Important Questions Class 8 Maths Part 2 Chapter 1 Fractions in Disguise

Fractions in Disguise is Class 8 Maths Part 2 Chapter 1 in Ganita Prakash. Important Questions Class 8 Maths Part 2 Chapter 1 cover fractions, decimals, percentages, profit and loss, discount, GST, simple interest, compound interest, and tricky percentage applications.

A 25% discount, 18% GST, 10% yearly interest, and 50% profit margin all use the same idea: a fraction written in another form. Class 8 Maths Chapter 1 Fractions in Disguise teaches students how to move between fractions, decimals, and percentages. The chapter becomes easier when students learn the base value first. Once the base is clear, profit, loss, discount, interest, and percentage increase questions become direct.

Key Takeaways

Class 8 Maths Chapter 1 Fractions in Disguise: Topics Covered

Before solving this chapter, identify the base quantity. In percentage questions, the same number can change meaning when the base changes.

1. Fractions as percentages
2. Decimals as percentages
3. FDP trio: fractions, decimals, percentages
4. Percentage of a quantity
5. Percentages greater than 100
6. Profit and loss
7. Discount and marked price
8. GST and taxes
9. Simple interest
10. Compound interest
11. Successive percentage changes
12. Tricky percentage questions

Important Questions for Class 8 Maths Chapter 1 with Answers

Important questions for Class 8 Maths Chapter 1 should be solved in order. Start with FDP conversions, then move to percentages, profit-loss, and interest.

These fraction in disguise class 8 important questions with answers are arranged by concept so students can practise every type of question clearly.

Fractions in Disguise Class 8 Questions on FDP Conversion

FDP means fractions, decimals, and percentages. This is the base skill for the full chapter.

If students can convert between these three forms quickly, most percentage class 8 maths questions become easier.

Fractions Decimals Percentages Class 8 Questions

Q1. Express the following fractions as percentages:
(i) 3/5
(ii) 7/14
(iii) 9/20
(iv) 72/150
(v) 1/3
(vi) 5/11

Ans.

(i) 3/5 × 100 = 60%
(ii) 7/14 = 1/2 = 50%
(iii) 9/20 × 100 = 45%
(iv) 72/150 × 100 = 48%
(v) 1/3 × 100 = 33.33% or 33⅓%
(vi) 5/11 × 100 = 45.45% approximately

Q2. Nandini has 25 marbles. 15 are white. What percentage of her marbles are white?
Ans.

Percentage = 15/25 × 100
Percentage = 60%

So, 60% of the marbles are white.

Q3. In a school, 15 of 80 students come by walking. What percentage of students walk to school?
Ans.

Percentage = 15/80 × 100
Percentage = 18.75%

Q4. Write the correct symbol: >, <, or =.

(i) 50% ___ 5%
(ii) 5/10 ___ 50%
(iii) 3/11 ___ 61%
(iv) 30% ___ 1/3

Ans.

(i) 50% > 5%
(ii) 5/10 = 50%
(iii) 3/11 = 27.27%, so 3/11 < 61%
(iv) 1/3 = 33.33%, so 30% < 1/3

Q5. Complete the FDP trio table.

Percentage Class 8 Maths Questions

Percentage questions depend on the word “of.” In maths, “of” usually means multiplication.

Find the base value first. Then convert the percentage into a fraction or decimal before multiplying.

Percentage of a Quantity Questions

Q1. Find the value of the following:

(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

Ans.

(i) 25% of 160 = 25/100 × 160 = 40
(ii) 16% of 250 = 16/100 × 250 = 40
(iii) 62% of 360 = 62/100 × 360 = 223.2
(iv) 140% of 40 = 140/100 × 40 = 56
(v) 1% of 1 hour = 1/100 × 60 minutes = 0.6 minutes
(vi) 7% of 10 kg = 7/100 × 10 = 0.7 kg

Q2. 30% of k is 70. Find 60%, 90%, and 120% of k.
Ans.

30% of k = 70

60% is double of 30%.
So, 60% of k = 140.

90% is three times 30%.
So, 90% of k = 210.

120% is four times 30%.
So, 120% of k = 280.

Q3. 100% of m is 215. Find 10%, 1%, and 6% of m.
Ans.

10% of m = 215 ÷ 10 = 21.5
1% of m = 215 ÷ 100 = 2.15
6% of m = 6 × 2.15 = 12.9

Q4. Fill in the blanks.

(i) 3 is ___% of 300.
(ii) ___ is 40% of 4.
(iii) 40 is 80% of ___.

Ans.

(i) 3/300 × 100 = 1%
(ii) 40/100 × 4 = 1.6
(iii) 40 = 80/100 × x, so x = 50

Q5. Is 10% of a day longer than 1% of a week?
Ans.

10% of a day = 10/100 × 24 hours = 2.4 hours

1% of a week = 1/100 × 7 × 24 hours = 1.68 hours

Since 2.4 > 1.68, 10% of a day is longer.

Q6. The maximum marks in a test are 75. Students scoring 80% or above get an A grade. What is the minimum score Zubin needs?
Ans.

Minimum score = 80% of 75
= 80/100 × 75
= 60 marks

Q7. A cyclist completes 40% of the Delhi-Agra journey and has covered 92 km. How many kilometres are left?
Ans.

40% of total distance = 92 km

Total distance = 92 × 100/40
= 230 km

Remaining distance = 230 - 92
= 138 km

Percentages Greater Than 100 Questions

A percentage greater than 100 means the new value is more than the original value.

In this section of ganita prakash class 8 part 2 chapter 1, students must avoid treating 150% as 150. It means 1.5 times the base.

Ganita Prakash Class 8 Part 2 Chapter 1 Questions

Q1. Kishanlal’s daily sales target is Rs. 5000. His Day 7 sales were 150% of target and Day 8 sales were 210% of target. Find both sales.
Ans.

Day 7 sales = 150/100 × 5000
= Rs. 7500

Day 8 sales = 210/100 × 5000
= Rs. 10,500

Q2. A farmer harvested 260 kg wheat last year and 650 kg this year. What percentage of last year’s harvest is this year’s harvest?
Ans.

Percentage = 650/260 × 100
= 250%

This year’s harvest is 250% of last year’s harvest.

Q3. The population of Bengaluru in 2025 is about 250% of its population in 2000. If the population in 2000 was 50 lakhs, find the 2025 population.
Ans.

Population in 2025 = 250/100 × 50 lakhs
= 125 lakhs

Q4. The price of a mobile phone is Rs. 8250. GST of 18% is added. Find the final price.
Ans.

GST = 18% of 8250
= 18/100 × 8250
= Rs. 1485

Final price = 8250 + 1485
= Rs. 9735

Profit and Loss Class 8 Questions

Profit and loss class 8 questions depend on cost price, selling price, marked price, discount, and percentage.

Always check the base. Profit and loss are calculated on cost price. Discount is calculated on marked price.

Profit, Loss and Discount Questions

Q1. A shopkeeper buys a geometry box for Rs. 75 and sells it for Rs. 110. Find the profit percentage.
Ans.

Profit = 110 - 75
= Rs. 35

Profit % = 35/75 × 100
= 46.67%

Q2. A carpenter makes chairs. Cost of materials is Rs. 475 and desired profit margin is 50%. Find the selling price.
Ans.

Profit = 50% of 475
= 50/100 × 475
= Rs. 237.50

Selling price = 475 + 237.50
= Rs. 712.50

Q3. A clothing shop offers 25% discount on a shirt marked Rs. 300. How much will Anwar pay?
Ans.

Discount = 25% of 300
= Rs. 75

Selling price = 300 - 75
= Rs. 225

Q4. Samson bought a car for Rs. 4,40,000 after a 15% discount. Find the original price.
Ans.

Selling price = 85% of marked price

4,40,000 = 85/100 × marked price

Marked price = 4,40,000 × 100/85
= Rs. 5,17,647 approximately

Q5. Rice was purchased at Rs. 35 per kg. Raghu sells 10 kg for Rs. 300. Find the loss percentage.
Ans.

Cost price of 10 kg = 35 × 10
= Rs. 350

Selling price = Rs. 300

Loss = 350 - 300
= Rs. 50

Loss % = 50/350 × 100
= 14.28% approximately

Q6. Shyamala bought vases at Rs. 2650 each. She sells one damaged vase at 18% loss. Find the selling price.
Ans.

Loss = 18% of 2650
= 18/100 × 2650
= Rs. 477

Selling price = 2650 - 477
= Rs. 2173

Q7. A shopkeeper sets a price with 35% profit margin and then offers 30% discount. Will he make profit or loss?
Ans.

Let cost price = Rs. 100.

Marked price after 35% profit = Rs. 135

Selling price after 30% discount = 70% of 135
= 70/100 × 135
= Rs. 94.50

Since selling price is less than cost price, he makes a loss of Rs. 5.50.

Loss % = 5.5%

Q8. A store offers two options: 20% discount or Rs. 50 discount on purchases above Rs. 150. Which is better for Rs. 180, Rs. 225, and Rs. 300?
Ans.

For Rs. 180:
20% discount = Rs. 36
Rs. 50 discount is better.

For Rs. 225:
20% discount = Rs. 45
Rs. 50 discount is better.

For Rs. 300:
20% discount = Rs. 60
20% discount is better.

Simple Interest and Compound Interest Class 8 Questions

Simple interest and compound interest class 8 questions test whether interest is calculated on the original principal or on the updated amount.

Simple interest uses the same principal every year. Compound interest adds interest to the principal before the next year begins.

Interest Questions from Fractions in Disguise Class 8 Solutions

Q1. A bank offers 10% per annum. Compare what one gets if Rs. 20,000 is deposited for 2 years with and without compounding.
Ans.

Without compounding:

SI = P × R × T / 100
= 20000 × 10 × 2 / 100
= Rs. 4000

Total amount = Rs. 24,000

With compounding:

Year 1 interest = 10% of 20,000
= Rs. 2000

Amount after Year 1 = Rs. 22,000

Year 2 interest = 10% of 22,000
= Rs. 2200

Amount after Year 2 = Rs. 24,200

Compound interest gives Rs. 200 more.

Q2. The population of a city rises by 3% every year. Current population is 1.5 crore. Find the expected population after 3 years.
Ans.

Population after 3 years = 1.5 × (1.03)³ crore

= 1.5 × 1.092727
= 1.639 crore approximately

Q3. Bacteria increase at 2.5% per hour. Initial count is 5,06,000. Find the count after 2 hours.
Ans.

Count after 2 hours = 5,06,000 × (1.025)²

= 5,06,000 × 1.050625
= 5,31,616 approximately

Q4. A TV bought for Rs. 21,000 depreciates by 5% after one year. Find its value.
Ans.

Value after 1 year = 21,000 × (1 - 0.05)

= 21,000 × 0.95
= Rs. 19,950

Q5. Jasmine invests amount p for 4 years at 6% per annum without compounding. Write the expression for total amount.
Ans.

Without compounding, interest = p × 0.06 × 4

Total amount = p + (p × 0.06 × 4)

Q6. Bus fare increased 3% last year and 4% this year. Find the overall increase.
Ans.

Overall factor = 1.03 × 1.04
= 1.0712

Overall increase = 1.0712 - 1
= 0.0712
= 7.12%

The increase is 7.12%, not 7%, because the changes are successive.

Fractions in Disguise Class 8 Extra Questions

Fractions in disguise class 8 extra questions test whether students understand the base value in each percentage.

Use small assumed values like 100 when the actual value is not given.

Class 8 Maths Extra Questions Chapter 1

Q1. Surbhi kept a 50% profit margin and then offered 50% discount. Did she make any loss?
Ans.

Let cost price = Rs. 100.

Marked price after 50% profit = Rs. 150

Selling price after 50% discount = Rs. 75

Since Rs. 75 is less than Rs. 100, she made a loss.

Loss = Rs. 25
Loss % = 25%

Q2. What is 5% of 40? What is 40% of 5? What do you notice?
Ans.

5% of 40 = 5/100 × 40 = 2

40% of 5 = 40/100 × 5 = 2

Both are equal.

In general, x% of y = y% of x.

Q3. If the length of a rectangle increases by 10% and area remains unchanged, by what percentage does breadth decrease?
Ans.

Let original length = L and breadth = B.

New length = 1.1L

Area unchanged:

1.1L × new breadth = L × B

New breadth = B / 1.1
= 10B/11

Decrease = B - 10B/11
= B/11

Percentage decrease = (B/11)/B × 100
= 100/11%
= 9 1/11%

Q4. In a room of 100 people, 99% are left-handed. How many left-handed people must leave to make it 98%?
Ans.

Currently, 99 out of 100 people are left-handed.

Let x left-handed people leave.

Left-handed people left = 99 - x
Total people left = 100 - x

(99 - x)/(100 - x) = 98/100

99 - x = 98 - 0.98x

1 = 0.02x

x = 50

So, 50 left-handed people must leave.

Q5. Three shops sell the same item at Rs. 100. Shop A gives buy 1 get 1 free. Shop B gives buy 2 get 1 free. Shop C gives buy 3 get 1 free. Find effective price and discount percentage.
Ans.

Shop A:
Pay Rs. 100 for 2 items.
Effective price = Rs. 50
Discount = 50%

Shop B:
Pay Rs. 200 for 3 items.
Effective price = Rs. 66.67
Discount = 33.33%

Shop C:
Pay Rs. 300 for 4 items.
Effective price = Rs. 75
Discount = 25%

Cheapest order: A, B, C.

Q6. A school excursion has 40% Grade 8 and 60% Grade 9 students. Among Grade 8 students, 60% are girls. Total students = 160. Find the percentage and number of Grade 8 girls.
Ans.

Grade 8 girls = 60% of 40%

= 0.6 × 0.4
= 0.24
= 24% of total students

Number of Grade 8 girls = 24% of 160
= 24/100 × 160
= 38.4

So, approximately 38 students are Grade 8 girls.

Fractions in Disguise Class 8 Worksheet Practice

A fractions in disguise class 8 worksheet should mix direct conversions, word problems, commercial maths, and tricky percentages.

Do not solve only formula-based questions. This chapter is application-heavy.

Worksheet-Style Maths Questions for Class 8 with Answers

Q1. Convert 0.35 into a percentage.
Ans. 0.35 × 100 = 35%

Q2. Convert 12.5% into a fraction.
Ans. 12.5% = 12.5/100 = 125/1000 = 1/8

Q3. Find 30% of Rs. 450.
Ans. 30/100 × 450 = Rs. 135

Q4. A jacket marked Rs. 2000 is sold at 15% discount. Find the selling price.
Ans.

Discount = 15% of 2000
= Rs. 300

Selling price = 2000 - 300
= Rs. 1700

Q5. A bag is sold for Rs. 840 after 20% loss. Find the cost price.
Ans.

Selling price = 80% of cost price

840 = 80/100 × CP

CP = 840 × 100/80
= Rs. 1050

Q6. Find the amount on Rs. 5000 at 8% simple interest for 2 years.
Ans.

SI = P × R × T / 100
= 5000 × 8 × 2 / 100
= Rs. 800

Amount = 5000 + 800
= Rs. 5800

Important MCQs for Class 8 Maths Chapter 1 Fractions in Disguise

MCQs in this chapter often test quick conversion and common traps.

Read the base carefully before choosing the option.

Q1. Percentages are fractions with denominator:
(a) 10
(b) 50
(c) 100
(d) 1000

Ans. (c) 100

Q2. 1% of 1 hour in minutes is:
(a) 1 minute
(b) 0.6 minutes
(c) 60 minutes
(d) 10 minutes

Ans. (b) 0.6 minutes

Q3. If selling price is more than cost price, the result is:
(a) Loss
(b) Profit
(c) Discount
(d) Tax

Ans. (b) Profit

Q4. Surbhi kept a 50% margin and then gave a 50% discount. She had:
(a) No loss
(b) 25% loss
(c) 50% profit
(d) 100% profit

Ans. (b) 25% loss

Q5. The population of elephants increased by 5%. If the population last decade was p, the population now is:
(i) p × 0.5
(ii) p × 0.05
(iii) p × 1.5
(iv) p × 1.05

Ans. (iv) p × 1.05

Q6. “The demand for cameras has fallen by 85%” means:
(i) Demand now is 85% of earlier demand
(ii) Demand now is 15% of earlier demand
(iii) Earlier demand was 185% of current demand

Ans. (ii) Demand now is 15% of earlier demand

Q7. A “30% + 20% discount” is:
(a) Same as 50%
(b) Less than 50%
(c) More than 50%
(d) Zero

Ans. (b) Less than 50%

Successive discount = 1 - (0.7 × 0.8)
= 1 - 0.56
= 44%

Ganita Prakash Part 2 Chapter 1 Solutions: Common Question Types

Ganita Prakash Part 2 Chapter 1 solutions often require more than direct formula use. Students must identify whether the question uses percentage increase, decrease, discount, interest, or profit.

This is why writing the base value is the first step in every solution.

Class 8 Maths Ganita Prakash Solutions Practice

Q1. How do you identify the base in a percentage question?
Ans. The base is the original quantity or the quantity after the word “of.”

For example, in 25% of 400, the base is 400.

Q2. Why are successive discounts not added directly?
Ans. Each discount is applied to a new reduced price.

A 30% discount followed by 20% discount is not 50%. It is 44%.

Q3. Why does compound interest grow faster than simple interest?
Ans. In compound interest, each year’s interest is added to the principal.

The next interest is calculated on the increased amount.

All Key Formulas from Class 8 Maths Chapter 1