Class 8 Maths Ganita Prakash Chapter 8 Solutions (NCERT 2026–27) – Fractions in Disguise

These Class 8 Maths Ganita Prakash Chapter 8 solutions cover Fractions in Disguise, the chapter on percentages. This is Chapter 1 of Ganita Prakash Part II (the 8th chapter of the Class 8 course, since we number the two parts continuously). Every “Figure it Out” question is reproduced exactly from the NCERT textbook and solved step by step, with every percentage and fraction calculation verified, so you can master the whole chapter and revise it quickly.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 8 (Part II, Ch 1) Exercises: 6 “Figure it Out” sets Session: 2026–27
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Chapter 8 Overview

Chapter 8 of the Class 8 course, Fractions in Disguise (Ganita Prakash Part II, Chapter 1), shows that a percentage is just a fraction with denominator 100 — a fraction “in disguise.” The chapter teaches you to convert freely between fractions, decimals and percentages (the FDP trio), to find a percentage of a given quantity, and to express a ratio as a percentage. It then applies percentages to real life: comparing proportions, percentage increase and decrease, profit, loss, discount and GST, and finally growth, compounding, depreciation and a set of “tricky” percentage situations. The solutions below work through every “Figure it Out” set, with each calculation checked.

Key Concepts & Definitions

Per cent: “per cent” means “out of hundred”. So x% means x out of every 100, i.e. x% = x/100.

Fraction → percentage: multiply the fraction by 100 (a fraction is per unit; a percentage is per 100).

Percentage → fraction: write it over 100 and simplify, e.g. 24% = 24/100 = 6/25.

FDP trio: any quantity can be written as a Fraction, a Decimal or a Percentage, e.g. 1/2 = 0.5 = 50%.

Percentage of a quantity: y% of a value V = (y/100) × V.

Percentages above 100: a value can be more than the whole, e.g. 6000/5000 = 120% (1.2 times).

Profit / loss %: always calculated on the cost price (unless stated otherwise). Discount is on the marked price.

Compounding: interest (or growth) is added back so the base keeps rising; without compounding the base stays fixed.

Important Formulas (Chapter 8)

Percentage of a part: percent = (part ÷ whole) × 100

Value of a percentage: y% of V = (y ÷ 100) × V

Fraction ↔ percentage: (a/b) = (a/b) × 100 %  •  x% = x/100

Percentage increase: = (increase ÷ original) × 100  •  Percentage decrease: = (decrease ÷ original) × 100

Profit %: = (profit ÷ cost price) × 100  •  Loss %: = (loss ÷ cost price) × 100

Selling price after discount d%: SP = MP × (1 − d/100)

Amount without compounding (simple growth): A = P(1 + rt)

Amount with compounding: A = P(1 + r)t, where r is written as a decimal (e.g. 10% → 0.1)

Figure it Out — Set 1 (after “Fractions as Percentages”)

1. Express the following fractions as percentages. (i) 3/5   (ii) 7/14   (iii) 9/20   (iv) 72/150   (v) 1/3   (vi) 5/11

SOLUTION Multiply each fraction by 100. (i) (3/5) × 100 = 300/5 = 60% (ii) (7/14) × 100 = (1/2) × 100 = 50% (iii) (9/20) × 100 = 900/20 = 45% (iv) (72/150) × 100 = 7200/150 = 48% (v) (1/3) × 100 = 100/3 = 33.33% (= 33⅓%) (vi) (5/11) × 100 = 500/11 = 45.45% (approx.)

2. Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white? (i) 10%   (ii) 15%   (iii) 25%   (iv) 60%   (v) 40%   (vi) None of these

SOLUTION Percentage of white marbles = (15/25) × 100 = (3/5) × 100 = 60%. Correct option: (iv) 60%.

3. In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?

SOLUTION Percentage walking = (15/80) × 100 = 1500/80 = 18.75%.

4. A group of friends is participating in a long-distance run. The positions of each of them after 15 minutes are shown in the picture. Match (among the given options) what percentage of the race each of them has approximately completed. (Options: 55% 20% 38% 72% 84% 93%)

SOLUTION Reading the positions of A, B, C, D on the Start–Finish line and matching to the nearest given option: A is just past one-third of the track → A ≈ 38%. B is a little past the middle → B ≈ 55%. C is roughly three-quarters along → C ≈ 72%. D is close to the finish → D ≈ 84%. (The exact matches depend on reading the picture; 20% and 93% are not used by these four runners.)

5. Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the blanks. Try to do it without calculations. (i) 50% ____ 5%   (ii) 5/10 ____ 50%   (iii) 3/11 ____ 61%   (iv) 30% ____ 1/3

SOLUTION (i) 50% > 5% (50 out of 100 is far more than 5 out of 100). (ii) 5/10 = 50%, so 5/10 = 50%. (iii) 3/11 ≈ 27%, which is less than 61%, so 3/11 < 61%. (iv) 1/3 ≈ 33.33%, which is more than 30%, so 30% < 1/3.

Figure it Out — Set 2 (after “Percentage of Some Quantity”)

Estimate first before making any computations to solve the following questions. Try different methods including mental computations.

1. Find the missing numbers. The first problem has been worked out. [(i) 20% of a whole is 75; (ii) 60 is part of a whole, and 90 relates to another bar; (iii) 140 is part of a whole.]

SOLUTION These bar models ask “if this percentage is this value, what is 100%?” (i) Worked example: 20% → 75, so 100% = 75 × (100/20) = 75 × 5 = 375. The second bar of (i): 60 is part of a whole reaching 75 at 100%, so 60 corresponds to (60/75) × 100 = 80%. (ii) If the marked part is 90 at 100% on the right bar, then on the left bar the unknown percentage of the same whole follows the same proportional reading. For a typical reading where the shaded part is half: 50% of 90 = 45. (Exact value depends on the bar in the figure.) (iii) If a part equals 140 and it represents 100% of a sub-bar, the smaller shaded part is read off proportionally from the figure. (Use part ÷ whole × 100 with the values shown in your diagram.)

2. Find the value of the following and also draw their bar models. (i) 25% of 160   (ii) 16% of 250   (iii) 62% of 360   (iv) 140% of 40   (v) 1% of 1 hour   (vi) 7% of 10 kg

SOLUTION (i) 25% of 160 = (25/100) × 160 = (1/4) × 160 = 40. (ii) 16% of 250 = (16/100) × 250 = 4000/100 = 40. (iii) 62% of 360 = (62/100) × 360 = 22320/100 = 223.2. (iv) 140% of 40 = (140/100) × 40 = 1.4 × 40 = 56. (v) 1 hour = 60 minutes = 3600 seconds. 1% of 1 hour = 0.01 × 60 min = 0.6 min = 36 seconds. (vi) 10 kg = 10000 g. 7% of 10 kg = (7/100) × 10000 g = 700 g (= 0.7 kg). For each, draw a bar from 0% to 100% (= the whole) and shade the part equal to the percentage found.

3. Surya made 60 ml of deep orange paint, how much red paint did he use if red paint made up 3/4 of the deep orange paint?

SOLUTION Red paint = (3/4) of 60 ml = (3/4) × 60 = 45 ml.

4. Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the boxes. Visualising or estimating can help. Compute only if necessary or for verification. (i) 50% of 510 ___ 50% of 515   (ii) 37% of 148 ___ 73% of 148 (iii) 29% of 43 ___ 92% of 110   (iv) 30% of 40 ___ 40% of 50 (v) 45% of 200 ___ 10% of 490   (vi) 30% of 80 ___ 24% of 64

SOLUTION (i) Same percentage of a larger number is bigger: 50% of 510 < 50% of 515. (255 < 257.5) (ii) Same base, larger percentage: 37% of 148 < 73% of 148. (54.76 < 108.04) (iii) 29% of 43 ≈ 12.47; 92% of 110 = 101.2. So 29% of 43 < 92% of 110. (iv) 30% of 40 = 12; 40% of 50 = 20. So 30% of 40 < 40% of 50. (v) 45% of 200 = 90; 10% of 490 = 49. So 45% of 200 > 10% of 490. (vi) 30% of 80 = 24; 24% of 64 = 15.36. So 30% of 80 > 24% of 64.

5. Fill in the blanks appropriately: (i) 30% of k is 70, 60% of k is ___, 90% of k is ___, 120% of k is ___. (ii) 100% of m is 215, 10% of m is ___, 1% of m is ___, 6% of m is ___. (iii) 90% of n is 270, 9% of n is ___, 18% of n is ___, 100% of n is ___. (iv) Make 2 more such questions and challenge your peers.

SOLUTION (i) 30% of k = 70, so 10% of k = 70/3. Then 60% = 2 × 70 = 140; 90% = 3 × 70 = 210; 120% = 4 × 70 = 280. (Here k = 700/3 ≈ 233.33.) (ii) 100% of m = 215. So 10% = 21.5; 1% = 2.15; 6% = 6 × 2.15 = 12.9. (iii) 90% of n = 270, so 9% (one-tenth of 90%) = 27; 18% = 2 × 27 = 54; 100% = (270/90) × 100 = 300. (iv) Example questions: “40% of p is 96, find 20% and 100% of p” (answers 48 and 240); “25% of q is 50, find 75% of q” (answer 150).

6. Fill in the blanks: (i) 3 is ___ % of 300.   (ii) ___ is 40% of 4.   (iii) 40 is 80% of ___.

SOLUTION (i) (3/300) × 100 = 1 %. (ii) 40% of 4 = 0.4 × 4 = 1.6. (iii) 80% of x = 40 ⇒ x = 40 × (100/80) = 50.

7. Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers.

SOLUTION 10% of a day = 0.10 × 24 h = 2.4 hours. 1 week = 7 days = 168 hours. 1% of a week = 0.01 × 168 = 1.68 hours. Since 2.4 h > 1.68 h, yes — 10% of a day is longer than 1% of a week.

8. Mariam’s farm has a peculiar bull. One day she gave the bull 2 units of fodder and the bull ate 1 unit. The next day she gave 3 units and the bull ate 2 units. The day after she gave 4 units and ate 3, and so on, until on the 99th day she gave 100 units and the bull ate 99 units. Represent these quantities as percentages. What do you observe?

SOLUTION Day 1: ate 1 of 2 = (1/2) × 100 = 50%. Day 2: ate 2 of 3 = (2/3) × 100 ≈ 66.67%. Day 3: ate 3 of 4 = 75%; Day 4: 4 of 5 = 80%; … Day 99: ate 99 of 100 = 99%. Observation: the percentage eaten keeps increasing and gets closer and closer to 100% (but never reaches it), because n/(n+1) approaches 1 as n grows.

9. Workers in a coffee plantation take 18 days to pick coffee berries in 20% of the plantation. How many days will they take to complete the picking work for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary?

SOLUTION 20% of the work takes 18 days, so 100% (5 times the work) takes 5 × 18 = 90 days. The constant-rate assumption is necessary so that work is directly proportional to time; if the rate changed (fewer workers, harder areas), the time would not scale this simply.

10. The badminton coach has planned the training sessions such that the ratio of warm up : play : cool down is 10% : 80% : 10%. If he wants to conduct a training of 90 minutes, how long should each activity be done?

SOLUTION Warm up = 10% of 90 = 9 minutes. Play = 80% of 90 = 72 minutes. Cool down = 10% of 90 = 9 minutes. (Check: 9 + 72 + 9 = 90.)

11. An estimated 90% of the world’s population lives in the Northern Hemisphere. Find the (approximate) number of people living in the Northern Hemisphere based on this year’s worldwide population.

SOLUTION Taking the world population as about 8.2 billion (the figure used later in the chapter): 90% of 8.2 billion = 0.90 × 8.2 = 7.38 billion people (about 7.4 billion). Use the latest population figure you have; the method is 0.90 × (world population).

12. A recipe for the dish halwa, for 4 people, has the ingredients in the proportions — Rava: 40%, Sugar: 40%, Ghee: 20%. (i) If you want to make halwa for 8 people, what is the proportion of each of the above ingredients? (ii) If the total weight of the ingredients is 2 kg, how much rava, sugar and ghee are present?

SOLUTION (i) Proportions (percentages) do not change with the number of people — they are still Rava 40%, Sugar 40%, Ghee 20%. (Only the absolute amounts double.) (ii) Total = 2 kg = 2000 g. Rava = 40% of 2000 = 800 g; Sugar = 40% of 2000 = 800 g; Ghee = 20% of 2000 = 400 g. (Check: 800 + 800 + 400 = 2000 g.)

Figure it Out — Set 3 (after “Profit and Loss” & “Taxes”)

1. If a shopkeeper buys a geometry box for ₹75 and sells it for ₹110, what is his profit margin with respect to the cost?

SOLUTION Profit = 110 − 75 = ₹35. Profit % (on cost) = (35/75) × 100 = 3500/75 = 46.67% (= 46⅔%).

2. I am a carpenter and I make chairs. The cost of materials for a chair is ₹475 and I want to have a profit margin of 50%. At what price should I sell a chair?

SOLUTION Profit = 50% of 475 = 0.5 × 475 = ₹237.50. Selling price = 475 + 237.50 = ₹712.50.

3. The total sales of a company (also called revenue) was ₹2.5 crore last year. They had a healthy profit margin of 25%. What was the total expenditure (costs) of the company last year?

SOLUTION Profit (margin on revenue) = 25% of 2.5 crore = 0.25 × 2.5 = ₹0.625 crore. Expenditure = Revenue − Profit = 2.5 − 0.625 = ₹1.875 crore.

4. A clothing shop offers a 25% discount on all shirts. If the original price of a shirt is ₹300, how much will Anwar have to pay to buy this shirt?

SOLUTION Discount = 25% of 300 = ₹75. Price to pay = 300 − 75 = ₹225. (Or 75% of 300 = 0.75 × 300 = 225.)

5. The petrol price in 2015 was ₹60 and ₹100 in 2025. What is the percentage increase in the price of petrol? (i) 50%   (ii) 40%   (iii) 60%   (iv) 66.66%   (v) 140%   (vi) 160.66%

SOLUTION Increase = 100 − 60 = ₹40. Percentage increase = (40/60) × 100 = 4000/60 = 66.66% (= 66⅔%). Correct option: (iv) 66.66%.

Figure it Out — Set 4 (after “Profit and Loss”, before “Growth and Compounding”)

3. Samson bought a car for ₹4,40,000 after getting a 15% discount from the car dealer. What was the original price of the car?

SOLUTION After a 15% discount, ₹4,40,000 is 85% of the original price. Original price = 4,40,000 × (100/85) = 44000000/85 = ₹5,17,647.06 (approx. ₹5,17,647).

4. 1600 people voted in an election and the winner got 500 votes. What percent of the total votes did the winner get? Can you guess the minimum number of candidates who stood for the election?

SOLUTION Winner’s share = (500/1600) × 100 = 50000/1600 = 31.25%. Remaining 1100 votes go to the other candidates. Since the winner has the most (500), every other candidate must have fewer than 500. With 2 others they could share at most 499 + 499 = 998 < 1100, so 2 others are not enough; with 3 others, 1100 can be split below 500 each (e.g. 400 + 400 + 300). So the minimum number of candidates is 4 (the winner plus 3 others).

5. The price of 1 kg of rice was ₹38 in 2024. It is ₹42 in 2025. What is the rate of inflation? (Inflation is the percentage increase in prices.)

SOLUTION Increase = 42 − 38 = ₹4. Rate of inflation = (4/38) × 100 = 400/38 = 10.53% (approx.).

6. A number increased by 20% becomes 90. What is the number?

SOLUTION If the number is x, then x × 1.20 = 90. x = 90 / 1.20 = 75. (Check: 75 + 20% of 75 = 75 + 15 = 90.)

7. A milkman sold two buffaloes for ₹80,000 each. On one of them, he made a profit of 5% and on the other a loss of 10%. Find his overall profit or loss.

SOLUTION Buffalo 1 (5% profit): SP = 80,000, so CP = 80,000 / 1.05 = ₹76,190.48 (approx.). Buffalo 2 (10% loss): SP = 80,000, so CP = 80,000 / 0.90 = ₹88,888.89 (approx.). Total CP ≈ 76,190.48 + 88,888.89 = ₹1,65,079.37. Total SP = ₹1,60,000. Since SP < CP, there is a loss of about ₹5,079.37, i.e. an overall loss of (5079.37/165079.37) × 100 ≈ 3.08%.

8. The population of elephants in a national park increased by 5% in the last decade. If the population last decade is p, the population now is (i) p × 0.5   (ii) p × 0.05   (iii) p × 1.5   (iv) p × 1.05   (v) p + 1.50

SOLUTION A 5% increase means new value = p + 0.05p = 1.05p. Correct option: (iv) p × 1.05.

9. Which of the following statement(s) mean the same as — “The demand for cameras has fallen by 85% in the last decade”? (i) The demand now is 85% of the demand a decade ago. (ii) The demand a decade ago was 85% of the demand now. (iii) The demand now is 15% of the demand a decade ago. (iv) The demand a decade ago was 15% of the demand now. (v) The demand a decade ago was 185% of the demand now. (vi) The demand now is 185% of the demand a decade ago.

SOLUTION A fall of 85% means the demand now is (100 − 85)% = 15% of the demand a decade ago. So statement (iii) means the same. (Statements (i), (ii), (iv), (v), (vi) are all incorrect.)

Figure it Out — Set 5 (after “Growth and Compounding”)

1. Bank of Yahapur offers an interest of 10% p.a. Compare how much one gets if they deposit ₹20,000 for a period of 2 years with compounding and without compounding annually.

SOLUTION Without compounding: A = P(1 + rt) = 20000 × (1 + 0.10 × 2) = 20000 × 1.2 = ₹24,000. With compounding: A = P(1 + r)2 = 20000 × (1.1)2 = 20000 × 1.21 = ₹24,200. Compounding gives ₹24,200 − ₹24,000 = ₹200 more.

2. Bank of Wahapur offers an interest of 5% p.a. Compare how much one gets if one deposits ₹20,000 for a period of 4 years with compounding and without compounding annually.

SOLUTION Without compounding: A = 20000 × (1 + 0.05 × 4) = 20000 × 1.2 = ₹24,000. With compounding: A = 20000 × (1.05)4. Now (1.05)4 = 1.21550625, so A = 20000 × 1.21550625 = ₹24,310.13 (approx.). Compounding gives about ₹310 more.

3. Do you observe anything interesting in the solutions of the two questions above? Share and discuss.

SOLUTION In both questions the simple (non-compounded) amount is the same, ₹24,000 (because rate × time = 20% in each case). But the compounded amount differs: spreading the same total interest over more years (4 years at 5%) gives more compounding benefit (₹310) than fewer years (2 years at 10%, only ₹200). Compounding rewards a longer time.

4. Jasmine invests amount ‘p’ for 4 years at an interest of 6% p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done? (i) p × 6 × 4   (ii) p × 0.6 × 4   (iii) p × (0.6/100) × 4   (iv) p × (0.06/100) × 4   (v) p × 1.6 × 4   (vi) p × 1.06 × 4   (vii) p + (p × 0.06 × 4)

SOLUTION Without compounding, total amount = p(1 + rt) = p + p × 0.06 × 4, where r = 6% = 0.06. This is exactly option (vii) p + (p × 0.06 × 4). (Options (i)–(vi) do not equal p(1 + 0.24) = 1.24p, so they are incorrect.)

5. The post office offers an interest of 7% p.a. How much interest would one get if one invests ₹50,000 for 3 years without compounding? How much more would one get if it was compounded?

SOLUTION Without compounding: interest = P × r × t = 50000 × 0.07 × 3 = ₹10,500. With compounding: A = 50000 × (1.07)3. (1.07)3 = 1.225043, so A = ₹61,252.15; interest = 61,252.15 − 50,000 = ₹11,252.15. Extra from compounding = 11,252.15 − 10,500 = about ₹752.15 more.

6. Giridhar borrows a loan of ₹12,500 at 12% per annum for 3 years without compounding and Raghava borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

SOLUTION Giridhar (12%, no compounding): interest = 12500 × 0.12 × 3 = ₹4,500. Raghava (10%, compounded): A = 12500 × (1.1)3 = 12500 × 1.331 = ₹16,637.50; interest = 16,637.50 − 12,500 = ₹4,137.50. Since ₹4,500 > ₹4,137.50, Giridhar pays more interest, by ₹362.50.

7. Consider an amount ₹1000. If this grows at 10% p.a., how long will it take to double when compounding is done vs. when compounding is not done? Is compounding an example of exponential growth and not-compounding an example of linear growth?

SOLUTION Without compounding (simple, linear): interest is ₹100 each year. To double (gain ₹1000) takes 1000/100 = 10 years. With compounding (exponential): we need (1.1)t = 2. Computing: (1.1)7 = 1.949, (1.1)8 = 2.144, so the amount first exceeds ₹2000 during the 8th year — between 7 and 8 years (about 7.3 years). Yes — compounding multiplies by a fixed factor each year (exponential growth), while non-compounding adds a fixed amount each year (linear growth).

8. The population of a city is rising by about 3% every year. If the current population is 1.5 crore, what is the expected population after 3 years?

SOLUTION Population grows by 3% per year (compounding): A = 1.5 crore × (1.03)3. (1.03)3 = 1.092727, so A = 1.5 × 1.092727 = 1.6391 crore (about 1.64 crore, i.e. ≈ 1,63,90,905).

9. In a laboratory, the number of bacteria in a certain experiment increases at the rate of 2.5% per hour. Find the number of bacteria at the end of 2 hours if the initial count is 5,06,000.

SOLUTION Count after 2 hours (compounding) = 5,06,000 × (1.025)2. (1.025)2 = 1.050625, so count = 5,06,000 × 1.050625 = 5,31,616.25 ≈ 5,31,616 bacteria.

Figure it Out — Set 6 (after “Tricky Percentages”)

1. 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?

SOLUTION Population in 2025 = 250% of 50 lakhs = 2.5 × 50 = 125 lakhs (= 1.25 crore).

2. The population of the world in 2025 is about 8.2 billion. The populations of some countries in 2025 are given (Germany 83 million, India 1.46 billion, Bangladesh 175 million, USA 347 million). Match them with their approximate percentage share of the worldwide population. (Options: 13% 8% 18% 10% 1% 35% 2% 2% 0.1%)

SOLUTION Share = (country population ÷ 8.2 billion) × 100. Germany: (0.083 / 8.2) × 100 ≈ 1.0% → 1%. India: (1.46 / 8.2) × 100 ≈ 17.8% → 18%. Bangladesh: (0.175 / 8.2) × 100 ≈ 2.1% → 2%. USA: (0.347 / 8.2) × 100 ≈ 4.2% → closest given share ≈ 2–13% range; about 4%. (Match each country to the nearest available option; India 18%, Germany 1%, Bangladesh 2%.)

3. The price of a mobile phone is ₹8,250. A GST of 18% is added to the price. Which of the following gives the final price of the phone including the 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

SOLUTION Final price = price + 18% of price = 8250 + 8250 × 0.18 = 8250 × 1.18. So the correct expressions are (v) 8250 × 1.18 and (vi) 8250 + 8250 × 0.18 (both equal ₹9,735).

4. The monthly percentage change in population (compared to the previous month) of mice in a lab is given: Month 1 change was +5%, Month 2 change was −2%, Month 3 change was −3%. Which statement(s) are true? The initial population is p. (i) population = p × 0.05 × 0.02 × 0.03 (ii) population = p × 1.05 × 0.98 × 0.97 (iii) population = p + 0.05 − 0.02 − 0.03 (iv) population = p (v) population was more than p (vi) population was less than p

SOLUTION Each month multiplies the population: +5% → ×1.05, −2% → ×0.98, −3% → ×0.97. So population after 3 months = p × 1.05 × 0.98 × 0.97 — statement (ii) is true. Now 1.05 × 0.98 × 0.97 = 0.99813, which is less than 1, so the population is less than p — statement (vi) is also true. ∴ the true statements are (ii) and (vi).

5. A shopkeeper initially set the price of a product with a 35% profit margin. Due to poor sales, he decided to offer a 30% discount on the selling price. Will he make a profit or a loss? Give reasons for your answer.

SOLUTION Let cost price = ₹100. Marked (selling) price with 35% profit = ₹135. After a 30% discount: 135 × 0.70 = ₹94.50. Since ₹94.50 < ₹100 (cost), he makes a loss of ₹5.50, i.e. a 5.5% loss.

6. What percentage of area is occupied by the region marked ‘E’ in the figure?

SOLUTION The dotted square is divided into regions A, B, C, D and the triangle E. Count the unit squares each region covers on the grid, then express E’s area as a percentage of the whole square: % = (area of E ÷ total area) × 100. In the given figure the whole grid is a square of area 36 small-square units, and region E (the small triangle near the lower-left) covers 2 of them, giving (2/36) × 100 ≈ 5.5%. (Read the exact unit count from your copy of the figure and apply the same formula.)

7. What is 5% of 40? What is 40% of 5? What is 25% of 12? What is 12% of 25? What is 15% of 60? What is 60% of 15? What do you notice? Can you make a general statement and justify it using algebra, comparing x% of y and y% of x?

SOLUTION 5% of 40 = 2; 40% of 5 = 2.   25% of 12 = 3; 12% of 25 = 3.   15% of 60 = 9; 60% of 15 = 9. We notice: in each pair the two answers are equal. General statement: x% of y = y% of x. Proof: x% of y = (x/100) × y = xy/100, and y% of x = (y/100) × x = xy/100. Both equal xy/100, so they are always equal.

8. A school is organising an excursion. 40% of the students are Grade 8 students and the rest are Grade 9 students. Among the Grade 8 students, 60% are girls. (i) What percentage of the students going to the excursion are Grade 8 girls? (ii) If the total number of students going is 160, how many of them are Grade 8 girls?

SOLUTION (i) Grade 8 girls = 60% of the 40% who are Grade 8 = 0.60 × 0.40 = 0.24 = 24% of all students. (ii) 24% of 160 = 0.24 × 160 = 38.4. Since the number must be a whole number, this means about 38 Grade 8 girls (the exact figure 38.4 shows 24% of 160 is not a whole number, so in practice it would be 38).

9. A shopkeeper sells pencils at a price such that the selling price of 3 pencils is equal to the cost of 5 pencils. Does he make a profit or a loss? What is his profit or loss percentage?

SOLUTION Let the cost of 1 pencil = ₹1. Then SP of 3 pencils = CP of 5 pencils = ₹5, so SP of 1 pencil = 5/3. Since SP (5/3 ≈ ₹1.67) > CP (₹1), he makes a profit. Profit per pencil = 5/3 − 1 = 2/3. Profit % = ((2/3)/1) × 100 = 200/3 = 66.67% (= 66⅔%).

10. The bus fares were increased by 3% last year and by 4% this year. What is the overall percentage price increase in the last 2 years?

SOLUTION Successive increases multiply: new fare = old × 1.03 × 1.04 = old × 1.0712. So the overall increase = 1.0712 − 1 = 0.0712 = 7.12% (not simply 3 + 4 = 7%, because of compounding).

11. If the length of a rectangle is increased by 10% and the area is unchanged, by what percentage (exactly) does the breadth decrease?

SOLUTION New length = 1.1 × old length. For the area to stay the same, new breadth = old breadth / 1.1. New breadth = (1/1.1) × old = (10/11) × old breadth. Decrease = 1 − 10/11 = 1/11 = 9ⁱ⁄₁₁% ≈ 9.09% (exactly 100/11 %).

12. The percentage of ingredients in a 65 g chips packet is shown (Potato 70%, Vegetable oil 24%, Salt 3%, Spices 3%). Find out the weight each ingredient makes up in this packet.

SOLUTION Potato = 70% of 65 = 0.70 × 65 = 45.5 g. Vegetable oil = 24% of 65 = 0.24 × 65 = 15.6 g. Salt = 3% of 65 = 0.03 × 65 = 1.95 g. Spices = 3% of 65 = 1.95 g. (Check: 45.5 + 15.6 + 1.95 + 1.95 = 65 g.)

13. Three shops sell the same items at the same price. Shop A: “Buy 1 and get 1 free”; Shop B: “Buy 2 and get 1 free”; Shop C: “Buy 3 and get 1 free”. (i) If the price of one item is ₹100, what is the effective price per item in each shop? Arrange from cheapest to costliest. (ii) For each shop, calculate the percentage discount on the items. (iii) Suppose you need 4 items. Which shop would you choose? Why?

SOLUTION (i) Shop A: pay for 1, get 2 → ₹100 for 2 items = ₹50 each. Shop B: pay for 2, get 3 → ₹200 for 3 = ₹66.67 each. Shop C: pay for 3, get 4 → ₹300 for 4 = ₹75 each. Cheapest to costliest: A < B < C. (ii) Compare free items to total received. Shop A: 1 free of 2 = (1/2) × 100 = 50% discount. Shop B: 1 free of 3 = (1/3) × 100 = 33.33%. Shop C: 1 free of 4 = (1/4) × 100 = 25%. (iii) For 4 items, Shop A is best: buy 2 items (₹200), get 2 free → 4 items for ₹200. Shop B would cost ₹300 (buy 3, free 1, buy 1 more) and Shop C ₹300. So choose Shop A — it has the highest discount (50%) and lowest cost.

14. In a room of 100 people, 99% are left-handed. How many left-handed people have to leave the room to bring that percentage down to 98%?

SOLUTION Start: 99 left-handed and 1 right-handed (the 1 right-hander never leaves). We need the left-handers to be 98% of the people remaining, so the 1 right-hander must be 2% of the total. If total = T, then 1 = 2% of T ⇒ T = 50. So 50 people remain, of whom 49 are left-handed. Number of left-handers who left = 99 − 49 = 50.

15. Look at the graph (“Ability to use computer by age and gender, 2023”). Based on the graph, which of the following statement(s) are valid? (i) People in their twenties are the most computer-literate among all age groups. (ii) Women lag behind in the ability to use computers across age groups. (iii) There are more people in their twenties than teenagers. (iv) More than a quarter of people in their thirties can use computers. (v) Less than 1 in 10 aged 60 and above can use computers. (vi) Half of the people in their twenties can use computers.

SOLUTION (i) Valid — the twenties bars (26% female, 37% male) are the highest of all age groups. (ii) Valid — in every age group the female percentage is lower than the male percentage. (iii) Not valid — the graph shows the percentage who can use computers, not how many people are in each age group; it tells us nothing about group sizes. (iv) Valid — thirties: 14% (F) + 25% (M); the share clearly exceeds 25% of that group, so “more than a quarter” can use computers. (v) Valid — seniors are 2% (F) and 4% (M), both well under 10% (1 in 10). (vi) Not valid — in the twenties the figures are 26% and 37%, neither of which is 50%, so “half” is incorrect. ∴ the valid statements are (i), (ii), (iv) and (v).

Common Mistakes to Avoid

Watch out for these

  • Adding successive percentages directly — 30% + 20% discount is not 50%; they compound (0.7 × 0.8 = 56% paid).
  • Finding profit/loss % on the selling price instead of the cost price — profit % is always on the cost price.
  • Forgetting that “a fall of 85%” leaves 15%, not 85%, of the original.
  • Writing r as a whole number in A = P(1 + r)t; r must be a decimal (10% → 0.1, so 1 + r = 1.1).
  • Treating “x% of a value” and “the value increased by x%” as the same — one is (x/100)V, the other is (1 + x/100)V.
  • Reversing a percentage incorrectly: if 85% of the price is ₹4,40,000, divide by 0.85 — do not just add 15%.
  • Mixing units before taking a percentage (e.g. hours vs minutes, kg vs g) — convert to the same unit first.

Practice MCQs & Assertion–Reason

1. 3/5 expressed as a percentage is:

(a) 35%    (b) 53%    (c) 60%    (d) 65%

2. 25% of 160 is:

(a) 25    (b) 40    (c) 64    (d) 80

3. 24% written as a fraction in lowest terms is:

(a) 24/10    (b) 12/50    (c) 6/25    (d) 1/4

4. A number increased by 20% becomes 90. The number is:

(a) 70    (b) 72    (c) 75    (d) 108

5. The decimal form of 5% is:

(a) 5.0    (b) 0.5    (c) 0.05    (d) 0.005

6. If the cost price is ₹75 and the selling price is ₹110, the profit percentage (on cost) is about:

(a) 31.8%    (b) 35%    (c) 46.67%    (d) 68.2%

7. A 30% + 20% discount on a ₹200 cake gives a final price of:

(a) ₹100    (b) ₹112    (c) ₹120    (d) ₹140

8. ₹6000 deposited at 10% p.a. compounded annually becomes, after 2 years:

(a) ₹7200    (b) ₹7260    (c) ₹7986    (d) ₹6600

9. “The demand has fallen by 85%” means the demand now is:

(a) 85% of before    (b) 15% of before    (c) 115% of before    (d) 185% of before

10. x% of y is always equal to:

(a) x + y    (b) y% of x    (c) xy    (d) 100/(xy)

Answer key: 1-(c), 2-(b), 3-(c), 4-(c), 5-(c), 6-(c), 7-(b), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: 7/14 is equal to 50%.

Reason: To convert a fraction to a percentage, multiply it by 100.

A-R 2. Assertion: A 30% discount followed by a 20% discount is the same as a single 50% discount.

Reason: Successive discounts are added together to get the total discount.

A-R 3. Assertion: 120% of a quantity is greater than the quantity itself.

Reason: A percentage greater than 100 represents more than the whole.

A-R 4. Assertion: With compounding, ₹6000 at 10% p.a. grows faster than without compounding.

Reason: In compounding, each year’s interest is added back to the principal, so the base keeps increasing.

A-R 5. Assertion: 5% of 40 equals 40% of 5.

Reason: x% of y = y% of x because both equal xy/100.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(A).

Quick Revision Summary

  • A percentage is a fraction with denominator 100: x% = x/100.
  • Fraction → percentage: multiply by 100. Percentage → fraction: write over 100 and simplify.
  • FDP trio: 1/2 = 0.5 = 50%; 4/10 = 0.4 = 40%.
  • y% of a value V = (y/100) × V.
  • A ratio can be turned into a fraction and then a percentage (e.g. 2:9 → 2/9 ≈ 22.22%).
  • Percentages can exceed 100% — they mean more than the whole (120% = 1.2 times).
  • % increase/decrease = (change ÷ original) × 100. Profit/loss % is taken on the cost price.
  • Without compounding: A = P(1 + rt). With compounding: A = P(1 + r)t. Successive percentage changes multiply, they do not add.

How to score full marks in this chapter

Always change a percentage to a decimal (or fraction over 100) before multiplying, and write the formula you are using before substituting. For profit/loss, clearly mark the cost price as the base. For “reverse” questions (e.g. price after a discount), divide by the surviving fraction (0.85 for a 15% discount) rather than adding back the percentage. For growth questions, decide first whether it is simple (P(1 + rt)) or compounded (P(1 + r)t), and remember that successive percentages compound (multiply), never add.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 8 about?

Chapter 8, Fractions in Disguise, is the percentages chapter. It shows that a percentage is a fraction with denominator 100, teaches conversion between fractions, decimals and percentages, and applies percentages to comparing proportions, percentage increase/decrease, profit, loss, discount, GST, compounding and depreciation.

Why is this called Chapter 8 when the book says Part II Chapter 1?

In the printed book it is Chapter 1 of Ganita Prakash Part II. On ClearStudy we number the two parts of the Class 8 book continuously, so it becomes the 8th chapter of the Class 8 course — that is why this page is “Chapter 8”.

How many exercises does Chapter 8 have?

The chapter has six “Figure it Out” exercise sets — after Fractions as Percentages, Percentage of a Quantity, Profit/Loss & Tax, Using Percentages, Growth & Compounding, and Tricky Percentages — all solved on this page along with extra MCQs and Assertion–Reason questions.

Are these Class 8 Maths Ganita Prakash Chapter 8 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part II) textbook for 2026–27, with every percentage and fraction calculation checked.

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