Official NCERT 2026-27 Solutions
NCERT Solutions for Class 8 Maths Chapter 8: Fractions in Disguise (Ganita Prakash Part II)
Authentic handwritten solutions in Topper Notebook style. Covers percentage fundamentals, successive discounts, GST calculations, compound growth, retail profit & loss, and data interpretation from Pages 28 to 31 of Ganita Prakash Part II.
Quick Chapter Navigation & Blueprint
- Part A: Core Percentage Models & Commercial Math (Questions 1 to 5) — Bengaluru population growth (250%), global population shares, 18% GST formulation, compounding lab mice population, and retail markup/discount net loss.
- Part B: Commutative Properties & Proportions (Questions 6 to 9) — Tangram region E percentage, proving x% of y = y% of x, school excursion Grade 8 girls ratio, and pencil cost-selling profit (66.7%).
- Part C: Successive Changes & Retail Deals (Questions 10 to 13) — Consecutive bus fare hikes (7.12%), constant area rectangle breadth contraction (9.09%), chips ingredients weight, and Buy-X-Get-Y-Free discount analysis.
- Part D: Mathematical Paradoxes & Graph Analytics (Questions 14 & 15) — The 99%-to-98% left-handed room puzzle (50 people leave) and National Statistics Office (NSO) computer literacy chart evaluation.
Official Textbook PDF Reference: Download NCERT Class 8 Maths Chapter 8 (hegp201.pdf)
Part A: Foundations, Taxes & Successive Changes (Pages 28–29)
Question 1 (Page 28):
The population of Bengaluru in 2025 is about 250% of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025?
The population of Bengaluru in 2025 is about 250% of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025?
Topper’s Step-by-Step Calculation:
- Base population in 2000 = 50 lakhs = 50,00,000.
- Population in 2025 = 250% of 50 lakhs.
- Converting percentage to fraction: 250% = 250 / 100 = 2.5.
- Population = 2.5 × 50 lakhs = 125 lakhs = 1.25 crore (1,25,00,000).
Final Answer: The population of Bengaluru in 2025 is 125 lakhs (1.25 crore or 12.5 million).
Question 2 (Page 28 – Global Population Shares):
The population of the world in 2025 is approximately 8.2 billion. Match the countries with their approximate percentage share:
India (1.46 billion), USA (347 million), Bangladesh (175 million), Germany (83 million).
Given percentage options: 18%, 4.2%, 2.1%, 1.0%.
The population of the world in 2025 is approximately 8.2 billion. Match the countries with their approximate percentage share:
India (1.46 billion), USA (347 million), Bangladesh (175 million), Germany (83 million).
Given percentage options: 18%, 4.2%, 2.1%, 1.0%.
Percentage Share Estimations:
- World Population = 8,200 million = 8.2 billion.
- India: (1.46 / 8.2) × 100 ≈ 17.8% ≈ 18%.
- USA: (347 / 8200) × 100 = (34.7 / 820) ≈ 4.23% ≈ 4.2%.
- Bangladesh: (175 / 8200) × 100 ≈ 2.13% ≈ 2.1%.
- Germany: (83 / 8200) × 100 ≈ 1.01% ≈ 1.0%.
Matches:
• India (1.46 B) → 18%
• USA (347 M) → 4.2%
• Bangladesh (175 M) → 2.1%
• Germany (83 M) → 1.0%
• India (1.46 B) → 18%
• USA (347 M) → 4.2%
• Bangladesh (175 M) → 2.1%
• Germany (83 M) → 1.0%
Question 3 (Page 28):
The price of a mobile phone is &inr;8,250. A GST of 18% is added. Which of the following gives the final price including GST?
(i) 8250 + 18 (ii) 8250 + 1800 (iii) 8250 + 18/100 (iv) 8250 × 18
(v) 8250 × 1.18 (vi) 8250 + 8250 × 0.18 (vii) 1.8 × 8250
The price of a mobile phone is &inr;8,250. A GST of 18% is added. Which of the following gives the final price including GST?
(i) 8250 + 18 (ii) 8250 + 1800 (iii) 8250 + 18/100 (iv) 8250 × 18
(v) 8250 × 1.18 (vi) 8250 + 8250 × 0.18 (vii) 1.8 × 8250
GST Algebraic Multiplier Analysis:
- Base Price = P = &inr;8,250.
- GST Amount = 18% of P = (18/100) × 8250 = 8250 × 0.18.
- Total Price = P + GST Amount = 8250 + 8250 × 0.18 → Matches (vi).
- Factoring out 8250: Total Price = 8250(1 + 0.18) = 8250 × 1.18 → Matches (v).
Correct Options: Both (v) 8250 × 1.18 and (vi) 8250 + 8250 × 0.18.
Question 4 (Page 28 – Compounded Monthly Change):
A lab mouse population changes over three months: Month 1: +5%, Month 2: −2%, Month 3: −3%. If initial population is p, which statements are true?
A lab mouse population changes over three months: Month 1: +5%, Month 2: −2%, Month 3: −3%. If initial population is p, which statements are true?
Compounding Growth Factors:
- After Month 1: Population = p(1 + 0.05) = p × 1.05.
- After Month 2: Population = (p × 1.05)(1 − 0.02) = p × 1.05 × 0.98.
- After Month 3: Population = p × 1.05 × 0.98 × (1 − 0.03) = p × 1.05 × 0.98 × 0.97 → Statement (ii) is TRUE.
- Multiplying factors: 1.05 × 0.98 × 0.97 = 1.029 × 0.97 = 0.99813.
- Since 0.99813 < 1, the final population is strictly less than p → Statement (vi) is TRUE.
Valid Statements: (ii) Population = p × 1.05 × 0.98 × 0.97 and (vi) Population after 3 months was less than p.
Question 5 (Page 28 – Markup vs. Discount):
A shopkeeper sets the price of a product with a 35% profit margin on cost. Due to poor sales, he offers a 30% discount on this selling price. Will he make a profit or a loss? Give reasons.
A shopkeeper sets the price of a product with a 35% profit margin on cost. Due to poor sales, he offers a 30% discount on this selling price. Will he make a profit or a loss? Give reasons.
Successive Multipliers Calculation:
- Let Cost Price (CP) = 100.
- Marked Selling Price = CP × (1 + 0.35) = 100 × 1.35 = 135.
- Discount of 30% is applied to the Marked Price (135):
Discount = 30% of 135 = 0.30 × 135 = 40.5.
Final Selling Price = 135 − 40.5 = 94.5. - Direct Multiplier Method: Final Price = CP × 1.35 × (1 − 0.30) = CP × 1.35 × 0.70 = CP × 0.945.
- Since 94.5 < 100 (CP), the shopkeeper incurs a LOSS.
- Loss Percentage = (100 − 94.5) / 100 = 5.5% loss.
Final Answer: The shopkeeper makes a LOSS of 5.5%.
Part B: Percentage Symmetries & Retail Margins (Pages 29–30)
Question 7 (Page 29 – Commutative Property of Percentages):
Evaluate: 5% of 40 and 40% of 5; 25% of 12 and 12% of 25; 15% of 60 and 60% of 15.
What do you notice? Generalise using algebra comparing x% of y and y% of x.
Evaluate: 5% of 40 and 40% of 5; 25% of 12 and 12% of 25; 15% of 60 and 60% of 15.
What do you notice? Generalise using algebra comparing x% of y and y% of x.
Algebraic Proof of Commutativity:
- 5% of 40 = (5/100) × 40 = 200/100 = 2. | 40% of 5 = (40/100) × 5 = 200/100 = 2.
- 25% of 12 = (1/4) × 12 = 3. | 12% of 25 = (12/100) × 25 = 300/100 = 3.
- 15% of 60 = 9. | 60% of 15 = 9.
- Algebraic Proof:
x% of y = (x / 100) × y = (x × y) / 100.
y% of x = (y / 100) × x = (y × x) / 100.
Since multiplication of real numbers is commutative (x × y = y × x):
x% of y = y% of x unconditionally!
General Rule: x% of y is always identically equal to y% of x. (e.g. Computing 16% of 50 is mentally identical to computing 50% of 16 = 8!).
Question 8 (Page 29):
In an excursion, 40% of students are Grade 8, and the rest Grade 9. Among Grade 8 students, 60% are girls.
(i) What percentage of all excursion students are Grade 8 girls?
(ii) If total students = 160, how many are Grade 8 girls?
In an excursion, 40% of students are Grade 8, and the rest Grade 9. Among Grade 8 students, 60% are girls.
(i) What percentage of all excursion students are Grade 8 girls?
(ii) If total students = 160, how many are Grade 8 girls?
Two-Tier Percentage Breakdown:
- (i) Fraction of Grade 8 students = 40% = 0.40.
- Fraction of Grade 8 girls = 60% of Grade 8 = 0.60 × 0.40 = 0.24 = 24%.
- (ii) For a total group of 160 students:
Number of Grade 8 girls = 24% of 160 = (24/100) × 160 = 0.24 × 160 = 38.4 ≈ 38 students.
Answers: (i) 24% of the total students are Grade 8 girls. (ii) For 160 students, there are 38 (or 38.4) Grade 8 girls.
Question 9 (Page 29):
A shopkeeper sells pencils such that the Selling Price (SP) of 3 pencils equals the Cost Price (CP) of 5 pencils. Does he make a profit or loss? Find the profit or loss percentage.
A shopkeeper sells pencils such that the Selling Price (SP) of 3 pencils equals the Cost Price (CP) of 5 pencils. Does he make a profit or loss? Find the profit or loss percentage.
Ratio Formulation:
- Given: 3 × SP = 5 × CP.
- Rearranging: SP / CP = 5 / 3.
- Since SP > CP (5 > 3), the shopkeeper makes a PROFIT.
- Let CP = &inr;3 per pencil → SP = &inr;5 per pencil.
- Profit per pencil = SP − CP = 5 − 3 = &inr;2.
- Profit Percentage = (Profit / CP) × 100 = (2 / 3) × 100 = 66.67% (or 66 ⅔%).
Final Answer: The shopkeeper makes a PROFIT of 66⅔% (66.67%).
Question 10 (Page 29):
Bus fares increased by 3% last year and by 4% this year. What is the overall percentage price increase in the last 2 years?
Bus fares increased by 3% last year and by 4% this year. What is the overall percentage price increase in the last 2 years?
Successive Compounded Percentage Formula:
- Let initial fare be 100.
- After Year 1 (+3%): Fare = 100 × 1.03 = 103.
- After Year 2 (+4% on 103): Fare = 103 × 1.04 = 107.12.
- Overall Multiplier Formula: (1 + r1)(1 + r2) = 1.03 × 1.04 = 1.0712.
- Net Percentage Increase = (1.0712 − 1) × 100 = 7.12%.
- (Notice: 7.12% is slightly greater than simple sum 3% + 4% = 7% due to compounding on the first year’s hike!).
Final Answer: The overall price increase is 7.12%.
Question 11 (Page 29 – Invariant Area):
If the length of a rectangle is increased by 10% and the area remains unchanged, by what percentage (exactly) must the breadth decrease?
If the length of a rectangle is increased by 10% and the area remains unchanged, by what percentage (exactly) must the breadth decrease?
Inverse Proportionality Derivation:
- Initial Area: A = L × B.
- New Length: L‘ = L(1 + 0.10) = 1.1L = (11/10)L.
- For area to remain unchanged: L‘ × B‘ = L × B.
- (11/10)L × B‘ = L × B → B‘ = (10/11)B.
- Decrease in Breadth = B − B‘ = B − (10/11)B = (1/11)B.
- Percentage Decrease = [(1/11)B / B] × 100 = 100 / 11 = 9 &frac11;% ≈ 9.09%.
Final Answer: The breadth must decrease by exactly 9 &frac11;% (approximately 9.09%).
Part C: Deals, Puzzles & Data Interpretation (Pages 29–31)
Question 13 (Page 29 – Retail Promotional Deals):
Three shops offer deals for items costing &inr;100 each:
• Shop A: “Buy 1 Get 1 Free”
• Shop B: “Buy 2 Get 1 Free”
• Shop C: “Buy 3 Get 1 Free”
(i) What is the effective price per item in each shop? Arrange from cheapest to costliest.
(ii) Calculate the percentage discount in each shop.
(iii) If you need exactly 4 items, which shop would you choose?
Three shops offer deals for items costing &inr;100 each:
• Shop A: “Buy 1 Get 1 Free”
• Shop B: “Buy 2 Get 1 Free”
• Shop C: “Buy 3 Get 1 Free”
(i) What is the effective price per item in each shop? Arrange from cheapest to costliest.
(ii) Calculate the percentage discount in each shop.
(iii) If you need exactly 4 items, which shop would you choose?
Unitary Cost & Discount Analysis:
- Shop A (Buy 1 Get 1 Free):
Pay for 1 item (&inr;100), receive 2 items.
Effective price per item = &inr;100 / 2 = &inr;50.
Discount = (1 free / 2 total) × 100 = 50% discount. - Shop B (Buy 2 Get 1 Free):
Pay for 2 items (&inr;200), receive 3 items.
Effective price per item = &inr;200 / 3 = &inr;66.67.
Discount = (1 free / 3 total) × 100 = 33.33% discount. - Shop C (Buy 3 Get 1 Free):
Pay for 3 items (&inr;300), receive 4 items.
Effective price per item = &inr;300 / 4 = &inr;75.
Discount = (1 free / 4 total) × 100 = 25% discount. - (i) Order (Cheapest to Costliest): Shop A (&inr;50) < Shop B (&inr;66.67) < Shop C (&inr;75).
- (iii) Buying 4 items:
• Shop A: Buy 2 get 2 free → pay for 2 items = &inr;200.
• Shop B: Buy 2 get 1 free (3 items for &inr;200), buy 1 more at &inr;100 → pay &inr;300.
• Shop C: Buy 3 get 1 free (4 items for pay of 3) → pay &inr;300.
Choose Shop A because you only pay &inr;200 for 4 items!
Summary:
(i) Effective Prices: Shop A: &inr;50 | Shop B: &inr;66.67 | Shop C: &inr;75.
(ii) Discounts: Shop A: 50% | Shop B: 33.33% | Shop C: 25%.
(iii) For 4 items: Shop A is the best deal (&inr;200 total).
(i) Effective Prices: Shop A: &inr;50 | Shop B: &inr;66.67 | Shop C: &inr;75.
(ii) Discounts: Shop A: 50% | Shop B: 33.33% | Shop C: 25%.
(iii) For 4 items: Shop A is the best deal (&inr;200 total).
Question 14 (Page 30 – The 99% to 98% Left-Handed Paradox):
In a room of 100 people, 99% are left-handed. How many left-handed people must leave the room to bring that percentage down to exactly 98%?
In a room of 100 people, 99% are left-handed. How many left-handed people must leave the room to bring that percentage down to exactly 98%?
Topper’s Constant Invariant Method:
- Initial Room Composition:
Total people = 100.
Left-handed = 99% of 100 = 99 people.
Right-handed = 100 − 99 = 1 person. - The Key Invariant: ONLY left-handed people leave the room. The number of right-handed people remains strictly 1!
- In the new group, left-handed people must make up 98%. This means the 1 right-handed person must represent:
100% − 98% = 2% of the new total! - Let the new total population be N:
2% of N = 1
(2 / 100) × N = 1
N = 100 / 2 = 50 people! - If the new total is 50 people, with 1 right-handed person, the number of left-handed people remaining is 49.
- Number of left-handed people that must leave:
99 (initial) − 49 (remaining) = 50 people!
Final Answer: Exactly 50 left-handed people must leave the room!
Exam Tip: Most students mistakenly guess that 1 person leaving reduces the percentage from 99% to 98%. Always focus on the invariant quantity (the 1 right-handed person) to find the correct total!