--- title: "NCERT Solutions for Class 8 Maths Chapter 8 Fractions in Disguise" url: https://www.swavid.com/maths/class/8/chapter/fractions-in-disguise/ncert-solutions dateModified: 2026-10-07T15:32:36+00:00 --- # NCERT Solutions for Class 8 Maths Chapter 8 Fractions in Disguise This chapter's questions cover fractions, percentages, their conversions, profit and loss, interest, and various word problems involving ratios and proportions. Free PDF (25 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-8/swavid-ncert-solutions-class-8-maths-chapter-8-fractions-in-disguise-5a405e8422.pdf ## Figure it Out ### Question 1 *3 marks · Short answer* Express the following fractions as percentages. (i) $\frac{3}{5}$ (ii) $\frac{7}{14}$ (iii) $\frac{9}{20}$ (iv) $\frac{72}{150}$ (v) $\frac{1}{3}$ (vi) $\frac{5}{11}$ **Part (i)** 1. Multiply the fraction by 100 to convert it into a percentage. 2. $\frac{3}{5} \times 100\% = 60\%$. Answer (i): 60% **Part (ii)** 1. Simplify the fraction and multiply by 100. 2. $\frac{7}{14} = \frac{1}{2}$, and $\frac{1}{2} \times 100\% = 50\%$. Answer (ii): 50% **Part (iii)** 1. Multiply the fraction by 100. 2. $\frac{9}{20} \times 100\% = 9 \times 5\% = 45\%$. Answer (iii): 45% **Part (iv)** 1. Multiply the fraction by 100. 2. $\frac{72}{150} \times 100\% = \frac{72}{3} \times 2\% = 24 \times 2\% = 48\%$. Answer (iv): 48% **Part (v)** 1. Multiply the fraction by 100. 2. $\frac{1}{3} \times 100\% = 33.33\%$. Answer (v): 33.33% **Part (vi)** 1. Multiply the fraction by 100. 2. $\frac{5}{11} \times 100\% = \frac{500}{11}\% = 45.45\%$. Answer (vi): 45.45% **Answer:** The percentages are (i) 60% (ii) 50% (iii) 45% (iv) 48% (v) 33.33% (vi) 45.45%. > Common mistake: Forgetting to multiply by 100 or incorrectly dividing instead. ### Question 2 *1 mark · MCQ* Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white? - 10% - 15% - 25% - 60% - 40% - None of these **Solution** 1. Find the fraction of white marbles: $\frac{15}{25}$. 2. Convert the fraction to a percentage by multiplying by 100: $\frac{15}{25} \times 100\% = 60\%$. **Answer:** (iv) 60% > Common mistake: Dividing 25 by 15 instead of 15 by 25. ### Question 3 *3 marks · Short answer* In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking? **Solution** 1. State the number of students walking and the total number of students: 15 out of 80. 2. Express the proportion as a fraction: $\frac{15}{80}$. 3. Multiply by 100 to find the percentage: $\frac{15}{80} \times 100\% = \frac{75}{4}\% = 18.75\%$. **Answer:** 18.75% of the students come by walking. > Common mistake: Placing the total in the numerator instead of the denominator. ### Question 4 *3 marks · Short answer* A group of friends is participating in a long-distance run. The positions of each of them after 15 minutes are shown in the following picture. Match (among the given options) what percentage of the race each of them has approximately completed. - 55% - 20% - 38% - 72% - 84% - 93% **Solution** 1. Observe the positions of A, B, C, and D on the number line between Start (0%) and Finish (100%). 2. Point A is slightly past halfway between 20% and 55%, closer to 38%. 3. Points B, C, and D correspond to the higher values on the line, matching 55%, 84%, and 93% respectively. **Answer:** A corresponds to 38%, B to 55%, C to 84%, and D to 93%. > Common mistake: Misreading the scale intervals on the bar model. ### Question 5 *3 marks · Short answer* Pairs of quantities are shown below. Identify and write appropriate symbols ʻ>ʼ, ʻ<ʼ, ʻ=ʼ in the blanks. Try to do it without calculations. (i) 50% ____ 5% (ii) $\frac{5}{10}$ ____ 50% (iii) $\frac{3}{11}$ _____ 61% (iv) 30% ____ $\frac{1}{3}$ **Part (i)** 1. Compare 50% and 5% directly by magnitude. 2. $50\% > 5\%$. Answer (i): > **Part (ii)** 1. Convert $\frac{5}{10}$ to a percentage: $\frac{1}{2} = 50\%$. 2. Compare with 50%. Answer (ii): = **Part (iii)** 1. Convert $\frac{3}{11}$ to a percentage: $\frac{300}{11}\% \approx 27.27\%$. 2. Compare 27.27% with 61%. Answer (iii): < **Part (iv)** 1. Convert $\frac{1}{3}$ to a percentage: $33.33\%$. 2. Compare 30% with 33.33%. Answer (iv): < **Answer:** The appropriate symbols are (i) >, (ii) =, (iii) <, (iv) <. > Common mistake: Comparing fractions and percentages directly without converting them to the same unit. ### Question 1 *3 marks · Short answer* Find the missing numbers. The first problem has been worked out. (i) (ii) **Part (i)** 1. Given that 20% corresponds to 15 (since 20% of 75 is 15). 2. The bar model is divided into five 20% parts, where each part is 15. Answer (i): 15 **Part (ii)** 1. Given that 60% corresponds to 54 or using proportion where 90 is 100%. 2. Calculating $\frac{60}{100} \times 90 = 54$. Answer (ii): 54 **Answer:** For (i), the missing number is 15. For (ii), the missing number is 45. > Common mistake: Incorrectly scaling the bar divisions. ### Question 2 *3 marks · Short answer* 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 **Part (i)** 1. $25\% \text{ of } 160 = \frac{25}{100} \times 160 = \frac{1}{4} \times 160 = 40$ Answer (i): 40 **Part (ii)** 1. $16\% \text{ of } 250 = \frac{16}{100} \times 250 = \frac{4}{25} \times 250 = 40$ Answer (ii): 40 **Part (iii)** 1. $62\% \text{ of } 360 = \frac{62}{100} \times 360 = 0.62 \times 360 = 223.2$ Answer (iii): 223.2 **Part (iv)** 1. $140\% \text{ of } 40 = \frac{140}{100} \times 40 = 1.4 \times 40 = 56$ Answer (iv): 56 **Part (v)** 1. $1\% \text{ of } 1 \text{ hour} = 1\% \text{ of } 60 \text{ minutes} = \frac{1}{100} \times 60 = 0.6 \text{ minutes} = 36 \text{ seconds}$ Answer (v): 36 seconds **Part (vi)** 1. $7\% \text{ of } 10 \text{ kg} = \frac{7}{100} \times 10 = \frac{7}{10} \text{ kg} = 0.7 \text{ kg}$ Answer (vi): 0.7 kg **Answer:** (i) 40, (ii) 40, (iii) 223.2, (iv) 56, (v) 36 seconds, (vi) 0.7 kg > Common mistake: Forgetting to convert units when finding percentages of quantities like hours or kilograms. ### Question 3 *3 marks · Short answer* Surya made 60 ml of deep orange paint, how much red paint did he use if red paint made up $\frac{3}{4}$ of the deep orange paint? **Solution** 1. Given total amount of paint is $60\text{ ml}$. 2. Red paint makes up $\frac{3}{4}$ of the deep orange paint. 3. Amount of red paint $= \frac{3}{4} \times 60 = 3 \times 15 = 45\text{ ml}$. **Answer:** 45 ml > Common mistake: Multiplying by the wrong fraction or confusing red and yellow parts. ### Question 4 *3 marks · Short answer* Pairs of quantities are shown below. Identify and write appropriate symbols ʻ>ʼ, ʻ<ʼ, ʻ=ʼ 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 **Part (i)** 1. Since $510 < 515$, $50\% \text{ of } 510 < 50\% \text{ of } 515$. Answer (i): < **Part (ii)** 1. Comparing $37\% \text{ of } 148$ and $73\% \text{ of } 148$, since $37 < 73$, the first value is smaller. Answer (ii): < **Part (iii)** 1. $29\% \text{ of } 43$ is much smaller than $92\% \text{ of } 110$. Answer (iii): < **Part (iv)** 1. $30\% \text{ of } 40 = 12$ and $40\% \text{ of } 50 = 20$, so $12 < 20$. Answer (iv): < **Part (v)** 1. $45\% \text{ of } 200 = 90$ and $10\% \text{ of } 490 = 49$, so $90 > 49$. Answer (v): > **Part (vi)** 1. $30\% \text{ of } 80 = 24$ and $24\% \text{ of } 64 = 15.36$, so $24 > 15.36$. Answer (vi): > **Answer:** (i) <, (ii) <, (iii) <, (iv) <, (v) >, (vi) > > Common mistake: Doing unnecessary full calculations instead of reasoning by magnitude. ### Question 5 *3 marks · Short answer* 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. **Part (i)** 1. Given $30\% \text{ of } k = 70$. 2. Therefore, $60\% \text{ of } k = 70 \times 2 = 140$, $90\% \text{ of } k = 70 \times 3 = 210$, and $120\% \text{ of } k = 70 \times 4 = 280$. Answer (i): 140, 210, 280 **Part (ii)** 1. Given $100\% \text{ of } m = 215$. 2. Therefore, $10\% \text{ of } m = 21.5$, $1\% \text{ of } m = 2.15$, and $6\% \text{ of } m = 2.15 \times 6 = 12.9$. Answer (ii): 21.5, 2.15, 12.9 **Part (iii)** 1. Given $90\% \text{ of } n = 270$, so $1\% \text{ of } n = 3$. 2. Thus, $9\% \text{ of } n = 27$, $18\% \text{ of } n = 54$, and $100\% \text{ of } n = 300$. Answer (iii): 30, 60, 300 **Answer:** (i) 140, 210, 280; (ii) 21.5, 2.15, 12.9; (iii) 30, 60, 300 > Common mistake: Multiplying or dividing incorrectly when scaling percentages. ### Question 6 *3 marks · Short answer* Fill in the blanks: (i) 3 is ____ % of 300. (ii) _____ is 40% of 4. (iii) 40 is 80% of _____. **Part (i)** 1. Let $3 = x\% \text{ of } 300$. 2. $\frac{x}{100} \times 300 = 3 \implies 3x = 3 \implies x = 1$. Answer (i): 1 **Part (ii)** 1. Let required value be $x$. 2. $x = 40\% \text{ of } 4 = \frac{40}{100} \times 4 = \frac{160}{100} = 1.6$. Answer (ii): 1.6 **Part (iii)** 1. Let $40 = 80\% \text{ of } x$. 2. $40 = \frac{80}{100} \times x \implies x = 40 \times \frac{100}{80} = 50$. Answer (iii): 50 **Answer:** (i) 1, (ii) 1.6, (iii) 50 > Common mistake: Setting up the fraction upside down when finding the base quantity. ### Question 7 *3 marks · Short answer* Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers. **Solution** 1. A day has $24\text{ hours}$, so $10\% \text{ of a day} = 10\% \text{ of } 24\text{ hours} = 2.4\text{ hours} = 144\text{ minutes}$. 2. A week has $7\text{ days}$, so $1\% \text{ of a week} = 1\% \text{ of } 168\text{ hours} = 1.68\text{ hours} = 100.8\text{ minutes}$. 3. Comparing both values, $144\text{ minutes} > 100.8\text{ minutes}$, hence $10\% \text{ of a day}$ is longer than $1\% \text{ of a week}$. **Answer:** Yes, 10% of a day (144 minutes) is longer than 1% of a week (100.8 minutes). > Common mistake: Comparing percentages directly without converting them to the same units of time. ### Question 8 *3 marks · Short answer* 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 the bull 3 units of fodder and the bull ate 2 units. The day after, she gave the bull 4 units and the bull ate 3 units. This continued, and on the 99th day she gave the bull 100 units and the bull ate 99 units. Represent these quantities as percentages. This task can be distributed among the class. What do you observe? **Solution** 1. On any given day $n$, the fodder given is $(n + 1)$ units and the fodder eaten is $n$ units. 2. The percentage of fodder eaten on day $n$ is calculated as $\frac{n}{n + 1} \times 100$. 3. On the 99th day, the percentage is $\frac{99}{100} \times 100 = 99\%$, and we observe that the percentage of fodder eaten increases closer to $100\%$ as days progress. **Answer:** On the 99th day, the bull eats 99% of the fodder given. > Common mistake: Confusing the numerator and denominator while calculating the percentage. ### Question 9 *3 marks · Short answer* 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** 1. Given: Workers take 18 days to pick berries in 20% of the plantation. 2. Let the total number of days required for the entire plantation ($100\%$) be $d$. 3. Using proportionality, $\frac{20}{100} = \frac{18}{d}$, which gives $d = \frac{18 \times 100}{20} = 90$ days. The assumption that the rate of work stays the same is necessary because worker efficiency or weather conditions might change over time. **Answer:** 90 days > Common mistake: Multiplying 18 by 5 without explaining the proportional relationship. ### Question 10 *3 marks · Short answer* 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** 1. Given total training time = 90 minutes and the ratio of activities is 10% : 80% : 10%. 2. Warm-up time = 10% of 90 minutes = $\frac{10}{100} \times 90 = 9$ minutes. 3. Play time = 80% of 90 minutes = $\frac{80}{100} \times 90 = 72$ minutes, and cool-down time = 10% of 90 minutes = 9 minutes. **Answer:** Warm-up: 9 minutes, Play: 72 minutes, Cool down: 9 minutes > Common mistake: Calculating percentages on an incorrect total time value. ### Question 11 *3 marks · Short answer* 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** 1. Given that approximately 90% of the world's population lives in the Northern Hemisphere. 2. Taking the world's population as approximately 8.2 billion. 3. Northern Hemisphere population = 90% of 8.2 billion = $\frac{90}{100} \times 8.2 = 7.38$ billion. **Answer:** Approximately 7.38 billion people > Common mistake: Using an outdated population figure instead of the current estimate. ### Question 12 *3 marks · Short answer* A recipe for the dish, halwa, for 4 people has the following ingredients in the given proportions — Rava: 40%, Sugar: 40%, and 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? **Part (i)** 1. The proportion of ingredients for a recipe does not change when the number of people served changes. 2. Therefore, the proportions remain: Rava: 40%, Sugar: 40%, and Ghee: 20%. Answer (i): Rava: 40%, Sugar: 40%, Ghee: 20% **Part (ii)** 1. Given total weight = 2 kg = 2000 g. 2. Rava = 40% of 2000 g = $\frac{40}{100} \times 2000 = 800$ g; Sugar = 40% of 2000 g = 800 g; Ghee = 20% of 2000 g = 400 g. Answer (ii): Rava: 800 g, Sugar: 800 g, Ghee: 400 g **Answer:** For part (i), proportions remain the same. For part (ii), rava = 800 g, sugar = 800 g, ghee = 400 g. > Common mistake: Scaling up the percentage values when doubling the number of people in part (i). ### Question 1 *3 marks · Short answer* 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** 1. Given: Cost Price (CP) = ₹75 and Selling Price (SP) = ₹110. 2. Profit amount = SP - CP = ₹110 - ₹75 = ₹35. 3. Profit margin with respect to cost = $\frac{\text{Profit}}{\text{Cost Price}} \times 100 = \frac{35}{75} \times 100 = \frac{140}{3} = 46.67\%$. **Answer:** 46.67% > Common mistake: Calculating profit percentage with respect to the selling price instead of the cost price. ### Question 2 *3 marks · Short answer* 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** 1. Given the cost price of a chair = ₹475 and profit margin = 50%. 2. Profit amount = $50\% \text{ of } 475 = \frac{50}{100} \times 475 = \text{₹}237.50$. 3. Selling price = Cost price + Profit = $475 + 237.50 = \text{₹}712.50$. **Answer:** ₹712.50 > Common mistake: Adding percentage directly instead of calculating 50% of the cost price. ### Question 3 *3 marks · Short answer* 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** 1. Given total sales (revenue) = ₹2.5 crore and profit margin = 25% of expenditure. 2. Let the total expenditure be $x$. Then sales = Expenditure + Profit = $x + 0.25x = 1.25x$. 3. $1.25x = 2.5 \text{ crore}$, which gives $x = \frac{2.5}{1.25} = 2 \text{ crore}$. **Answer:** ₹2 crore > Common mistake: Taking 25% of the total sales as profit instead of 25% of the expenditure. ### Question 4 *3 marks · Short answer* 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** 1. Given original price of the shirt = ₹300 and discount = 25%. 2. Discount amount = $25\% \text{ of } 300 = \frac{25}{100} \times 300 = \text{₹}75$. 3. Selling price (amount Anwar has to pay) = Original price - Discount = $300 - 75 = \text{₹}225$. **Answer:** ₹225 > Common mistake: Adding the discount amount to the original price. ### Question 5 *1 mark · MCQ* The petrol price in 2015 was ₹60 and ₹100 in 2025. What is the percentage increase in the price of petrol? - 50% - 40% - 60% - 66.66% - 140% - 160.66% **Solution** 1. Increase in petrol price = $100 - 60 = \text{₹}40$. 2. Percentage increase = $\frac{\text{Increase}}{\text{Original price}} \times 100 = \frac{40}{60} \times 100 = 66.66\%$. **Answer:** (iv) 66.66% > Common mistake: Dividing the increase by the final price instead of the original price. ### Question 3 *3 marks · Short answer* 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** 1. Given selling price after 15% discount = ₹4,40,000. 2. Let the original price be $p$. Then selling price = $p - 15\% \text{ of } p = 85\% \text{ of } p$. 3. $0.85p = 440000$, which gives $p = \frac{440000}{0.85} = \text{₹}5,17,647.06$. **Answer:** ₹5,17,647.06 > Common mistake: Calculating 15% of the selling price and adding it directly. ### Question 4 *3 marks · Short answer* 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** 1. Total votes = 1600, winner's votes = 500. 2. Winner's percentage = $\frac{500}{1600} \times 100 = 31.25\%.$ 3. For the winner to get 500 votes out of 1600 while securing the highest share among candidates, the minimum number of candidates must be 4 so that remaining votes can be distributed among others. **Answer:** 31.25%, minimum 4 candidates > Common mistake: Dividing total votes by winner votes instead of winner votes by total votes. ### Question 5 *3 marks · Short answer* 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** 1. Given: original price in 2024 = ₹38, new price in 2025 = ₹42 2. Amount of increase = ₹42 - ₹38 = ₹4 3. Rate of inflation = $\frac{\text{Amount of increase}}{\text{Original price}} \times 100$ 4. Rate of inflation = $\frac{4}{38} \times 100 = \frac{200}{19}\% = 10.53\%$ **Answer:** 10.53% > Common mistake: Dividing the increase by the new price instead of the original price. ### Question 6 *3 marks · Short answer* A number increased by 20% becomes 90. What is the number? **Solution** 1. Let the required number be x. 2. According to the question, the number increased by 20% becomes 90, so $x + \frac{20}{100}x = 90$. 3. Simplifying the equation gives $1.2x = 90$, which means $x = \frac{90}{1.2} = 75$. **Answer:** 75 > Common mistake: Subtracting 20% of 90 from 90 instead of setting up the equation with the original number. ### Question 7 *3 marks · Short answer* 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** 1. Selling price of each buffalo = ₹80,000, so total selling price for two buffaloes = ₹1,60,000. 2. For the first buffalo with a 5% profit, cost price = $\frac{80,000}{1.05} = ₹76,190.48$. 3. For the second buffalo with a 10% loss, cost price = $\frac{80,000}{0.9} = ₹88,888.89$, making the total cost price ₹1,65,079.37, resulting in an overall loss of ₹5,079.37. **Answer:** Overall loss of ₹5,079.37 > Common mistake: Assuming that a 5% gain and a 10% loss on equal selling prices cancel out to a 5% loss. ### Question 8 *1 mark · MCQ* The population of elephants in a national park increased by 5% in the last decade. If the population of the elephants last decade is p, the population now is - p × 0.5 - p × 0.05 - p × 1.5 - p × 1.05 - p + 1.50 **Solution** 1. An increase of 5% on population p means the new population is $p + 5\% \text{ of } p$. 2. This can be written as $p + 0.05p = 1.05p$, which is $p \times 1.05$. 3. Therefore, the correct option is (iv). **Answer:** (iv) p × 1.05 > Common mistake: Selecting p × 0.05, which represents only the increase amount and not the new total population. ### Question 9 *1 mark · MCQ* Which of the following statement(s) mean the same as — “The demand for cameras has fallen by 85% in the last decade”? - The demand now is 85% of the demand a decade ago. - The demand a decade ago was 85% of the demand now. - The demand now is 15% of the demand a decade ago. - The demand a decade ago was 15% of the demand now. - The demand a decade ago was 185% of the demand now. - The demand now is 185% of the demand a decade ago. **Solution** 1. A fall of 85% in demand means the current demand is 100% - 85% = 15% of the demand a decade ago. 2. Thus, the demand now is 15% of the demand a decade ago, which corresponds to statement (iii). **Answer:** (iii) The demand now is 15% of the demand a decade ago. > Common mistake: Choosing option (i) by confusing the percentage decrease with the remaining percentage. ### Question 1 *3 marks · Short answer* 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** 1. Given principal $P = \text{₹}20,000$, rate $r = 10\%$ p.a., and time $t = 2$ years. 2. Without compounding (simple interest), total amount = $P(1 + rt) = 20000(1 + 0.10 \times 2) = 20000(1.2) = \text{₹}24,000$. 3. With compounding annually, total amount = $P(1 + r)^t = 20000(1 + 0.10)^2 = 20000(1.21) = \text{₹}24,200$. **Answer:** Without compounding: ₹24,000; With compounding: ₹24,200 > Common mistake: Applying compound interest formula for simple interest calculations. ### Question 2 *3 marks · Short answer* 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** 1. Given: Principal $P = \text{₹}20,000$, rate $r = 5\% = 0.05$ per annum, time $t = 4$ years. 2. Without compounding (Simple Interest), total amount = $p(1 + rt) = 20000 \times (1 + 0.05 \times 4) = 20000 \times 1.20 = \text{₹}24,000$. 3. With compounding annually, total amount = $p(1 + r)^t = 20000 \times (1.05)^4 = 20000 \times 1.21550625 = \text{₹}24,310.13$. **Answer:** Without compounding, the total amount is ₹24,000. With compounding, the total amount is ₹24,310.13. > Common mistake: Multiplying the interest rate directly for all years in compound interest. ### Question 3 *3 marks · Short answer* Do you observe anything interesting in the solutions of the two questions above? Share and discuss. **Solution** 1. Compare the total amounts obtained from both options in the previous two questions. 2. Observe that for shorter time periods and lower rates, the difference between simple and compound interest is relatively small. 3. As the time period increases, the compound interest amount grows much faster than the simple interest amount due to interest being earned on interest. **Answer:** Compounding yields a higher total amount than simple interest, and the difference increases with time. > Common mistake: Assuming simple and compound interest give the same returns. ### Question 4 *1 mark · MCQ* 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? - p × 6 × 4 - p × 0.6 × 4 - p × \frac{0.6}{100} × 4 - p × \frac{0.06}{100} × 4 - p × 1.6 × 4 - p × 1.06 × 4 - p + (p × 0.06 × 4) **Solution** 1. The total amount with simple interest for $p$, rate $r$, and time $t$ is given by $p(1 + rt)$. 2. Substituting $r = 6\% = 0.06$ and $t = 4$ gives $p + (p \times 0.06 \times 4)$. 3. Therefore, option (vii) is the correct expression. **Answer:** (vii) $p + (p \times 0.06 \times 4)$ > Common mistake: Selecting compound interest expressions for simple interest. ### Question 5 *3 marks · Short answer* 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** 1. Given: Principal $P = \text{₹}50,000$, rate $r = 7\% = 0.07$ per annum, time $t = 3$ years. 2. Interest without compounding = $P \times r \times t = 50000 \times 0.07 \times 3 = \text{₹}10,500$. 3. Amount with compounding = $P(1 + r)^t = 50000 \times (1.07)^3 = 50000 \times 1.225043 = \text{₹}61,252.15$, so compound interest = $\text{₹}11,252.15$. 4. Difference in interest = $11,252.15 - 10,500 = \text{₹}752.15$. **Answer:** Interest without compounding is ₹10,500, and compounding yields ₹752.15 more. > Common mistake: Calculating only the total amount instead of the interest earned. ### Question 6 *3 marks · Short answer* 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** 1. For Giridhar (without compounding): Principal $P = 12500$, rate $r = 0.12$, time $t = 3$. Interest = $12500 \times 0.12 \times 3 = \text{₹}4,500$. 2. For Raghava (with compounding): Principal $P = 12500$, rate $r = 0.10$, time $t = 3$. Amount = $12500 \times (1.1)^3 = 12500 \times 1.331 = \text{₹}16,637.50$, so interest = $16637.50 - 12500 = \text{₹}4,137.50$. 3. Comparing the interests, Giridhar pays more interest by $4500 - 4137.50 = \text{₹}362.50$. **Answer:** Giridhar pays more interest by ₹362.50. > Common mistake: Comparing total amounts directly instead of the interest amounts paid. ### Question 7 *3 marks · Short answer* 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** 1. Given Principal $P = \text{₹}1000$ growing at $10\%$ p.a. 2. Without compounding, interest per year is constant at $1000 \times 0.10 = \text{₹}100$. To double (reach ₹2000, i.e., ₹1000 interest), it takes $1000 / 100 = 10$ years. 3. With compounding, amount after $t$ years is $1000 \times (1.1)^t$. Setting this to 2000 gives $(1.1)^t = 2$, which takes about $7.27$ years ($t = \frac{\log 2}{\log 1.1}$). Compounding is exponential growth, while simple interest is linear growth. **Answer:** It takes 10 years without compounding and about 7.27 years with compounding. Compounding is exponential growth and simple interest is linear growth. > Common mistake: Assuming doubling time is the same for both methods. ### Question 8 *3 marks · Short answer* 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** 1. Given current population $P = 1.5$ crore, rate of growth $r = 3\%$ per year, and time $t = 3$ years. 2. Using the formula for population growth, expected population = $P \times (1 + r)^t$. 3. Expected population = $1.5 \times (1 + 0.03)^3 = 1.5 \times (1.03)^3 = 1.5 \times 1.092727 = 1.6390905$ crore. **Answer:** 1.639 crore (or approximately 1,63,90,905) > Common mistake: Multiplying by $1 + 3 \times 0.03$ instead of using compounding $(1.03)^3$. ### Question 9 *3 marks · Short answer* 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** 1. Given initial bacterial count = $5,06,000$, growth rate = $2.5\%$ per hour, and time = $2$ hours. 2. Population after $2$ hours = $5,06,000 \times \left(1 + \frac{2.5}{100}\right)^2$. 3. Calculating the value: $5,06,000 \times (1.025)^2 = 5,06,000 \times 1.050625 = 5,31,616.25$. **Answer:** 5,31,616 bacteria (approximately) > Common mistake: Calculating simple interest style growth instead of compound growth. ### Question 1 *3 marks · Short answer* 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** 1. Given population in 2000 = $50$ lakhs, and population in 2025 = $250\%$ of population in 2000. 2. Expected population in 2025 = $250\% \text{ of } 50 \text{ lakhs} = \frac{250}{100} \times 50$. 3. Calculating the value: $2.5 \times 50 = 125$ lakhs. **Answer:** 125 lakhs (or 1.25 crore) > Common mistake: Dividing by 100 incorrectly or multiplying 50 by 2.5 wrongly. ### Question 2 *3 marks · Short answer* The population of the world in 2025 is about 8.2 billion. The populations of some countries in 2025 are given. Match them with their approximate percentage share of the worldwide population. [Hint: Writing these numbers in the standard form and estimating can help]. **Part Germany** 1. Calculate the percentage share: $\frac{83 \text{ million}}{8.2 \text{ billion}} \times 100 = \frac{83 \times 10^6}{8.2 \times 10^9} \times 100$ 2. Simplify the expression to get approximately $1\%$. Answer Germany: $1\%$ **Part India** 1. Calculate the percentage share: $\frac{1.46 \text{ billion}}{8.2 \text{ billion}} \times 100 = \frac{1.46}{8.2} \times 100$ 2. Simplify the expression to get approximately $18\%$. Answer India: $18\%$ **Part Bangladesh** 1. Calculate the percentage share: $\frac{175 \text{ million}}{8.2 \text{ billion}} \times 100 = \frac{175 \times 10^6}{8.2 \times 10^9} \times 100$ 2. Simplify the expression to get approximately $2\%$. Answer Bangladesh: $2\%$ **Part USA** 1. Calculate the percentage share: $\frac{347 \text{ million}}{8.2 \text{ billion}} \times 100 = \frac{347 \times 10^6}{8.2 \times 10^9} \times 100$ 2. Simplify the expression to get approximately $4\%$. Answer USA: $4\%$ **Answer:** Germany: $1\%$, India: $18\%$, Bangladesh: $2\%$, USA: $4\%$ > Common mistake: Mixing up millions and billions when writing out the ratio. ### Question 3 *1 mark · MCQ* 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? - 8250 + 18 - 8250 + 1800 - 8250 + \frac{18}{100} - 8250 × 18 - 8250 × 1.18 - 8250 + 8250 × 0.18 - 1.8 × 8250 **Solution** 1. The final price including GST is the original price plus the GST amount. 2. GST amount is $18\%$ of ₹8,250, which is $8250 \times 0.18$ or $8250 \times \frac{18}{100}$. 3. Therefore, the correct options representing this are (v) $8250 \times 1.18$, (vi) $8250 + 8250 \times 0.18$, etc. Selecting the standard textbook equivalent expression. **Answer:** (v) $8250 \times 1.18$ (also matches (vi) and (vii) depending on multiple correct selections) > Common mistake: Adding just 18 rupees instead of 18 percent. ### Question 4 *1 mark · MCQ* 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%, and Month 3 change was –3%. Which of the following statement(s) are true? The initial population is p. - The population after three months was p × 0.05 × 0.02 × 0.03. - The population after three months was p × 1.05 × 0.98 × 0.97. - The population after three months was p + 0.05 – 0.02 – 0.03. - The population after three months was p. - The population after three months was more than p. - The population after three months was less than p. **Solution** 1. Successive percentage changes involve multiplying the initial value by the respective multiplier for each period. 2. A +5% change gives factor 1.05, -2% gives 0.98, and -3% gives 0.97. 3. Thus, the population after three months is $p \times 1.05 \times 0.98 \times 0.97$, making statement (ii) true. **Answer:** (ii) The population after three months was $p \times 1.05 \times 0.98 \times 0.97$. > Common mistake: Adding percentages directly instead of multiplying their decimal factors. ### Question 5 *3 marks · Short answer* 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** 1. Let the cost price of the product be $x$. 2. With a 35% profit margin, the selling price before discount is $1.35x$. 3. After a 30% discount on this selling price, the final selling price is $1.35x - 0.30 \times (1.35x) = 1.35x \times 0.70 = 0.945x$. 4. Since the final selling price ($0.945x$) is less than the cost price ($x$), the shopkeeper makes a loss. **Answer:** The shopkeeper makes a loss. > Common mistake: Adding and subtracting percentages directly without accounting for the base change. ### Question 6 *3 marks · Short answer* What percentage of area is occupied by the region marked ‘E’ in the figure? **Solution** 1. Count the total number of grid squares in the figure, which is $10 \times 10 = 100$ squares. 2. Count the number of squares occupied by the region marked 'E'. 3. Region 'E' occupies 15 squares out of 100, which is $15\%$. **Answer:** $15\%$ > Common mistake: Miscounting the grid squares for region E. ### Question 7 *3 marks · Short answer* 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** 1. $5\%$ of $40 = 2$ and $40\%$ of $5 = 2$. 2. $25\%$ of $12 = 3$ and $12\%$ of $25 = 3$. 3. $15\%$ of $60 = 9$ and $60\%$ of $15 = 9$. 4. We notice that $x\%$ of $y$ is always equal to $y\%$ of $x$. 5. General statement: $x\% \text{ of } y = \frac{x}{100} \times y = \frac{xy}{100} = \frac{y}{100} \times x = y\% \text{ of } x$. **Answer:** x% of y is equal to y% of x. > Common mistake: Assuming the percentage value depends on the order of numbers. ### Question 8 *3 marks · Short answer* A school is organising an excursion for its students. 40% of them are Grade 8 students and the rest are Grade 9 students. Among these Grade 8 students, 60% are girls. [Hint: Drawing a rough diagram can help]. (i) What percentage of the students going to the excursion are Grade 8 girls? (ii) If the total number of students going to the excursion is 160, how many of them are Grade 8 girls? **Part (i)** 1. Grade 8 students make up $40\%$ of the total students. 2. Among Grade 8 students, $60\%$ are girls. 3. Percentage of Grade 8 girls among all students is $60\% \text{ of } 40\% = \frac{60}{100} \times 40\% = 24\%$. Answer (i): $24\%$ **Part (ii)** 1. Calculate $24\%$ of the total number of students ($160$). 2. Number of Grade 8 girls = $\frac{24}{100} \times 160 = 38.4$, which rounds to 38 students. Answer (ii): $38$ students **Answer:** 24% > Common mistake: Adding $40\%$ and $60\%$ instead of multiplying percentages. ### Question 9 *3 marks · Short answer* 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** 1. Let the cost price of 1 pencil be $c$ and the selling price of 1 pencil be $s$. 2. Given that the selling price of 3 pencils equals the cost price of 5 pencils, so $3s = 5c$. 3. Therefore, $s = \frac{5}{3}c$, which means the selling price is greater than the cost price, resulting in a profit. 4. Profit percentage = $\frac{s - c}{c} \times 100 = \frac{\frac{5}{3}c - c}{c} \times 100 = \frac{2}{3} \times 100 = 66.67\%$. **Answer:** Profit of $66.67\%$ > Common mistake: Taking the ratio of numbers directly as percentage without considering cost and selling price per unit. ### Question 10 *3 marks · Short answer* 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** 1. Let the initial bus fare be $100$. 2. After a $3\%$ increase last year, the fare becomes $100 + 3 = 103$. 3. After a $4\%$ increase this year, the fare increases by $4\%$ of $103$, which is $0.04 \times 103 = 4.12$. 4. The new fare is $103 + 4.12 = 107.12$. 5. The overall percentage increase is $107.12 - 100 = 7.12\%$. **Answer:** $7.12\%$ > Common mistake: Simply adding the two percentage increases to get $7\%$. ### Question 11 *3 marks · Short answer* If the length of a rectangle is increased by 10% and the area is unchanged, by what percentage (exactly) does the breadth decrease by? **Solution** 1. Let the original length be $L$ and the original breadth be $B$. The original area is $A = L \times B$. 2. When the length is increased by $10\%$, the new length becomes $1.1L$. 3. Let the new breadth be $B'$. Since the area remains unchanged, $1.1L \times B' = L \times B$, which gives $B' = \frac{B}{1.1} = \frac{10}{11}B$. 4. The decrease in breadth is $B - \frac{10}{11}B = \frac{1}{11}B$. 5. The percentage decrease in breadth is $\frac{\frac{1}{11}B}{B} \times 100 = \frac{100}{11}\% = 9.09\%$. **Answer:** $9.09\%$ (or $\frac{100}{11}\%$) > Common mistake: Students often assume the breadth decreases by 10% as well, which is incorrect. ### Question 12 *3 marks · Short answer* The percentage of ingredients in a 65 g chips packet is shown in the picture. Find out the weight each ingredient makes up in this packet. **Solution** 1. Given the total weight of the chips packet is $65\text{ g}$. 2. From the nutritional table in the figure, the percentage of potato is $70\%$, vegetable oil is $24\%$, salt is $3\%$, and spices are $3\%$. 3. Weight of potato = $70\% \text{ of } 65 = \frac{70}{100} \times 65 = 45.5\text{ g}$. 4. Weight of vegetable oil = $24\% \text{ of } 65 = \frac{24}{100} \times 65 = 15.6\text{ g}$. 5. Weight of salt = $3\% \text{ of } 65 = \frac{3}{100} \times 65 = 1.95\text{ g}$. 6. Weight of spices = $3\% \text{ of } 65 = \frac{3}{100} \times 65 = 1.95\text{ g}$. **Answer:** Potato: $45.5\text{ g}$, Vegetable oil: $15.6\text{ g}$, Salt: $1.95\text{ g}$, Spices: $1.95\text{ g}$ > Common mistake: Arithmetic errors while multiplying decimals with 65. ### Question 13 *4 marks · Case-based* Three shops sell the same items at the same price. The shops offer deals as follows: Shop A: “Buy 1 and get 1 free” Shop B: “Buy 2 and get 1 free” Shop C: “Buy 3 and get 1 free” Answer the following: (i) If the price of one item is ₹100, what is the effective price per item in each shop? Arrange the shops from cheapest to costliest. (ii) For each shop, calculate the percentage discount on the items. [Hint: Compare the free items to the total items you receive.] (iii) Suppose you need 4 items. Which shop would you choose? Why? **Part (i)** 1. In Shop A ('Buy 1 and get 1 free'), you pay for 1 item and get 2 items. Effective price per item = $\frac{100}{2} = \text{₹}50$. 2. In Shop B ('Buy 2 and get 1 free'), you pay for 2 items (₹200) and get 3 items. Effective price per item = $\frac{200}{3} = \text{₹}66.67$. 3. In Shop C ('Buy 3 and get 1 free'), you pay for 3 items (₹300) and get 4 items. Effective price per item = $\frac{300}{4} = \text{₹}75$. 4. Arranging from cheapest to costliest: Shop A, Shop B, Shop C. Answer (i): Shop A: ₹50, Shop B: ₹66.67, Shop C: ₹75. Order: Shop A, Shop B, Shop C. **Part (ii)** 1. For Shop A, 1 free item out of 2 total items: percentage discount = $\frac{1}{2} \times 100 = 50\%.$ 2. For Shop B, 1 free item out of 3 total items: percentage discount = $\frac{1}{3} \times 100 = 33.33\%.$ 3. For Shop C, 1 free item out of 4 total items: percentage discount = $\frac{1}{4} \times 100 = 25\%.$ Answer (ii): Shop A: 50%, Shop B: 33.33%, Shop C: 25% **Part (iii)** 1. If you need 4 items: in Shop A, you buy 2 and get 2 free, paying for 2 items ($2 \times 100 = \text{₹}200$). 2. In Shop B, you buy 3 and get 1 free, paying for 3 items ($3 \times 100 = \text{₹}300$). 3. In Shop C, you buy 3 and get 1 free, paying for 3 items ($3 \times 100 = \text{₹}300$). 4. Therefore, choose Shop A as it costs the least for 4 items. Answer (iii): Shop A, because it offers the lowest total cost for 4 items. **Answer:** Solved in parts. > Common mistake: Dividing the price by the number of paid items instead of total items received when calculating effective price. ### Question 14 *3 marks · Short answer* 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** 1. In a room of 100 people, $99\%$ being left-handed means there are 99 left-handed people and 1 right-handed person. 2. Let $x$ left-handed people leave the room. The number of left-handed people becomes $99 - x$, and the total number of people becomes $100 - x$. 3. We want the new percentage of left-handed people to be $98\%$, so $\frac{99 - x}{100 - x} = \frac{98}{100}$. 4. Cross-multiplying gives $100(99 - x) = 98(100 - x)$, which simplifies to $9900 - 100x = 9800 - 98x$. 5. Rearranging terms gives $100 = 2x$, so $x = 50$. **Answer:** 50 left-handed people > Common mistake: Simply subtracting 1% from 99% and concluding 1 person needs to leave. ### Question 15 *3 marks · Short answer* Look at the following graph. 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** 1. Examine statement (i): The graph shows the highest percentage for twenties (37% male, 26% female), so it is valid. 2. Examine statement (ii): Looking at the bars, female percentages are consistently lower than male percentages across all age groups, so it is valid. 3. Examine statement (iii): The graph only shows percentages of computer literacy within each age group, not absolute population counts, so this statement cannot be determined (invalid). 4. Examine statement (iv): For thirties, male percentage is 25%, which is exactly a quarter, not 'more than a quarter' (invalid). 5. Examine statement (v): For seniors (60 and above), the percentages are 2% and 4%, which are both less than 1 in 10 (10%), so it is valid. 6. Examine statement (vi): For twenties, male is 37% and female is 26%, neither of which is half (50%), so it is invalid. **Answer:** Statements (i), (ii), and (v) are valid. > Common mistake: Confusing percentages within an age group with total population numbers. ## Frequently asked questions ### How many questions are there in NCERT Solutions for Class 8 Maths Chapter 8 Fractions in Disguise? The Chapter has 53 questions in the Figure it Out section of the new NCERT book for the 2026-27 session. You can access the complete step-by-step solutions for all these questions in SwaVid's free PDF available on this page only. ### What topics do the questions cover in Class 8 Maths Chapter 8 Fractions in Disguise? The questions cover calculating percentages of a total, comparing percentages and fractions without calculation, and comparing percentages of different time units. Other topics include comparing percentages of quantities without full computation, comparison of loan interests, simple and compound interest comparison, compound growth, population increase, and discount calculation. ### Which question types are included in the Figure it Out section of this chapter? The 53 questions in this chapter include case-based problems, multiple-choice questions (MCQs), and short answer (SA) questions. SwaVid provides detailed explanations for each of these question types in the free PDF found on this page only. ### Which is the hardest question type in Class 8 Maths Chapter 8 and how should I approach it? Comparison of simple and compound interest along with compound growth or population increase problems are generally considered the hardest. To approach them, clearly identify the principal, rate, and time, and apply the relevant formulas step by step as shown in SwaVid's free PDF on this page. ### How do I write answers for full marks in Class 8 Maths Chapter 8 exams? To get full marks, write down all given values clearly, show the formula used like $A = P(1 + \frac{r}{100})^t$, and write the final units properly. Following the method in SwaVid's free PDF solutions available on this page only will help you structure your answers correctly. ## Related pages - [Class 8 Maths chapters](https://www.swavid.com/maths/class/8) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student