NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 3 Proportional Reasoning - 2

The purpose of Chapter 3, Proportional Reasoning - 2, is to allow for the further comprehension of the relationship between and the scaling of quantities. This chapter will expand on previous material that utilized ratio notation by introducing students to the relationship of direct and inverse proportion (e.g., when two quantities increase together or decrease in the opposite direction) and the handling of multi-part ratios and technique of cross-multiplication to verify the proportionality of a number of different things (recipes, distance, etc.).

By learning about these topics, students also improve their logical thinking abilities and prediction abilities for everyday scenarios involving real-world outcomes (for example, changing the number of ingredients used in a recipe, or calculating how long it will take to drive from one location to another). NCERT Solutions for Class 8 gives students an organized way to improve their essential scaling skills with concrete examples of successful uses of scaling methods.

1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 3 Solutions

Refine your scaling and reasoning skills with our comprehensive PDF guide. This chapter covers advanced ratio sharing, constant quotients in direct proportion, and constant products in inverse proportion.

Chapter 3 : Proportional Reasoning - 2

2.0Key Concepts covered in Class 8 Maths : Proportional Reasoning - 2

This chapter provides the mathematical framework for analyzing how different variables interact and scale in various contexts.

Multi-part Ratios: Learning to divide quantities into several parts using ratios like a : b : c : d, and calculating the value of each specific part.
Direct Proportion: Understanding that two quantities are directly proportional when their quotient remains constant as they change by the same factor.
Inverse Proportion: An inverse proportional relationship describes how the value of one item goes up as the value of the other item goes down, and how both items multiply to equal the same number.
Verifying Proportionality: To verify whether or not two ratios are equal by using a method known as cross multiplication.
Graphical Representation: Visualizing proportional relationships through charts and diagrams to better understand the rate of change.

3.0NCERT Solutions for Class 8 Maths Chapter 3 : All Exercises

Exercise 3.1

This section provides a recap and extension of direct proportion. Students solve problems where two quantities increase together at a constant rate, such as cost-to-quantity relationships and recipes requiring specific ingredient scaling.

Exercise 3.2

Students learn to understand & apply the concept of the inverse proportion as part of their study in Exercise 3.2. Through exercises that include solving for the travel time of 2 different rates of speed and determining how long supplies would last for either an increase or decrease in the number of consumers consuming those supplies; this will provide students with valuable learning experiences.

Exercise 3.3

In Exercise 3.3 students are given a quantity of items (money, food, land) to distribute among different individuals according to a multi-part ratio with each individual receiving a different part based on the complex ratio made up by the individual parts.

Exercise 3.4

In this last exercise of the chapter, students will use proportional reasoning to solve complicated percentage and data problems. The students will solve multi-step word problems about 3-dimensional scaling (compound scaling), converting units, and determining proportionality using the cross-multiplication method in various situations with the same kinds of problems required in the previous activities.

4.0Detailed Class 8 Maths Chapter 3 Proportional Reasoning - NCERT Solutions

Figure it out-01

1. A cricket coach schedules practice sessions that include different activities in a specific ratio - time for warm-up/cooldown : time for batting : time for bowling : time for fielding $:: 3 : 4 : 3 : 5$ . If each session is 150 minutes long, how much time is spent on each activity?

Sol. Given, time for warm-up/cool-down : time for batting : Time for bowling : time for fielding $:: 3 : 4 : 3 : 5$

Total number of parts $= 3 + 4 + 3 + 5 = 15$ Total time of each session = 150 minutes So, time for warm-up/cool-down $= \frac{3}{15} \times 150 = 30$ minutes Time for batting $= \frac{4}{15} \times 150 = 40$ minutes Time for bowling $= \frac{3}{15} \times 150 = 30$ minutes

Time for fielding $= \frac{5}{15} \times 150 = 50$ minutes

2. A school library has books in different languages in the following ratio no. of Odiya books : no. of Hindi books : no. of English books : $3 : 2 : 1$ .

If the library has 288 Odiya books, how many Hindi and English books does it have?

Sol. Given, No. of Odiya books : No. of Hindi books : No. of English books : $3 : 2 : 1$ . Let x be the total number of books.

No. of Odiya books $= \frac{3}{6} \times x$ $\Rightarrow 288 = \frac{3}{6} \times x$ $\Rightarrow x = 576$ $∴$ No. of Hindi books $= \frac{2}{6} \times 576 = 192$ and Number of English books $= \frac{1}{6} \times 576 = 96$

3. I have 100 coins in the ratio no. of ₹10 coins : no. of ₹5 coins : no. of ₹2 coins : no. of ₹1 coins $:: 4 : 3 : 2 : 1$ . How much money do I have in coins?

Sol. Given no. of ₹ 10 coins : no. of ₹ 5 coins : no. of ₹ 2 coins : no. of ₹ 1 coins $:: 4 : 3 : 2 : 1$ .

Total number of coins $= 100$ Total number of ratio parts $= 4 + 3 + 2 + 1 = 10$ Number of ₹ 10 coins $= \frac{4}{10} \times 100 = 40$ Number of ₹ 5 coins $= \frac{3}{10} \times 100 = 30$ Number of ₹ 2 coins $= \frac{2}{10} \times 100 = 20$ Number of ₹ 1 coins $= \frac{1}{10} \times 100 = 10$ Total money $= 40 \times 10 + 30 \times 5 + 20 \times 2 + 10 \times 1$ $= 400 + 150 + 40 + 10 = ₹600$

4. Construct a triangle with sidelengths in the ratio $3 : 4 : 5$ . Will all the triangles drawn with this ratio of sidelengths be congruent to each other? Why or why not?

Sol. We can construct triangles with sides in the ratio $3 : 4 : 5$ .

They will not be congruent to each other. Reason: Let's consider some triangles having sides in the ratio $3 : 4 : 5$ .

Triangle 1: Let sides be $3 cm, 4 cm, 5 cm$ Triangle 2: Let sides be $6 cm, 8 cm, 10 cm$ Triangle 3: Let sides be $9 cm, 12 cm, 15 cm$ Though all these triangles have the same ratio ( $3 : 4 : 5$ ), their actual sizes are different.

Congruent triangles must have the same shape and size.

These triangles have the same shape (they are similar) but different sizes.

Hence, they are not congruent.

5. Can you construct a triangle with side lengths in the ratio $1 : 3 : 5$ ? Why or why not? Sol. For a triangle to exist, it must satisfy the triangle inequality theorem, which states: The sum of any two sides of a triangle must be greater than the third side.

Let's take Side $1 = 1 cm$ Side 2 $= 3 cm$ Side $3 = 5 cm$

$1 + 3 = 4 \Rightarrow 4 > 5$ (No)
$1 + 5 = 6 \Rightarrow 6 > 3$ (Yes)
$3 + 5 = 8 \Rightarrow 8 > 1$ (Yes)

Since the first condition fails ( $1 + 3 = 4 < 5$ ) For a triangle to be formed, the sum of any two sides must be greater than the third side. We can't construct a triangle with these side lengths.

Figure it out-02

1. A group of 360 people were asked to vote for their favourite season from the three seasons -rainy, winter and summer. 90 liked the summer season, 120 liked the rainy season, and the rest liked the winter. Draw a pie chart to show this information.

Sol. Given, total number of people $= 360$ 90 people liked the summer season. 120 people liked the rainy season. $∴$ People like winter season

$= 360 - (120 + 90) = 150$

So, angle for summer season

$= \frac{90}{360} \times 36 0^{\circ} = 9 0^{\circ}$

Angle for rainy season

$= \frac{120}{360} \times 36 0^{\circ} = 12 0^{\circ}$

Angle for winter season

$= \frac{150}{360} \times 36 0^{\circ} = 15 0^{\circ}$

2. Draw a pie chart based on the following information about viewers' of favourite type of TV channel: Entertainment-50%, Sports 25%, News-15%, Information-10%. Sol. Given, Entertainment

= 50%

Sports

= 25%

News

= 15%

Information = 10% Angle for entertainment

= 50%

36 0^{\circ}

= \frac{50}{100} \times 36 0^{\circ} = 18 0^{\circ}

Angle for sports

= 25%

36 0^{\circ}

= \frac{25}{100} \times 36 0^{\circ} = 9 0^{\circ}

Angle for news

= 15%

36 0^{\circ}

= \frac{15}{100} \times 36 0^{\circ} = 5 4^{\circ}

Angle for information

= 10%

36 0^{\circ}

= \frac{10}{100} \times 36 0^{\circ} = 3 6^{\circ}

3. Prepare a pie chart that shows the favourite subjects of the students in your class. You can collect the data of the number of students for each subject shown in the table (each student should choose only one subject). Then write these numbers in the table and construct a pie chart:

Subject	Number of students
Language
Arts Education
Vocational Education
Social Science
Physical Education
Maths
Science

Sol.

Subject	Number of students
Language	4
Arts Education	6
Vocational Education	9
Social Science	3
Physical Education	10
Maths	12
Science	16

Angle for Language $= \frac{4}{60} \times 36 0^{\circ} = 2 4^{\circ}$ Angle for Arts Education $= \frac{6}{60} \times 36 0^{\circ} = 3 6^{\circ}$ Angle for Vocational Education $= \frac{9}{60} \times 36 0^{\circ} = 5 4^{\circ}$ Angle for Social Science $= \frac{3}{60} \times 36 0^{\circ} = 1 8^{\circ}$ Angle for Physical Education $= \frac{10}{60} \times 36 0^{\circ} = 6 0^{\circ}$ Angle for Maths $= \frac{12}{60} \times 36 0^{\circ} = 7 2^{\circ}$ Angle for Science $= \frac{16}{60} \times 36 0^{\circ} = 9 6^{\circ}$

Science
Language
Arts Education
Vocational education
Social Science
Physical Education
Maths

Figure it out-03

1. Which of these are inverse proportion? (i)

x	40	80	25	16
y	20	10	32	50

(ii)

x	40	80	25	16
y	20	10	12.5	8

(iii)

x	30	90	150	10
y	15	5	3	45

Sol. (i) $x_{1} = 40, x_{2} = 80, x_{3} = 25, x_{4} = 16$

$y_{1} = 20, y_{2} = 10, y_{3} = 32, y_{4} = 50 x_{1} y_{1} = 40 \times 20 = 800 x_{2} y_{2} = 80 \times 10 = 800 x_{3} y_{3} = 25 \times 32 = 800 x_{4} y_{4} = 16 \times 50 = 800$

So, $x_{1} y_{1} = x_{2} y_{2} = x_{3} y_{3} = x_{4} y_{4} = 800$ $∴ x$ and y are in inverse proportion. (ii) $x_{1} = 40, x_{2} = 80, x_{3} = 25, x_{4} = 16$

$y_{1} = 20, y_{2} = 10, y_{3} = 12.5, y_{4} = 8 x_{1} y_{1} = 40 \times 20 = 800 x_{2} y_{2} = 80 \times 10 = 800 x_{3} y_{3} = 25 \times 12.5 = 312.5 x_{4} y_{4} = 16 \times 8 = 128$

So, $x_{1} y_{1} = x_{2} y_{2} \neq = x_{3} y_{3} \neq = x_{4} y_{4}$ $∴ x$ and $y$ are not in inverse proportion. (iii) $x_{1} = 30, x_{2} = 90, x_{3} = 150, x_{4} = 10$

$y_{1} = 15, y_{2} = 5, y_{3} = 3, y_{4} = 45 x_{1} y_{1} = 30 \times 15 = 450 x_{2} y_{2} = 90 \times 5 = 450 x_{3} y_{3} = 150 \times 3 = 450 x_{4} y_{4} = 10 \times 45 = 450$

So, $x_{1} y_{1} = x_{2} y_{2} = x_{3} y_{3} = x_{4} y_{4} = 450$ $∴ x$ and $y$ are in inverse proportion.

2. Fill in the empty cells if x and y are in inverse proportion.

x	16	12		36
y	9		48

Sol.

x	16	12	$X_{3}$	36
y	9	$y_{2}$	48	$y_{4}$

$∵ x$ and y are in inverse proportion. $∴ 16 \times 9 = 12 \times y_{2}$ $\Rightarrow y_{2} = \frac{16 \times 9}{12} = 12$ $∴ 16 \times 9 = x 3 \times 48$ $\Rightarrow x_{3} = \frac{16 \times 9}{48} = 3$ and $16 \times 9 = 36 \times y_{4}$ $\Rightarrow y_{4} = \frac{16 \times 9}{36} = 4$ $\Rightarrow y_{4} = 4$

Figure it out-04

1. Which of the following pairs of quantities are in inverse proportion? (i) The number of taps filling a water tank and the time taken to fill it. (ii) The number of painters hired and the days needed to paint a wall of fixed size. (iii) The distance a car can travel and the amount of petrol in the tank. (iv) The speed of a cyclist and the time taken to cover a fixed route. (v) The length of cloth bought and the price paid at a fixed rate per metre. (vi) The number of pages in a book and the time required to read it at a fixed reading speed.

Sol. (i) If the number of taps increases, then the time taken to fill the tank decreases. If the number of taps decreases, then the time taken to fill the tank increases. The quantities change in opposite directions by the same factor. If we double the no. of taps, the time taken becomes half.

Hence, they are in inverse proportion. (ii) If the number of painters increases, then the number of days needed decreases.

If the number of painters decreases, then the number of days needed increases.

If we double the no. of painters, the work gets done in half the time.

Hence, they are in inverse proportion. (iii) If the distance increases, then the amount of petrol also increases. If the distance decreases, then the amount of petrol also increases. So, they are not in inverse proportion. (iv) If the speed increases, then the time taken decreases. If the speed decreases, then the time taken increases. For a fixed distance, if speed doubles, time becomes half. Hence, they are in inverse proportion. (v) If the amount of cloth increases, then the price to be paid increases. If the amount of cloth decreases, then the price to be paid decreases. Both quantities decrease together and increase together, so they are in direct proportion. (vi) If the number of pages increases, then the time taken to read increases. If the number of pages decreases, then the time taken to read decreases. Both quantities decrease together and increase together, so they are in direct proportion.

2. If 24 pencils cost ₹ 120 , how much will 20 such pencils cost?

Sol. The number of pencils and the cost of pencils are in direct proportion. If x is the required cost, then $\frac{24}{20} = \frac{120}{x}$ $\Rightarrow x \times 24 = 120 \times 20$ $\Rightarrow x = 100$ So, the cost of 20 such pencils, is ₹ 100

3. A tank on a building has enough water to supply 20 families living there for 6 days. If 10 more families move in there, how long will the water last? What assumptions do you need to make to work out this problem?

Sol. The number of families and the number of days is in inverse proportion.

Assumptions needed (i) All families consume equal amounts of water. (ii) The daily water consumption per family remains fixed. (iii) No additional water is supplied to the tank.

Let the water last for x days. So, $20 \times 6 = 30 \times x$ $\Rightarrow x = 4$ So, the water will last for 4 days.

4. Fill in the average number of hours each living being sleeps in a day by looking at the charts. Select the appropriate hours from this list: 15, 2.5, 20, 8, 3.5, 13, 10.5, 18 .

Sol. Common Sleep Patterns: Average no. of hours a giraffe sleeps = 2.5 hours

Average number of hours an elephant sleeps $= 3.5$ hours

Average number of hours a boy sleeps = 8 hours

Average number of hours a dog sleeps = 10.5 hours

Average no. of hours a cat sleeps = 13 hours Average no. of hours a squirrel sleeps = 15 hours

Average no. of hours a snake sleeps $= 18$ hours

Average no. of hours a bat sleeps $= 20$ hours

5. The pie chart on the right shows the result of a survey carried out to find the modes of transport used by children to go to school. Study the pie chart and answer the following questions. (i) What is the most common mode of transport? (ii) What fraction of children travel by car? (iii) If 18 children travel by car, how many children took part in the survey? How many children use taxis to travel to school? (iv) By which two modes of transport are equal numbers of children travelling? Sol. (i) The largest angle is $12 0^{\circ}$ , which corresponds to the Bus. (ii) Angle corresponding to car $= 36 0^{\circ} - (9 0^{\circ} + 12 0^{\circ} + 6 0^{\circ} + 6 0^{\circ}) = 3 0^{\circ}$ The fraction of children who travel by $car \frac{36}{360} = \frac{1}{12}$ (iii) Let x be the total number of children who took part in the survey. Then $18 = \frac{1}{12} \times x$ $\Rightarrow x = 18 \times 12 = 216$ The number of children using taxis is 0 , as taxis are not a category in this chart. (iv) Cycle and two-wheeler ( $6 0^{\circ}$ each)

6. Three workers can paint a fence in 4 days. If one more worker joins the team, how many days will it take them to finish the work? What are the assumptions you need to make?

Sol. When the number of workers increases, the number of days needed to paint the fence decreases.

Car30

^{\circ}

Assumptions Needed

(i) All workers work at the same speed/rate. (ii) The work is uniformly distributed among all workers. (iii) All workers work for the same number of hours each day. So, the number of workers and the number of days, is in inverse proportion. Let x be the no. of days taken. $3 \times 4 = 4 \times x \to x = 3$

So, they will take 3 days to finish the work.

7. It takes 6 hours to fill 2 tanks of the same size with a pump. How long will it take to fill 5 such tanks with the same pump?

Sol. No. of hours and no. of tanks are in direct proportion. Let 5 such tanks take x hours. $\frac{6}{2} = \frac{x}{5}$ $\Rightarrow x = 15$ So, 5 tanks will take 15 hours.

8. A given set of chairs are arranged in 25 rows, with 12 chairs in each row. If the chairs are rearranged with 20 chairs in each row, how many rows does this new arrangement have?

Sol. No. of rows and no. of chairs in each row are in inverse proportion.

Let the new arrangement has $x$ rows. $25 \times 12 = x \times 20$ $\Rightarrow x = 15$ So, the new arrangement has 15 rows.

9. A school has 8 periods a day, each of 45 minutes duration. How long is each period, if the school has 9 periods a day, assuming that the number of school hours per day stays the same?

Sol. No. of periods and the duration of each period are in inverse proportion. Let each period be of $x$ minutes. $8 \times 45 = 9 \times x$ $\Rightarrow x = 40$ So, each period is 40 minutes.

10. A small pump can fill a tank in 3 hours, while a large pump can fill the same tank in 2 hours. If both pumps are used together, how long will the tank take to fill?

Sol. Consider the work done to fill the tank as 1 unit of work. Let us find the work done by each pump in 1 hour.

Small pump fills the tank in 3 hours, so in 1 hour it does $\frac{1}{3}$ unit of work.
Large pump fills the tank in 2 hours, so in 1 hour it does $\frac{1}{2}$ unit of work.

So, the work done by both pumps in 1 hour is $\frac{1}{3} + \frac{1}{2} = \frac{5}{6}$ units of work.

Therefore, to complete $\frac{5}{6}$ units of work, it takes 1 hour when both work together. How much time will it take to complete 1 unit of work?

The quantity of work and time taken are directly proportional.

So, $\frac{5}{6} : 1 :: 1 : x$ Since it is a direct proportion,

$\frac{5}{6} \times x = 1 x = \frac{6}{5}$

Therefore, both pumps together will fill the tank in $\frac{6}{5}$ hours (or 1.2 hours).

11. A factory requires 42 machines to produce a given number of toys in 63 days. How many machines are required to produce the same number of toys in 54 days?

Sol. Let the number of machines be x

Number of Machines	42	X
Number of Days	63	54

Number of machines and no. of days are in inverse proportion. So, $42 \times 63 = x \times 54$ $\Rightarrow x = \frac{42 \times 63}{54} = 7 \times 7$ or $x = 49$ So, 49 machines are required.

12. A car takes 2 hours to reach a destination, travelling at a speed of $60 km / h$ . How long will the car take if it travels at a speed of $80 km / h$ ?

Sol. Let the car take $t$ hours.

Speed $(km / hr)$	60	$80 ↑$
Time $(hrs)$	2	$t ↓$

The speed of the car and the time taken are in inverse proportion. So, $2 \times 60 = t \times 80$ $t = \frac{2 \times 60}{30} = \frac{3}{2}$ $\Rightarrow t = 1.5$ hours So, the car will take 1.5 hours.

5.0Key Features and Benefits of Class 8 Maths Ganith Prakash 2 Chapter 3

Practical Problem-Solving: The chapter uses relatable examples, such as mixing idli batter or adjusting speeds for travel, making the abstract concept of proportionality easy to grasp.
Algorithmic Mastery: Students learn to use systematic formulas to solve for unknown values in proportional sets, ensuring accuracy in both school and competitive exams.
Emphasis on Logical Connections: By contrasting direct and inverse proportions, the solutions help students identify which mathematical rule to apply based on the real-world context of the problem.
Skill for Data Interpretation: Learning to handle multi-part ratios is a vital skill for reading and interpreting modern data sets, financial reports, and scientific mixtures.
Foundation for Advanced Physics: The concept of inverse proportion is a cornerstone for understanding physical laws like speed vs. time or pressure vs. volume in higher grades.
Academic Excellence: These solutions are aligned with the latest CBSE curriculum, providing a solid foundation for high performance in school assessments and national-level Olympiads.

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NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 3 Proportional Reasoning - 2

1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 3 Solutions

2.0Key Concepts covered in Class 8 Maths : Proportional Reasoning - 2

3.0NCERT Solutions for Class 8 Maths Chapter 3 : All Exercises