Class 10 Mathematics: Pair of Linear Equations in Two Variables important questions
Pair of Linear Equations in Two Variables is a chapter of Class 10 Mathematics for the CBSE. These 15 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.
Practice · 1 marks
1. If the system 3x + y = 1 and (2k − 1)x + (k − 1)y = 2k + 1 is inconsistent, then k = (a) −1 (b) 0 (c) 1 (d) 2
(d) 2 3/(2k − 1) = 1/(k − 1) gives 3k − 3 = 2k − 1, so k = 2 (and then 1/(k − 1) = 1 ≠ 1/5).
Important · 2 marks
2. If 49x + 51y = 499 and 51x + 49y = 501, find x and y.
Adding: 100x + 100y = 1000, so x + y = 10. Subtracting: −2x + 2y = −2, so x − y = 1. x = 11/2 and y = 9/2.
Important · 3 marks
3. A train covered a distance at a uniform speed. Had it been 6 km/h faster, it would have taken 4 hours less than scheduled; had it been 6 km/h slower, it would have taken 6 hours more. Find the length of the journey.
Let the speed be x km/h and time y hours; distance = xy. (x + 6)(y − 4) = xy gives −2x + 3y = 12 … (i) (x − 6)(y + 6) = xy gives x − y = 6 … (ii) Solving: x = 30, y = 24. Distance = 30 × 24 = 720 km.
Important · 3 marks
4. Anuj divided his chocolates into lots A and B. He sold lot A at ₹2 for 3 chocolates and lot B at ₹1 per chocolate, collecting ₹400. Had he sold lot A at ₹1 per chocolate and lot B at ₹4 for 5 chocolates, he would have collected ₹460. Find the total number of chocolates.
Let the lots have x and y chocolates. (2/3)x + y = 400, i.e. 2x + 3y = 1200 … (i) x + (4/5)y = 460, i.e. 5x + 4y = 2300 … (ii) Solving: x = 300, y = 200. Total = 500 chocolates.
Practice · 1 marks
5. The lines representing the pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are non-intersecting (parallel). Which statement is true? (a) a₁/a₂ = b₁/b₂ = c₁/c₂ (b) a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (c) a₁/a₂ ≠ b₁/b₂ = c₁/c₂ (d) a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
(b) a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the condition for parallel lines (no solution).
Important · 3 marks
6. The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits differ by 2, find the number. How many such numbers are there?
Let the number be 10x + y. (10x + y) + (10y + x) = 66, so x + y = 6. If x − y = 2: x = 4, y = 2, giving 42. If y − x = 2: x = 2, y = 4, giving 24. There are two such numbers: 42 and 24.
Important · 3 marks
7. Solve: 2/√x + 3/√y = 2 and 4/√x − 9/√y = −1 (x, y > 0).
Let p = 1/√x and q = 1/√y: 2p + 3q = 2 and 4p − 9q = −1. Multiply the first by 3 and add: 10p = 5, so p = 1/2 and q = 1/3. √x = 2 and √y = 3, so x = 4 and y = 9.
Practice · 1 marks
8. The value of k for which the system 3x − ky = 7 and 6x + 10y = 3 is inconsistent is: (A) −10 (B) −5 (C) 5 (D) 7
(B) −5 Inconsistent means a₁/a₂ = b₁/b₂ ≠ c₁/c₂: 3/6 = −k/10, so k = −5 (and 7/3 ≠ 1/2).
Very Important · 5 marks
9. Solve graphically: x + 2y = 3 and 2x − 3y + 8 = 0.
x + 2y = 3 passes through (3, 0), (1, 1) and (−1, 2). 2x − 3y + 8 = 0 passes through (−4, 0), (2, 4) and (−1, 2). The lines intersect at (−1, 2), so x = −1 and y = 2.
Very Important · 5 marks
10. Places A and B are 180 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction at different speeds, they meet in 9 hours. If they travel towards each other, they meet in 1 hour. Find the speeds of the two cars.
Let the speeds be x and y km/h (x > y). Same direction: 9x − 9y = 180, so x − y = 20 … (i) Towards each other: x + y = 180 … (ii) Solving: x = 100 km/h and y = 80 km/h.
Practice · 1 marks
11. If the lines 3x + 2ky = 2 and 2x + 5y + 1 = 0 are not parallel, then k must be: (A) 15/4 (B) ≠ 15/4 (C) any rational number (D) any rational number with 4 as denominator
(B) k ≠ 15/4 Not parallel means a₁/a₂ ≠ b₁/b₂: 3/2 ≠ 2k/5, so k ≠ 15/4.
Important · 3 marks
12. The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their expenditures in the ratio 5 : 7. If each saves ₹15,000 a month, find their incomes.
Let incomes be 3x, 4x and expenditures 5y, 7y. 3x − 5y = 15000 and 4x − 7y = 15000. Solving gives x = 30000 (and y = 15000). Incomes: Aryan ₹90,000 and Babban ₹1,20,000.
Important · 3 marks
13. Solve graphically: 2x + y = 6 and 2x − y − 2 = 0. Find the area of the triangle formed by the two lines and the x-axis.
2x + y = 6 passes through (3, 0) and (0, 6); 2x − y = 2 passes through (1, 0) and (0, −2). They meet at (2, 2), so x = 2, y = 2. The triangle has vertices (1, 0), (3, 0), (2, 2): base 2 and height 2, so area = ½ × 2 × 2 = 2 square units.
Practice · 1 marks
14. Tarun solved a pair of linear equations and found their only point of intersection as (5, −1). One line was x − y = 6. Which could have been the other line? I: 3x − 3y = 18 II: 2x − 3y = 13 III: 2x − 3y = 16 (A) I only (B) II only (C) I and II (D) II and III
(B) II only (5, −1) also satisfies 3x − 3y = 18, but that line is the same as x − y = 6 (infinitely many solutions). Only 2x − 3y = 13 passes through (5, −1) and gives exactly one point of intersection.
Very Important · 4 marks
15. A student company sells sanitizer in small bottles at ₹10 and large bottles at ₹15. In June it sold 1000 bottles for ₹12,750. It then raised both prices by ₹2, and in July sold 2500 bottles for ₹34,250. (i) Form a linear equation for June's sales. (ii) Form a linear equation for July's sales. (iii) (A) How many bottles of each size were sold in June? OR (B) How many of each size were sold in July?
Let x small and y large bottles be sold. (i) 10x + 15y = 12750 (or 2x + 3y = 2550). (ii) 12x + 17y = 34250. (iii) (A) x + y = 1000 and 10x + 15y = 12750 give x = 450, y = 550. OR (B) x + y = 2500 and 12x + 17y = 34250 give x = 1650, y = 850.
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