Chapter 1 Fractions in Disguise

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NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 1 Fractions in Disguise

The first chapter, titled "Fractions in Disguise", introduces students to the idea of percentages, which can be defined as fractions with denominators of 100. This allows students to use these three number types (fractions, decimals and percentages) for everyday problem solving (proportionality, increasing/decreasing amounts, and money).

By having a strong command over these topics, students are able to develop good numerical reasoning skills which will allow them to quickly estimate mentally and compare items in real-world situations, including how to analyze discounts or health statistics. In addition, students can use NCERT Solutions for Class 8 as a source of step-by-step instructions and alternative solutions to help reinforce their understanding of content and apply it successfully.

1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 1 Solutions

Students can learn using percentages by reading the detailed PDF of this chapter on Percentage Basics. These solutions teaches you how to compare and convert number types to solve real-world problems that are more complex than they may seem. Download the free PDF of the NCERT Solutions from below:

Chapter 1 : Fractions in Disguise

2.0Key Concepts covered in Class 8 Maths : Fractions in Disguise

This chapter shifts from basic fractional parts to the universal language of percentages, providing tools to analyze data and financial growth effectively.

Understanding Percentages: Learning that "per cent" means "by the hundred" and represents a proportion out of a whole of 100.
Conversions (The FDP Trio): Mastering the fluid movement between Fractions, Decimals, and Percentages for efficient calculation.
Proportional Comparison: Show you can compare two numbers (e.g. exam scores or the ratios of recipes) using %s when they have different bases.
Percentage Change: Calculating the rate of change through percentage increase or decrease relative to an original base value.
Financial Applications: Applying percentage concepts to understand profit, loss, and the mechanics of compounding interest in transactions.

3.0NCERT Solutions for Class 8 Maths Chapter 1 : All Exercises

Exercise 1.1 The purpose of this exercise is to help the student convert fractions and decimals to a percentage of a quantity. They do this by seeing an example of how to express multiple values as part of 100 (one hundred), then solving problems related to mixing proportions and identifying basic % (percentage) values in real situations.

Exercise 1.2 Activities in this section include determining the percentage of a given amount (quantity). Example of determining the shaded portion of an object; Analyzing a sample of survey data and calculating the proportional amount of a certain item based upon a particular percentage and also the total number.

Exercise 1.3 The exercise covers increases and decreases (percentages). Students will calculate the rate of change (rise or fall) of a variety of items like price, population, measurement so students will be able to interpret both how much an item has increased/decreased when compared to its original base.

Exercise 1.4 Focused on financial transactions, this section teaches profit and loss calculations. Students determine percentage gains or losses on sales, analyze overhead costs, and solve problems related to buying and selling various commercial items.

Exercise 1.5 Students work with the concept of simple interest and basic growth. They solve problems involving the amount of interest accrued over time; finding the principal; and how the fixed rates affect certain total amounts, including several savings.

Exercise 1.6 This section introduces the power of compounding. Questions guide students through multi-step calculations where interest is added back to the principal, demonstrating how money or populations grow exponentially over several consecutive time periods.

Exercise 1.7 The last task includes a variety of multi-step problems and real-world problems. Students will analyze health statistics and nutrition labels using the various types of percentage calculations they have learned to be able to compare data and make logical conclusions based on complex data.

4.0Detailed Class 8 Maths Chapter 1 Fractions in Disguise - NCERT Solutions

Figure it out-01

1. 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) $\frac{5}{11}$

Sol. (i) $\frac{3}{5} = \frac{3}{5} \times 100% = 60%$ (ii) $\frac{7}{14} = \frac{7}{14} \times 100% = 50%$ (iii) $\frac{9}{20} = \frac{9}{20} \times 100% = 45%$ (iv) $\frac{72}{150} = \frac{72}{150} \times 100% = 48%$ (v) $\frac{1}{3} = \frac{1}{3} \times 100% = 33 \frac{1}{3} %$ (vi) $\frac{5}{11} = \frac{5}{11} \times 100% = \frac{500}{11} % = 45 \frac{5}{11} %$

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

Sol. (iv) 60% Percentage of white marbles $= \frac{15}{25} \times 100% = 60%$

3. In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking? Sol. Percentage of students coming by walking $= \frac{15}{80} \times 100% = 18.75%$

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 following picture. Match (among the given options) what percentage of the race each of them has approximately completed.

Sol. $A = 38%$ $B = 55%$ $C = 72%$ D = 93%

5. A pair of quantities is shown below. Identify and write appropriate symbols '>', '<', '=' in the blanks. Try to do it without calculations. (i) $50%$ $____$ 5% (ii) $\frac{5}{10} - 50%$ (iii) $\frac{3}{11} - 61%$ (iv) $30% - \frac{1}{3}$

Sol. (i) $50% > 5%$ (ii) Since, $\frac{5}{10} = \frac{1}{2} \Rightarrow \frac{1}{2} \times 100 = 50%$

$∴ \frac{5}{10} = 50%$

(iii) Since $\frac{3}{11} \times 100$

$\Rightarrow \frac{300}{11} = 27.27% ∴ \frac{3}{11} < 61%$

(iv) Since, $\frac{1}{3} \times 100 = 33.33%$

$∴ 30% < \frac{1}{3}$

Figure it out-02

1. Find the missing numbers. The first problem has been worked out. (i)

(ii)

(iii)

Sol. (i) 20%; 60 (ii)

Ten Parts repesent 100% One part represnts 10%

Total $= 90$ $60%$ of $90 = \frac{60}{100} \times 90 = 54$ So, 10%; 54 (iii)

\frac{1}{4}

100% = \frac{1}{4} \times 100 = 25%

\frac{3}{4}

140 = \frac{3}{4} \times 140 = 105

So, 25%; 105

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

Sol. (i) $25%$ of $160 = \frac{25}{100} \times 160 = 40$

(ii)

16%

250 = \frac{16}{100} \times 250 = 40

\frac{16}{100} = \frac{4}{25}

(iii)

62%

360 = \frac{62}{100} \times 360 = 223.2

62% = \frac{62}{100} = \frac{31}{50}

(iv)

140%

40 = \frac{140}{100} \times 40 = 56

140% = \frac{140}{100} = \frac{7}{5}

1 hour = 60 minutes (v)

1%

of 1 hour

= 1%

of 60 min

= 1%

of 3600 sec .

(\frac{3}{300} \times 100% = 1%) = 36

sec . (vi)

7%

10 kg = \frac{7}{100} \times 10 kg = 0.7 kg

3. 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?

Sol. Quantity of orange paint $= 60 ml$ Quantity of red paint $= \frac{3}{4} \times 60 ml = 45 ml$

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

Sol. (i) $50%$ of $510 < 50%$ of $515 (510 < 515)$ (ii) $37%$ of $148 < 73%$ of $148 (37 < 73)$ (iii) $29%$ of $43 < 92%$ of 110

$(29 \times 92 < 43 \times 110)$

(iv) $30%$ of $40 < 40%$ of 50

$(30 < 40; 40 < 50)$

(v) $45%$ of $200 > 10%$ of $490 (90 > 49)$ (vi) $30%$ of $80 > 24%$ of 64 ( $24 > 15.36$ )

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.

Sol. (i) $(30%$ of k is 70 $) \times 2$ $= 60%$ of $k = 140$ ( $30%$ of k is 70 ) $\times 3$ $= 90%$ of $k = 210$ $(30%$ of k is 70 $) \times 4$ $= 120%$ of $k = 280$ (ii) $(100%$ of m is 215 $) \div 10$ $10%$ of m is 21.5 $(100%$ of m is 215 $) \div 100$ $1%$ of m is 2.15 $(1%$ of $m$ is 2.15 $) \times 6$ $= 6%$ of m is 12.9 (iii) $(90%$ of $n$ is 270 $) \div 10$ $9%$ of $n$ is 27 ( $9%$ of $n$ is 27 ) $\times 2$ $18%$ of $n$ is 54 . (iv) $25%$ of $x$ is $50.50%$ of $x$ is $____$ , $75%$ of x is $____$ , $200%$ of x is $____$ . $80%$ of $y$ is $160.10%$ of $y$ is $____$ , $5%$ of $y$ is $____$ , $100% y$ is $____$ .

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

Sol. (i) $\frac{x}{100} \times 300 = 3$ or $x = 1$ (ii) $40 = \frac{40}{100} \times 4$ or $x = 1.6$ (iii) $40 = \frac{80}{100} \times x$ or $x = \frac{40 \times 100}{80} = 5 0^{\circ}$

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

Sol. Yes, $10%$ of a day $= \frac{10}{100} \times 24 hrs = 2.4 hr$ $1%$ of week $= \frac{10}{100} \times (24 \times 7) hrs = 1.68 hrs$ Other questions: Is $10%$ of a day $> 1%$ of a week. Is 10% of a month (30 days) < 50% of a week?

Is 50% of a dozen (12) > 10% of a score (20)? Is 80% of a century < 45% of a double century?

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 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?

Sol. $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{98}{99}, \frac{99}{100}$ , $50%, 66 \frac{2}{3} %, 75%, 80%, \dots 98 \frac{98}{99} %, 99%$ . It follows the pattern $(\frac{n}{n + 1}) < 100$ where n is the day number.

As n increases, the percentage gets closer and closer to $100%$ , but never quite reaches it.

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? Sol. ( $20%$ work in 18 days) $\times 5$ $= 100%$ work in 90 days. The work will be completed in 90 days. Necessary Assumptions

Weather conditions might change.
Workers might get tired over time. This reduces their efficiency.
Some workers might take leave or breaks.

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 last?

Sol. Warm-up time $= 10%$ of 90 min $\frac{10}{100} \times 90 = 9 min$ Play time $= 80%$ of 90 min $\frac{80}{100} \times 90 = 72 min$ Cool down time $= 10%$ of $90 min = 9 min$ .

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.

Sol. World population as of Jan 1, 2026 $= 8.3$ billion $90%$ of 8.3 billion $= \frac{90}{100} \times 8.3$ billion $= 7.47$ Hence, 7.47 billion people live in the northern hemisphere.

12. 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?

Sol. (i) Proportion remains the same. (ii) Rava $= \frac{40}{100} \times 2 kg = 0.8 kg$

$Sugar = \frac{40}{100} \times 2 kg = 0.8 kg Ghee = \frac{20}{100} \times 2 kg = 0.4 kg$

Figure it out-03

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

Sol. Profit $= ₹110 - ₹75 = ₹35$ Profit % = $\frac{35}{75} \times 100$ $= 0.4667 \times 100 = 46.67%$

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?

Sol. Cost of material $=$ ₹475 Profit $= 50%$ of 475 $= \frac{50}{100} \times 475 = ₹237.50$ Sale price $= ₹475 + ₹237.50 = ₹712.50$

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

Sol. Let the total expenditure (cost) be x Then profit $= ₹ \frac{25 x}{100}$ or $₹0.25 x$ $∴$ Revenue $= x + 0.25 x = 2.5$ crores $\Rightarrow 1.25 x = 2.5$ $\Rightarrow x = \frac{2.5}{1.25} = \frac{250}{125} = 2$ ∴ The total expenditure of the company is ₹2 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?

Sol. Marked price $= ₹300$ Discount $= ₹ \frac{25}{100} \times 300 = ₹75$ Sale price = ₹300 - ₹75 = ₹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) $160.66%$

Sol. Increase in price $= ₹100 - ₹60 = ₹40$ Increase $% = \frac{40}{60} \times 100% = 66 \frac{2}{3} %$ or 66.66%

6. 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?

Sol. Let the marked price of the car be ₹x Discount $= \frac{15}{100} x = 0.15 x$ Sale price $= x - 0.15 x = 0.85 x$ Now $0.85 x = 4, 40, 000$ $x = \frac{4 , 40 , 000}{0.85} = 517647$ Marked price of the car is ₹5,17,647.

7. 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?

Sol. Vote $%$ (winner) $= \times 100% = 31.25%$ and $100 \div 31.25 = 3.2$ ∴ In all, there were at least 4 candidates. This means at least 3 more candidates.

8. 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.) Sol. Increase in price $= ₹42 - ₹38 = ₹4$ Rate of inflation $= \frac{4}{38} \times 100% = 10.52%$

9. A number increased by $20%$ becomes 90. What is the number? Sol. 120% of a number is 90 $∴ 1%$ of the number is $\frac{90}{120}$ $100%$ of the number $= \frac{90}{120} \times 100 = 75$ Therefore, the number was 75.

10. 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. Sol. SP of $1^{st}$ buffalo $= ₹80, 000$ Profit% = 5% $∴ CP = ₹ \frac{100}{100 + 5} \times 80, 000 = ₹76, 190$ SP of $2^{nd}$ buffalo $= ₹80, 000$ Loss % $= 10$ % $∴ CP = ₹ \frac{100}{100 - 10} \times 80, 000 = ₹88, 889$ Total CP = ₹76,190 + ₹88,889 = ₹1,65,079 Total SP = ₹80,000 + ₹80,000 = ₹1,60,000 Loss $=$ ₹1,65,079 - ₹1,60,000 = ₹5,079 $Loss % = \frac{5079}{165079} \times 100% = 3%$

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

Sol. (iv) $p \times 1.05$ Population 10 years ago = p Increase in population $= \frac{5}{100} P = 0.05 p$ $∴$ Current population $= p + 0.05 p = 1.05 p$

12. Which of the following statements(s) means 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. Sol. Statement: The demand for cameras has fallen by 85% in last decade. Only (iii) means the same.

Figure it out-04

1. The Bank of Yahapur offers an interest rate of $10%$ p.a. Compare the amount one earns by depositing ₹20,000 for a period of 2 years with and without compounding annually. Sol. Without compounding Amount $= P (1 + \frac{r t}{100})$ $= 20, 000 \times (1 + \frac{10 \times 2}{100}) = 20, 000 (1 + 0.20)$ $= 20, 000 \times 1.20 = ₹24, 000$ With compounding Amount $= P (1 + r)^{t} = 20, 000 \times (1 + \frac{10}{100})^{2}$ $= 20, 000 \times 1.21 = ₹24, 200$ Comparison: Without compounding = ₹24,000 With compounding $=$ ₹24,200 Difference $=$ ₹24,200 - ₹24,000 = ₹200 Hence, with compounding, one gets ₹200 more than without compounding.

2. The Bank of Wahapur offers an interest rate of $5%$ p.a. Compare the amount one earns by depositing ₹20,000 for a period of 4 years with and without compounding annually. Sol. $P = ₹20, 000; t = 4, r = 5$ $S I = \frac{20000 \times 5 \times 4}{100} = ₹4, 000$ $A = ₹20, 000 \times (1 + \frac{5}{100})^{4}$ $= ₹20, 000 \times (\frac{21}{20})^{4}$ = ₹ $20, 000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} = ₹24310.13$ $CI = ₹24, 310.13 - ₹20, 000 = ₹4, 310.13$ SI (without compounding) < CI (with compounding)

3. Do you observe anything interesting in the solutions of the two questions above? Share and discuss. Sol. If the rate percent and time is same, the interest received with compounding is more than the interest received without compounding.

Figure it out-05

1. Jasmine invests an 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) $p \times 6 \times 4$ (ii) $p \times 0.6 \times 4$ (iii) $p \times \frac{0.6}{100} \times 4$ (iv) $p \times \frac{0.06}{100} \times 4$ (v) $p \times 1.6 \times 4$ (vi) $p \times 1.06 \times 4$ (vii) $p + (p \times 0.06 \times 4)$

Sol. $P = p, R = 6, T = 4$ Amount $= p + I = p + \frac{p \times 6 \times 4}{100} = p + (p \times 0.06 \times 4)$ $= p + 0.24 p = 1.24 p$ Hence (vii) is correct.

2. 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 were compounded? Sol. Without compounding $P = ₹50, 000; R = 7% pa; T = 3$ years $I = ₹ \frac{50 , 000 \times 7 \times 3}{100} = ₹10, 500$ Amount $= ₹50, 000 + ₹10, 500 = ₹60, 500$ With compounding $A = 50, 000 (1.07)^{3} = 61252.15$ Difference $= 61252.15 - 50, 000 = 11252.15$ Extra interest $= 11252.15 - 10500 =$ ₹752.15

3. 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? Sol. Interest (Giridhar) $= (1 + \frac{10}{100})^{3} = ₹4, 500$ For Raghava with compounding $A = 12, 500 \times (1 + \frac{10}{100})^{3} = 12, 500 \times \frac{1331}{1000}$ = ₹16637.5 (Raghava) Interest $=$ ₹16,637.5 - ₹12,500 $=$ ₹4137.50 Difference $=$ ₹4500 - ₹4137.50 = ₹362.50 Giridhar pays ₹362.5 more than Raghava.

4. Consider an amount of ₹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?

Sol. ₹1000 becomes ₹2,000 Interest = ₹1000 Without compounding $1000 = \frac{1000 \times 10 \times t}{100}$ $t = 10$ years With compounding $1000 (1 + \frac{10}{100})^{n} = 2000$ $(1.1)^{n} = 2$ This can be done by hit and trial $1. 1^{2} = 1.21$ $1. 1^{3} = 1.331$ $1. 1^{4} = 1.4641$ $1. 1^{5} = 1.6$ $1. 1^{6} = 1.77$ $1. 1^{7} = 1.94$ $1. 1^{8} = 2.14$ 1.94 < 2 < 2.14

Time would be between 7 and 8 years (The nearest answer is 7.2 years)

5. 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? Sol. Population after 3 years $= 1.5 \times (1 + \frac{3}{100})^{3}$ crores $= 1.5 \times (1.03)^{3}$ crores $= 1.639$ crores

6. 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. Sol. Total bacteria after 2 hours $= 5, 06, 000 \times (1 + \frac{25}{100})^{2}$ $= 506000 \times (1.025)^{2} = 5, 31, 616$

Figure it out-06

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?

Sol. Population in $2000 = 50$ lakhs Population in 2025 $= 50$ lakhs $+ 250%$ of 50 lakhs $= 50$ lakhs $+ 2.5 \times 50$ lakhs $= 50$ lakhs + 125 lakhs $= 175$ lakhs or 1.75 crores

2. 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 Germany 83 million $13%8%$ $13%8%$ 18% 18% 10% 1% 35% 2% 2% Sol. Germany $= \frac{83}{8200} \times 100% \sim 1%$ India $= \frac{1.46}{8.20} \times 100% \sim 18%$ Bangladesh $= \frac{175}{8200} \times 100% \sim 2%$ USA $= \frac{347}{8200} \times 100% \sim 4%$

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? of the worldwide population.

Hint: Writing these numbers in the standard form and estimating can help. (i) $8250 + 18$ (ii) $8250 + 1800$ (iii) $8250 + \frac{18}{100}$ (iv) $8250 \times 18$ (v) $8250 \times 1.18$ (vi) $8250 + 8250 \times 0.18$ (vii) $1.8 \times 8250$

Sol. Price of mobile phone $= ₹8, 250$ GST @ $18% = \frac{8250 \times 18}{100}$ $= ₹8250 \times 0.18$ Total cost $= ₹ (8250 + 8250 \times 0.18)$ Options (v) and (vi) are correct.

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 $- 2%$ , and Month 3 change was $- 3%$ . Which of the following statements are true? The initial population is p. (i) The population after three months was $p \times 0.05 \times 0.02 \times 0.03$ . (ii) The population after three months was $p \times 1.05 \times 0.98 \times 0.97$ . (iii) The population after three months was $p + 0.05 - 0.02 - 0.03$ . (iv) The population after three months was p . (v) The population after three months was more than p. (vi) The population after three months was less than p. Sol. Population after 3 months $= p (1 + \frac{5}{100}) (1 - \frac{2}{100}) (1 - \frac{3}{100})$ $= p \times 1.05 \times 0.98 \times 0.97 = 0.99813 p$ Option (ii) and (vi) are correct.

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 loss? Give reasons for your answer. Sol. Let CP be ₹100. $P % = 35%$ $P = ₹35$ MP = ₹135 Discount $= 30%$ of $135 = ₹40.50$ New MP $=$ ₹ $135 -$ ₹ $40.50 =$ ₹ 94.50 New MP < CP ∴ the shopkeeper makes a loss. Reason: Although he initially added a 35% profit margin, the $30%$ discount is calculated on the increased selling price (not the cost price), which results in a larger absolute discount amount that exceeds the original profit.

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

Sol. Total area $= 8 \times 8 = 64$ sq. units And area of $E = 8$ sq. units ∴ Required $% = \frac{8}{64} \times 100% = 12.5%$

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 ?

Sol. $5%$ of $40 = \frac{5}{100} \times 40 = 2$ $40%$ of $5 = \frac{40}{100} \times 5 = 2$ $25%$ of $12 = \frac{25}{100} \times 12 = 3$ $12%$ of $25 = \frac{12}{100} \times 25 = 3$ $15%$ of $60 = \frac{15}{100} \times 60 = 9$ $60%$ of $15 = \frac{60}{100} \times 15 = 9$ Justification $x %$ of $y = y %$ of x Since Multiplication is commutative $\frac{x}{100} \times y = \frac{y}{100} \times x$

8. 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? Sol. Let no. of students be 100. Then, no. of students of grade 8 $= \frac{40}{100} \times 100 = 40$ No. of students of grade $9 = 100 - 40 = 60$ (i) No. of grade 8 girls $= \frac{60}{100} \times 40 = 24%$ (ii) $100 : 24 :: 160 : x$ $100 x = 24 \times 160$ $x = 38.4$ No. of grade 8 girls is 38.4

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?

Sol. SP of 3 pencils = CP of 5 pencils Let SP of 3 pencils = CP of 5 pencils $= 3 \times 5 = 15$ Then SP = ₹5; CP = ₹3 Profit $= ₹2$ Profit $% = \frac{2}{3} \times 100% = 66 \frac{2}{3} % (\sim 67%)$

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?

Sol. Let the bus fare 2 years ago be ₹ 100 Present bus fare $= ₹100 \times 1.03 \times 1.04 = ₹107.12$ Increase = ₹7.12 Increase $% = \frac{7.12}{100} \times 100% = 7.12%$

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

Sol.

$l (l + \frac{10}{100} l) (b - \frac{x}{100} b) = A$

$1.1 l \times \frac{( 100 - x ) b}{100} = A$ $1.1 l \times \frac{100 - x}{100} b = l \times b$ $1.1 l \times (\frac{100 - x}{100}) = 1 \Rightarrow 100 - x = \frac{1 \times 100}{1.1}$ $x = 100 - \frac{1000}{11} \Rightarrow x = \frac{1100 - 1000}{11}$ $\frac{100}{11} = 9 \frac{1}{11}$ Breadth is decreased by $9 \frac{1}{11} %$ .

12. 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.

Sol. Potato $= \frac{70}{100} \times 65 g = 45.5 g$ Veg Oil $= \frac{24}{100} \times 65 g = 15.6 g$ Salt $= 390$ of $65 g = 1.95 g$ Spice $= \frac{3}{100} \times 65 g = 1.95 g$ Verification: $45.5 + 15.6 + 1.95 + 1.95 = 65.05 g$

13. 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?

Sol. (i) Effective price per item at shop A $= ₹ \frac{100}{2} = ₹50$ Effective price per item at shop B $= ₹ \frac{200}{3} = ₹66 \frac{2}{3}$ Effective price per item at shop C $= ₹ \frac{300}{4} = ₹75$ Cheapest to costliest: Shop A< Shop B < Shop C (ii) Discount at shop $A = \frac{1}{2} \times 100% = 50%$

Discount at shop B $= \frac{1}{3} \times 100%$ $= \frac{100}{3} % = 33.3%$ Discount at shop C $= \frac{1}{4} \times 100% = 25%$ (iii) To buy 4 items, we choose shop A. Pay for 2 , and the other 2 are free.

14. In a room of 100 people, $99%$ are lefthanded. How many left-handed people have to leave the room to bring that percentage down to $98%$ ? Sol. No. of people in room $= 100$ No. of left handers $= 99%$ of $100 = 99$ No. of right-handers = 1 Let x people leave. No. of left-handers is 98%. and no. of right-handers is $2%$ . Then $\frac{99 - x}{100 - x} = \frac{98}{100}$ $\Rightarrow 100 (99 - x) = 98 (100 - x)$ $\Rightarrow 9900 - 100 x = 9800 - 98 x$ $\Rightarrow 9900 - 9800 = 100 x - 98 x$ $\Rightarrow 2 x = 100$ $\Rightarrow x = 50$ $∴ 50$ left-handers must leave the room.

15. Look at the following graph.

Ability to use computer by age and gender (2023)

The ability to use computers is highest among those is their twenties and teenagers.

Based on the graph, which of the following statements are valid? (i) People in their twenties are the most computer-literate among all age groups. (ii) Women lag 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.

Sol. (i) True (ii) True (iii) True (iv) False (v) True (vi) False

5.0Key Features and Benefits of Class 8 Maths Ganith Prakash 2 Chapter 1

Diverse Problem-Solving Methods: The chapter encourages students to explore multiple ways to solve a single problem—such as proportional reasoning, decimal multiplication, or bar models—enriching their mathematical flexibility.
Emphasis on Mental Estimation: By practicing "free-hand" and mental computations, students improve their number sense and learn to make quick, accurate estimates without relying on a calculator.
Real-World Contextual Learning: Lessons are grounded in practical examples like food label reading (KYC - Know Your Contents), global physical activity rates, and historical interest rates from Kautilya’s Arthaśhāstra.
Visual Representation Tools: The use of bar models and rough diagrams is highlighted as a key feature to help students better visualize and understand complex word problems.
Clarification of Complex Concepts: The chapter explicitly addresses potential confusion, such as why percentages can exceed 100% and why certain comparisons between percentages require a common base.
Preparation for Competitive Success: These foundational skills in percentages and financial math are highly relevant for the CBSE curriculum and provide a significant advantage in Olympiads and other competitive examinations.

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Frequently Asked Questions

What concepts are explained in NCERT Solutions for Class 8 Maths Chapter 1 Fractions in Disguise?

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Why is Chapter 1 Fractions in Disguise important for Class 8 students?

What are equivalent fractions explained in this chapter?

How can students improve their fraction-solving skills?

Do NCERT Solutions for Class 8 include answers to all textbook exercises?

How are fractions used in real life?

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NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 1 Fractions in Disguise

1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 1 Solutions

2.0Key Concepts covered in Class 8 Maths : Fractions in Disguise

3.0NCERT Solutions for Class 8 Maths Chapter 1 : All Exercises

4.0Detailed Class 8 Maths Chapter 1 Fractions in Disguise - NCERT Solutions

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5.0Key Features and Benefits of Class 8 Maths Ganith Prakash 2 Chapter 1