Official NCERT 2026-27 Solutions
NCERT Solutions for Class 8 Maths Chapter 7: Proportional Reasoning-1 (Ganita Prakash)
Authentic handwritten solutions in Topper Notebook style. Features complete ratio sharing, mixture modifications, unitary method, density comparisons, ancient Lilavati problems, and rates of work for all 19 exercises from Pages 170 to 178 of Ganita Prakash Part I.
Quick Chapter Navigation & Blueprint
- Part A: Direct Proportion & Construction Scales (Page 170) — Earth’s orbital speed in km/week and house masonry brick estimation (108 ft walls).
- Part B: Ratio Division & Mixture Transformations (Page 175) — Dividing &inr;4,500 in 2:3, acid-water solutions, paint color shade lightening, idli batter ratio, and double-bucket paint mixing.
- Part C: Real-World Comparisons & Historical Math (Pages 176–177) — Fruit juice ratios, school bus fleet scheduling, Delhi vs. Mumbai population density, crane anatomy scaling, Lilavati saffron problem, and sibling age ratio equations.
- Part D: Rates, Alloys & Agricultural Optimization (Pages 177–178) — Gold density, farm manure acreage calculation, tap flow rate, land valuation, tractor vs. oxen ploughing times, and &inr;10 coin alloy metal valuation.
Official Textbook PDF Reference: Download NCERT Class 8 Maths Chapter 7 (hegp107.pdf)
Part A: Direct Proportions & Scale Estimation (Page 170)
Question 1 (Page 170 – Earth’s Orbit):
The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week? (Take 1 year = 52 weeks).
The Earth travels approximately 940 million kilometres around the Sun in a year. How many kilometres will it travel in a week? (Take 1 year = 52 weeks).
Topper’s Unitary Conversion & Division:
- 1 million = 10,00,000 (10 lakh) = 106.
- Total distance in 1 year = 940 × 106 km = 940,000,000 km.
- 1 year = 52 weeks.
- Distance traveled per week = 940,000,000 ÷ 52.
- Dividing:
940,000,000 ÷ 52 = 18,076,923.076… km.
Rounding to the nearest whole kilometre = 18,076,923 km.
Final Answer: The Earth travels approximately 18,076,923 km in a week.
Question 2 (Page 170 – House Masonry Bricks):
A mason needs to build both the outer perimeter walls and the interior partition wall of a house. To build 10 feet of wall, he requires approximately 1450 bricks. How many bricks would he need to build the entire house if total wall lengths are: 12 + 12 + 12 + 15 + 9 + 15 + 9 + 9 + 9 + 6 = 108 feet?
A mason needs to build both the outer perimeter walls and the interior partition wall of a house. To build 10 feet of wall, he requires approximately 1450 bricks. How many bricks would he need to build the entire house if total wall lengths are: 12 + 12 + 12 + 15 + 9 + 15 + 9 + 9 + 9 + 6 = 108 feet?
Topper’s Direct Proportion Formulation:
- Total length of wall to be constructed = 108 feet.
- Given: 10 feet of wall requires 1,450 bricks.
- Rate of bricks per foot = 1450 / 10 = 145 bricks per foot.
- Setting up direct proportion: 10 : 1450 = 108 : x.
- x = (1450 × 108) / 10 = 145 × 108.
145 × 108 = 145 × (100 + 8) = 14500 + 1160 = 15,660 bricks.
Final Answer: The mason will need 15,660 bricks to build the house.
Part B: Ratio Divisions & Mixture Transformations (Page 175)
Question 1 (Page 175):
Divide &inr;4,500 into two parts in the ratio 2 : 3.
Divide &inr;4,500 into two parts in the ratio 2 : 3.
Step-by-Step Ratio Sharing:
- Total number of ratio parts = 2 + 3 = 5 parts.
- Value of 1 part = &inr;4,500 / 5 = &inr;900.
- First Part (2 parts) = 2 × &inr;900 = &inr;1,800.
- Second Part (3 parts) = 3 × &inr;900 = &inr;2,700.
- Check: 1800 + 2700 = 4500 ✓.
The two parts are &inr;1,800 and &inr;2,700.
Question 2 (Page 175):
In a science lab, acid and water are mixed in the ratio 1 : 5 to make a solution. In a bottle containing 240 mL of the solution, how much acid and water does it contain?
In a science lab, acid and water are mixed in the ratio 1 : 5 to make a solution. In a bottle containing 240 mL of the solution, how much acid and water does it contain?
Solution Breakdown:
- Total ratio parts = 1 + 5 = 6 parts.
- Volume of 1 part = 240 mL / 6 = 40 mL.
- Volume of Acid (1 part) = 1 × 40 mL = 40 mL.
- Volume of Water (5 parts) = 5 × 40 mL = 200 mL.
The bottle contains 40 mL of acid and 200 mL of water.
Question 3 (Page 175 – Paint Lightening):
Blue and yellow paints are mixed in the ratio 3 : 5 to produce green paint. To produce 40 mL of green paint, how much of each colour is needed? If 20 mL of yellow paint is added to make a lighter shade, what is the new ratio of blue to yellow?
Blue and yellow paints are mixed in the ratio 3 : 5 to produce green paint. To produce 40 mL of green paint, how much of each colour is needed? If 20 mL of yellow paint is added to make a lighter shade, what is the new ratio of blue to yellow?
Step-by-Step Two-Stage Mixture Analysis:
- Stage 1 (Initial 40 mL green paint):
Total ratio parts = 3 + 5 = 8 parts.
Volume of 1 part = 40 mL / 8 = 5 mL.
• Blue paint = 3 × 5 mL = 15 mL.
• Yellow paint = 5 × 5 mL = 25 mL. - Stage 2 (Adding 20 mL yellow paint):
• New volume of blue paint = 15 mL (unchanged).
• New volume of yellow paint = 25 mL + 20 mL = 45 mL. - New Ratio:
Blue : Yellow = 15 : 45 = (15 ÷ 15) : (45 ÷ 15) = 1 : 3.
Answers:
Initial amounts: 15 mL of blue and 25 mL of yellow.
New ratio after addition: 1 : 3.
Initial amounts: 15 mL of blue and 25 mL of yellow.
New ratio after addition: 1 : 3.
Question 4 (Page 175):
To make soft idlis, you need to mix rice and urad dal in the ratio 2 : 1. If you need 6 cups of this mixture, how many cups of rice and urad dal will you need?
To make soft idlis, you need to mix rice and urad dal in the ratio 2 : 1. If you need 6 cups of this mixture, how many cups of rice and urad dal will you need?
Direct Ratio Calculation:
- Total parts = 2 + 1 = 3 parts.
- 1 part = 6 cups / 3 = 2 cups.
- Cups of rice = 2 × 2 = 4 cups.
- Cups of urad dal = 1 × 2 = 2 cups.
You will need 4 cups of rice and 2 cups of urad dal.
Question 5 (Page 175 – Paint Bucket Addition):
I have one bucket of orange paint made by mixing red and yellow paints in the ratio 3 : 5. I added another full bucket of yellow paint to this mixture. What is the ratio of red paint to yellow paint in the new mixture?
I have one bucket of orange paint made by mixing red and yellow paints in the ratio 3 : 5. I added another full bucket of yellow paint to this mixture. What is the ratio of red paint to yellow paint in the new mixture?
Fractional Part Arithmetic:
- In the 1st bucket of orange paint (ratio 3 : 5):
• Fraction of red paint = 3/8 bucket.
• Fraction of yellow paint = 5/8 bucket. - A second full bucket (1 whole bucket = 8/8) of pure yellow paint is added:
• New red paint quantity = 3/8 bucket.
• New yellow paint quantity = 5/8 + 1 = 5/8 + 8/8 = 13/8 bucket. - New Ratio of Red to Yellow:
Red : Yellow = (3/8) : (13/8) = 3 : 13.
Final Ratio: The ratio of red paint to yellow paint in the new mixture is 3 : 13.
Part C: Density, Scheduling & Historical Algebra (Pages 176–177)
Question 2 (Page 176 – School Bus Scheduling):
Last year, 3 buses were hired for a school trip carrying 162 students and teachers, and all buses were full. This year, 204 students and teachers are going. How many buses will be needed? Will all buses be full?
Last year, 3 buses were hired for a school trip carrying 162 students and teachers, and all buses were full. This year, 204 students and teachers are going. How many buses will be needed? Will all buses be full?
Capacity & Remainder Analysis:
- Capacity of 1 bus = 162 / 3 = 54 passengers.
- For 204 passengers:
Number of buses needed = 204 / 54 = 3 with a remainder of 42. - Since 3 buses can only carry 3 × 54 = 162 people, 42 people would be left behind. Therefore, 4 buses are needed!
- Total capacity of 4 buses = 4 × 54 = 216 seats.
- Empty/vacant seats = 216 − 204 = 12 seats.
Answers: 4 buses are needed. No, they will not all be full — exactly 12 seats will remain vacant.
Question 3 (Page 176 – Population Density):
Delhi: Area = 1,484 sq. km, Population ≈ 30 million.
Mumbai: Area = 550 sq. km, Population ≈ 20 million.
Which city is more crowded? Why?
Delhi: Area = 1,484 sq. km, Population ≈ 30 million.
Mumbai: Area = 550 sq. km, Population ≈ 20 million.
Which city is more crowded? Why?
Population Density Calculations:
- Population density = Total Population ÷ Total Land Area (persons per sq. km).
- Delhi Density:
30,000,000 / 1,484 ≈ 20,216 persons/sq. km. - Mumbai Density:
20,000,000 / 550 ≈ 36,364 persons/sq. km. - Comparing: 36,364 > 20,216. Mumbai has over 16,000 more people squeezed into every single square kilometre!
Final Answer: Mumbai is significantly more crowded because its population density (36,364 people/sq. km) is substantially higher than Delhi’s (20,216 people/sq. km).
Question 5 (Page 176 – Ancient Problem from Bhaskara’s Lilavati):
“If 2½ palas of saffron costs 37 niskas, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?”
“If 2½ palas of saffron costs 37 niskas, O expert businessman! tell me quickly what quantity of saffron can be bought for 9 niskas?”
Topper’s Rule of Three (Trairasika):
- Given: 37 niskas buys 2½ = 5/2 = 2.5 palas of saffron.
- Quantity of saffron bought for 1 niska = 2.5 / 37 palas.
- Quantity of saffron bought for 9 niskas:
Q = (2.5 × 9) / 37 = 22.5 / 37 = 45 / 74 palas ≈ 0.608 palas.
(Note: In ancient inverse problems, if price varies inversely: 37 × 2.5 / 9 = 92.5 / 9 ≈ 10.28 palas). - Direct proportional value: 45/74 palas (or ~0.61 palas).
Final Answer: Exactly 45/74 palas (approximately 0.61 palas) of saffron can be bought for 9 niskas.
Question 6 (Page 176 – Sibling Age Problem):
Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain’s age when the ratio of her age to her brother’s age is 1 : 2?
Harmain is a 1-year-old girl. Her elder brother is 5 years old. What will be Harmain’s age when the ratio of her age to her brother’s age is 1 : 2?
Algebraic Linear Equation Formulation:
- Let the number of years that pass be x.
- Harmain’s future age = 1 + x.
- Her brother’s future age = 5 + x.
- Setting up the ratio:
(1 + x) / (5 + x) = 1 / 2 - Cross-multiplying:
2(1 + x) = 1(5 + x)
2 + 2x = 5 + x
2x − x = 5 − 2
x = 3 years. - Harmain’s age after 3 years = 1 + 3 = 4 years old.
(Check: Brother’s age = 5 + 3 = 8 years; ratio = 4/8 = 1/2 ✓).
Final Answer: Harmain will be 4 years old when the ratio of their ages becomes 1 : 2.
Part D: Densities, Rates & Alloy Valuations (Pages 177–178)
Question 7 (Page 177 – Density of Gold):
The mass of equal volumes of gold and water are in the ratio 37 : 2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold?
The mass of equal volumes of gold and water are in the ratio 37 : 2. If 1 litre of water is 1 kg in mass, what is the mass of 1 litre of gold?
Mass Proportion Calculation:
- Mass of Gold : Mass of Water = 37 : 2.
- Mass of 1 litre of water = 1 kg.
- Let mass of 1 litre of gold be M kg:
M / 1 = 37 / 2
M = 18.5 kg.
Final Answer: The mass of 1 litre of gold is 18.5 kg.
Question 9 (Page 177 – Tap Flow Rate):
A tap takes 15 seconds to fill a 500 mL mug of water. How much time does the same tap take to fill a bucket of 10-litre capacity?
A tap takes 15 seconds to fill a 500 mL mug of water. How much time does the same tap take to fill a bucket of 10-litre capacity?
Unit Conversion & Direct Time Scaling:
- 10 litres = 10 × 1,000 mL = 10,000 mL.
- Number of 500 mL mugs in the bucket = 10,000 / 500 = 20 mugs.
- Time required = 20 × 15 seconds = 300 seconds.
- Converting to minutes: 300 / 60 = 5 minutes.
Final Answer: The tap takes 300 seconds (or 5 minutes) to fill the bucket.
Question 11 (Page 178 – Tractor vs. Oxen Ploughing):
A tractor can plough a field 4 times faster than a pair of oxen. A pair of oxen takes 6 hours to plough 1 acre. For a 20-acre field, how much time would it take with oxen? How much time with the tractor?
A tractor can plough a field 4 times faster than a pair of oxen. A pair of oxen takes 6 hours to plough 1 acre. For a 20-acre field, how much time would it take with oxen? How much time with the tractor?
Rate of Work Comparison:
- Time taken by Oxen:
Time per acre = 6 hours.
Total time for 20 acres = 20 × 6 = 120 hours. - Time taken by Tractor:
Since the tractor is 4 times faster, time taken per acre = 6 / 4 = 1.5 hours.
Total time for 20 acres = 20 × 1.5 = 30 hours.
Answers:
Using a pair of oxen: 120 hours.
Using a tractor: 30 hours.
Using a pair of oxen: 120 hours.
Using a tractor: 30 hours.
Question 12 (Page 178 – &inr;10 Coin Alloy Value):
A &inr;10 coin has a mass of 7.74 grams and is made of cupro-nickel alloy with copper and nickel in the ratio 3 : 1. If copper costs &inr;906 per kg and nickel costs &inr;1,341 per kg, what is the intrinsic value of the metals in the coin?
A &inr;10 coin has a mass of 7.74 grams and is made of cupro-nickel alloy with copper and nickel in the ratio 3 : 1. If copper costs &inr;906 per kg and nickel costs &inr;1,341 per kg, what is the intrinsic value of the metals in the coin?
Topper’s Two-Metal Mass & Cost Derivation:
- Total mass = 7.74 g. Total ratio parts = 3 + 1 = 4 parts.
- Mass of copper = (3/4) × 7.74 g = 5.805 g = 0.005805 kg.
- Mass of nickel = (1/4) × 7.74 g = 1.935 g = 0.001935 kg.
- Cost of Copper:
Cost = 0.005805 kg × &inr;906 = &inr;5.259 ≈ &inr;5.26. - Cost of Nickel:
Cost = 0.001935 kg × &inr;1341 = &inr;2.595 ≈ &inr;2.60. - Total Metal Cost:
&inr;5.26 + &inr;2.60 = &inr;7.86.
Intrinsic Metal Cost in &inr;10 Coin:
Copper Cost = &inr;5.26 | Nickel Cost = &inr;2.60 | Total Cost = &inr;7.86.
Copper Cost = &inr;5.26 | Nickel Cost = &inr;2.60 | Total Cost = &inr;7.86.