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CBSE · NCERT
MATHEMATICS – CHAPTER 3 : PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
Chapter-wise Question Bank (CBSE Board + Competitive · Moderate to HOTS Level)
General Instructions
- This question bank is based on NCERT Class X Mathematics, Chapter 3 (Pair of Linear Equations in Two Variables), and is set in the CBSE board pattern with competitive (HOTS) questions mixed in.
- Section A: Q.1 to Q.50 are MCQs of 1 mark each (50 marks). Section B: Q.51 to Q.60 are Assertion–Reason questions of 1 mark each (10 marks).
- Section C: Q.61 to Q.70, Very Short Answer, 2 marks each (20 marks). Section D: Q.71 to Q.80, Short Answer, 3 marks each (30 marks).
- Section E: Q.81 to Q.84, Case-Based, 4 marks each (16 marks). Section F: Q.85 to Q.89, Long Answer, 5 marks each (25 marks).
- Show all steps for Sections C to F. Complete solutions are given after the questions.
SECTION – A (Multiple Choice Questions · 1 Mark each)
Q.1 – Q.50 Choose the correct option.50 × 1 = 50
1. The pair of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution if –
(a) a1/a2 ≠ b1/b2(b) a1/a2 = b1/b2 ≠ c1/c2(c) a1/a2 = b1/b2 = c1/c2(d) a1/a2 = c1/c2 ≠ b1/b2
2. The pair has infinitely many solutions (coincident lines) if –
(a) a1/a2 ≠ b1/b2(b) a1/a2 = b1/b2 = c1/c2(c) a1/a2 = b1/b2 ≠ c1/c2(d) a1/a2 ≠ c1/c2
3. The pair has no solution (parallel lines) if –
(a) a1/a2 ≠ b1/b2(b) a1/a2 = b1/b2 ≠ c1/c2(c) a1/a2 = b1/b2 = c1/c2(d) b1/b2 ≠ c1/c2 = a1/a2
4. The solution of the pair 2x + 3y = 12 and x – y = 1 is –
(a) x = 2, y = 3(b) x = 4, y = 1(c) x = 1, y = 0(d) x = 3, y = 2
5. The pair x + y = 5 and 2x + 2y = 10 has –
(a) a unique solution(b) infinitely many solutions(c) no solution(d) exactly two solutions
6. The pair 2x + 3y = 5 and 4x + 6y = 11 has –
(a) a unique solution(b) no solution(c) infinitely many solutions(d) exactly two solutions
7. The solution of x – 2y = 0 and 3x + 4y – 20 = 0 is –
(a) (2, 4)(b) (4, 2)(c) (5, 1)(d) (–4, –2)
8. The value of k for which the pair kx + 3y = 3 and 12x + ky = 6 has no solution is –
(a) 6(b) –6(c) ±6(d) 12
9. The pair kx + 2y = 5 and 3x + y = 1 has a unique solution if –
(a) k ≠ 6(b) k = 6(c) k ≠ 2(d) k = 3
10. The value of k for which 2x + 3y = 7 and (k – 1)x + (k + 2)y = 3k have infinitely many solutions is –
(a) 3(b) 5(c) –7(d) 7
11. The graphs of the equations x = 2 and y = 3 intersect at –
(a) (3, 2)(b) (0, 0)(c) (2, 0)(d) (2, 3)
12. The pair x + 2y = 4 and 2x + 4y = 12 is –
(a) inconsistent (parallel lines)(b) consistent with a unique solution(c) dependent(d) consistent with two solutions
13. The solution of x + y = 14 and x – y = 4 is –
(a) x = 9, y = 5(b) x = 5, y = 9(c) x = 10, y = 4(d) x = 8, y = 6
14. The solution of 3x + 4y = 10 and 2x – 2y = 2 is –
(a) x = 1, y = 2(b) x = 2, y = 1(c) x = 3, y = 0.25(d) x = 0, y = 2.5
15. The sum of two numbers is 50 and their difference is 10. The larger number is –
(a) 20(b) 40(c) 25(d) 30
16. A father is three times as old as his son. After 12 years he will be twice as old as his son. The present age of the son is –
(a) 12 years(b) 10 years(c) 14 years(d) 36 years
17. 5 pencils and 7 pens together cost ₹50, while 7 pencils and 5 pens cost ₹46. The cost of one pen is –
(a) ₹3(b) ₹5(c) ₹4(d) ₹7
18. The sum of the digits of a two-digit number is 9 and the number exceeds the number obtained by reversing its digits by 27. The number is –
(a) 36(b) 63(c) 72(d) 81
19. A boat goes 12 km downstream in 1 hour and returns upstream in 2 hours. The speed of the boat in still water is –
(a) 6 km/h(b) 3 km/h(c) 12 km/h(d) 9 km/h
20. The perimeter of a rectangle is 26 cm and its length exceeds its breadth by 3 cm. Its area is –
(a) 30 cm2(b) 42 cm2(c) 40 cm2(d) 36 cm2
21. The lines 2x + y = 6 and 2x – y = 2 intersect at the point –
(a) (1, 4)(b) (3, 0)(c) (2, 1)(d) (2, 2)
22. The area of the triangle formed by the line 2x + 3y = 12 and the coordinate axes is –
(a) 24 square units(b) 6 square units(c) 12 square units(d) 8 square units
23. If x = 2, y = –1 is a solution of 2x + 3y = k, then k equals –
(a) 1(b) 7(c) –1(d) 5
24. If x = 2, y = 3 is a solution of 3x + ky = 12, then k equals –
(a) 3(b) –2(c) 2(d) 6
25. A pair of linear equations which has no solution is called –
(a) an inconsistent pair(b) a consistent pair(c) a dependent pair(d) a quadratic pair
26. A consistent pair of linear equations in two variables can have –
(a) either a unique solution or infinitely many solutions(b) only a unique solution(c) only infinitely many solutions(d) exactly two solutions
27. Two lines are parallel if –
(a) a1/a2 ≠ b1/b2(b) a1/a2 = b1/b2 ≠ c1/c2(c) a1/a2 = b1/b2 = c1/c2(d) a1/a2 = c1/c2
28. A pair of dependent linear equations is always –
(a) inconsistent(b) having a unique solution(c) having no solution(d) consistent
29. The pair 5x – 15y = 8 and 3x – 9y = 24/5 has –
(a) infinitely many solutions(b) a unique solution(c) no solution(d) exactly two solutions
30. The pair 2x – 3y = 8 and 4x – 6y = 9 has –
(a) a unique solution(b) infinitely many solutions(c) no solution(d) two solutions
31. The value of k for which 4x + 6y – 1 = 0 and 2x + ky – 7 = 0 represent parallel lines is –
(a) 6(b) 2(c) 3(d) 12
32. The line 3x + 2y = 12 meets the y-axis at the point –
(a) (4, 0)(b) (6, 0)(c) (0, 6)(d) (0, 4)
33. The solution of 99x + 101y = 499 and 101x + 99y = 501 is –
(a) x = 3, y = 2(b) x = 2, y = 3(c) x = 5, y = 0(d) x = 4, y = 1
34. The solution of 37x + 43y = 123 and 43x + 37y = 117 is –
(a) x = 2, y = 1(b) x = 3, y = 0(c) x = 0, y = 3(d) x = 1, y = 2
35. If 2x + 3y = 11 and 2x – 4y = –24, and y = mx + 3, the value of m is –
(a) 1(b) 2(c) –1(d) –2
36. The incomes of two persons are in the ratio 9 : 7 and their expenditures are in the ratio 4 : 3. If each saves ₹2000, the income of the first person is –
(a) ₹14000(b) ₹16000(c) ₹12000(d) ₹18000
37. The solution of x + y = a + b and ax – by = a2 – b2 is –
(a) x = a, y = b(b) x = b, y = a(c) x = a + b, y = 0(d) x = a – b, y = b
38. The graphs of y = 0 and y = –7 are –
(a) intersecting at (0, –7)(b) parallel lines, so the pair has no solution(c) coincident lines(d) perpendicular lines
39. If the lines of a pair of linear equations intersect at a point, the pair is –
(a) inconsistent(b) dependent(c) consistent with a unique solution(d) having infinitely many solutions
40. The value of k for which kx + 3y = k – 3 and 12x + ky = k have infinitely many solutions is –
(a) –6(b) 12(c) 6(d) 3
41. If 2x + y = 5 and 3x – y = 5, then x + y equals –
(a) 4(b) 2(c) 3(d) 5
42. The area of the triangle formed by the lines x + y = 6, x – y = 2 and the x-axis is –
(a) 8 square units(b) 6 square units(c) 2 square units(d) 4 square units
43. The pair 3x – 5y = 20 and 6x – 10y = 40 represents –
(a) coincident lines(b) parallel lines(c) intersecting lines(d) perpendicular lines
44. A taxi charges a fixed amount plus a fixed rate per km. The fare for 10 km is ₹105 and for 15 km is ₹155. The fixed charge is –
(a) ₹10(b) ₹15(c) ₹5(d) ₹50
45. A two-digit number is 4 times the sum of its digits, and when 18 is added to it, the digits are reversed. The number is –
(a) 36(b) 48(c) 42(d) 24
46. If the lines 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then k equals –
(a) 4/15(b) 15/4(c) 2/5(d) –15/4
47. The line 2x + 3y = 12 meets the x-axis at the point –
(a) (0, 4)(b) (6, 0)(c) (4, 0)(d) (0, 6)
48. The cost of 2 kg of apples and 1 kg of grapes is ₹160 and the cost of 4 kg of apples and 2 kg of grapes is ₹300. Geometrically, the pair of equations represents –
(a) intersecting lines(b) coincident lines(c) perpendicular lines(d) parallel lines
49. If a1/a2 = b1/b2 = c1/c2, the two lines represented by the pair are –
(a) parallel(b) intersecting(c) coincident(d) perpendicular
50. For which value of p does the pair px + 3y = p – 3 and 12x + py = p have no solution?
(a) –6(b) 6(c) 12(d) –12
SECTION – B (Assertion (A) – Reason (R) · 1 Mark each)
Q.51 – Q.60 Choose: (a) Both A and R true, R explains A; (b) Both true, R does not explain A; (c) A true, R false; (d) A false, R true.10 × 1 = 10
51. Assertion (A): The pair x + 2y – 3 = 0 and 2x + 4y – 6 = 0 has infinitely many solutions.
Reason (R): For the pair, a1/a2 = b1/b2 = c1/c2.
Reason (R): For the pair, a1/a2 = b1/b2 = c1/c2.
52. Assertion (A): The pair 2x + 3y = 5 and 4x + 6y = 10 has infinitely many solutions.
Reason (R): Any two lines in a plane intersect in exactly one point.
Reason (R): Any two lines in a plane intersect in exactly one point.
53. Assertion (A): The graphs of x + y = 4 and 2x + 2y = 8 coincide.
Reason (R): Every linear equation in two variables represents a straight line.
Reason (R): Every linear equation in two variables represents a straight line.
54. Assertion (A): The pair 2x + 3y = 6 and 4x + 6y = 10 has a unique solution.
Reason (R): Two lines with a1/a2 = b1/b2 ≠ c1/c2 are parallel.
Reason (R): Two lines with a1/a2 = b1/b2 ≠ c1/c2 are parallel.
55. Assertion (A): The pair 2x – 3y = 5 and 4x – 6y = 7 is inconsistent.
Reason (R): For this pair a1/a2 = b1/b2 ≠ c1/c2.
Reason (R): For this pair a1/a2 = b1/b2 ≠ c1/c2.
56. Assertion (A): The graphs of x = 3 and y = –2 intersect at (3, –2).
Reason (R): The lines x = 3 and y = –2 are parallel.
Reason (R): The lines x = 3 and y = –2 are parallel.
57. Assertion (A): A consistent pair of linear equations always has a unique solution.
Reason (R): A consistent pair can represent coincident lines.
Reason (R): A consistent pair can represent coincident lines.
58. Assertion (A): The solution of x + y = 5 and x – y = 1 is x = 3, y = 2.
Reason (R): Adding the two equations gives 2x = 6.
Reason (R): Adding the two equations gives 2x = 6.
59. Assertion (A): The pair x + y = 1 and 2x + 2y = 2 is a dependent pair.
Reason (R): A dependent pair is always consistent.
Reason (R): A dependent pair is always consistent.
60. Assertion (A): The solution of x – 2y = 0 and 3x + 4y – 20 = 0 is (4, 2).
Reason (R): The two lines intersect at two points.
Reason (R): The two lines intersect at two points.
SECTION – C (Very Short Answer · 2 Marks each)
61. Check whether the pair 2x + 3y = 9 and 4x + 6y = 18 is consistent. If so, state the number of solutions.(2)
62. Solve: x + y = 14, x – y = 4.(2)
63. For what value of k will the pair kx + 2y = 5 and 3x + y = 1 have a unique solution?(2)
64. Without solving, find whether 3x + 2y = 5 and 2x – 3y = 7 has a unique solution, no solution or infinitely many solutions.(2)
65. Find the point of intersection of the lines 2x + y = 6 and 2x – y = 2.(2)
66. If x = 2, y = –1 is a solution of 2x + 3y = k, find k.(2)
67. The sum of two numbers is 34 and their difference is 14. Find the numbers.(2)
68. 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Form the pair of equations and find the cost of a pencil and of a pen.(2)
69. For what value of k are the lines 4x + 6y – 1 = 0 and 2x + ky – 7 = 0 parallel?(2)
70. Find the coordinates of the points where the line 2x + 3y = 12 meets the x-axis and the y-axis.(2)
SECTION – D (Short Answer · 3 Marks each)
71. Solve graphically: 2x + y = 6 and 2x – y = 2. Also find the area of the triangle formed by these lines and the x-axis.(3)
72. Solve by the substitution method: x + 2y = 7 and 2x – y = 4.(3)
73. Solve by the elimination method: 2x + 3y = 13 and 5x – 2y = 4.(3)
74. Find the value of k for which the pair (k – 3)x + 3y = k and kx + ky = 12 has infinitely many solutions.(3)
75. Five years ago Nuri was thrice as old as Sonu. Ten years later Nuri will be twice as old as Sonu. Find the present ages of Nuri and Sonu.(3)
76. A man travels 300 km partly by train and partly by car. If he covers 60 km by train and the rest by car, it takes 4 hours; if he covers 100 km by train and the rest by car, it takes 10 minutes more. Find the speeds of the train and the car.(3)
77. The numerator and denominator of a fraction are such that if 1 is added to the numerator and subtracted from the denominator, the fraction becomes 1; if 1 is added to the denominator, the fraction becomes 1/2. Find the fraction.(3)
78. The sum of the digits of a two-digit number is 9. Also, 9 times this number is twice the number obtained by reversing the digits. Find the number.(3)
79. Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. She got 25 notes in all. Find how many notes of each denomination she received.(3)
80. ABCD is a cyclic quadrilateral with ∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°. Find x, y and the four angles.(3)
SECTION – E (Case-Based Questions · 4 Marks each)
81. Read the passage and answer the questions that follow.(4)
A taxi service charges a fixed amount for the ride plus a fixed amount for each km travelled. The fare for a journey of 10 km is ₹105 and for a journey of 15 km is ₹155. Let the fixed charge be ₹x and the charge per km be ₹y.
(i) Write the pair of linear equations that represents this situation.(1)
(ii) Find the fixed charge and the charge per km.(1)
(iii) Find the fare for 25 km, and the distance travelled if the fare is ₹355.(2)
82. Read the passage and answer the questions that follow.(4)
The sum of the present ages of Anil and his father is 70 years. After 10 years the father will be twice as old as Anil will be then. Let Anil's present age be x years and his father's y years.
(i) Write the two linear equations.(1)
(ii) Find the present ages of Anil and his father.(1)
(iii) How many years ago was the father exactly three times as old as Anil? What were their ages then?(2)
83. Read the passage and answer the questions that follow.(4)
A boat goes 24 km upstream and 28 km downstream in 6 hours, and 30 km upstream and 21 km downstream in 6½ hours. Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
(i) Form the equations in a = 1/(x – y) and b = 1/(x + y).(1)
(ii) Solve them to find x – y and x + y.(1)
(iii) Find the speed of the boat in still water and the speed of the stream.(2)
84. Read the passage and answer the questions that follow.(4)
Rajan drew the graphs of the equations x + y = 6 and 2x – y = 0 on the same graph paper.
(i) Where does the line x + y = 6 meet the two axes?(1)
(ii) Find the point of intersection of the two lines.(1)
(iii) Find the area of the triangle formed by the two lines and the y-axis.(2)
SECTION – F (Long Answer · 5 Marks each)
85. Solve graphically x – y + 1 = 0 and 3x + 2y – 12 = 0. Also find the area of the triangle formed by these lines and the x-axis.(5)
86. Solve: 2/x + 3/y = 13 and 5/x – 4/y = –2 (x ≠ 0, y ≠ 0).(5)
87. A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If 3 is added to both, it becomes 5/6. Find the fraction.(5)
88. The places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. Find the speeds of the two cars.(5)
89. 2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and by 1 man alone to finish the work.(5)
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CBSE · NCERT
MATHEMATICS – CHAPTER 3 : COMPLETE SOLUTIONS
Pair of Linear Equations in Two Variables · Answer Key with Step-by-Step Solutions
SECTION – A (MCQ Solutions)
Quick Answer Key (Q.1 – Q.50)
| 1. (a) | 2. (b) | 3. (b) | 4. (d) | 5. (b) | 6. (b) | 7. (b) | 8. (b) | 9. (a) | 10. (d) |
| 11. (d) | 12. (a) | 13. (a) | 14. (b) | 15. (d) | 16. (a) | 17. (b) | 18. (b) | 19. (d) | 20. (c) |
| 21. (d) | 22. (c) | 23. (a) | 24. (c) | 25. (a) | 26. (a) | 27. (b) | 28. (d) | 29. (a) | 30. (c) |
| 31. (c) | 32. (c) | 33. (a) | 34. (d) | 35. (c) | 36. (d) | 37. (a) | 38. (b) | 39. (c) | 40. (c) |
| 41. (c) | 42. (d) | 43. (a) | 44. (c) | 45. (d) | 46. (b) | 47. (b) | 48. (d) | 49. (c) | 50. (a) |
Solutions with Explanation
1. Ans. (a) a1/a2 ≠ b1/b2
If the ratios of the x- and y-coefficients differ, the lines have different slopes and intersect at exactly one point.
2. Ans. (b) a1/a2 = b1/b2 = c1/c2
All three ratios equal means one equation is a multiple of the other – the two lines coincide.
3. Ans. (b) a1/a2 = b1/b2 ≠ c1/c2
Same slope but different intercepts – the lines never meet.
4. Ans. (d) x = 3, y = 2
From x – y = 1, x = y + 1. Then 2(y + 1) + 3y = 12 ⇒ 5y = 10 ⇒ y = 2, x = 3.
5. Ans. (b) infinitely many solutions
1/2 = 1/2 = 5/10 – the lines coincide.
6. Ans. (b) no solution
2/4 = 3/6 = 1/2 but 5/11 ≠ 1/2 – parallel lines, so no solution.
7. Ans. (b) (4, 2)
x = 2y ⇒ 6y + 4y = 20 ⇒ y = 2, x = 4.
8. Ans. (b) –6
No solution needs k/12 = 3/k ≠ 3/6 ⇒ k2 = 36. For k = 6, k/12 = 1/2 = 3/6 (coincident); for k = –6, –1/2 ≠ 1/2. So k = –6.
9. Ans. (a) k ≠ 6
Unique solution ⇔ k/3 ≠ 2/1 ⇒ k ≠ 6.
10. Ans. (d) 7
2/(k – 1) = 3/(k + 2) ⇒ 2k + 4 = 3k – 3 ⇒ k = 7. Check: 2/6 = 3/9 = 7/21 = 1/3 ✓.
11. Ans. (d) (2, 3)
x = 2 is a vertical line, y = 3 a horizontal line; they meet at (2, 3).
12. Ans. (a) inconsistent (parallel lines)
1/2 = 2/4 ≠ 4/12 ⇒ parallel lines, no solution – the pair is inconsistent.
13. Ans. (a) x = 9, y = 5
Add: 2x = 18 ⇒ x = 9; y = 14 – 9 = 5.
14. Ans. (b) x = 2, y = 1
Multiply the second by 2: 4x – 4y = 4. Add to the first: 7x = 14 ⇒ x = 2, y = 1.
15. Ans. (d) 30
x + y = 50, x – y = 10 ⇒ x = 30, y = 20.
16. Ans. (a) 12 years
F = 3S; F + 12 = 2(S + 12) ⇒ 3S + 12 = 2S + 24 ⇒ S = 12 (F = 36).
17. Ans. (b) ₹5
5p + 7q = 50 and 7p + 5q = 46. Adding: p + q = 8; subtracting: q – p = 2 ⇒ q = 5, p = 3.
18. Ans. (b) 63
x + y = 9 and (10x + y) – (10y + x) = 27 ⇒ x – y = 3 ⇒ x = 6, y = 3 ⇒ 63.
19. Ans. (d) 9 km/h
Downstream speed = 12 km/h, upstream = 6 km/h. Still water = (12 + 6)/2 = 9 km/h; stream = 3 km/h.
20. Ans. (c) 40 cm2
l + b = 13 and l – b = 3 ⇒ l = 8, b = 5 ⇒ area = 40 cm2.
21. Ans. (d) (2, 2)
Add: 4x = 8 ⇒ x = 2; y = 6 – 4 = 2.
22. Ans. (c) 12 square units
Intercepts: (6, 0) and (0, 4). Area = ½ × 6 × 4 = 12.
23. Ans. (a) 1
k = 2(2) + 3(–1) = 4 – 3 = 1.
24. Ans. (c) 2
6 + 3k = 12 ⇒ k = 2.
25. Ans. (a) an inconsistent pair
No common solution ⇒ inconsistent (the lines are parallel).
26. Ans. (a) either a unique solution or infinitely many solutions
Consistent means at least one solution: intersecting lines (one) or coincident lines (infinitely many).
27. Ans. (b) a1/a2 = b1/b2 ≠ c1/c2
Equal slopes but different intercepts.
28. Ans. (d) consistent
Dependent (coincident) pairs have infinitely many solutions, so they are consistent.
29. Ans. (a) infinitely many solutions
5/3 = –15/–9 = 5/3 and 8/(24/5) = 40/24 = 5/3 – all three ratios are equal.
30. Ans. (c) no solution
2/4 = –3/–6 = 1/2 but 8/9 ≠ 1/2 ⇒ parallel lines.
31. Ans. (c) 3
4/2 = 6/k ⇒ k = 3. Also –1/–7 = 1/7 ≠ 2, so the lines are parallel and distinct.
32. Ans. (c) (0, 6)
Put x = 0: 2y = 12 ⇒ y = 6.
33. Ans. (a) x = 3, y = 2
Add: 200(x + y) = 1000 ⇒ x + y = 5. Subtract: –2x + 2y = –2 ⇒ x – y = 1. So x = 3, y = 2.
34. Ans. (d) x = 1, y = 2
Add: 80(x + y) = 240 ⇒ x + y = 3. Subtract: –6x + 6y = 6 ⇒ y – x = 1 ⇒ x = 1, y = 2.
35. Ans. (c) –1
Subtract: 7y = 35 ⇒ y = 5; 2x = 11 – 15 = –4 ⇒ x = –2. Then 5 = –2m + 3 ⇒ m = –1.
36. Ans. (d) ₹18000
9x – 4y = 2000 and 7x – 3y = 2000 ⇒ x = 2000, y = 4000. Income of the first = 9x = ₹18000 (second = ₹14000).
37. Ans. (a) x = a, y = b
Substituting x = a, y = b: a + b ✓ and a·a – b·b = a2 – b2 ✓ (uniqueness since a/1 ≠ –b/1 when a + b ≠ 0).
38. Ans. (b) parallel lines, so the pair has no solution
Both are horizontal lines at different heights – they never meet.
39. Ans. (c) consistent with a unique solution
Intersecting lines have exactly one common point.
40. Ans. (c) 6
k/12 = 3/k = (k – 3)/k ⇒ k2 = 36 and k – 3 = 3 ⇒ k = 6 (k = –6 fails the third ratio).
41. Ans. (c) 3
Add: 5x = 10 ⇒ x = 2; y = 1. So x + y = 3.
42. Ans. (d) 4 square units
The lines meet at (4, 2); they cut the x-axis at (6, 0) and (2, 0). Area = ½ × 4 × 2 = 4.
43. Ans. (a) coincident lines
3/6 = –5/–10 = 20/40 = 1/2 – all ratios equal.
44. Ans. (c) ₹5
x + 10y = 105, x + 15y = 155 ⇒ 5y = 50 ⇒ y = 10, x = 5.
45. Ans. (d) 24
10x + y = 4(x + y) ⇒ y = 2x; 10x + y + 18 = 10y + x ⇒ y – x = 2 ⇒ x = 2, y = 4 ⇒ 24.
46. Ans. (b) 15/4
3/2 = 2k/5 ⇒ 4k = 15 ⇒ k = 15/4. (c-ratio 2/1 ≠ 3/2, so the lines are distinct.)
47. Ans. (b) (6, 0)
Put y = 0: 2x = 12 ⇒ x = 6.
48. Ans. (d) parallel lines
2a + g = 160 and 4a + 2g = 300: 2/4 = 1/2 ≠ 160/300 ⇒ parallel lines (no solution).
49. Ans. (c) coincident
Equal ratios mean one equation is a constant multiple of the other.
50. Ans. (a) –6
p/12 = 3/p ⇒ p = ±6. p = 6 gives (p – 3)/p = 1/2 = p/12 (infinitely many); p = –6 gives 3/2 ≠ –1/2 ⇒ no solution.
SECTION – B (Assertion – Reason)
51. Ans. (a)
Both true; 1/2 = 2/4 = –3/–6, so the lines coincide.
52. Ans. (c)
A is true: 2/4 = 3/6 = 5/10, so the lines coincide. R is false: two lines may be parallel (no common point) or coincident (infinitely many).
53. Ans. (b)
Both true, but R does not explain why these two lines coincide.
54. Ans. (d)
A is false: 2/4 = 3/6 ≠ 6/10 – the lines are parallel (no solution). R is true.
55. Ans. (a)
Both true: 1/2 = 1/2 ≠ 5/7 – parallel lines, no common solution.
56. Ans. (c)
A is true. R is false: x = 3 is vertical and y = –2 is horizontal, so they are perpendicular.
57. Ans. (d)
A is false (coincident lines give infinitely many solutions). R is true.
58. Ans. (a)
Both true; adding eliminates y and gives x = 3, then y = 2.
59. Ans. (b)
Both true, but R does not explain why this pair is dependent.
60. Ans. (c)
A is true. R is false: two distinct lines intersect in at most one point.
SECTION – C (2 Marks each)
61. Ans.
a1/a2 = 2/4 = 1/2; b1/b2 = 3/6 = 1/2; c1/c2 = 9/18 = 1/2. [1] All ratios are equal: the lines coincide, so the pair is consistent with infinitely many solutions. [1]
62. Ans.
Adding: 2x = 18 ⇒ x = 9. [1] Then y = 14 – 9 = 5. Solution: x = 9, y = 5. [1]
63. Ans.
For a unique solution a1/a2 ≠ b1/b2. [1] k/3 ≠ 2/1 ⇒ k ≠ 6. [1]
64. Ans.
a1/a2 = 3/2 and b1/b2 = 2/(–3) = –2/3. [1] Since 3/2 ≠ –2/3, the pair has a unique solution (the lines intersect). [1]
65. Ans.
Adding: 4x = 8 ⇒ x = 2. [1] From the first, y = 6 – 4 = 2. The lines meet at (2, 2). [1]
66. Ans.
Substitute: k = 2(2) + 3(–1). [1] k = 4 – 3 = 1. [1]
67. Ans.
x + y = 34 and x – y = 14. [1] Adding: 2x = 48 ⇒ x = 24, y = 10. The numbers are 24 and 10. [1]
68. Ans.
Let a pencil cost ₹x and a pen ₹y: 5x + 7y = 50 and 7x + 5y = 46. [1] Adding: x + y = 8; subtracting: y – x = 2. So y = 5 and x = 3: pencil ₹3, pen ₹5. [1]
69. Ans.
Parallel lines: a1/a2 = b1/b2 ≠ c1/c2. [1] 4/2 = 6/k ⇒ k = 3, and c1/c2 = 1/7 ≠ 2 ✓. [1]
70. Ans.
On the x-axis y = 0: 2x = 12 ⇒ x = 6, the point (6, 0). [1] On the y-axis x = 0: 3y = 12 ⇒ y = 4, the point (0, 4). [1]
SECTION – D (3 Marks each)
71. Ans.
2x + y = 6: (0, 6), (1, 4), (2, 2), (3, 0). 2x – y = 2: (0, –2), (1, 0), (2, 2), (3, 4). [1] Plotting, the lines intersect at (2, 2), so x = 2, y = 2. [1] The lines meet the x-axis at (3, 0) and (1, 0): base = 2, height = 2 ⇒ area = ½ × 2 × 2 = 2 square units. [1]
72. Ans.
From the first, x = 7 – 2y. [1] Substitute in the second: 2(7 – 2y) – y = 4 ⇒ 14 – 5y = 4 ⇒ y = 2. [1] Then x = 7 – 4 = 3. x = 3, y = 2. [1]
73. Ans.
Multiply the first by 2 and the second by 3: 4x + 6y = 26 and 15x – 6y = 12. [1] Adding: 19x = 38 ⇒ x = 2. [1] Then 4 + 3y = 13 ⇒ y = 3. x = 2, y = 3. [1]
74. Ans.
Infinitely many solutions: (k – 3)/k = 3/k = k/12. [1] 3/k = k/12 ⇒ k2 = 36 ⇒ k = ±6. [1] (k – 3)/k = 3/k ⇒ k – 3 = 3 ⇒ k = 6. Hence k = 6 (k = –6 fails). [1]
75. Ans.
Let Nuri's present age = x and Sonu's = y. x – 5 = 3(y – 5) ⇒ x – 3y = –10 …(i) [1] x + 10 = 2(y + 10) ⇒ x – 2y = 10 …(ii) [1] (ii) – (i): y = 20; x = 50. Nuri 50 years, Sonu 20 years. [1]
76. Ans.
Let the train speed = u km/h and car speed = v km/h. 60/u + 240/v = 4 and 100/u + 200/v = 25/6. [1] Put 1/u = a, 1/v = b: 60a + 240b = 4 and 100a + 200b = 25/6. Solving: a = 1/60, b = 1/80. [1] Train: 60 km/h; car: 80 km/h. (Check: 1 + 3 = 4 ✓; 5/3 + 5/2 = 25/6 ✓.) [1]
77. Ans.
Let the fraction be x/y. (x + 1)/(y – 1) = 1 ⇒ y = x + 2. [1] x/(y + 1) = 1/2 ⇒ 2x = y + 1 = x + 3 ⇒ x = 3. [1] y = 5. The fraction is 3/5. [1]
78. Ans.
Let the digits be x (tens) and y (units). x + y = 9 …(i) [½] 9(10x + y) = 2(10y + x) ⇒ 88x = 11y ⇒ y = 8x …(ii) [1½] From (i): x + 8x = 9 ⇒ x = 1, y = 8. The number is 18. [1]
79. Ans.
Let ₹50 notes = x and ₹100 notes = y. x + y = 25 …(i) and 50x + 100y = 2000 ⇒ x + 2y = 40 …(ii). [1½] (ii) – (i): y = 15; x = 10. [1] 10 notes of ₹50 and 15 notes of ₹100. (Check: 500 + 1500 = 2000 ✓.) [½]
80. Ans.
Opposite angles of a cyclic quadrilateral are supplementary. ∠A + ∠C = 180°: 2x + 2y – 20 = 180 ⇒ x + y = 100. [1] ∠B + ∠D = 180°: y + 20 + x + y = 180 ⇒ x + 2y = 160 ⇒ y = 60, x = 40. [1] ∠A = 110°, ∠B = 80°, ∠C = 70°, ∠D = 100° (sum = 360° ✓). [1]
SECTION – E (Case-Based · 4 Marks each)
81. Ans.
(i) x + 10y = 105 and x + 15y = 155. [1]
(ii) Subtract: 5y = 50 ⇒ y = 10; x = 105 – 100 = 5. Fixed charge ₹5; ₹10 per km. [1]
(iii) Fare for 25 km = 5 + 10 × 25 = ₹255. [1] 5 + 10d = 355 ⇒ d = 35 km. [1] [2]
82. Ans.
(i) x + y = 70 and y + 10 = 2(x + 10), i.e. y – 2x = 10. [1]
(ii) y = 70 – x ⇒ 70 – x – 2x = 10 ⇒ x = 20, y = 50. Anil 20 years; father 50 years. [1]
(iii) Let it be t years ago: 50 – t = 3(20 – t) ⇒ 50 – t = 60 – 3t ⇒ t = 5. [1] 5 years ago; ages were 15 and 45 (45 = 3 × 15 ✓). [1] [2]
83. Ans.
(i) 24a + 28b = 6 and 30a + 21b = 13/2. [1]
(ii) Multiply the first by 3 and the second by 4: 72a + 84b = 18 and 120a + 84b = 26 ⇒ 48a = 8 ⇒ a = 1/6; then 28b = 2 ⇒ b = 1/14. So x – y = 6 and x + y = 14. [1]
(iii) x = (6 + 14)/2 = 10 km/h. [1] y = (14 – 6)/2 = 4 km/h. [1] [2]
84. Ans.
(i) At (6, 0) on the x-axis and (0, 6) on the y-axis. [1]
(ii) Adding: 3x = 6 ⇒ x = 2; y = 4. The lines meet at (2, 4). [1]
(iii) 2x – y = 0 passes through the origin and (2, 4); x + y = 6 meets the y-axis at (0, 6). The vertices are (0, 0), (0, 6) and (2, 4). [1] Base on the y-axis = 6, height = 2 ⇒ area = ½ × 6 × 2 = 6 square units. [1] [2]
SECTION – F (Long Answer · 5 Marks each)
85. Ans.
Table for x – y + 1 = 0: (–1, 0), (0, 1), (1, 2), (2, 3). [1] Table for 3x + 2y = 12: (0, 6), (2, 3), (4, 0). [1] Plotting both lines, they intersect at (2, 3), so the solution is x = 2, y = 3. [1] (Algebraic check: x = y – 1 ⇒ 3y – 3 + 2y = 12 ⇒ y = 3, x = 2 ✓.) [½]
The lines meet the x-axis at (–1, 0) and (4, 0): base = 5 units; the third vertex is (2, 3): height = 3. [1] Area = ½ × 5 × 3 = 7.5 square units. [½]
The lines meet the x-axis at (–1, 0) and (4, 0): base = 5 units; the third vertex is (2, 3): height = 3. [1] Area = ½ × 5 × 3 = 7.5 square units. [½]
86. Ans.
Put 1/x = a and 1/y = b: 2a + 3b = 13 …(i) and 5a – 4b = –2 …(ii). [1] Multiply (i) by 4 and (ii) by 3: 8a + 12b = 52 and 15a – 12b = –6. [1] Adding: 23a = 46 ⇒ a = 2. [1] From (i): 4 + 3b = 13 ⇒ b = 3. [1] Hence x = 1/a = 1/2 and y = 1/b = 1/3. (Check: 4 + 9 = 13 ✓; 10 – 12 = –2 ✓.) [1]
87. Ans.
Let the fraction be x/y. (x + 2)/(y + 2) = 9/11 ⇒ 11x + 22 = 9y + 18 ⇒ 11x – 9y = –4 …(i). [1½] (x + 3)/(y + 3) = 5/6 ⇒ 6x + 18 = 5y + 15 ⇒ 6x – 5y = –3 …(ii). [1½] Multiply (i) by 5 and (ii) by 9: 55x – 45y = –20 and 54x – 45y = –27. Subtracting: x = 7. [1] From (ii): 42 – 5y = –3 ⇒ y = 9. The fraction is 7/9. (Check: 9/11 ✓, 10/12 = 5/6 ✓.) [1]
88. Ans.
Let the speed of the car from A be x km/h and that from B be y km/h (x > y). [1] Same direction: the car from A gains 100 km in 5 hours: 5x – 5y = 100 ⇒ x – y = 20 …(i). [1½] Towards each other: x + y = 100 …(ii). [1] Adding: 2x = 120 ⇒ x = 60; y = 40. [1] The speeds are 60 km/h and 40 km/h. [½]
89. Ans.
Let one woman take x days and one man take y days alone; their one-day work = 1/x and 1/y. Put a = 1/x, b = 1/y. [1] 4(2a + 5b) = 1 ⇒ 8a + 20b = 1 …(i); 3(3a + 6b) = 1 ⇒ 9a + 18b = 1 …(ii). [1½] Multiply (i) by 9 and (ii) by 8: 72a + 180b = 9 and 72a + 144b = 8 ⇒ 36b = 1 ⇒ b = 1/36. [1] From (i): 8a = 1 – 20/36 = 16/36 ⇒ a = 1/18. [1] 1 woman alone: 18 days; 1 man alone: 36 days. [½]
QUICK REVISION · KEY POINTS (CBSE / COMPETITIVE)
- Standard form: a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Each equation is a straight line.
- a1/a2 ≠ b1/b2: intersecting lines – unique solution (consistent).
- a1/a2 = b1/b2 = c1/c2: coincident lines – infinitely many solutions (consistent, dependent).
- a1/a2 = b1/b2 ≠ c1/c2: parallel lines – no solution (inconsistent).
- Substitution: express one variable in terms of the other and substitute. Elimination: make the coefficients of one variable equal (and opposite), then add/subtract.
- Reducible equations: put 1/x = a and 1/y = b to get a linear pair; for speed problems use downstream = x + y and upstream = x – y.
- Word problems: define two variables, translate the two conditions into equations, solve and check; in digit problems the number is 10x + y and the reversed number is 10y + x.
- A line ax + by = c meets the x-axis at (c/a, 0) and the y-axis at (0, c/b); the area of the triangle formed with the axes = ½ × |c/a| × |c/b|.
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