CBSE Class 8 Mathematics Part II Chapter 1 Questions and Answers

Class 8 Maths

Fractions In Disguise Class 8 Maths Ganita Prakash Part 2 Chapter 1 NCERT Solutions

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Figure it Out (Page 3)

1. Express the following fractions as percentages.

Solution:

(i) 3/5 = (3/5) × 100% = 3 × 20% = 60%.

(ii) 7/4 = (7/4) × 100% = 7 × 25% = 175%.

(iii) 9/20 = (9/20) × 100% = 9 × 5% = 45%.

(iv) 72/150 = (72/150) × 100% = (72/15) × 10% = (24/5) × 10% = 24 × 2% = 48%.

(v) 1/3 = (1/3) × 100% = (100/3)% = 33 1/3%.

(vi) 5/11 = (5/11) × 100% = (500/11)% = 45 5/11%.

2. Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?

Solution:
Total marbles = 25
White marbles = 15
Percentage of white marbles = (15/25) × 100% = 15 × 4% = 60%.
Therefore, (iv) 60% is the correct answer.

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

Solution:
Total students = 80
Students coming to school by walking = 15
Percentage of students coming to school by walking = (15/80) × 100% = (15/4) × 5% = (75/4)% = 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.

Solution:
A is clearly well before halfway → 38%
B is approximately at the halfway mark → 55%
C is around three-fourths of the distance → 72%
D is very near the finish → 93%

5. Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the blanks. Try to do it without calculations.

Solution:
(i) 50% > 5%
Clearly, 50% is greater than 5%.

(ii) 5/10 ___ 50%
(5/10) × 100% = 50%
∴ 5/10 = 50%

(iii) 3/11 ___ 61%
(3/11) × 100% = (300/11)% = 27.27%
∴ 3/11 < 61%

(iv) 30% ____ 1/3
(1/3) × 100% = (100/3)% = 33.3%
∴ 30% < 1/3

Figure it Out (Page 12 – 14)

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

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

Solution:

2. Find the value of the following and also draw their bar models.

Solution:
(i) 25% of 160 = (25/100) × 160 = (1/4) × 160 = 160/4 = 40.
(ii) 16% of 250 = (16/100) × 250 = (16/10) × 25 = (16/2) × 5 = 8 × 5 = 40.
(iii) 62% of 360 = (62/100) × 360 = (62/10) × 36 = (62/5) × 18 = 1116/5 = 223.2.
(iv) 140% of 40 = (140/100) × 40 = (14/10) × 40 = 14 × 4 = 56.
(v) 1% of 1 hour = (1/100) × 60 min = (1/100) × 3600 sec = 36 sec.
(vi) 7% of 10 kg = (7/100) × 10000 g = 7 × 100 g = 700 g.

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:
Quantity of orange paint = 60 ml
Quantity of red paint = (3/4) × 60 ml = 3 × 15 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

Solution:
(i) Since 510 < 515 ∴ 50% of 510 < 50% of 515
(ii) Since 37% < 73% ∴ 37% of 148 < 73% of 148
(iii) 29% of 43 = (29/100) × 43 = 12.47; 92% of 110 = (92/100) × 110 = 101.2. Since 12.47 < 101.2 ∴ 29% of 43 < 92% of 110
(iv) 30% of 40 = 12; 40% of 50 = 20. Since 12 < 20 ∴ 30% of 40 < 40% of 50
(v) 45% of 200 = 90; 10% of 490 = 49. Since 90 > 49 ∴ 45% of 200 > 10% of 490
(vi) 30% of 80 = 24; 24% of 64 = 15.36. Since 24 > 15.36 ∴ 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 is 70 → 60% of k = 140, 90% of k = 210, 120% of k = 280.
(ii) 100% of m is 215 (m = 215) → 10% of m = 21.5, 1% of m = 2.15, 6% of m = 12.9.
(iii) 90% of n is 270 (n = 300) → 9% of n = 27, 18% of n = 54, 100% of n = 300.
(iv) Example challenge questions:
- 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% of y is ___.

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

Solution:
(i) 3 is 1 % of 300.
(ii) 1.6 is 40% of 4.
(iii) 40 is 80% of 50.

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

Solution:
1 day = 24 hours → 10% of a day = (10/100) × 24 = 2.4 hours.
1 week = 7 days = 168 hours → 1% of a week = (1/100) × 168 = 1.68 hours.
Since 2.4 > 1.68, 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 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. What do you observe?

Solution:
Day 1 = (1/2) × 100% = 50%
Day 2 = (2/3) × 100% = 66.67%
Day 3 = (3/4) × 100% = 75%
Day 4 = (4/5) × 100% = 80%
...
Day 99 = (99/100) × 100% = 99%
The fraction follows the pattern n / (n + 1) which approaches 1 (or 100%) as n increases. The percentage of fodder eaten increases every day and gets closer and closer to 100%, but never actually reaches 100%.

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:
Days taken for 20% of the plantation = 18.
Days taken for 100% (5 × 20%) = 5 × 18 = 90 days.
Therefore, workers will take 90 days to complete the entire plantation.

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 time = 10% of 90 min = (10/100) × 90 = 9 min.
Play time = 80% of 90 min = (80/100) × 90 = 72 min.
Cool down time = 10% of 90 min = (10/100) × 90 = 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.

Solution:
World population ≈ 8.3 billion
90% of 8.3 billion = (90/100) × 8.3 billion = 7.47 billion people.

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?

Solution:
(i) The proportions remain the same in the recipe: Rava: 40%, Sugar: 40%, Ghee: 20%.
(ii) Total weight = 2 kg (2000 g):
Rava = 40% of 2000 g = 800 g (0.8 kg)
Sugar = 40% of 2000 g = 800 g (0.8 kg)
Ghee = 20% of 2000 g = 400 g (0.4 kg)

Figure it Out (Page 19 – 20)

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 percentage = (35/75) × 100% = (7/15) × 100% = 46.67% (or 46.77%).

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:
Cost of material = ₹475
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:
Let total expenditure be x.
Revenue = Expenditure + Profit
x + 0.25x = 2.5 crore
1.25x = 2.5 crore
x = 2.5 / 1.25 = 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?

Solution:
Discount = 25% of 300 = (25/100) × 300 = ₹75.
Final 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?

Solution:
Increase in price = ₹100 – ₹60 = ₹40.
Percentage increase = (40/60) × 100% = (2/3) × 100% = 66.67%.

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?

Solution:
Let marked price be y.
Sale price = y – 0.15y = 0.85y = 4,40,000.
y = 4,40,000 / 0.85 ≈ ₹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?

Solution:
Percentage of votes received by winner = (500/1600) × 100% = 31.25%.

8. The price of 1 kg of rice was ₹ 38 in 2024. It is ₹42 in 2025. What is the rate of inflation?

Solution:
Increase = ₹42 – ₹38 = ₹4.
Rate of inflation = (4/38) × 100% = 10.52%.

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

Solution:
y + 0.20y = 90
1.2y = 90
y = 90 / 1.2 = 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.

Solution:
C.P. of 1st buffalo = (100 / 105) × 80,000 ≈ ₹76,190
C.P. of 2nd buffalo = (100 / 90) × 80,000 ≈ ₹88,889
Total S.P. = ₹80,000 + ₹80,000 = ₹1,60,000
Total C.P. = ₹76,190 + ₹88,889 = ₹1,65,079
Loss = ₹1,65,079 – ₹1,60,000 = ₹5,079
Loss percentage = (5,079 / 1,65,079) × 100% ≈ 3%.

11. 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:

Solution:
New population = p + 0.05p = 1.05p.
Therefore, (iv) p × 1.05 is the correct option.

12. Which of the following statement(s) mean the same as — “The demand for cameras has fallen by 85% in the last decade”?

Solution:
(iii) The demand now is 15% of the demand a decade ago.

Figure it Out (Page 22 – 23)

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:
Principal = ₹20,000; Rate = 10%; Time = 2 years
(i) Without compounding (Simple): Amount = 20,000 × (1 + 0.10 × 2) = ₹24,000.
(ii) With compounding: Amount = 20,000 × (1 + 0.10)² = 20,000 × 1.21 = ₹24,200.
Difference = ₹24,200 – ₹24,000 = ₹200 more with compounding.

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:
(i) Without compounding: Amount = 20,000 × (1 + 0.05 × 4) = 20,000 × 1.2 = ₹24,000.
(ii) With compounding: Amount = 20,000 × (1 + 0.05)⁴ = 20,000 × (1.05)⁴ ≈ ₹24,310.13.
With compounding, the final amount is greater.

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

Solution:
The interest earned with compounding is always greater than the interest earned without compounding.

Figure it Out (Page 24)

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?

Solution:
Total amount = p + (p × 0.06 × 4).
Therefore, (vii) p + (p × 0.06 × 4) is the correct expression.

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:
(i) Without compounding: Amount = 50,000 × (1 + 0.07 × 3) = ₹60,500 (Interest = ₹10,500).
(ii) With compounding: Amount = 50,000 × (1.07)³ ≈ ₹61,252.15 (Interest = ₹11,252.15).
Extra interest due to compounding = 11,252.15 − 10,500 = ₹752.15.

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:
(i) Giridhar's interest = 12,500 × 0.12 × 3 = ₹4,500.
(ii) Raghava's amount = 12,500 × (1.1)³ = ₹16,637.5 → Interest = ₹4,137.5.
Difference = ₹4,500 – ₹4,137.5 = ₹362.50.
Giridhar paid ₹362.50 more than Raghava.

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

Solution:
(i) Without compounding: 2000 = 1000 × (1 + 0.10t) → 2 = 1 + 0.10t → t = 10 years.
(ii) With compounding: 2000 = 1000 × (1.1)^t → 2 = (1.1)^t → t ≈ 7.27 years.
Compounding represents exponential growth, while non-compounding represents 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 = 1.5 × (1.03)³ ≈ 1.639 crore.

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:
Total count = 5,06,000 × (1.025)² = 5,31,616.

Try it Out (Page 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 = 2.5 × 50 lakh = 125 lakh (1.25 crore).
(Note: If computed as an increase by 250%: 50 + 125 = 175 lakh or 1.75 crore).

2. The population of the world in 2025 is about 8.2 billion. Match the countries with their approximate percentage share of the worldwide population.

Solution:
Germany: 83 million / 8.2 billion ≈ 1%
India: 1.46 billion / 8.2 billion ≈ 18%
Bangladesh: 175 million / 8.2 billion ≈ 2%
USA: 347 million / 8.2 billion ≈ 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?

Solution:
Correct options are: (v) 8250 × 1.18 and (vi) 8250 + 8250 × 0.18.

4. The monthly percentage change in population of mice in a lab is given: Month 1 (+5%), Month 2 (–2%), Month 3 (–3%). Initial population is p. Which statement(s) are true?

Solution:
Population = p × 1.05 × 0.98 × 0.97 = 0.9987p.
Correct statements: (ii) The population after three months was p × 1.05 × 0.98 × 0.97 and (vi) The population after three months was less than p.

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 Price = ₹135.
Discount = 30% of 135 = ₹40.50.
Final Price = 135 − 40.50 = ₹94.50.
Loss = 100 − 94.50 = ₹5.50.
The shopkeeper makes a loss because the 30% discount is applied to the higher marked price.

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

Solution:
Total area = 144 sq. units.
Area of region E = 18 sq. units.
Percentage = (18 / 144) × 100% = 12.5% (or 12.25%).

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?

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
Observation: x% of y = y% of x.
Justification: (x/100) × y = (y/100) × x because multiplication is commutative.

8. A school is organising an excursion. 40% are Grade 8 students (the rest are Grade 9). Among 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 is 160, how many of them are Grade 8 girls?

Solution:
(i) Percentage of Grade 8 girls = 60% of 40% = 24%.
(ii) Number of Grade 8 girls = 24% of 160 = 38.4 ≈ 38 girls.

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 CP of 1 pencil = ₹1 → CP of 3 pencils = ₹3.
SP of 3 pencils = ₹5.
Profit on 3 pencils = ₹2.
Profit percentage = (2/3) × 100% = 66 2/3% profit.

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:
1.03 × 1.04 = 1.0712.
Overall percentage increase = 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?

Solution:
New breadth = B / 1.10 = 10B / 11.
Decrease = B − 10B/11 = B/11.
Percentage decrease = (1/11) × 100% = 9 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.

Solution:
Potato (70%) = 0.70 × 65 = 45.5 g
Veg oil (24%) = 0.24 × 65 = 15.6 g
Salt (3%) = 0.03 × 65 = 1.95 g
Spice (3%) = 0.03 × 65 = 1.95 g

13. Three shops sell the same items at the same price (₹100 each):
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) What is the effective price per item in each shop? Arrange from cheapest to costliest.
(ii) Calculate the percentage discount on the items for each shop.
(iii) Suppose you need 4 items. Which shop would you choose? Why?

Solution:
(i) Effective price: Shop A = ₹50, Shop B = ₹66.67, Shop C = ₹75. Order: Shop A < Shop B < Shop C.
(ii) Discounts: Shop A = 50%, Shop B = 33.3%, Shop C = 25%.
(iii) Choose Shop A because buying 2 items gives 2 free, fulfilling the 4 items for only ₹200.

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:
Right-handed persons = 1 (constant).
For 1 right-handed person to be 2% of the total: Total = 1 / 0.02 = 50 people.
Number of left-handed people who must leave = 100 − 50 = 50 people.

15. 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. → True
(ii) Women lag behind in the ability to use computers across age groups. → True
(iii) There are more people in their twenties than teenagers. → True
(iv) More than a quarter of people in their thirties can use computers. → False
(v) Less than 1 in 10 aged 60 and above can use computers. → True
(vi) Half of the people in their twenties can use computers. → False