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). Follow the textbook questions with worked steps, percentage bar models and original teaching diagrams. Where the printed choices or data are inconsistent, the solution explains the issue instead of forcing an incorrect answer.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 8 (Part II, Ch 1) Solutions: 6 sections with textbook page references Session: 2026–27
On this page

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 %: retail questions usually use cost price. A company’s profit margin on revenue uses revenue as its base. Read the stated reference amount. Discount is on the marked price.

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

A ten by ten grid has sixty of its one hundred squares shaded blue, showing three fifths equals sixty percent.
Fractions, decimals and percentages describe the same share: 3/5 = 0.6 = 60%.

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 — pages 3–4

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% = 33⅓% ≈ 33.33% (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%)

SOLUTIONRead distance from Start toward Finish and choose the nearest offered percentage: A: 38%, B: 55%, C: 72%, D: 93%.D is very close to the finish. The unused choices are 20% and 84%. These are approximate matches to the picture, not exact measurements.
Approximate race positions A B C D are matched to the options thirty-eight, fifty-five, seventy-two and ninety-three percent respectively.
Page 4, Question 4: positions are redrawn approximately from the textbook; answers are the closest offered percentages.

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 — pages 12–14

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

1. Find the missing numbers in the bar diagrams. The first problem has been worked out. See the redrawn models below.

SOLUTION(i) Five equal parts make 100%, so one part is 20% (the worked example). In the right bar, the whole is 75 and four parts are 4/5 × 75 = 60, or 80% of the whole.(ii) One of ten equal parts is 10%. Six of ten parts of 90 give 6/10 × 90 = 54.(iii) One of four equal parts is 25%. Three of four parts of 140 give 3/4 × 140 = 105.
Three rows show one fifth is twenty percent and four fifths of seventy-five is sixty; one tenth is ten percent and six tenths of ninety is fifty-four; one quarter is twenty-five percent and three quarters of one hundred forty is one hundred five.
Pages 12–13, Question 1: the whole values are 75, 90 and 140. The first left-hand example is already 20%; 75 is not the value of that 20%.

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). The models below shade the required share. For 140%, extend the diagram beyond the 100% whole; for 1% of an hour and 7% of 10 kg, convert units first.
Bars shade twenty-five percent of one hundred sixty, sixteen percent of two hundred fifty, and sixty-two percent of three hundred sixty, giving forty, forty, and two hundred twenty-three point two.
Page 13, Question 2(i–iii): the full bar is the given whole, not the answer.
The first bar extends to one hundred forty percent, giving fifty-six from a base of forty. The next bars shade one percent of three thousand six hundred seconds and seven percent of ten thousand grams.
Question 2(iv–vi): 140% needs a whole bar plus 40% of another. Small shares are drawn to scale.

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.)
The ninety-minute training bar gives ten percent or nine minutes to warm up, eighty percent or seventy-two minutes to play, and ten percent or nine minutes to cool down.
Question 10: all three activities use the same total of 90 minutes as their base.

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.

SOLUTIONFor a 2026 estimate, use approximately 8.30 billion people worldwide. Then 90% × 8.30 billion = 7.47 billion people in the Northern Hemisphere, using the question’s 90% assumption.Source: Worldometer’s 2026 population table lists 8,300,678,395 for 1 July 2026, based on UN population projections. This is an estimate, not an exact census count.If using the chapter’s 2025 rounded base of 8.2 billion instead, the answer is 7.38 billion. State the year and base you choose.

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 — page 19

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 = 46⅔% ≈ 46.67%.

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.
Two worked cards contrast profit divided by cost with profit divided by revenue. Thirty-five rupees on cost seventy-five gives about forty-six point six seven percent; costs are seventy-five percent of two point five crore revenue.
Pages 18–19: the company question explicitly uses revenue as the base; retail markup questions here use cost.

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%

SOLUTIONUsing the question’s prices, increase = ₹100 − ₹60 = ₹40.Percentage increase = (40/60) × 100% = 66⅔% ≈ 66.67%.The intended option is (iv) 66.66%, which truncates the repeating decimal. Correct rounding to two decimal places gives 66.67%.

Figure it Out — Set 4 — page 20

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 — pages 22–24

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.
A comparison shows twenty thousand rupees initially, twenty-two thousand after one year, and after two years twenty-four thousand without compounding or twenty-four thousand two hundred with compounding.
Question 1: the extra ₹200 is interest on the first year’s ₹2000 interest.

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?

SOLUTIONWithout compounding: ₹100 interest is added each year. ₹1000 of extra interest takes 10 years.With annual compounding: after 7 years, ₹1000 × 1.1⁷ ≈ ₹1948.72; after 8 years, ₹1000 × 1.1⁸ ≈ ₹2143.59. The balance first reaches or exceeds double at the end of year 8, when that year’s interest is credited.Compounding is exponential growth because the amount is multiplied by 1.1 each year. Without compounding, adding a fixed ₹100 each year gives 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 — pages 28–30

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%)

SOLUTIONUse the supplied rounded 2025 values, with 8.2 billion = 8200 million as the whole.Germany: 83/8200 × 100 ≈ 1.01% → 1%.India: 1460/8200 × 100 ≈ 17.80% → 18%.Bangladesh: 175/8200 × 100 ≈ 2.13% → 2%.USA: 347/8200 × 100 ≈ 4.23%, or about 4%. The printed list has no suitable 4% option; it contains 2% twice. Do not force the USA into an incorrect listed percentage.

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?

SOLUTIONThe full square has 8 × 8 = 64 square units. Its lower-left portion is a 4 × 4 square divided by a diagonal into equal triangles D and E.Area of E = ½ × 4 × 4 = 8 square units.Percentage occupied by E = (8/64) × 100 = 12.5%.
An eight by eight square is split into A B C D E. The lower-left four by four square is bisected diagonally; triangle E has area eight out of sixty-four, or twelve point five percent.
Page 29, Question 6: count eight intervals along the full side, not the number of dots. E is half of a 4 × 4 square.

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 are 60% of the 40% in Grade 8: 0.60 × 0.40 = 0.24 = 24% of all students.(ii) 24% of 160 = 38.4. A student count must be an integer, so the stated total and exact percentages are inconsistent. There is no whole-number answer satisfying all of them.Only if the percentages were approximate could one report about 38 girls. Rounding to 38 is not an exact solution to the printed data.
A total-student bar separates twenty-four percent Grade Eight girls, sixteen percent other Grade Eight students, and sixty percent Grade Nine students. Twenty-four percent of one hundred sixty gives a non-integer count.
Question 8: 38.4 is the arithmetic result. No whole-number answer satisfies all the stated exact data; rounding does not repair the inconsistency.

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% = 66⅔% ≈ 66.67%.

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?

SOLUTIONLet the original length and breadth be L and B. New length = 1.1L. Keeping LB unchanged requires new breadth = B/1.1 = (10/11)B.The decrease is B − (10/11)B = B/11, which is 1/11 of the original breadth.Percentage decrease = (1/11) × 100% = 100/11% = 9 1/11% ≈ 9.09%.
Two equal-area rectangles have dimensions ten by eleven and eleven by ten. Length increases ten percent while breadth decreases by one eleventh, or nine and one eleventh percent.
Question 11: the dimensions are a convenient example. The exact fractional change is 1/11; the percentage change is 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 2 and get 1 free, then buy 1 more) and Shop C ₹300. So choose Shop A — it has the highest discount (50%) and lowest cost.
Shop A offers one paid and one free item for fifty percent discount, shop B two paid and one free for thirty-three and one third percent, and shop C three paid and one free for twenty-five percent.
Question 13: for exactly four items, A costs ₹200; B and C cost ₹300.

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 in the textbook comparison: the twenties have the highest female percentage (26%) and highest male percentage (37%). Exact combined age-group percentages would also require the group weights.(ii) Valid for the survey shown: each female bar is lower than its corresponding male bar. This describes the reported 2023 survey, not an inherent difference in ability.(iii) Not valid: percentages do not reveal the total number of people in the age groups.(iv) Not valid: the thirties values are 14% and 25%. Their combined percentage is a weighted average between 14% and 25%, so it cannot be more than a quarter. Adding 14% + 25% is incorrect because the bases differ.(v) Valid: the senior percentages, 2% and 4%, are both below 10%, so their combined share is also below one in ten.(vi) Not valid: a combined percentage between 26% and 37% cannot be 50%.The intended valid statements are (i), (ii) and (v).
Paired horizontal bars show female and male percentages: children four and about five, teenagers twenty-four and twenty-nine, twenties twenty-six and thirty-seven, thirties fourteen and twenty-five, forties seven and fourteen, fifties four and nine, seniors two and four.
Page 30: original redraw of the NSS 79th-round survey chart reproduced in NCERT. The unlabelled children’s male bar is approximate; other values follow printed labels.

Textbook activities and worked examples

Percentage practice (pages 7–12)

Percentage10020050801035287
25%255012.5202.58.7571.75
10%10205813.528.7
20%204010162757.4
5%5102.540.51.7514.35
40%40802032414114.8
15%15307.5121.55.2543.05

For example, 15% = 10% + 5%. Find these two shares mentally and add them.

FractionDecimalPercentage
1/20.550%
11100%
1/40.2525%
3/40.7575%
1/100.110%
1/1000.011%
1/200.055%
43/1000.4343%
9/100.990%
11/101.1110%
22200%
5/22.5250%
3/200.1515%
173/1001.73173%
179/503.58358%
289/10000.28928.9%
61/203.05305%

On a number line from 0 to 4, mark each decimal at its matching point. For example, 110% is 1.1, just after 1; 358% is 3.58, between 3.5 and 3.6.

Mixtures and reverse percentages (pages 9–10)

A 2:7 millet-to-water mixture contains 9 parts in total. Millet is 2/9 ≈ 22.22%; for 500 mL, use 500 × 2/9 ≈ 111.11 mL millet.

A mixture bar has two orange millet parts and seven blue water parts. Millet is two ninths of the total, approximately twenty-two point two two percent.
Page 9: keep 2/9 exact while calculating; round only the final amount.

The cyclist has completed 92 km, which is 40% of the journey. Total = 92/0.4 = 230 km; remaining distance = 138 km.

A journey bar shows the completed forty percent as ninety-two kilometres and the remaining sixty percent as one hundred thirty-eight kilometres. The total is two hundred thirty kilometres.
Page 10: 40% of the journey is already complete, so the remaining share is 60%.

Sales targets (page 11)

For a ₹5000 daily target: day 5 sales of ₹7800 achieve 156%; day 6 sales of ₹9550 achieve 191%. A 150% day means ₹7500, and a 210% day means ₹10,500.

Prices and the sample bill (pages 17–19)

A ₹36 notebook sold at 20% profit costs the buyer ₹43.20. A crayon box sold for ₹50 at 25% profit had cost ₹40. Strawberries sold at ₹80 with 12% loss had cost 80/0.88 ≈ ₹90.91. A ₹1800 cooker after 35% discount sells for ₹1170; with cost ₹900, profit is ₹270 or 30%.

The textbook bill is correct: 3 × ₹150 = ₹450; each 9% tax is ₹40.50; total = 450 + 40.50 + 40.50 = ₹531. These rates are the exercise data.

Depreciation and offers (pages 25–27)

₹21,000 depreciated by 5% becomes ₹19,950. A village population of 1250 declining by 10% for three years is modelled by 1250 × 0.9³ = 911.25, or about 911 people.

Compare 20% off with ₹50 off: for ₹180, choose ₹50 off and pay ₹130; for ₹225, choose ₹50 off and pay ₹175; for ₹300, choose 20% off and pay ₹240. They are equal at ₹250.

A 50% markup followed by a 50% discount gives 1.5 × 0.5 = 0.75 times cost: a 25% loss. If the final selling price is ₹12,000, cost was ₹16,000 and loss ₹4000. A 50% markup permits a break-even discount of 33⅓%.

If Ariba’s amount is 120% of Arun’s, Arun’s amount is 1/1.2 = 5/6 = 83⅓% of Ariba’s amount.

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).
  • Choosing the wrong base: retail markup uses cost, while a stated profit margin on revenue uses revenue.
  • 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: For successive discounts, multiply the fractions of the price that remain after each discount.

A-R 3. Assertion: 120% of a positive 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. Retail profit/loss % uses cost; a stated revenue margin uses revenue.
  • 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, identify whether the question uses cost or revenue 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?

This page organises the textbook solutions into six sections, with page references. It also includes in-text activities and original MCQs and Assertion–Reason practice.

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.

Textbook reference: NCERT Ganita Prakash, Chapter 8 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

← Chapter 7 All Ganita Prakash Chapters Chapter 9 →
Scroll to Top