Q10)
Answer: 168 km. Explanation: Speed = 14/25 km/min = 33.6 km/h. In 5 h: 168 km.
Q10. Which in inverse? Workers and time. (Ex 11.2 Q1 i)
Answer: Yes. Explanation: More workers, less time; product constant.
Q11. Prize money table for winners. (Ex 11.2 Q2)
Answer: Inversely proportional. Table: 25000,20000,12500,10000,5000. Explanation: Product 100000 constant.
Q12. Spokes table, angle for 15, spokes for 40°. (Ex 11.2 Q3)
Answer: i) Yes inverse. ii) 24°. iii) 9. Explanation: Angle = 360/spokes.
Q13. Sweets if children reduced by 4. (Ex 11.2 Q4)
Answer: 6. Explanation: 24 × 5 / 20 = 6. Inverse.
Q14. Food for 20 animals last 6 days, for 30? (Ex 11.2 Q5)
Answer: 4 days. Explanation: 20 × 6 / 30 = 4. Inverse.
Q15. 3 persons rewire in 4 days, time for 4? (Ex 11.2 Q6)
Answer: 3 days. Explanation: 3 × 4 / 4 = 3. Inverse.
