Exercise 3.1 Class 10 Effective Solutions
This post provides detailed solutions to Exercise 3.1 Class 10 Mathematics, which covers key concepts of ratios, proportions, and variations. In this exercise, students encounter a range of problems that test their understanding of how to compare quantities, simplify ratios, and solve for unknowns in proportional relationships. We’ve worked through each question step-by-step, explaining the methods used to find accurate and simplified answers. These solutions are designed to help you grasp the methods and confidently tackle similar problems on your own.
Question 1. Express the following as a ratio (a : b) and as a fraction in its simplest (lowest) form.
(i) Rs. 750 : Rs. 1250
\( \frac{\text{Rs.750}}{\text{Rs.1250}} = \frac{75 \div 25}{125 \div 25} = \frac{3}{5} = 3:5 \)
(ii) 450 cm : 3 m \( (\because 1 \text{m} = 100 \text{cm}) \)
\( 450 \text{cm} : 300 \text{cm} \)
\( \frac{450}{300} = \frac{45 \div 15}{30 \div 15} = \frac{3}{2} = 3:2 \)
(iii) 4kg : 2kg 750g \( (\because 1 \text{kg} = 1000 \text{g}) \)
\( = \frac{4000g}{(2000g + 750g)} = \frac{4000g}{2750g} \)
\( \frac{400 \div 25}{275 \div 25} = \frac{16}{11} = 16:11 \)
(iv) 27min. 30s : 1h
\( 1 \text{h} = 60 \text{min} = 60 \times 60 \text{s} = 3600 \text{s} \)
\( (27 \times 60 \text{s} + 30 \text{s}) : 3600 \text{s} \)
\( = 1620 \text{s} : 3600 \text{s} \)
\( = \frac{1650}{3600} = \frac{165 \div 15}{360 \div 15} = \frac{11}{24} = 11:24 \)
(v) \(75^\circ : 225^\circ\)
\( \frac{75 \times 1^\circ}{225 \times 1^\circ} = \frac{75 \div 25}{225 \div 25} = \frac{3}{9} = \frac{1}{3} = 1:3 \)
Question 2. In a class of 60 students, 25 students are girls and the remaining students are boys. Compute the ratio of:
(i) Boys to total students
\( \frac{35}{60} = \frac{35 \div 5}{60 \div 5} = \frac{7}{12} = 7:12 \)
(ii) Boys to girls
\( \frac{35}{25} = \frac{35 \div 5}{25 \div 5} = \frac{7}{5} = 7:5 \)
Question 3. If (3(4x – 5y) = 2x – 7y), find the ratio (x : y).
Solution:
\( 3(4x – 5y) = 2x – 7y \)
\( 12x – 15y = 2x – 7y \)
\( 12x – 2x = 15y – 7y \)
\( 10x = 8y \)
\( \frac{x}{y} = \frac{8}{10} = \frac{4}{5} \)
Thus, \( x : y = 4:5 \).
Question 4. Find the value of (p), if the ratios (2p + 5 : 3p + 4) and (3:4) are equal.
Solution:
First ratio = \(2p + 5 : 3p + 4\)
Second ratio = \(3 : 4\)
According to the given condition:
\( \frac{2p + 5}{3p + 4} = \frac{3}{4} \)
Cross-multiplying:
\( 4(2p + 5) = 3(3p + 4) \)
\( 8p + 20 = 9p + 12 \)
\( 20 – 12 = 9p – 8p \)
\( p = 8 \)
Question 5 Exercise 3.1 Class 10. If the ratios \(3x + 1 : 6 + 4x\) and \(2 : 5\) are equal. Find the value of \(x\).
Solution: \(1^{\text{st}}\) Ratio = \(3x + 1 : 6 + 4x\)
\(2^{text{nd}}\) Ratio = \(2 : 5\)
According to the given condition:
\( \frac{3x + 1}{6 + 4x} = \frac{2}{5} \)
Cross-multiplying:
\( 5(3x + 1) = 2(6 + 4x) \)
Expanding:
\( 15x + 5 = 12 + 8x \)
Simplifying:
\( 15x – 8x = 12 – 5 \)
\( 7x = 7 \)
\( x = \frac{7}{7} = 1 \)
Thus, \(x = 1\).
Question 6 Exercise 3.1 Class 10. Two numbers are in the ratio (5 : 8). If 9 is added to each number, we get a new ratio (8 : 11). Find the numbers.
Solution: Ratio between the two numbers = \(5 : 8\)
Number to be added = 9
New ratio = \(8 : 11\)
Let \(1^{\text{st}}\) Number = \(5x\)
\(2^{\text{nd}}\) Number = \(8x\)
By condition:
\( \frac{5x + 9}{8x + 9} = \frac{8}{11} \)
Cross-multiplying:
\( 11(5x + 9) = 8(8x + 9) \)
Expanding:
\( 55x + 99 = 64x + 72 \)
Simplifying:
\( 99 – 72 = 64x – 55x \)
\( 27 = 9x \)
\( x = \frac{27}{9} = 3 \)
Thus, \(1^{\text{st}}\) number = \(5 \times 3 = 15\)
\(2^{\text{nd}}\) number = \(8 \times 3 = 24\)
Required numbers are 15 and 24.
Question 7 Exercise 3.1 Class 10. If 10 is added in each number of the ratio (4 : 13), we get a new ratio (1 : 2). What are the numbers?
Solution: Ratio of two numbers = \(4 : 13\)
Number to be added = 10
New ratio = \(1 : 2\)
Let \(1^{\text{st}}\) number = \(4x\)
\(2^{\text{nd}} number\) = \(13x\)
By condition:
\( \frac{4x + 10}{13x + 10} = \frac{1}{2} \)
Cross-multiplying:
\( 2(4x + 10) = 1(13x + 10) \)
Expanding:
\( 8x + 20 = 13x + 10 \)
Simplifying:
\( 20 – 10 = 13x – 8x \)
\( 10 = 5x \)
\( x = \frac{10}{5} = 2 \)
Thus, \(1^{\text{st}}\) number = \(4 \times 2 = 8\)
\(2^{\text{nd}}\) number = \(13 \times 2 = 26\)
Required numbers are 8 and 26.
Question 8 of Exercise 3.1 Class 10. Find the cost of 8 kg of mangoes, if 5 kg of mangoes cost Rs. 250.
Solution:
Mangoes = 5 kg
Price = Rs. 250
Mangoes = 8 kg
Let Price = \(x\) (for 8 kg)
We know that
\( \frac{5 \text{kg}}{8 \text{kg}} = \frac{250}{x} \)
Cross-multiplying:
\( 5x = 8 \times 250 \)
\( x = \frac{8 \times 250}{5} \)
\( x = 8 \times 50 \)
\( x = \text{Rs.400} \)
Question 9 of Exercise 3.1 Class 10. If \(a : b = 7 : 6\), find the value of \(3a + 5b : 7b – 5a\).
Solution: \(a : b = 7 : 6\)
\( \frac{a}{b} = \frac{7}{6} \)
\( \frac{3a + 5b}{7b – 5a} = \frac{3a + 5b}{7b – 5a} \)
Dividing numerator and denominator by \(b\):
\( \frac{3a + 5b}{7b – 5a} = \frac{\frac{b}{b}}{\frac{b}{b}} = \frac{3 \times \frac{a}{b} + 5}{7 – 5 \times \frac{a}{b}} \)
Substituting \( \frac{a}{b} = \frac{7}{6} \):
\( \frac{3 \times \frac{7}{6} + 5}{7 – 5 \times \frac{7}{6}} = \frac{\frac{21}{6} + 5}{7 – \frac{35}{6}} = \frac{\frac{21 + 30}{6}}{\frac{42 – 35}{6}} = \frac{\frac{51}{6}}{\frac{7}{6}} = \frac{51}{7} \)
Thus, \( 3a + 5b : 7b – 5a = 51 : 7 \).
Question 10 of Exercise 3.1 Class 10. Complete the following
If \( \frac{24}{7} = \frac{6}{x} \), then \( 4x = \dots \)
(i)
Solution:
\( 24 \times x = 7 \times 6 \)
\( x = \frac{7 \times 6}{24} = \frac{42}{24} = 7 \)
Thus, \( 4x = 7 \times 6 \).
\(\textbf{Q.10 (ii) Complete the following:}\)
If \( \frac{5a}{3x} = \frac{15b}{y} \), then \( ay = \dots \)
Solution:
\( 5a \times y = 15b \times 3x \)
\( ay = 45bx \)
Thus, \( ay = 45bx \).
\(\textbf{Q.10 (iii) Complete the following:}\)
If \( \frac{9pq}{2 \ell m} = \frac{18p}{5m} \), then \( 5q = \dots \)
Solution:
\( 5m \times (9pq) = 18p \times 2 \ell m \)
Simplifying:
\( 45pqm = 36 p \ell m \)
Cancelling common terms:
\( 5q = 4 \ell \)
Thus, \( 5q = 4 \ell \).
Question 11 of Exercise 3.1 Class 10.
Find (x) in the following proportions
(i) \( 3x – 2 : 4 :: 2x + 3 : 7 \)
Solution:
Product of Extremes = Product of Means
\( 7(3x – 2) = 4(2x + 3) \)
Expanding:
\( 21x – 14 = 8x + 12 \)
Simplifying:
\( 21x – 8x = 12 + 14 \)
\( 13x = 26 \)
\( x = \frac{26}{13} = 2 \)
Thus, \( x = 2 \).
\(\textbf{(ii) \(3x – 1 : 7 :: \frac{3}{5} : 2x : \frac{7}{5}\)}\)
Solution:
Product of Extremes = Product of Means
\( \frac{3x – 1}{7} \times \frac{7}{5} = \frac{3}{5} \times 2x \)
\( \frac{3x – 1}{5} = \frac{2x}{5} \)
Multiplying both sides by 5:
\( 3x – 1 = 2x \)
Simplifying:
\( 3x – 2x = 1 \)
Thus, \(x = 1\).
\(\textbf{(iii) \( \frac{x – 3}{2} : \frac{5}{x – 1} :: \frac{x – 1}{3} : \frac{4}{x + 4} \)}\)
Solution:
Product of Extremes = Product of Means
\( \frac{x – 3}{2} \times \frac{4}{x + 4} = \frac{5}{x – 1} \times \frac{x – 1}{3} \)
\( \frac{4(x – 3)}{2(x + 4)} = \frac{5(x – 1)}{3(x – 1)} \)
Simplifying:
\( 4(x – 3) \times 3 = 5(x – 1) \times 2(x + 4) \)
Expanding:
\( 12(x – 3) = 10(x – 1) \)
Simplifying:
\( 12x – 36 = 10x + 40 \)
Solving for \(x\):
\( 2x = 76 \)
\( x = \frac{76}{2} = 38 \)
Thus, \(x = 38\).
\(\textbf{(iv) \( p^2 + pq + q^2 : x :: p^3 – q^3 : (p – q) \)}\)
Solution:
Product of Extremes = Product of Means
\( (p^2 + pq + q^2)(p – q) = x(p^3 – q^3) \)
Expanding and simplifying:
\( x = \frac{(p^2 + pq + q^2)(p – q)}{p^3 – q^3} \)
Simplifying the cubic terms:
\( x = p – q \)
Thus, \( x = p – q \).
We hope that these solutions to Exercise 3.1 Class 10 have helped you understand the concepts of ratios, proportions, and variations in depth. Our step-by-step explanations aim to make each problem easier to follow and reinforce your understanding of the methods used. All the important formulas of this unit are here do check them before solving questions. Let’s move on to the notes of exercise 3.2 .For a comprehensive explanation of direct and inverse variations with detailed examples, visit BYJU’S website here For further clarification and to deepen your understanding, we’ve included a video lecture of this exercise below. Watch the video to review key points and clear any remaining doubts. Keep practicing, and you’ll master these essential concepts in no time!