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Fractions in Disguise - Exercise 1.2 Class 8 Maths - Chapter 1 NCERT Solutions

Class 8 Maths NCERT Ganita Prakash Term 2 Chapter 1 Fractions in Disguise Exercise 1.2 detailed solutions and benefits are provided on this page, which would be helpful for students to grasp the ideas covered in the exercise. The solutions are provided by the ALLEN expert faculty in simple words, which can be useful for students while preparing for their final exams and related competitive examinations.

The exercise is about finding a percentage of a quantity, comparing percentages when the quantities are different, and using proportional reasoning. A key focus of this particular exercise is the connection between fractions, decimals, and percentages, which is explored through different percentage-based questions

1.0Download Class 8 Maths Chapter 1 Ex 1.2 NCERT Solutions PDF

Download class 8 maths chapter 1 fraction in disguise exercise 1.2 solutions pdf from here to learn the detailed solutions

Chapter 1 : Fractions in Disguise Exercise 1.2

2.0Detailed NCERT Solutions on Fractions in Disguise Exercise 1.2 Class 8 Maths (Ganita Prakashan Term 2)

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}

of

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

\frac{3}{4}

of

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%

of

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

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

(iii)

62%

of

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

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

(iv)

140%

of

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%

of

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 $= \frac{40}{100} \times 2 kg = 0.8 kg$

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

3.0Key Concepts of Chapter 1- Fractions in Disguise Exercise 1.2

Finding a percentage of a given quantity using different methods.
Using fractions, decimals, and proportional reasoning to calculate percentages.
Representing percentage values using bar models.
Comparing percentages of different quantities.
Finding related percentage values when one percentage value is known.
Understanding and calculating percentages greater than 100%.
Identifying percentage patterns and relationships in quantity-based situations.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 1 Exercise 1.2

Explains how to calculate percentages of quantities using the methods introduced in Fractions in Disguise.
Shows the relationship between percentage, fraction, and decimal forms while solving Exercise 1.2 questions.
Breaks down comparisons involving different percentages and different base quantities.
Explains how to work with percentage values greater than 100% in quantity-based questions.
Demonstrates the use of bar models for representing and finding percentage quantities.
Provides the working for questions that use a known percentage to determine other percentage values.
Helps interpret the percentage patterns and relationships presented in the exercise.

NCERT Class 8 Maths Ch. 1 Fractions in Disguise Other Exercises:-

Exercise 1.2

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 : Fractions in Disguise

Chapter 2 : The Baudhayana - The Pythagoras Theorem

Chapter 3 : Proportional Reasoning - 2

Chapter 4 : Exploring Some Geometric Themes

Chapter 5 : Tales by Dots and Lines

Chapter 6 : Algebra Play

Chapter 7 : Area

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Fractions in Disguise - Exercise 1.2 Class 8 Maths - Chapter 1 NCERT Solutions

Class 8 Maths NCERT Ganita Prakash Term 2 Chapter 1 Fractions in Disguise Exercise 1.2 detailed solutions and benefits are provided on this page, which would be helpful for students to grasp the ideas covered in the exercise. The solutions are provided by the ALLEN expert faculty in simple words, which can be useful for students while preparing for their final exams and related competitive examinations.

The exercise is about finding a percentage of a quantity, comparing percentages when the quantities are different, and using proportional reasoning. A key focus of this particular exercise is the connection between fractions, decimals, and percentages, which is explored through different percentage-based questions

1.0Download Class 8 Maths Chapter 1 Ex 1.2 NCERT Solutions PDF

Download class 8 maths chapter 1 fraction in disguise exercise 1.2 solutions pdf from here to learn the detailed solutions

Chapter 1 : Fractions in Disguise Exercise 1.2

2.0Detailed NCERT Solutions on Fractions in Disguise Exercise 1.2 Class 8 Maths (Ganita Prakashan Term 2)

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}

of

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

\frac{3}{4}

of

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%

of

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

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

(iii)

62%

of

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

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

(iv)

140%

of

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%

of

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}