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Proportional Reasoning-2 Exercise 3.1 Class 8 Maths - Chapter 3 NCERT Solutions

NCERT Class 8 Maths Chapter 3 Proportional Reasoning-2 Exercise 3.1 NCERT Solutions provides detailed, step-by-step solutions to the questions given in the exercise. The solutions are curated by the ALLEN expert team with clear explanations aligned with the updated syllabus, making the concepts easier to understand and the answers easier to follow.Exercise 3.1: Ratios and proportional reasoning for solving different problems. It deals with the division of quantities in a given ratio, calculation of unknown values, finding total values and construction of triangles from given side ratios. It also covers tasks on triangle inequality and on determining if given lengths can form a triangle.

1.0Download Class 8 Maths Chapter 3 Ex 3.1 NCERT Solutions PDF

Download the free PDF of Class 8 Maths Chapter 3 Exercise 3.1 NCERT Solutions to review the solved questions, practise the methods used in the exercise, and revise the concepts covered in Proportional Reasoning-2.

Chapter 3 : Proportional Reasoning - 2

2.0Detailed NCERT Solutions on Proportional Reasoning-2 Exercise 3.1 Class 8 Maths (Ganita Prakash Term 2)

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.

3.0Key Concepts of Chapter 3 Proportional Reasoning-2 Exercise 3.1

Dividing quantities into parts according to a given ratio.
Using a known quantity and its ratio to determine the other quantities.
Calculating the value of different denominations distributed in a given ratio.
Constructing triangles with side lengths in a specified ratio.
Understanding the difference between similar and congruent triangles when their side ratios are the same.
Applying the triangle inequality theorem to determine whether given side lengths can form a triangle.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 3 Exercise 3.1

Explains how to divide a total time into different activities using a given ratio.
Shows how to calculate the number of books in different languages when their ratio and one quantity are known.
Helps solve problems involving different coin denominations distributed in a given ratio and calculate their total value.
Provides clear steps for constructing triangles when their side lengths are given in a ratio.
Explains why triangles having the same side ratio can be similar but not congruent when their sizes are different.
Shows how to use the triangle inequality to check whether three lengths in a given ratio can form a triangle.

NCERT Class 8 Maths Ch. 3 Proportional Reasoning - 2 Other Exercises:-

Exercise 3.1

Exercise 3.2