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Maharashtra SSC Class 10 Mathematics Chapter 1 Linear Equations in Two Variables: textbook solutions

Solutions written by Shishya, checked, not the board's official answers

Step-by-step working for the practice-set questions of Chapter 1, “Linear Equations in Two Variables”, in Mathematics Part I (Algebra), Standard X (Maharashtra SSC, English medium; the chapter is on pages 1-29 of the printed book). Each question is in our own short words and cited by its set and number; open the textbook for the exact wording and figures.

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How these were checked: drafted with AI on 28 Oct 2026. A solution is shown only if a calculator check recomputed every final value from the question's own numbers and every arithmetic step in the working agreed. 58 of 62 questions in this chapter passed; the others are listed below as not solved here. These are not the board's official answers: the textbook and your teacher are the authority, and if something here disagrees with the book, follow the book and tell us.

Textbook PDF read on 28 Oct 2026 from the publisher's official site.

Practice Set 1.1

Practice Set 1.1, Q1
Solve by adding the equations: 5x + 3y = 9 and 2x − 3y = 12.
1. Add the two equations: (5x + 3y) + (2x − 3y) = 9 + 12
2. Combine like terms: 7x = 21
3. Divide both sides by 7: x = 21 ÷ 7
4. x = 3
5. Substitute x = 3 into the first equation: 5(3) + 3y = 9
6. Simplify: 15 + 3y = 9
7. Subtract 15 from both sides: 3y = 9 − 15
8. 3y = −6
9. Divide both sides by 3: y = −6 ÷ 3
10. y = −2
Answer: x = 3, y = -2
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Practice Set 1.1, Q2(1)
Solve: 3a + 5b = 26 and a + 5b = 22.
1. Given equations: 3a + 5b = 26 and a + 5b = 22.
2. Subtract the second equation from the first: (3a + 5b) − (a + 5b) = 26 − 22.
3. Simplify the left side: 3a − a + 5b − 5b = 2a.
4. Simplify the right side: 26 − 22 = 4.
5. So, 2a = 4.
6. Divide both sides by 2: a = 4 ÷ 2.
7. a = 2.
8. Substitute a = 2 into the second equation: 2 + 5b = 22.
9. Subtract 2 from both sides: 5b = 22 − 2.
10. 5b = 20.
11. Divide both sides by 5: b = 20 ÷ 5.
12. b = 4.
Answer: a = 2, b = 4
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Practice Set 1.1, Q2(2)
Solve: x + 7y = 10 and 3x − 2y = 7.
1. Given the equations: x + 7y = 10 and 3x − 2y = 7.
2. Multiply the first equation by 3 to make the coefficient of x the same in both equations.
3. 3(x + 7y) = 3(10) gives 3x + 21y = 30.
4. Subtract the second equation (3x − 2y = 7) from this new equation.
5. (3x + 21y) − (3x − 2y) = 30 − 7.
6. 23y = 23.
7. y = 1.
8. Substitute y = 1 into the first original equation x + 7y = 10.
9. x + 7(1) = 10.
10. x + 7 = 10.
11. x = 3.
Answer: x = 3, y = 1
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Practice Set 1.1, Q2(3)
Solve: 2x − 3y = 9 and 2x + y = 13.
1. Given equations: 2x − 3y = 9 and 2x + y = 13.
2. Subtract the first equation from the second: (2x + y) − (2x − 3y) = 13 − 9.
3. Simplify the left side: 2x + y − 2x + 3y = 4y.
4. Simplify the right side: 13 − 9 = 4.
5. So, 4y = 4.
6. Divide both sides by 4: y = 4 ÷ 4.
7. y = 1.
8. Substitute y = 1 into the second equation: 2x + 1 = 13.
9. Subtract 1 from both sides: 2x = 13 − 1.
10. 2x = 12.
11. Divide both sides by 2: x = 12 ÷ 2.
12. x = 6.
Answer: x = 6, y = 1
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Practice Set 1.1, Q2(4)
Solve: 5m − 3n = 19 and m − 6n = −7.
1. Given equations: 5m − 3n = 19 and m − 6n = −7.
2. Multiply the first equation by 2 to match the coefficient of n: 2(5m − 3n) = 2(19).
3. This gives 10m − 6n = 38.
4. Subtract the second equation from this new equation: (10m − 6n) − (m − 6n) = 38 − (−7).
5. Simplify the left side: 10m − m = 9m.
6. Simplify the right side: 38 + 7 = 45.
7. So, 9m = 45.
8. Divide by 9: m = 45 ÷ 9.
9. m = 5.
10. Substitute m = 5 into the second equation: 5 − 6n = −7.
11. Move 5 to the right side: −6n = −7 − 5.
12. −6n = −12.
13. Divide by −6: n = −12 ÷ (−6).
14. n = 2.
Answer: m = 5, n = 2
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Practice Set 1.1, Q2(5)
Solve: 5x + 2y = −3 and x + 5y = 4.
1. 5x + 2y = −3 … (i)
2. x + 5y = 4 … (ii)
3. Multiply equation (ii) by 5 to get 5x + 25y = 20 … (iii)
4. Subtract equation (i) from equation (iii): (5x + 25y) − (5x + 2y) = 20 − (−3)
5. 23y = 23
6. y = 23 ÷ 23
7. y = 1
8. Substitute y = 1 into equation (ii): x + 5(1) = 4
9. x + 5 = 4
10. x = 4 − 5
11. x = −1
Answer: x = −1, y = 1
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Practice Set 1.1, Q2(6)
Solve: (1/3)x + y = 10/3 and 2x + (1/4)y = 11/4.
1. Multiply the first equation by 3 to clear the denominator: x + 3y = 10.
2. Multiply the second equation by 4 to clear the denominator: 8x + y = 11.
3. From the second equation, express y in terms of x: y = 11 - 8x.
4. Substitute y = 11 - 8x into the first equation: x + 3(11 - 8x) = 10.
5. Simplify the equation: x + 33 - 24x = 10.
6. Combine like terms: -23x + 33 = 10.
7. Isolate the x term: -23x = 10 - 33.
8. Calculate the constant: -23x = -23.
9. Solve for x: x = -23 / -23.
10. x = 1.
11. Substitute x = 1 into y = 11 - 8x to find y: y = 11 - 8(1).
12. y = 11 - 8.
13. y = 3.
Answer: x = 1, y = 3
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Practice Set 1.1, Q2(7)
Solve: 99x + 101y = 499 and 101x + 99y = 501.
1. Add the two equations.
2. 99x + 101y + 101x + 99y = 499 + 501
3. 200x + 200y = 1000
4. Divide the entire equation by 200.
5. x + y = 5
6. Multiply this equation by 99 to match the coefficient of x in the first original equation.
7. 99x + 99y = 495
8. Subtract this new equation from the first original equation (99x + 101y = 499).
9. (99x + 101y) - (99x + 99y) = 499 - 495
10. 2y = 4
11. y = 2
12. Substitute y = 2 into the equation x + y = 5.
13. x + 2 = 5
14. x = 3
Answer: x = 3, y = 2
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Practice Set 1.1, Q2(8)
Solve: 49x − 57y = 172 and 57x − 49y = 252.
1. Write the given equations as 49x − 57y = 172 and 57x − 49y = 252.
2. Add the two equations to get 106x − 106y = 424.
3. Divide the sum by 106 to obtain x − y = 4.
4. Subtract the second equation from the first to get −8x − 8y = −80.
5. Divide by −8 to obtain x + y = 10.
6. Add x − y = 4 and x + y = 10: 2x = 14.
7. x = 14 ÷ 2 = 7.
8. y = 10 − 7 = 3.
Answer: x = 7, y = 3
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Practice Set 1.2

Practice Set 1.2, Q1
Complete the tables of points for drawing the graphs: (I) x + y = 3 with x = 3, y = 5 and y = 3 given; (II) x − y = 4 with y = 0, x = −1 and x = 0 given.
1. (I) For x + y = 3, when x = 3, y = 3 - 3 = 0, so the point is (3, 0).
2. (I) For x + y = 3, when y = 5, x = 3 - 5 = -2, so the point is (-2, 5).
3. (I) For x + y = 3, when y = 3, x = 3 - 3 = 0, so the point is (0, 3).
4. (II) For x - y = 4, when y = 0, x = 4 + 0 = 4, so the point is (4, 0).
5. (II) For x - y = 4, when x = -1, y = -1 - 4 = -5, so the point is (-1, -5).
6. (II) For x - y = 4, when x = 0, y = 0 - 4 = -4, so the point is (0, -4).
Answer: (I) (3, 0), (-2, 5), (0, 3); (II) (4, 0), (-1, -5), (0, -4)
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Practice Set 1.2, Q2(1)
Solve graphically: x + y = 6 and x − y = 4.
1. For the equation x + y = 6, set x = 0 to get y = 6, giving point (0, 6).
2. For the equation x + y = 6, set y = 0 to get x = 6, giving point (6, 0).
3. Draw a straight line through (0, 6) and (6, 0).
4. For the equation x − y = 4, set x = 0 to get −y = 4, so y = −4, giving point (0, −4).
5. For the equation x − y = 4, set y = 0 to get x = 4, giving point (4, 0).
6. Draw a straight line through (0, −4) and (4, 0).
7. The two lines intersect at the point (5, 1).
8. Therefore, the solution is x = 5 and y = 1.
Answer: x = 5, y = 1
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Practice Set 1.2, Q2(2)
Solve graphically: x + y = 5 and x − y = 3.
1. Write the first equation as x + y = 5.
2. Find two points for the first line: when x = 0, y = 5, and when y = 0, x = 5.
3. Write the second equation as x − y = 3.
4. Find two points for the second line: when x = 0, y = −3, and when y = 0, x = 3.
5. Draw both lines on the same coordinate plane.
6. Observe that the lines intersect at the point (4, 1).
7. Verify the solution in the first equation: 4 + 1 = 5.
8. Verify the solution in the second equation: 4 − 1 = 3.
Answer: x = 4, y = 1
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Practice Set 1.2, Q2(3)
Solve graphically: x + y = 0 and 2x − y = 9.
1. For the equation x + y = 0, rearrange it to y = -x.
2. Find two points for the first line: when x = 0, y = 0 and when x = 1, y = -1.
3. For the equation 2x - y = 9, rearrange it to y = 2x - 9.
4. Find two points for the second line: when x = 0, y = -9 and when x = 5, y = 1.
5. Plot the points (0, 0) and (1, -1) on the graph and draw the line for x + y = 0.
6. Plot the points (0, -9) and (5, 1) on the graph and draw the line for 2x - y = 9.
7. Observe that the two lines intersect at the point (3, -3).
Answer: x = 3, y = -3
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Practice Set 1.2, Q2(4)
Solve graphically: 3x − y = 2 and 2x − y = 3.
1. For the equation 3x − y = 2, when x = 0, y = −2, giving point (0, −2).
2. For the equation 3x − y = 2, when y = 0, x = 2/3, giving point (2/3, 0).
3. For the equation 2x − y = 3, when x = 0, y = −3, giving point (0, −3).
4. For the equation 2x − y = 3, when y = 0, x = 3/2, giving point (3/2, 0).
5. Draw the line through (0, −2) and (2/3, 0) for 3x − y = 2.
6. Draw the line through (0, −3) and (3/2, 0) for 2x − y = 3.
7. The two lines intersect at the point (−1, −5).
8. Therefore, the solution is x = −1 and y = −5.
Answer: x = -1, y = -5
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Practice Set 1.2, Q2(5)
Solve graphically: 3x − 4y = −7 and 5x − 2y = 0.
1. Given equations are 3x − 4y = −7 and 5x − 2y = 0.
2. For 3x − 4y = −7, when x = 1, 3(1) − 4y = −7, so −4y = −10 and y = 5/2.
3. For 3x − 4y = −7, when x = −1, 3(−1) − 4y = −7, so −4y = −4 and y = 1.
4. For 5x − 2y = 0, when x = 0, 5(0) − 2y = 0, so y = 0.
5. For 5x − 2y = 0, when x = 2, 5(2) − 2y = 0, so 2y = 10 and y = 5.
6. Plot the points (1, 5/2) and (−1, 1) for the first equation and draw the line.
7. Plot the points (0, 0) and (2, 5) for the second equation and draw the line.
8. The two lines intersect at the point (1, 5/2).
9. Therefore, the solution is x = 1 and y = 5/2.
Answer: x = 1, y = 5/2
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Practice Set 1.2, Q2(6)
Solve graphically: 2x − 3y = 4 and 3y − x = 4.
1. Given equations are 2x − 3y = 4 and 3y − x = 4.
2. Rearrange the first equation as 3y = 2x − 4, so y = (2x − 4)/3.
3. Rearrange the second equation as x = 3y − 4, so y = (x + 4)/3.
4. For the first line, when x = 2, y = (2 × 2 − 4)/3 = 0, giving point (2, 0).
5. For the first line, when x = 5, y = (2 × 5 − 4)/3 = 2, giving point (5, 2).
6. For the second line, when x = 1, y = (1 + 4)/3 = 5/3, giving point (1, 5/3).
7. For the second line, when x = 4, y = (4 + 4)/3 = 8/3, giving point (4, 8/3).
8. Plot the points (2, 0) and (5, 2) on the graph and draw a straight line through them.
9. Plot the points (1, 5/3) and (4, 8/3) on the graph and draw a straight line through them.
10. The two lines intersect at the point (8, 4).
11. Therefore, the solution is x = 8 and y = 4.
Answer: x = 8, y = 4
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Practice Set 1.3

Not solved here yet: Q2 (see the textbook).

Practice Set 1.3, Q1
Find the value of the determinant with rows (3, 2) and (4, 5).
1. The determinant has rows (3, 2) and (4, 5).
2. For a determinant with rows (a, b) and (c, d), the value is ad − bc.
3. Here a = 3, b = 2, c = 4, d = 5
4. Substitute the values: 3 × 5 − 2 × 4
5. Calculate the products: 15 − 8
6. Subtract to find the result: 7
Answer: 7
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Practice Set 1.3, Q3(1)
Solve by Cramer's rule: 3x − 4y = 10 and 4x + 3y = 5.
1. Write the system of equations as 3x − 4y = 10 and 4x + 3y = 5.
2. Identify the coefficients: a₁ = 3, b₁ = −4, c₁ = 10, a₂ = 4, b₂ = 3, c₂ = 5.
3. Calculate the determinant D = |3 −4; 4 3| = (3 × 3) − (−4 × 4) = 9 − (−16) = 25.
4. Calculate the determinant Dₓ = |10 −4; 5 3| = (10 × 3) − (−4 × 5) = 30 − (−20) = 50.
5. Calculate the determinant Dᵧ = |3 10; 4 5| = (3 × 5) − (10 × 4) = 15 − 40 = −25.
6. Find x using Cramer's rule: x = Dₓ / D = 50 / 25 = 2.
7. Find y using Cramer's rule: y = Dᵧ / D = −25 / 25 = −1.
Answer: x = 2, y = -1
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Practice Set 1.3, Q3(2)
Solve by Cramer's rule: 4x + 3y − 4 = 0 and 6x = 8 − 5y.
1. Rewrite the equations in the form ax + by = c: 4x + 3y = 4 and 6x + 5y = 8.
2. Calculate the determinant D of the coefficient matrix: D = 4 × 5 − 6 × 3.
3. D = 20 − 18.
4. D = 2.
5. Calculate the determinant Dx by replacing the x-column with the constants: Dx = 4 × 5 − 3 × 8.
6. Dx = 20 − 24.
7. Dx = −4.
8. Calculate the determinant Dy by replacing the y-column with constants: Dy = 4 × 8 − 4 × 6.
9. Dy = 32 − 24.
10. Dy = 8.
11. Find x using Cramer's rule: x = Dx ÷ D.
12. x = −4 ÷ 2.
13. x = −2.
14. Find y using Cramer's rule: y = Dy ÷ D.
15. y = 8 ÷ 2.
16. y = 4.
Answer: x = -2, y = 4
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Practice Set 1.3, Q3(3)
Solve by Cramer's rule: x + 2y = −1 and 2x − 3y = 12.
1. Write the equations in the form ax + by = c: x + 2y = -1 and 2x - 3y = 12.
2. Identify the coefficients: a = 1, b = 2, c = -1, d = 2, e = -3, f = 12.
3. Calculate the determinant D = a × e − b × d = 1 × (-3) − 2 × 2 = -3 − 4 = -7.
4. Calculate the determinant Dx = c × e − b × f = (-1) × (-3) − 2 × 12 = 3 − 24 = -21.
5. Calculate the determinant Dy = a × f − c × d = 1 × 12 − (-1) × 2 = 12 + 2 = 14.
6. Find x using Cramer's rule: x = Dx ÷ D = -21 ÷ -7 = 3.
7. Find y using Cramer's rule: y = Dy ÷ D = 14 ÷ -7 = -2.
Answer: x = 3, y = -2
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Practice Set 1.3, Q3(4)
Solve by Cramer's rule: 6x − 4y = −12 and 8x − 3y = −2.
1. Write the equations as 6x − 4y = −12 and 8x − 3y = −2.
2. Calculate the determinant D = |6 −4; 8 −3| = 6(−3) − (−4)(8) = −18 + 32 = 14.
3. Calculate the determinant Dx = |−12 −4; −2 −3| = (−12)(−3) − (−4)(−2) = 36 − 8 = 28.
4. Calculate the determinant Dy = |6 −12; 8 −2| = 6(−2) − (−12)(8) = −12 + 96 = 84.
5. Find x using Cramer's rule: x = Dx / D = 28 / 14 = 2.
6. Find y using Cramer's rule: y = Dy / D = 84 / 14 = 6.
Answer: x = 2, y = 6
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Practice Set 1.3, Q3(5)
Solve by Cramer's rule: 4m + 6n = 54 and 3m + 2n = 28.
1. Write the system as 4m + 6n = 54 and 3m + 2n = 28.
2. Calculate the determinant D of the coefficient matrix: D = 4 × 2 − 6 × 3.
3. D = 8 − 18 = −10.
4. Calculate the determinant Dm by replacing the first column with constants: Dm = 54 × 2 − 6 × 28.
5. Dm = 108 − 168 = −60.
6. Calculate the determinant Dn by replacing the second column with constants: Dn = 4 × 28 − 54 × 3.
7. Dn = 112 − 162 = −50.
8. Find m using Cramer's rule: m = Dm ÷ D.
9. m = −60 ÷ −10 = 6.
10. Find n using Cramer's rule: n = Dn ÷ D.
11. n = −50 ÷ −10 = 5.
Answer: m = 6, n = 5
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Practice Set 1.3, Q3(6)
Solve by Cramer's rule: 2x + 3y = 2 and x − y/2 = 1/2.
1. Write the equations in standard form: 2x + 3y = 2 and x − (1/2)y = 1/2.
2. Identify coefficients: a₁ = 2, b₁ = 3, c₁ = 2; a₂ = 1, b₂ = −1/2, c₂ = 1/2.
3. Calculate the determinant D = (2)(−1/2) − (3)(1) = −1 − 3 = −4.
4. Calculate the determinant Dₓ = (2)(−1/2) − (3)(1/2) = −1 − 3/2 = −5/2.
5. Calculate the determinant Dᵧ = (2)(1/2) − (2)(1) = 1 − 2 = −1.
6. Find x using Cramer's rule: x = Dₓ / D = (−5/2) / (−4) = 5/8.
7. Find y using Cramer's rule: y = Dᵧ / D = (−1) / (−4) = 1/4.
Answer: x = 5/8, y = 1/4
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Practice Set 1.4

Practice Set 1.4, Q1(1)
Solve: 2/x − 3/y = 15 and 8/x + 5/y = 77.
1. Let 1/x = u and 1/y = v.
2. The equations become 2u − 3v = 15 and 8u + 5v = 77.
3. Multiply the first equation by 4 to get 8u − 12v = 60.
4. Subtract this new equation from the second equation: (8u + 5v) − (8u − 12v) = 77 − 60.
5. Simplify to find 17v = 17.
6. Solve for v to get v = 1.
7. Substitute v = 1 into the first equation: 2u − 3(1) = 15.
8. Simplify to get 2u = 18.
9. Solve for u to get u = 9.
10. Since u = 1/x, we have 1/x = 9, so x = 1/9.
11. Since v = 1/y, we have 1/y = 1, so y = 1.
Answer: x = 1/9, y = 1
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Practice Set 1.4, Q1(2)
Solve: 10/(x + y) + 2/(x − y) = 4 and 15/(x + y) − 5/(x − y) = −2.
1. Let 1/(x + y) = a and 1/(x − y) = b.
2. The first equation becomes 10a + 2b = 4.
3. The second equation becomes 15a − 5b = −2.
4. Multiply the first equation by 5 to get 50a + 10b = 20.
5. Multiply the second equation by 2 to get 30a − 10b = −4.
6. Add the two new equations to get 80a = 16.
7. Solve for a to get a = 16/80 = 1/5.
8. Substitute a = 1/5 into 10a + 2b = 4 to get 10(1/5) + 2b = 4.
9. Simplify to get 2 + 2b = 4, so 2b = 2.
10. Solve for b to get b = 1.
11. Since a = 1/(x + y) = 1/5, we have x + y = 5.
12. Since b = 1/(x − y) = 1, we have x − y = 1.
13. Add the equations x + y = 5 and x − y = 1 to get 2x = 6.
14. Solve for x to get x = 3.
15. Substitute x = 3 into x + y = 5 to get 3 + y = 5.
16. Solve for y to get y = 2.
Answer: x = 3, y = 2
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Practice Set 1.4, Q1(3)
Solve: 27/(x − 2) + 31/(y + 3) = 85 and 31/(x − 2) + 27/(y + 3) = 89.
1. Let 1/(x − 2) = u and 1/(y + 3) = v.
2. The equations become 27u + 31v = 85 and 31u + 27v = 89.
3. Multiply the first equation by 31 to get 837u + 961v = 2635.
4. Multiply the second equation by 27 to get 837u + 729v = 2403.
5. Subtract the second result from the first to get 232v = 232.
6. Solve for v to get v = 1.
7. Substitute v = 1 into 27u + 31v = 85 to get 27u + 31 = 85.
8. Solve for u to get 27u = 54, so u = 2.
9. Since u = 1/(x − 2) = 2, we have x − 2 = 1/2, so x = 5/2.
10. Since v = 1/(y + 3) = 1, we have y + 3 = 1, so y = −2.
Answer: x = 5/2, y = -2
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Practice Set 1.4, Q1(4)
Solve: 1/(3x + y) + 1/(3x − y) = 3/4 and 1/(2(3x + y)) − 1/(2(3x − y)) = −1/8.
1. Let 1/(3x + y) = a and 1/(3x − y) = b.
2. The first equation becomes a + b = 3/4.
3. The second equation becomes (1/2)a − (1/2)b = −1/8.
4. Multiply the second equation by 2 to get a − b = −1/4.
5. Add the equations a + b = 3/4 and a − b = −1/4 to get 2a = 1/2.
6. Solve for a to get a = 1/4.
7. Substitute a = 1/4 into a + b = 3/4 to get 1/4 + b = 3/4.
8. Solve for b to get b = 1/2.
9. Since a = 1/(3x + y) = 1/4, we have 3x + y = 4.
10. Since b = 1/(3x − y) = 1/2, we have 3x − y = 2.
11. Add the equations 3x + y = 4 and 3x − y = 2 to get 6x = 6.
12. Solve for x to get x = 1.
13. Substitute x = 1 into 3x + y = 4 to get 3(1) + y = 4.
14. Solve for y to get y = 1.
Answer: x = 1, y = 1
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Practice Set 1.5

Practice Set 1.5, Q1
Two numbers differ by 3. Twice the smaller plus three times the greater is 19. Find the numbers.
1. Let the smaller number be x and the greater number be y.
2. Since the numbers differ by 3, we have y - x = 3.
3. Twice the smaller plus three times the greater is 19, so 2x + 3y = 19.
4. From the first equation, y = x + 3.
5. Substitute y = x + 3 into the second equation: 2x + 3(x + 3) = 19.
6. Simplify the equation: 2x + 3x + 9 = 19.
7. Combine like terms: 5x + 9 = 19.
8. Subtract 9 from both sides: 5x = 10.
9. Divide by 5: x = 2.
10. Substitute x = 2 into y = x + 3 to find y: y = 2 + 3.
11. Calculate y: y = 5.
Answer: 5 and 2
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Practice Set 1.5, Q2
A rectangle has opposite sides 2x + y + 8 and 4x − y, and the other pair of opposite sides 2y and x + 4. Find x and y, then the rectangle's perimeter and area.
1. Opposite sides of a rectangle are equal, so 2x + y + 8 = 4x − y.
2. Rearranging the terms gives 2y − 2x = −8.
3. Dividing by 2 gives y − x = −4, or x − y = 4.
4. The other pair of opposite sides are equal, so 2y = x + 4.
5. Substitute x = y + 4 into the second equation.
6. 2y = (y + 4) + 4.
7. 2y = y + 8.
8. y = 8.
9. Substitute y = 8 into x = y + 4.
10. x = 8 + 4.
11. x = 12.
12. Calculate the length of the first pair of sides: 2x + y + 8 = 2(12) + 8 + 8 = 40.
13. Calculate the length of the second pair of sides: 2y = 2(8) = 16.
14. Perimeter = 2 × (40 + 16).
15. Perimeter = 2 × 56 = 112.
16. Area = 40 × 16.
17. Area = 640.
Answer: x = 12, y = 8, perimeter 112, area 640
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Practice Set 1.5, Q3
A father's age plus twice his son's age is 70; twice the father's age plus the son's age is 95. Find their present ages.
1. Let the father's present age be x years and the son's present age be y years.
2. From the first condition, x + 2y = 70.
3. From the second condition, 2x + y = 95.
4. Multiply the first equation by 2 to get 2x + 4y = 140.
5. Subtract the second equation from this new equation: (2x + 4y) - (2x + y) = 140 - 95.
6. Simplify the subtraction to find 3y = 45.
7. Divide by 3 to get y = 15.
8. Substitute y = 15 into the first equation: x + 2(15) = 70.
9. Calculate x + 30 = 70.
10. Subtract 30 from both sides to get x = 40.
Answer: son 15 years, father 40 years
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Practice Set 1.5, Q4
A fraction's denominator is 4 more than twice its numerator. If 6 is taken from both, the denominator becomes 12 times the numerator. Find the fraction.
1. Let the numerator of the fraction be x.
2. The denominator is 4 more than twice the numerator, so the denominator is 2x + 4.
3. The original fraction is x / (2x + 4).
4. If 6 is subtracted from the numerator, it becomes x - 6.
5. If 6 is subtracted from the denominator, it becomes (2x + 4) - 6 = 2x - 2.
6. According to the condition, the new denominator is 12 times the new numerator.
7. So, 2x - 2 = 12(x - 6).
8. Expanding the right side gives 2x - 2 = 12x - 72.
9. Rearranging terms gives 72 - 2 = 12x - 2x.
10. Simplifying the equation gives 70 = 10x.
11. Solving for x gives x = 70 / 10.
12. Therefore, x = 7.
13. The numerator is 7.
14. The denominator is 2(7) + 4 = 14 + 4 = 18.
15. The fraction is 7/18.
Answer: 7/18
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Practice Set 1.5, Q5
A truck holds 10 tonnes. It is full with 150 boxes of type A and 100 of type B, and also full with 260 of type A and 40 of type B. Find the weight of each type of box.
1. Let the weight of one box of type A be x kg and the weight of one box of type B be y kg.
2. The truck holds 10 tonnes, which is equal to 10000 kg.
3. From the first condition, 150 boxes of type A and 100 boxes of type B fill the truck, so 150x + 100y = 10000.
4. Dividing the first equation by 50 gives 3x + 2y = 200.
5. From the second condition, 260 boxes of type A and 40 boxes of type B fill the truck, so 260x + 40y = 10000.
6. Dividing the second equation by 20 gives 13x + 2y = 500.
7. Subtracting the simplified first equation from the simplified second equation gives (13x + 2y) - (3x + 2y) = 500 - 200.
8. This simplifies to 10x = 300.
9. Solving for x gives x = 30.
10. Substituting x = 30 into 3x + 2y = 200 gives 3(30) + 2y = 200.
11. This becomes 90 + 2y = 200.
12. Subtracting 90 from both sides gives 2y = 110.
13. Solving for y gives y = 55.
Answer: A 30 kg; B 55 kg
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Practice Set 1.5, Q6
Vishal travels 1900 km, partly by bus at 60 km/h and partly by aeroplane at 700 km/h, in 5 hours in all. Find the distance he travels by bus.
1. Let the distance travelled by bus be x km.
2. Then, the distance travelled by aeroplane is (1900 − x) km.
3. Time taken by bus = x ÷ 60 hours.
4. Time taken by aeroplane = (1900 − x) ÷ 700 hours.
5. Total time = 5 hours, so x ÷ 60 + (1900 − x) ÷ 700 = 5.
6. Multiply the entire equation by 4200 (LCM of 60 and 700) to clear denominators.
7. 70x + 6(1900 − x) = 21000.
8. 70x + 11400 − 6x = 21000.
9. 64x + 11400 = 21000.
10. 64x = 21000 − 11400.
11. 64x = 9600.
12. x = 9600 ÷ 64.
13. x = 150.
Answer: 150 km
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Problem Set 1

Not solved here yet: Q1(5), Q3(3), Q5(5) (see the textbook).

Problem Set 1, Q1(1)
Choose the correct option: for 4x + 5y = 19, y when x = 1 is (A) 4 (B) 3 (C) 2 (D) −3.
1. Given equation: 4x + 5y = 19
2. Substitute x = 1 into the equation: 4(1) + 5y = 19
3. 4 + 5y = 19
4. 5y = 19 − 4
5. 5y = 15
6. y = 15 ÷ 5
7. y = 3
8. The value 3 corresponds to option (B)
Answer: (B)
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Problem Set 1, Q1(2)
Choose the correct option: Dx = 49, Dy = −63 and D = 7; x = (A) 7 (B) −7 (C) 1/7 (D) −1/7.
1. The value of x is given by the formula x = Dx ÷ D.
2. Substitute the given values into the formula: x = 49 ÷ 7.
3. x = 7.
4. This matches option (A).
Answer: (A)
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Problem Set 1, Q1(3)
Choose the correct option: the determinant with rows (5, 3) and (−7, −4) is (A) −1 (B) −41 (C) 41 (D) 1.
1. The determinant of a 2x2 matrix with rows (5, 3) and (−7, −4) is calculated as (5 × −4) − (3 × −7).
2. 5 × −4 = −20
3. 3 × −7 = −21
4. −20 − (−21) = −20 + 21
5. −20 + 21 = 1
6. The calculated value is 1, which corresponds to option (D).
Answer: (D)
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Problem Set 1, Q1(4)
Choose the correct option: for x + y = 3 and 3x − 2y − 4 = 0, D = (A) 5 (B) 1 (C) −5 (D) −1.
1. The given equations are x + y = 3 and 3x − 2y − 4 = 0.
2. Rewrite the first equation as x + y − 3 = 0.
3. The coefficients are a₁ = 1, b₁ = 1, c₁ = −3 and a₂ = 3, b₂ = −2, c₂ = −4.
4. The determinant D is calculated as a₁b₂ − a₂b₁.
5. D = (1)(−2) − (3)(1).
6. D = −2 − 3.
7. D = −5.
8. The value −5 corresponds to option (C).
Answer: (C)
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Problem Set 1, Q2
Complete the table of points for the graph of 2x − 6y = 3: x = −5 (find y) and y = 0 (find x).
1. The given equation is 2x − 6y = 3.
2. For the first point, substitute x = −5 into the equation.
3. 2(−5) − 6y = 3
4. −10 − 6y = 3
5. −6y = 3 + 10
6. −6y = 13
7. y = 13 ÷ (−6)
8. y = −13/6
9. For the second point, substitute y = 0 into the equation.
10. 2x − 6(0) = 3
11. 2x − 0 = 3
12. 2x = 3
13. x = 3 ÷ 2
14. x = 3/2
Answer: y = -13/6; x = 3/2
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Problem Set 1, Q3(1)
Solve graphically: 2x + 3y = 12 and x − y = 1.
1. Write the first equation as 2x + 3y = 12.
2. Find two points for the first line: when x = 0, y = 4, and when y = 0, x = 6.
3. Write the second equation as x − y = 1.
4. Find two points for the second line: when x = 0, y = −1, and when y = 0, x = 1.
5. Plot the points (0, 4) and (6, 0) and draw the line for 2x + 3y = 12.
6. Plot the points (0, −1) and (1, 0) and draw the line for x − y = 1.
7. Observe that the two lines intersect at the point (3, 2).
Answer: x = 3, y = 2
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Problem Set 1, Q3(2)
Solve graphically: x − 3y = 1 and 3x − 2y + 4 = 0.
1. For the equation x − 3y = 1, rearrange to y = (x − 1)/3.
2. Find points for the first line: when x = 1, y = 0; when x = 4, y = 1.
3. For the equation 3x − 2y + 4 = 0, rearrange to y = (3x + 4)/2.
4. Find points for the second line: when x = 0, y = 2; when x = −2, y = −1.
5. Plot the points (1, 0) and (4, 1) and draw the first line.
6. Plot the points (0, 2) and (−2, −1) and draw the second line.
7. Observe that the two lines intersect at the point (−2, −1).
8. The solution is x = −2 and y = −1.
Answer: x = -2, y = -1
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Problem Set 1, Q3(4)
Solve graphically: 3x − y − 2 = 0 and 2x + y = 8.
1. For the equation 3x − y − 2 = 0, rearrange it as y = 3x − 2.
2. Find points for the first line: when x = 0, y = −2; when x = 1, y = 1; when x = 2, y = 4.
3. For the equation 2x + y = 8, rearrange it as y = 8 − 2x.
4. Find points for the second line: when x = 0, y = 8; when x = 2, y = 4; when x = 4, y = 0.
5. Plot the points (0, −2), (1, 1), (2, 4) and draw the first line.
6. Plot the points (0, 8), (2, 4), (4, 0) and draw the second line.
7. Observe that both lines intersect at the point (2, 4).
8. The intersection point gives the solution to the system of equations.
9. Therefore, x = 2 and y = 4.
Answer: x = 2, y = 4
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Problem Set 1, Q3(5)
Solve graphically: 3x + y = 10 and x − y = 2.
1. For the equation 3x + y = 10, set x = 0 to get y = 10, giving point (0, 10).
2. For the equation 3x + y = 10, set y = 0 to get 3x = 10, so x = 10/3, giving point (10/3, 0).
3. For the equation x − y = 2, set x = 0 to get −y = 2, so y = −2, giving point (0, −2).
4. For the equation x − y = 2, set y = 0 to get x = 2, giving point (2, 0).
5. Plot the points (0, 10) and (10/3, 0) and draw the line for 3x + y = 10.
6. Plot the points (0, −2) and (2, 0) and draw the line for x − y = 2.
7. The two lines intersect at the point (3, 1).
8. Therefore, the solution is x = 3 and y = 1.
Answer: x = 3, y = 1
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Problem Set 1, Q4
Find the determinants with rows: (1) (4, 3) and (2, 7) (2) (5, −2) and (−3, 1) (3) (3, −1) and (1, 4)
1. (1) The determinant has rows (4, 3) and (2, 7).
2. (1) Multiply the main diagonal: 4 × 7 = 28
3. (1) Multiply the other diagonal: 3 × 2 = 6
4. (1) Subtract the second product from the first: 28 − 6 = 22
5. (2) The determinant has rows (5, −2) and (−3, 1).
6. (2) Multiply the main diagonal: 5 × 1 = 5
7. (2) Multiply the other diagonal: −2 × −3 = 6
8. (2) Subtract the second product from the first: 5 − 6 = −1
9. (3) The determinant has rows (3, −1) and (1, 4).
10. (3) Multiply the main diagonal: 3 × 4 = 12
11. (3) Multiply the other diagonal: −1 × 1 = −1
12. (3) Subtract the second product from the first: 12 − (−1) = 12 + 1 = 13
Answer: (1) 22; (2) -1; (3) 13
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Problem Set 1, Q5(1)
Solve by Cramer's method: 6x − 3y = −10 and 3x + 5y − 8 = 0.
1. Write the equations in the form ax + by = c: 6x − 3y = −10 and 3x + 5y = 8.
2. Identify the coefficients: a₁ = 6, b₁ = −3, c₁ = −10; a₂ = 3, b₂ = 5, c₂ = 8.
3. Calculate the determinant D = |6 −3; 3 5| = (6 × 5) − (−3 × 3) = 30 − (−9) = 30 + 9 = 39.
4. Calculate the determinant Dx = |−10 −3; 8 5| = (−10 × 5) − (−3 × 8) = −50 − (−24) = −50 + 24 = −26.
5. Calculate the determinant Dy = |6 −10; 3 8| = (6 × 8) − (−10 × 3) = 48 − (−30) = 48 + 30 = 78.
6. Find x using Cramer's rule: x = Dx / D = −26 / 39 = −2/3.
7. Find y using Cramer's rule: y = Dy / D = 78 / 39 = 2.
Answer: x = -2/3, y = 2
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Problem Set 1, Q5(2)
Solve by Cramer's method: 4m − 2n = −4 and 4m + 3n = 16.
1. Write the system of equations: 4m − 2n = −4 and 4m + 3n = 16.
2. Calculate the determinant D = 4 × 3 − (−2) × 4.
3. D = 12 + 8.
4. D = 20.
5. Calculate the determinant Dm by replacing the first column with constants: Dm = −4 × 3 − (−2) × 16.
6. Dm = −12 + 32.
7. Dm = 20.
8. Calculate the determinant Dn by replacing the second column with constants: Dn = 4 × 16 − (−4) × 4.
9. Dn = 64 + 16.
10. Dn = 80.
11. Find m using Cramer's rule: m = Dm ÷ D.
12. m = 20 ÷ 20.
13. m = 1.
14. Find n using Cramer's rule: n = Dn ÷ D.
15. n = 80 ÷ 20.
16. n = 4.
Answer: m = 1, n = 4
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Problem Set 1, Q5(3)
Solve by Cramer's method: 3x − 2y = 5/2 and (1/3)x + 3y = −4/3.
1. Write the equations in the form ax + by = c: 3x − 2y = 5/2 and (1/3)x + 3y = −4/3.
2. Calculate the determinant D = |3 -2; 1/3 3| = 3(3) − (−2)(1/3) = 9 + 2/3 = 29/3.
3. Calculate the determinant Dx = |5/2 -2; -4/3 3| = (5/2)(3) − (−2)(−4/3) = 15/2 − 8/3 = 29/6.
4. Calculate the determinant Dy = |3 5/2; 1/3 -4/3| = 3(−4/3) − (5/2)(1/3) = −4 − 5/6 = −29/6.
5. Find x using Cramer's rule: x = Dx / D = (29/6) / (29/3) = 1/2.
6. Find y using Cramer's rule: y = Dy / D = (−29/6) / (29/3) = −1/2.
Answer: x = 1/2, y = -1/2
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Problem Set 1, Q5(4)
Solve by Cramer's method: 7x + 3y = 15 and 12y − 5x = 39.
1. Write the equations in standard form: 7x + 3y = 15 and -5x + 12y = 39.
2. Calculate the determinant D: D = 7(12) - 3(-5) = 84 + 15 = 99.
3. Calculate the determinant Dx: Dx = 15(12) - 3(39) = 180 - 117 = 63.
4. Calculate the determinant Dy: Dy = 7(39) - 15(-5) = 273 + 75 = 348.
5. Find x using Cramer's rule: x = Dx / D = 63 / 99 = 7/11.
6. Find y using Cramer's rule: y = Dy / D = 348 / 99 = 116/33.
Answer: x = 7/11, y = 116/33
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Problem Set 1, Q6(1)
Solve: 2/x + 2/(3y) = 1/6 and 3/x + 2/y = 0.
1. Let 1/x = u and 1/y = v.
2. The first equation becomes 2u + 2/3 v = 1/6.
3. Multiply by 6 to clear denominators: 12u + 4v = 1.
4. The second equation becomes 3u + 2v = 0.
5. Multiply the second equation by 2: 6u + 4v = 0.
6. Subtract this from the first modified equation: (12u + 4v) - (6u + 4v) = 1 - 0.
7. Simplify to find u: 6u = 1, so u = 1/6.
8. Substitute u = 1/6 into 3u + 2v = 0: 3(1/6) + 2v = 0.
9. Simplify: 1/2 + 2v = 0, so 2v = -1/2.
10. Solve for v: v = -1/4.
11. Since u = 1/x, we have 1/x = 1/6, so x = 6.
12. Since v = 1/y, we have 1/y = -1/4, so y = -4.
Answer: x = 6, y = -4
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Problem Set 1, Q6(2)
Solve: 7/(2x + 1) + 13/(y + 2) = 27 and 13/(2x + 1) + 7/(y + 2) = 33.
1. Let 1/(2x + 1) = a and 1/(y + 2) = b.
2. The equations become 7a + 13b = 27 and 13a + 7b = 33.
3. Multiply the first equation by 13: 91a + 169b = 351.
4. Multiply the second equation by 7: 91a + 49b = 231.
5. Subtract the second result from the first: 120b = 120.
6. Solve for b: b = 120/120 = 1.
7. Substitute b = 1 into 7a + 13b = 27: 7a + 13 = 27.
8. Solve for a: 7a = 14, so a = 2.
9. Since a = 1/(2x + 1) = 2, we have 2x + 1 = 1/2.
10. Solve for x: 2x = 1/2 - 1 = -1/2, so x = -1/4.
11. Since b = 1/(y + 2) = 1, we have y + 2 = 1.
12. Solve for y: y = 1 - 2 = -1.
Answer: x = -1/4, y = -1
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Problem Set 1, Q6(3)
Solve: 148/x + 231/y = 527/(xy) and 231/x + 148/y = 610/(xy).
1. Multiply both equations by xy to clear denominators: 148y + 231x = 527 and 231y + 148x = 610.
2. Label the first equation as (1) and the second as (2).
3. Multiply equation (1) by 231: 34188y + 53361x = 121737.
4. Multiply equation (2) by 148: 34188y + 21904x = 90280.
5. Subtract the second result from the first: (53361 - 21904)x = 121737 - 90280.
6. 31457x = 31457.
7. x = 31457 ÷ 31457.
8. x = 1.
9. Substitute x = 1 into equation (1): 148y + 231(1) = 527.
10. 148y + 231 = 527.
11. 148y = 527 - 231.
12. 148y = 296.
13. y = 296 ÷ 148.
14. y = 2.
Answer: x = 1, y = 2
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Problem Set 1, Q6(4)
Solve: (7x − 2y)/(xy) = 5 and (8x + 7y)/(xy) = 15.
1. Given equations are (7x − 2y)/(xy) = 5 and (8x + 7y)/(xy) = 15.
2. Split the first: 7x/(xy) − 2y/(xy) = 7/y − 2/x = 5.
3. Split the second: 8x/(xy) + 7y/(xy) = 8/y + 7/x = 15.
4. Let 1/x = u and 1/y = v, so the equations become −2u + 7v = 5 and 7u + 8v = 15.
5. Multiply the first equation by 7: −14u + 49v = 35.
6. Multiply the second equation by 2: 14u + 16v = 30.
7. Add the two new equations: 65v = 65, so v = 1.
8. Substitute v = 1 into −2u + 7v = 5: −2u + 7 = 5, so u = 1.
9. Since u = 1/x, we have 1/x = 1, so x = 1.
10. Since v = 1/y, we have 1/y = 1, so y = 1.
Answer: x = 1, y = 1
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Problem Set 1, Q6(5)
Solve: 1/(2(3x + 4y)) + 1/(5(2x − 3y)) = 1/4 and 5/(3x + 4y) − 2/(2x − 3y) = −3/2.
1. Let 1/(3x + 4y) = u and 1/(2x − 3y) = v.
2. The first equation becomes (1/2)u + (1/5)v = 1/4.
3. The second equation becomes 5u − 2v = −3/2.
4. Multiply the first equation by 20 to get 10u + 4v = 5.
5. Multiply the second equation by 2 to get 10u − 4v = −3.
6. Add the two new equations: 20u = 2, so u = 1/10.
7. Substitute u = 1/10 into 10u + 4v = 5 to get 1 + 4v = 5, so v = 1.
8. Since u = 1/(3x + 4y) = 1/10, we have 3x + 4y = 10.
9. Since v = 1/(2x − 3y) = 1, we have 2x − 3y = 1.
10. Multiply the first linear equation by 3 and the second by 4 to get 9x + 12y = 30 and 8x − 12y = 4.
11. Add these equations: 17x = 34, so x = 2.
12. Substitute x = 2 into 2x − 3y = 1 to get 4 − 3y = 1, so y = 1.
Answer: x = 2, y = 1
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Problem Set 1, Q7(1)
A two-digit number and the number with its digits swapped add up to 143. The units digit is 3 more than the tens digit. Find the number.
1. Let the tens digit be x and the units digit be y.
2. The number is 10x + y and the swapped number is 10y + x.
3. Given that the sum of the numbers is 143, so (10x + y) + (10y + x) = 143.
4. Simplifying the equation gives 11x + 11y = 143.
5. Dividing by 11 gives x + y = 13.
6. Given that the units digit is 3 more than the tens digit, so y = x + 3.
7. Substitute y = x + 3 into x + y = 13 to get x + (x + 3) = 13.
8. Solving 2x + 3 = 13 gives 2x = 10.
9. Therefore x = 5.
10. Substitute x = 5 into y = x + 3 to get y = 5 + 3 = 8.
11. The number is 10(5) + 8 = 58.
Answer: 58
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Problem Set 1, Q7(2)
1.5 kg tea and 5 kg sugar plus ₹50 rickshaw fare cost ₹700 in all; 2 kg tea and 7 kg sugar ordered online with free delivery cost ₹880. Find the price per kg of tea and of sugar.
1. Let the price of 1 kg tea be ₹x and the price of 1 kg sugar be ₹y.
2. From the first condition, 1.5x + 5y + 50 = 700.
3. Subtracting 50 from both sides gives 1.5x + 5y = 650.
4. Multiplying the entire equation by 2 to remove the decimal gives 3x + 10y = 1300.
5. From the second condition, 2x + 7y = 880.
6. Multiplying the second equation by 3 gives 6x + 21y = 2640.
7. Multiplying the first derived equation by 2 gives 6x + 20y = 2600.
8. Subtracting the two new equations gives (6x + 21y) − (6x + 20y) = 2640 − 2600.
9. This simplifies to y = 40.
10. Substituting y = 40 into 2x + 7y = 880 gives 2x + 7(40) = 880.
11. Calculating 7 × 40 gives 2x + 280 = 880.
12. Subtracting 280 from both sides gives 2x = 600.
13. Dividing by 2 gives x = 300.
Answer: tea ₹300, sugar ₹40
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Problem Set 1, Q7(3)
Anushka got ₹2500 in ₹100 and ₹50 notes. With the numbers of the two kinds of notes swapped, she would have got ₹500 less. How many notes of each kind did she get?
1. Let the number of ₹100 notes be x and the number of ₹50 notes be y.
2. The total amount received is ₹2500, so 100x + 50y = 2500.
3. Dividing the equation by 50 gives 2x + y = 50.
4. If the numbers of notes are swapped, the amount is ₹500 less than ₹2500, which is ₹2000.
5. The equation for the swapped notes is 100y + 50x = 2000.
6. Dividing this equation by 50 gives 2y + x = 40.
7. From the first simplified equation, y = 50 − 2x.
8. Substitute y = 50 − 2x into the second simplified equation: x + 2(50 − 2x) = 40.
9. Simplify to get x + 100 − 4x = 40.
10. Combine like terms to get −3x + 100 = 40.
11. Subtract 100 from both sides to get −3x = −60.
12. Divide by −3 to find x = 20.
13. Substitute x = 20 into y = 50 − 2x to get y = 50 − 40.
14. Calculate y = 10.
Answer: 20 notes of ₹100; 10 notes of ₹50
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Problem Set 1, Q7(4)
Manish's and Savita's present ages add up to 31. Three years ago Manish was 4 times as old as Savita. Find their present ages.
1. Let the present age of Manish be x years and the present age of Savita be y years.
2. According to the first condition, the sum of their present ages is 31, so x + y = 31.
3. Three years ago, Manish's age was x - 3 and Savita's age was y - 3.
4. According to the second condition, three years ago Manish was 4 times as old as Savita, so x - 3 = 4(y - 3).
5. Simplify the second equation to get x - 3 = 4y - 12.
6. Rearrange the equation to x - 4y = -9.
7. Subtract the simplified second equation from the first equation: (x + y) - (x - 4y) = 31 - (-9).
8. This gives 5y = 40.
9. Solving for y, we get y = 8.
10. Substitute y = 8 into the first equation: x + 8 = 31.
11. Solving for x, we get x = 23.
Answer: Manish 23, Savita 8
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Problem Set 1, Q7(5)
Skilled and unskilled workers' daily wages are in the ratio 5 : 3, and one of each together earn ₹720 a day. Find each daily wage.
1. Let the daily wage of a skilled worker be 5x and that of an unskilled worker be 3x.
2. According to the problem, the sum of their wages is 720.
3. 5x + 3x = 720
4. 8x = 720
5. x = 720 ÷ 8
6. x = 90
7. Daily wage of a skilled worker = 5 × 90 = 450
8. Daily wage of an unskilled worker = 3 × 90 = 270
Answer: skilled ₹450, unskilled ₹270
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Problem Set 1, Q7(6)
A and B are 30 km apart on a straight road. Hamid rides from A and Joseph from B at the same time; riding towards each other they meet after 20 minutes; riding in the same direction (Joseph away from A) Hamid catches him after 3 hours. Find their speeds.
1. Let Hamid's speed be x km/h and Joseph's speed be y km/h.
2. Time taken to meet when riding towards each other is 20 minutes, which is 20/60 = 1/3 hour.
3. Distance covered by Hamid in 1/3 hour is x × 1/3 = x/3 km.
4. Distance covered by Joseph in 1/3 hour is y × 1/3 = y/3 km.
5. Since they meet, the sum of their distances equals the total distance between A and B, so x/3 + y/3 = 30.
6. Multiplying the equation by 3 gives x + y = 90.
7. Time taken for Hamid to catch Joseph when riding in the same direction is 3 hours.
8. Relative speed when moving in the same direction is x - y km/h.
9. Distance covered by Hamid relative to Joseph in 3 hours is 3(x - y) km.
10. Since Hamid catches Joseph, this relative distance equals the initial separation of 30 km, so 3(x - y) = 30.
11. Dividing by 3 gives x - y = 10.
12. Adding the equations x + y = 90 and x - y = 10 gives 2x = 100.
13. Solving for x gives x = 50.
14. Substituting x = 50 into x + y = 90 gives 50 + y = 90.
15. Solving for y gives y = 40.
Answer: Hamid 50 km/h, Joseph 40 km/h
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Other chapters solved for this book

Chapter 2: Quadratic Equations — 73 questions solved
Chapter 3: Arithmetic Progression — 47 questions solved
Chapter 5: Probability — 25 questions solved

← All state-board textbook solutions

Maharashtra SSC Class 10 Mathematics Chapter 1 Linear Equations in Two Variables: textbook solutions

Practice Set 1.1

Practice Set 1.1, Q1

Practice Set 1.1, Q2(1)

Practice Set 1.1, Q2(2)

Practice Set 1.1, Q2(3)

Practice Set 1.1, Q2(4)

Practice Set 1.1, Q2(5)

Practice Set 1.1, Q2(6)

Practice Set 1.1, Q2(7)

Practice Set 1.1, Q2(8)

Practice Set 1.2

Practice Set 1.2, Q1

Practice Set 1.2, Q2(1)

Practice Set 1.2, Q2(2)

Practice Set 1.2, Q2(3)

Practice Set 1.2, Q2(4)

Practice Set 1.2, Q2(5)

Practice Set 1.2, Q2(6)

Practice Set 1.3

Practice Set 1.3, Q1

Practice Set 1.3, Q3(1)

Practice Set 1.3, Q3(2)

Practice Set 1.3, Q3(3)

Practice Set 1.3, Q3(4)

Practice Set 1.3, Q3(5)

Practice Set 1.3, Q3(6)

Practice Set 1.4

Practice Set 1.4, Q1(1)

Practice Set 1.4, Q1(2)

Practice Set 1.4, Q1(3)

Practice Set 1.4, Q1(4)

Practice Set 1.5

Practice Set 1.5, Q1

Practice Set 1.5, Q2

Practice Set 1.5, Q3

Practice Set 1.5, Q4

Practice Set 1.5, Q5

Practice Set 1.5, Q6

Problem Set 1

Problem Set 1, Q1(1)

Problem Set 1, Q1(2)

Problem Set 1, Q1(3)

Problem Set 1, Q1(4)

Problem Set 1, Q2

Problem Set 1, Q3(1)

Problem Set 1, Q3(2)

Problem Set 1, Q3(4)

Problem Set 1, Q3(5)

Problem Set 1, Q4

Problem Set 1, Q5(1)

Problem Set 1, Q5(2)

Problem Set 1, Q5(3)

Problem Set 1, Q5(4)

Problem Set 1, Q6(1)

Problem Set 1, Q6(2)

Problem Set 1, Q6(3)

Problem Set 1, Q6(4)

Problem Set 1, Q6(5)

Problem Set 1, Q7(1)

Problem Set 1, Q7(2)

Problem Set 1, Q7(3)

Problem Set 1, Q7(4)

Problem Set 1, Q7(5)

Problem Set 1, Q7(6)