Fractions in Disguise - Chapter 1 Class 8 (Ganita Prakash II)

Master Fractions in Disguise - Chapter 1 Class 8 (Ganita Prakash II) with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.

NCERT Solutions

Fractions in Disguise - Chapter 1 Class 8 (Ganita Prakash II) – NCERT Solutions

Each question below opens its complete step-by-step Teachoo solution.

Figure it out - Page 3, 4

5 questions

Question 1

Express the following fractions as percentages. (i) 3/53/5 = 𝟑/𝟓 × 100%
= 3 × 20%
= 60 %
Question 1 Express the following fractions as percentages. (ii) 7/147/14 = 𝟕/𝟏𝟒 × 100%
= 1/2 × 100%
= 50 %
Question 1 Express the following fractions as percentages. (iii) 9/209/20 = 𝟗/𝟐𝟎 × 100%
= 9 × 5%
= 45 %
Question 1 Express the following fractions as percentages. (iv) 72/15072/150 = 𝟕𝟐/𝟏𝟓𝟎 × 100%
= 72 × 100/150 %
= 72 × 𝟐/𝟑 %
= 72/3 × 2%
= 24 × 2 %
= 48 %
Question 1 Express the following fractions as percentages. (v) 1/31/3 = 𝟏/𝟑 × 100%
= 100/3 %
= 33 𝟏/𝟑 %
Question 1 Express the following fractions as percentages. (vi) 5/115/11 = 𝟓/𝟏𝟏 × 100%
= 500/11 %
= 45 𝟓/𝟏𝟏 %

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Question 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 Let’s first find Fraction of marbles which are white
And, then convert fraction into percentage

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Question 3

In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?Let’s first find Fraction of marbles which are white
And, then convert fraction into percentage

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Question 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.We need to match the dots (A, B, C, D) to the percentages: 55%, 20%, 38%, 72%, 84%, 93%.
Dot A
It is near the start, definitely less than a quarter (25%).
Best match: 20%

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Question 5

Pairs of quantities are shown below. Identify and write appropriate symbols ‘>’, ‘<’, ‘=’ in the blanks. Try to do it without calculations. (i) 50% ____ 5%Since both are in percentages, we just compare the numbers

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Figure it out - Page 12, 13, 14

12 questions

Question 1

Find the missing numbers. The first problem has been worked outLet’s check the Maths
Left Side
We see 5 blocks make up 100%.
This means each block is 𝟏𝟎𝟎÷𝟓=𝟐𝟎%.

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Question 2

Find the value of the following and also draw their bar models. (i) 25% of 160Now,
25% of 160 = 25% × 160
= 𝟐𝟓/𝟏𝟎𝟎 × 160
= 1/4 × 160
= 40

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Question 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?Given that red paint makes up 3/4 of this mixture
We need to find 3/4 of 60 ml

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Question 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 515Since the percentage is the same, we compare the values

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Question 5

Fill in the blanks appropriately: (i) 30% of k is 70, 60% of k is _____, 90% of k is _____, 120% of k is ______.Now,
60% of k
60% of k = 2 × 30% of k
= 2 × 70
= 140
90% of k
90% of k = 3 × 30% of k
= 3 × 70
= 210

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Question 6

Fill in the blanks: (i) 3 is ____ % of 300.Let blank be x

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Question 7

Is 10% of a day longer than 1% of a week? Create such questions and challenge your peers.Let’s convert both days & weeks into hours

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Question 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?Let’s find the percentage quantity each day
Day 1
She gave 2, ate 1
Percentage quantity bull ate = (𝑈𝑛𝑖𝑡𝑠 𝑎𝑡𝑒)/(𝑈𝑛𝑖𝑡 𝑔𝑖𝑣𝑒𝑛) × 100 %
= 1/2 × 100 %
= 50%

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Question 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?Let’s try making a bar diagram of this

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Question 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?We can find 10% of total,
and then find 80% by doing 80% = 8 × 10%

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Question 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.As of 2024-2025, the world population is approximately 8.2 Billion.

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Question 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?Given proportions are for 4 people.

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Figure it out - Page 19, 20

12 questions

Question 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?Given
Cost price = ₹ 75
Selling Price = ₹ 110

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Question 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?Given
Cost Price = ₹ 475
Profit margin = Profit percentage = 50%

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Question 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?Given
Revenue = ₹ 2.5 crore
Profit margin = 25%

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Question 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?Now,
Amount he has to pay = Original Price – Discount

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Question 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% We need to find Percentage increase
Now,
Percentage increase = 𝑪𝒉𝒂𝒏𝒈𝒆/(𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒑𝒓𝒊𝒄𝒆) × 100

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Question 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?Here,
Amount paid = Original Price – Discount

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Question 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?Given
Total votes = 1600
Winner votes = 500

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Question 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.)Here,
Rate of inflation = Percentage increase in prices
= 𝑪𝒉𝒂𝒏𝒈𝒆/(𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒑𝒓𝒊𝒄𝒆) × 100

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Question 9

A number increased by 20% becomes 90. What is the number?Now,
Percentage increase = 𝑪𝒉𝒂𝒏𝒈𝒆/(𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓) × 100

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Question 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.Given,
Selling Price of buffalos = ₹ 80,000 each
Total Selling price (SP) = 2 × Selling price of 1 buffalo
= 2 × 80,000
= ₹ 1,60,000

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Question 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 (i) p × 0.5 (ii) p × 0.05 (iii) p × 1.5 (iv) p × 1.05 (v) p + 1.50 Given
Percentage increase = 5%
Original population = p

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Question 12

Which of the following statement(s) mean the same as — “The demand for cameras has fallen by 85% in the last decade”? (i) The demand now is 85% of the demand a decade ago. (ii) The demand a decade ago was 85% of the demand now. (iii) The demand now is 15% of the demand a decade ago. (iv) The demand a decade ago was 15% of the demand now. (v) The demand a decade ago was 185% of the demand now. (vi) The demand now is 185% of the demand a decade ago.Given
Demand has fallen by 85%

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Figure it out - Page 22, 23, 24

9 questions

Question 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.Here, we have to consider both examples
Simple Interest
Compound Interest

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Question 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.Here, we have to consider both examples
Simple Interest
Compound Interest

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Question 3

Do you observe anything interesting in the solutions of the two questions above? Share and discuss. Let us try to generalise the pattern observed in each of the options.Yes, this is a brilliant mathematical pattern!
Notice that the Simple Interest in both Question 1 and Question 2 is exactly the same (₹24,000). That's because ( 10%×2 years) is mathematically the same as ( 5%×4 years).
However, look at the Compound Interest. Even though the simple math looks equal, compounding over more years ( 4 years instead of 2 ) gave us more money (₹ 24,310 vs ₹ 24,200 ), even at a lower interest rate! The more times the money "snowballs" (compounds), the bigger it gets.

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Question 4

Jasmine invests amount ‘p’ for 4 years at an interest of 6% p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done? (i) p × 6 × 4 (ii) p × 0.6 × 4 (iii) p × 0.6/100 × 4 (iv) p × 0.06/100 × 4 (v) p × 1.6 × 4 (vi) p × 1.06 × 4 (vii) p + (p × 0.06 × 4) So, we use Simple Interest
Now, given that
Principal = p
Rate = R = 6% per year
= 6/100 = 0.06
Time = 4 years

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Question 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?We have to find Interest (not Amount) in both cases
Simple Interest
Compound Interest

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Question 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?Let’s find one-by-one
Giridhar
This is without compounding – simple interest
Given
Principal = P = ₹ 12,500
Rate = R = 12% per year
= 12/100
Time = 3 years

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Question 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?To double our money, we need our total amount to reach ₹2000.
Let’s do both examples

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Question 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?Given,
Population = P = 1.5 crore
Rate = R = 3% per year
= 𝟑/𝟏𝟎𝟎
Time = t = 3 years

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Question 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.Given,
Initial count of bacteria = 506000
It is increasing at the rate of 2.5% per hour
Here, 2.5% is the compounded Rate.

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Figure it out - Page 28, 29, 30

15 questions

Question 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?Given
Population in 2000 = 50 lakhs

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Question 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 of the worldwide population. [Hint: Writing these numbers in the standard form and estimating can help].Since most of the population is given in million
We write Population in millions

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Question 3

The price of a mobile phone is ₹8,250. A GST of 18% is added to the price. Which of the following gives the final price of the phone including the GST? (i) 8250 + 18 (ii) 8250 + 1800 (iii) 8250 + 18/100 (iv) 8250 × 18 (v) 8250 × 1.18 (vi) 8250 + 8250 × 0.18 (vii) 1.8 × 8250Now,
Final price = Listed price + Tax (GST)

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Question 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 statement(s) are true? The initial population is p . (i) The population after three months was p × 0.05 × 0.02 × 0.03. (ii) The population after three months was p × 1.05 × 0.98 × 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.Given
Initial Population = p

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Question 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.To make our calculation (and life) easier,
we assume Cost Price = ₹ 100

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Question 6

What percentage of area is occupied by the region marked ‘E’ in the figure?We can write Percentage of Area as
Percentage of Area occupied by E = (𝑨𝒓𝒆𝒂 𝒐𝒇 𝑬)/(𝑻𝒐𝒕𝒂𝒍 𝑨𝒓𝒆𝒂) × 100

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Question 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?Let’s find 5% of 40 and 40% of 5
5% of 40
5% of 40 = 5/100 × 40
= 200/100
= 2

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Question 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?Given
Total number of students = 160

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Question 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?Given that
Selling Price of 3 pencils = Cost price of 5 pencils
3 × Selling Price of 1 pencil = 5 × Cost price of 1 pencil
Selling Price of 1 pencil = 𝟓/𝟑 × Cost price of 1 pencil

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Question 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?Let
Initial Bus fare = b
Note: We can assume 100 and solve also – it will be easier that way

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Question 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?Let
Original Length = L
Original Breadth = B

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Question 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.Let’s find it one-by-one
Potato
Here,
Percentage of Potato = (𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝑷𝒐𝒕𝒂𝒕𝒐)/(𝑻𝒐𝒕𝒂𝒍 𝑾𝒆𝒊𝒈𝒉𝒕) × 100
70 = (𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑃𝑜𝑡𝑎𝑡𝑜)/65 × 100
70 × 65/100 = Weight of Potato=
Weight of Potato = 70 × 𝟔𝟓/𝟏𝟎𝟎
Weight of Potato = 7 × 65/10
Weight of Potato = 455/10
Weight of Potato = 45.5 g

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Question 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.Here,
Price per item = (𝑨𝒎𝒐𝒖𝒏𝒕 𝒑𝒂𝒊𝒅 𝒇𝒐𝒓 𝒕𝒉𝒐𝒔𝒆 𝒊𝒕𝒆𝒎𝒔)/(𝑻𝒐𝒕𝒂𝒍 𝒊𝒕𝒆𝒎𝒔)

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Question 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%?Since 99% are left handed in a group of 100
Number of left handed people = 99% × Total
= 99/100 × 100
= 99

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Question 15

Look at the following graph. Based on the graph, which of the following statement(s) are valid?
Ability to use computer by age and gender (2023)
The ability to use computers is highest among those in their twenties and teenagers.
Source: NSS Round 79, Comprehensive Annual Modular Survey, National Statistics Office
Data For IndiaQuestion 15 (i) (i) People in their twenties are the most computer-literate among all age groups.Since Twenties bar is longest for males and females both
Thus, the given statement is Valid
Question 15 Based on the graph, which of the following statement(s) are valid? (ii) Women lag behind in the ability to use computers across age groups.The blue bar is shorter than the yellow bar in every single category.
Thus, the given statement is Valid
Question 15 Based on the graph, which of the following statement(s) are valid? (iii) There are more people in their twenties than teenagers.The graph shows percentages (ability), not the total number of people (population size).
Thus, the given statement is Invalid
Question 15 Based on the graph, which of the following statement(s) are valid? (iv) More than a quarter of people in their thirties can use computers.Men are 25% (exactly a quarter), but women are 14%.
The average of the two will be less than 25%.
Thus, the given statement is Invalid
Question 15 Based on the graph, which of the following statement(s) are valid? (v) Less than 1 in 10 aged 60 and above can use computers.1 in 10 means 1/10
And, in percentage = 1/10 × 100 = 10%

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Why Learn This With Teachoo?

Fractions in Disguise is Chapter 1 of NCERT Class 8 Ganita Prakash Part 2. It develops percentages as another way to express fractions and decimals and applies them to comparison, increase and decrease, profit and loss, taxes, growth, compounding and depreciation. Teachoo explains each calculation in context and provides step-by-step solutions to the chapter’s Figure it out questions.

Percentages as equivalent forms

Percentage means “per hundred.” A percentage is therefore a fraction with denominator 100, though its value may be less than, equal to or greater than 100%. Students convert among fractions, decimals and percentages and use convenient benchmarks to calculate percentages mentally.

Finding a percentage of a quantity means multiplying that quantity by the corresponding fraction or decimal. Conversely, a part can be expressed as a percentage of a whole by forming part ÷ whole × 100%. The base or whole must be identified correctly; the same numerical change can represent different percentages relative to different starting values.

Percentages greater than 100 describe quantities larger than the reference whole. Comparing proportions using percentages allows ratios with different totals to be evaluated on a common scale.

Change, commerce and compounding

Percentage increase or decrease compares a change with the original value. Profit and loss use cost price as the base under standard definitions, while profit or loss is the difference between selling and cost prices. Tax is calculated on the specified taxable amount and then added where appropriate.

Growth and compounding apply successive percentage changes. Each new change is calculated on the updated amount, not repeatedly on the original amount. Depreciation uses the same multiplicative structure for decrease. This explains why equal percentage increases and decreases do not cancel: a 20% rise followed by a 20% fall acts on different bases.

Tricky Percentages and Finding Mistakes in Percentages develop error detection and reasoning. The Figure it out sets on pages 3–4, 12–14, 19–20, 22–24 and 28–30 cover the full range of applications.

Topics covered on Teachoo

Teachoo covers:

  • definition and uses of percentages;

  • percentage of a quantity;

  • quick percentage calculation;

  • fractions, decimals and percentages;

  • percentages greater than 100;

  • comparison of proportions;

  • percentage increase and decrease;

  • profit and loss;

  • taxes;

  • growth and compounding;

  • depreciation;

  • tricky percentages and error analysis; and

  • Figure it out solutions for all listed pages.

Learning outcomes

Students should be able to convert equivalent forms, calculate a percentage of a quantity and identify the correct base in comparison problems. They should solve profit, loss, tax, growth and depreciation questions, apply successive changes multiplicatively and explain why reverse percentage changes may differ. They should also spot misleading or incorrect percentage reasoning.

Why is this chapter important?

Percentages appear in discounts, exam scores, interest, inflation, business reports, tax and data. The chapter builds financial and statistical literacy while preparing students for compound interest, growth models and algebraic multipliers.

How Teachoo helps you study

Teachoo separates the direct conversion skills from commercial and growth applications. Write every percentage as a multiplier when several changes occur: a 10% increase multiplies by 1.10, while a 10% decrease multiplies by 0.90. This prevents the mistaken addition or cancellation of successive rates.

For commerce questions, label cost price, selling price, tax rate and base amount before calculating. Estimate whether the final amount should rise or fall, and use Teachoo’s worked answer to check the interpretation as well as the arithmetic.

Common mistakes to avoid

Always identify “percentage of what?” Do not use selling price as the base for standard profit percentage unless the problem defines it that way. Successive changes are not normally added, and equal rise and fall percentages do not restore the original value. Distinguish percentage points from percent change when comparing two rates.

Quick revision checklist

Convert common fractions and decimals to percentages without a calculator, calculate a percentage of a quantity and work backwards to find the whole. Solve one profit-or-loss and one tax problem with every base labelled. Apply two successive changes using multipliers and compare the result with simply adding the rates. Finally, explain a misleading percentage statement by identifying the omitted or incorrectly chosen reference value.

Deeper reasoning and concept connections

Study Fractions in Disguise (Ganita Prakash Part 2) through comparison and justification. Place two related examples side by side, identify the decisive difference and explain why one method works in each case. Then create a new example and a deliberate non-example. This forces the definition to do real work and exposes gaps that passive reading hides.

Students should also practise reversing questions. After solving for an answer, ask what question could have produced it, whether more than one answer is possible and which extra condition would make the result unique. Reverse reasoning develops flexibility and is especially useful for missing-value, assertion–reason and error-analysis questions. The goal is to understand the network of ideas, not merely the order of a textbook solution.

How to solve unfamiliar and competency-based questions

Begin by separating facts from conclusions. Facts are given by the question or a known property; conclusions must be derived. Draw or rewrite the problem so each fact has a visible place. If several methods are possible, prefer the one with fewer assumptions and an easy final check. Record intermediate results rather than doing everything mentally.

Competency questions often change context without changing mathematics. Replace names and story details with variables, shapes, sets or data values. After solving, restore the context and check feasibility: counts should be whole where required, lengths and areas should have suitable units, probabilities should lie between 0 and 1, and constructed figures should satisfy every stated condition.

What complete mastery looks like

For Fractions in Disguise (Ganita Prakash Part 2), a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.

Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.

Additional frequently asked questions

What should a student know before starting Fractions in Disguise (Ganita Prakash Part 2)?

Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.

How can a student check an answer in Fractions in Disguise (Ganita Prakash Part 2)?

Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.

How many questions are enough for strong preparation?

There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.

How should Teachoo solutions be used without becoming dependent on them?

Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.

Frequently asked questions

Can a percentage be greater than 100?

Yes. It means the quantity is greater than the reference whole; for example, 150% equals 1.5 times the base.

Why do equal percentage increase and decrease not cancel?

The decrease is calculated on the changed amount, so the two operations use different bases.

How is depreciation calculated?

For a depreciation rate r%, multiply the current value by 1 − r/100 for each period.

Treat every percentage as a relationship between a part, a base and a rate. Correctly identifying the base solves half the problem.