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Questions
Section A: Objective and Very Short Answer (1 mark each)
Q1. Find a so that the point (2, 9) lies on the line ax − 3y = 5. [1 mark]
View Answer & Explanation
Answer: Substituting: 2a − 27 = 5, so 2a = 32 and a = 16.
Teacher's Note:
1. A point on a line satisfies its equation.
2. Replace x by 2 and y by 9.
Q2. Find k so that the lines 2x − 3y = 9 and kx − 9y = 18 are parallel. [1 mark]
View Answer & Explanation
Answer: Write as 2x − 3y − 9 = 0 and kx − 9y − 18 = 0. Parallel needs 2/k = 3/9 ≠ 9/18, i.e. 2/k = 1/3 ≠ 1/2.
So k = 6.
Teacher's Note:
1. Parallel lines need equal ratios of x and y coefficients but a different ratio for constants.
2. Check the third ratio is different, which it is.
Q3. For what value of k does the pair x + 2y = 5 and 3x + ky + 15 = 0 have no solution? [1 mark]
View Answer & Explanation
Answer: No solution needs 1/3 = 2/k ≠ −5/15. So k = 6, and 1/3 ≠ −1/3 holds. k = 6.
Teacher's Note:
1. No solution means parallel lines.
2. Verify the constant-term ratio is different.
Q4. Determine the values of a for which 3x + 2y − 4 = 0 and ax − y − 3 = 0 have a unique solution. [1 mark]
View Answer & Explanation
Answer: Unique solution needs 3/a ≠ 2/(−1) = −2, so a ≠ −3/2. Hence a ≠ −3/2.
Teacher's Note:
1. Use a₁/a₂ ≠ b₁/b₂ for intersecting lines.
2. Keep the signs of the coefficients.
Q5. Write any one linear equation that is parallel to 2x − 3y = 5. [1 mark]
View Answer & Explanation
Answer: Keep the ratio of coefficients and change the constant: for example 4x − 6y = 7.
Teacher's Note:
1. Multiply the x and y coefficients by the same number.
2. Make sure the constant does not scale in the same way.
Q6. Find the point where the line −3x + 7y = 3 meets the x-axis. [1 mark]
View Answer & Explanation
Answer: On the x-axis y = 0, so −3x = 3 and x = −1. The point is (−1, 0).
Teacher's Note:
1. Every point on the x-axis has y = 0.
2. Substitute and solve for x.
Q7. Find k for which the pair (k − 3)x + 3y = k and k(x + y) = 12 has infinitely many solutions. [1 mark]
View Answer & Explanation
Answer: Need (k − 3)/k = 3/k = k/12. From 3/k = k/12, k² = 36. From (k − 3)/k = 3/k, k − 3 = 3 so k = 6.
Hence k = 6, which also satisfies k² = 36.
Teacher's Note:
1. Infinite solutions mean coincident lines with all three ratios equal.
2. Use two ratios and check the answer in the third.
Q8. For what value of k do x + 5y − 7 = 0 and 4x + 20y + k = 0 represent coincident lines? [1 mark]
View Answer & Explanation
Answer: Need 1/4 = 5/20 = −7/k, so k = −28.
The answer is k = −28.
Teacher's Note:
1. All three ratios must be equal.
2. The first two already equal 1/4.
Q9. Does the pair 2x + 3y = 5 and 4x + 6y = 10 have a solution? What kind of pair is it? [1 mark]
View Answer & Explanation
Answer: The ratios 2/4 = 3/6 = 5/10 are all equal, so the lines are coincident, the pair is consistent and dependent, and it has infinitely many solutions.
Teacher's Note:
1. Compare all three ratios.
2. Equal ratios mean the lines coincide.
Q10. The pair of equations x + 2y − 4 = 0 and 2x + 4y − 12 = 0 represents [1 mark]
(a) intersecting lines
(b) coincident lines
(c) parallel lines
(d) perpendicular lines
View Answer & Explanation
Answer: (c) parallel lines. Since 1/2 = 2/4 ≠ 4/12, the lines never meet.
Teacher's Note:
1. Compare a₁/a₂, b₁/b₂ and c₁/c₂.
2. The first two ratios agree, the third does not.
Section B: Short Answer (2 and 3 marks)
Q11. Five pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen. [3 marks]
View Answer & Explanation
Answer: 5x + 7y = 50 and 7x + 5y = 46.
Adding: 12(x + y) = 96, so x + y = 8. Subtracting: 2y − 2x = 4, so y − x = 2.
Hence pencil = ₹3, pen = ₹5.
Teacher's Note:
1. Adding and subtracting the equations simplifies the work here.
2. Check: 15 + 35 = 50 and 21 + 25 = 46.
Q12. Solve: 3x − y = 3 and 7x + 2y = 20. [2 marks]
View Answer & Explanation
Answer: Multiply the first by 2: 6x − 2y = 6. Add to the second: 13x = 26, so x = 2.
Then y = 3(2) − 3 = 3. x = 2, y = 3.
Teacher's Note:
1. Make the coefficient of y equal in size.
2. Substitute back to find the other variable.
Q13. Solve by substitution: 3x − 5y = −1 and x − y = −1. [3 marks]
View Answer & Explanation
Answer: From the second, x = y − 1. Then 3(y − 1) − 5y = −1, so −2y = 2 and y = −1.
x = −2. x = −2, y = −1.
Teacher's Note:
1. Express one variable from the simpler equation.
2. Check in 3x − 5y: −6 + 5 = −1.
Q14. Solve: px + qy = p − q and qx − py = p + q. [3 marks]
View Answer & Explanation
Answer: Multiply the first by p and the second by q: p²x + pqy = p² − pq and q²x − pqy = pq + q².
Adding: (p² + q²)x = p² + q², so x = 1. Then p + qy = p − q gives y = −1.
x = 1, y = −1.
Teacher's Note:
1. Multiply so that the y terms cancel on adding.
2. Assume p² + q² ≠ 0.
Q15. Solve for x and y: x + y = a + b and ax − by = a² − b². [3 marks]
View Answer & Explanation
Answer: From the first, x = a + b − y. Substituting: a(a + b − y) − by = a² − b², so ab + b² = (a + b)y, giving y = b.
Then x = a. x = a, y = b.
Teacher's Note:
1. Substitute x from the first equation into the second.
2. Factor (a + b) out on the left.
Q16. Find the fraction which becomes 2/3 when the numerator is increased by 2, and 4/7 when the denominator is increased by 4. [3 marks]
View Answer & Explanation
Answer: Let the fraction be x/y. Then (x + 2)/y = 2/3 gives 3x − 2y = −6, and x/(y + 4) = 4/7 gives 7x − 4y = 16.
Solving: x = 28, y = 45. The fraction is \( \dfrac{28}{45} \).
Teacher's Note:
1. Form two linear equations from the two conditions.
2. Check: 30/45 = 2/3 and 28/49 = 4/7.
Q17. The sum of the digits of a two-digit number is 9. If 27 is subtracted from the number, the digits get reversed. Find the number. [3 marks]
View Answer & Explanation
Answer: Let the tens digit be x and the units digit y. Then x + y = 9 and (10x + y) − 27 = 10y + x, so x − y = 3.
Solving: x = 6, y = 3. The number is 63.
Teacher's Note:
1. Write the number as 10x + y.
2. Reversing the digits gives 10y + x.
Q18. In a taxi, the fare is a fixed charge plus a charge per kilometre. For 12 km the fare is ₹89 and for 20 km it is ₹145. Find the fare for 30 km. [3 marks]
View Answer & Explanation
Answer: Let fixed charge = x and rate = y. x + 12y = 89 and x + 20y = 145. Subtracting: 8y = 56, so y = 7 and x = 5.
Fare for 30 km = 5 + 30 × 7 = ₹215.
Teacher's Note:
1. Subtract the equations to remove the fixed charge.
2. Then substitute to find it.
Q19. Find the vertices of the triangle formed by the lines 4x − y = 4 and 4x + y = 12 with the x-axis, and find its area. [3 marks]
View Answer & Explanation
Answer: Adding the equations: 8x = 16, so x = 2 and y = 4: vertex (2, 4).
With y = 0: 4x = 4 gives (1, 0), and 4x = 12 gives (3, 0).
Area = ½ × base 2 × height 4 = 4 square units.
Teacher's Note:
1. Find the meeting point of the lines and the x-intercepts of each.
2. The base lies on the x-axis.
Q20. Find the point of intersection of x − y = −1 and 3x + 2y = 12, and the area of the triangle bounded by these lines and the x-axis. [3 marks]
View Answer & Explanation
Answer: Solving: 2(x − y) = −2 plus 3x + 2y = 12 gives 5x = 10, x = 2, y = 3.
x-intercepts: (−1, 0) and (4, 0). Base = 5, height = 3.
Area = ½ × 5 × 3 = 7.5 square units.
Teacher's Note:
1. Find where each line cuts the x-axis.
2. The height equals the y-coordinate of the meeting point.
Section C: Long Answer (4 marks)
Q21. Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. [4 marks]
View Answer & Explanation
Answer: Downstream speed = 20/2 = 10 km/h = x + y. Upstream speed = 4/2 = 2 km/h = x − y.
Adding: 2x = 12, so x = 6; then y = 4.
Still water: 6 km/h; current: 4 km/h.
Teacher's Note:
1. Downstream speed is boat speed plus current, upstream is the difference.
2. Half the sum of the two speeds gives the still-water speed.
Q22. In a triangle ABC, ∠C = 3∠B = 2(∠A + ∠B). Find the three angles. [4 marks]
View Answer & Explanation
Answer: From 3B = 2(A + B): B = 2A. And C = 3B = 6A.
A + B + C = 9A = 180°, so A = 20°, B = 40°, C = 120°.
Teacher's Note:
1. Express all angles in terms of one angle.
2. Use the angle sum of 180°.
Q23. 8 men and 12 boys can finish a piece of work in 10 days, while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and by one boy alone. [4 marks]
View Answer & Explanation
Answer: Let one man do 1/x of the work per day and one boy 1/y. 8/x + 12/y = 1/10 and 6/x + 8/y = 1/14.
With m = 1/x and b = 1/y: 80m + 120b = 1 and 84m + 112b = 1. Subtracting: 4m = 8b, so m = 2b.
Then 280b = 1, b = 1/280, m = 1/140.
One man: 140 days; one boy: 280 days.
Teacher's Note:
1. Take work per day as the unknown.
2. Scale the equations to get equal right-hand sides.
Q24. Find k for which the pair 2x + 5y = 3 and (k + 1)x + 2(k + 2)y = 2k has infinitely many solutions. [4 marks]
View Answer & Explanation
Answer: Need 2/(k + 1) = 5/(2(k + 2)) = 3/(2k). From 2/(k + 1) = 3/(2k): 4k = 3k + 3, so k = 3.
Check: 2/4 = 5/10 = 3/6 = 1/2. k = 3.
Teacher's Note:
1. Equate the first and third ratios, then verify the middle one.
2. All three ratios must equal 1/2.
Q25. Students of a class are made to stand in rows. If one student is extra in a row there would be 2 rows less, and if one student is less in a row there would be 3 rows more. Find the number of students in the class. [4 marks]
View Answer & Explanation
Answer: Let there be x students per row and y rows. (x + 1)(y − 2) = xy gives y − 2x = 2. (x − 1)(y + 3) = xy gives 3x − y = 3.
Adding: x = 5, so y = 12.
Students = 5 × 12 = 60.
Teacher's Note:
1. Total students equals students per row times rows.
2. Expand each condition and simplify to a linear equation.
Q26. Aman travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes 4 hours; if he covers 130 km by train and the rest by car, it takes 18 minutes longer. Find the speeds of the train and the car. [4 marks]
View Answer & Explanation
Answer: Let train speed = u and car speed = v km/h. 250/u + 120/v = 4 and 130/u + 240/v = 4.3.
With a = 1/u, b = 1/v: 250a + 120b = 4 and 130a + 240b = 4.3. Doubling the first and subtracting the second: 370a = 3.7, so a = 0.01.
Then b = 0.0125. Train: 100 km/h; car: 80 km/h.
Teacher's Note:
1. 18 minutes equals 0.3 hours.
2. Use reciprocals of the speeds as the new unknowns.
Q27. Riya travels 300 km partly by train and partly by bus. She takes 4 hours if she travels 60 km by bus and the rest by train, and 10 minutes more if she travels 100 km by bus and the rest by train. Find the speed of the train and the bus. [4 marks]
View Answer & Explanation
Answer: Let train speed = x, bus speed = y. 240/x + 60/y = 4 and 200/x + 100/y = 25/6.
Let a = 1/x, b = 1/y: 240a + 60b = 4 and 200a + 100b = 25/6. Solving gives a = 1/80 and b = 1/60.
Train: 80 km/h; bus: 60 km/h.
Teacher's Note:
1. 10 minutes equals 1/6 hour, so the total is 25/6 hours.
2. Check: 3 + 1 = 4 and 2.5 + 5/3 = 25/6.
Q28. Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the pair represents (i) intersecting lines and (ii) parallel lines. [4 marks]
View Answer & Explanation
Answer: (i) For intersecting lines the coefficient ratios must differ: for example 4x − 3y − 8 = 0 (2/4 ≠ 3/(−3)).
(ii) For parallel lines: for example 4x + 6y + 16 = 0 (2/4 = 3/6 ≠ −8/16).
Other correct equations are possible.
Teacher's Note:
1. Choose coefficients using the ratio conditions.
2. Parallel needs the first two ratios equal and the third different.
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Useful Resources and Notes for Class 10 Mathematics Chapter 03 Pair of Linear Equations in Two Variables
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