NCERT Exemplar Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Solutions (CBSE 2026-27)
This post covers in-syllabus NCERT Exemplar questions strictly based on the rationalised curriculum: graphical method of solution, conditions for consistency and inconsistency, and algebraic solutions via Substitution and Elimination methods, including situational daily-life word problems. Outdated syllabus topics (Cross-Multiplication method and non-linear reducible equations) have been omitted.
Section A: Multiple Choice Questions (MCQs)
Comparing the given pair with standard form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\):
\[\frac{a_1}{a_2} = \frac{5}{3}\] \[\frac{b_1}{b_2} = \frac{-15}{-9} = \frac{5}{3}\] \[\frac{c_1}{c_2} = \frac{-8}{-\frac{24}{5}} = \frac{8 \times 5}{24} = \frac{40}{24} = \frac{5}{3}\]Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{5}{3}\), the lines are coincident and the pair has infinitely many solutions.
Let the tens digit be \(x\) and units digit be \(y\). The original number is \(10x + y\).
Given that the sum of the digits is \(9\):
\[x + y = 9 \quad \text{— (1)}\]When \(27\) is added, digits are reversed (\(10y + x\)):
\[(10x + y) + 27 = 10y + x\] \[9x – 9y = -27 \implies x – y = -3 \quad \text{— (2)}\]Adding (1) and (2):
\[2x = 6 \implies x = 3\]Substituting \(x = 3\) in (1):
\[3 + y = 9 \implies y = 6\]Hence, the number is \(10(3) + 6 = 36\).
Comparing coefficients:
\[\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}\]Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel and possess no common solution.
Ratios of coefficients:
\[\frac{a_1}{a_2} = \frac{1}{-3} = -\frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{2}{-6} = -\frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{5}{1} = 5\]Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the pair represents parallel lines with no solution.
A pair of linear equations is defined as consistent if it has at least one solution. Intersecting lines have exactly one solution, while coincident lines have infinitely many solutions. Hence, the lines are intersecting or coincident.
\(y = 0\) represents the x-axis, and \(y = -7\) represents a straight line parallel to the x-axis lying \(7\) units below it. Being parallel, they never intersect, resulting in no solution.
The graph of \(x = a\) is a vertical line parallel to the y-axis, and \(y = b\) is a horizontal line parallel to the x-axis. They intersect perpendicularly at the unique point \((a, b)\).
Rewriting both equations in standard form:
\[3x – y + 8 = 0\] \[6x – ky + 16 = 0\]For coincident lines: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
\[\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16} \implies \frac{1}{2} = \frac{1}{k} \implies k = 2\]For parallel lines: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
\[\frac{3}{2} = \frac{2k}{5} \implies 4k = 15 \implies k = \frac{15}{4}\]Check constant ratio: \(\frac{c_1}{c_2} = \frac{-2}{1} = -2 \neq \frac{3}{2}\). Hence, \(k = \frac{15}{4}\).
For infinitely many solutions: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
\[\frac{c}{6} = \frac{-1}{-2} = \frac{2}{3} \implies \frac{c}{6} = \frac{1}{2} = \frac{2}{3}\]Since \(\frac{1}{2} \neq \frac{2}{3}\), this equality is contradictory. Therefore, no such value of \(c\) exists.
A dependent linear equation is obtained by multiplying \(-5x + 7y – 2 = 0\) by a non-zero constant \(k\). Multiplying the entire equation by \(k = -2\):
\[-2(-5x + 7y – 2) = 0 \implies 10x – 14y + 4 = 0 \implies 10x – 14y = -4\]Adding the two equations:
\[(x – y) + (x + y) = 2 + 4 \implies 2x = 6 \implies x = 3\]Substitute \(x = 3\) into \(x + y = 4\):
\[3 + y = 4 \implies y = 1\]Therefore, \(a = 3\) and \(b = 1\).
Let the number of \(\text{Re } 1\) coins be \(x\) and \(\text{Rs } 2\) coins be \(y\):
\[x + y = 50 \quad \text{— (1)}\] \[1(x) + 2(y) = 75 \implies x + 2y = 75 \quad \text{— (2)}\]Subtracting (1) from (2):
\[(x + 2y) – (x + y) = 75 – 50 \implies y = 25\]Substituting \(y = 25\) into (1):
\[x + 25 = 50 \implies x = 25\]Hence, the number of coins are \(25\) and \(25\).
Let the son’s present age be \(x\) years and father’s present age be \(y\) years.
\[y = 6x \quad \text{— (1)}\]Four years hence, son’s age \(= x + 4\), father’s age \(= y + 4\):
\[y + 4 = 4(x + 4) \implies y + 4 = 4x + 16 \implies y – 4x = 12 \quad \text{— (2)}\]Substitute \(y = 6x\) from (1) into (2):
\[6x – 4x = 12 \implies 2x = 12 \implies x = 6\]Present age of father: \(y = 6(6) = 36\) years. Ages are \(6\) and \(36\).
Section B: Short Answer Questions with Reasoning
Ratios of coefficients:
\[\frac{a_1}{a_2} = \frac{-1}{1/2} = -2\] \[\frac{b_1}{b_2} = \frac{2}{-1/4} = -8\]Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) (\(-2 \neq -8\)), the pair has a unique solution.
Rewrite equations as:
\[4x + 3y = 6\] \[12x + 9y = 15\]Comparing ratios:
\[\frac{a_1}{a_2} = \frac{4}{12} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{6}{15} = \frac{2}{5}\]Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), they represent parallel lines, not coincident lines.
For infinitely many solutions:
\[\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\]Ratios from the given equations:
\[\frac{\lambda}{2} = \frac{3}{6} = \frac{-7}{14}\] \[\frac{3}{6} = \frac{1}{2}, \quad \text{while } \frac{-7}{14} = -\frac{1}{2}\]Since \(\frac{1}{2} \neq -\frac{1}{2}\), \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\). Thus, there is no real value of \(\lambda\) for which the system can have infinitely many solutions.
Comparing coefficients:
\[\frac{a_1}{a_2} = \frac{1}{5}, \quad \frac{b_1}{b_2} = \frac{-2}{-10} = \frac{1}{5}\]Because \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\), the system can never have a unique solution for any value of \(c\) (it has either no solution if \(c \neq 40\) or infinitely many solutions if \(c = 40\)).
Every point on the line \(x = 7\) has an x-coordinate of \(7\). This line is vertical and perpendicular to the x-axis, which means it is parallel to the y-axis, not the x-axis.
Section C: Algebraic Solutions & Consistency Conditions
For infinitely many solutions:
\[\frac{4}{2p + 7q} = \frac{5}{p + 8q} = \frac{2}{2q – p + 1}\]Equating the first two ratios:
\[4(p + 8q) = 5(2p + 7q) \implies 4p + 32q = 10p + 35q \implies 6p + 3q = 0 \implies q = -2p \quad \text{— (1)}\]Equating the first and third ratios:
\[\frac{4}{2p + 7q} = \frac{2}{2q – p + 1} \implies 2(2q – p + 1) = 2p + 7q\] \[4q – 2p + 2 = 2p + 7q \implies 4p + 3q = 2 \quad \text{— (2)}\]Substituting (1) into (2):
\[4p + 3(-2p) = 2 \implies -2p = 2 \implies p = -1\]Substituting \(p = -1\) into (1):
\[q = -2(-1) = 2\]Answer: \(p = -1, \, q = 2\).
Add the two equations:
\[(21x + 47x) + (47y + 21y) = 110 + 162\] \[68x + 68y = 272 \implies x + y = 4 \quad \text{— (1)}\]Subtract the first equation from the second:
\[(47x – 21x) + (21y – 47y) = 162 – 110\] \[26x – 26y = 52 \implies x – y = 2 \quad \text{— (2)}\]Add (1) and (2):
\[2x = 6 \implies x = 3\]Substitute \(x = 3\) into (1):
\[3 + y = 4 \implies y = 1\]Answer: \(x = 3, \, y = 1\).
(i) no solution? (ii) infinitely many solutions? (iii) a unique solution?
Comparing with standard form: \(\frac{a_1}{a_2} = \frac{\lambda}{1}\), \(\frac{b_1}{b_2} = \frac{1}{\lambda}\), \(\frac{c_1}{c_2} = \frac{\lambda^2}{1}\).
(i) No solution: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
\[\lambda = \frac{1}{\lambda} \implies \lambda^2 = 1 \implies \lambda = \pm 1\]If \(\lambda = 1\), \(\frac{c_1}{c_2} = 1 = \lambda\), giving infinite solutions. If \(\lambda = -1\), \(\frac{a_1}{a_2} = -1\), but \(\frac{c_1}{c_2} = 1 \neq -1\). Thus, \(\lambda = -1\).
(ii) Infinitely many solutions: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \mathbf{\lambda = 1}\).
(iii) Unique solution: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies \lambda \neq \frac{1}{\lambda} \implies \mathbf{\lambda \neq \pm 1}\).
Adding the two given equations:
\[(2x + y) + (4x – y) = 23 + 19 \implies 6x = 42 \implies x = 7\]Substitute \(x = 7\) into the first equation:
\[2(7) + y = 23 \implies 14 + y = 23 \implies y = 9\]Evaluating required expressions:
\[5y – 2x = 5(9) – 2(7) = 45 – 14 = 31\] \[\frac{y}{x} – 2 = \frac{9}{7} – 2 = \frac{9 – 14}{7} = -\frac{5}{7}\]Opposite sides of a rectangle are equal in length:
\[x + 3y = 13 \quad \text{— (1)}\] \[3x + y = 7 \quad \text{— (2)}\]Multiplying (1) by 3:
\[3x + 9y = 39 \quad \text{— (3)}\]Subtracting (2) from (3):
\[(3x + 9y) – (3x + y) = 39 – 7 \implies 8y = 32 \implies y = 4\]Substitute \(y = 4\) into (1):
\[x + 3(4) = 13 \implies x + 12 = 13 \implies x = 1\]Answer: \(x = 1, \, y = 4\).
Clear denominators by multiplying the first equation by 10 and the second by 24:
\[x + 2y = 10 \quad \text{— (1)}\] \[3x + 4y = 360 \quad \text{— (2)}\]Multiplying (1) by 2:
\[2x + 4y = 20 \quad \text{— (3)}\]Subtracting (3) from (2):
\[x = 340\]Substitute \(x = 340\) into (1):
\[340 + 2y = 10 \implies 2y = -330 \implies y = -165\]Now, finding \(\lambda\) using \(y = \lambda x + 5\):
\[-165 = \lambda(340) + 5 \implies 340\lambda = -170 \implies \lambda = -\frac{170}{340} = -\frac{1}{2}\]Answer: \(x = 340, \, y = -165, \, \lambda = -\frac{1}{2}\).
Find the intercepts for both straight lines:
- For \(2x + y = 4\): when \(x = 0 \implies y = 4\) (Point \(A(0, 4)\)); when \(y = 0 \implies x = 2\) (Point \(C(2, 0)\)).
- For \(2x – y = 4\): when \(x = 0 \implies y = -4\) (Point \(B(0, -4)\)); when \(y = 0 \implies x = 2\) (Point \(C(2, 0)\)).
Both lines intersect each other at \(C(2, 0)\).
Vertices of the triangle: \(A(0, 4)\), \(B(0, -4)\), and \(C(2, 0)\).
Area Calculation: The base lies along the y-axis between \(A(0, 4)\) and \(B(0, -4)\):
\[\text{Base} = 4 – (-4) = 8 \text{ units}\] \[\text{Altitude} = \text{x-coordinate of } C = 2 \text{ units}\] \[\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 8 \times 2 = 8 \text{ sq. units}\]Section D: Real-Life Contextual Word Problems
Let CP of the table be \(\text{Rs } x\) and CP of the chair be \(\text{Rs } y\).
From the first scenario:
\[\frac{110}{100}x + \frac{125}{100}y = 1050 \implies 110x + 125y = 105000 \quad \text{— (1)}\]From the second scenario:
\[\frac{125}{100}x + \frac{110}{100}y = 1065 \implies 125x + 110y = 106500 \quad \text{— (2)}\]Adding (1) and (2):
\[235x + 235y = 211500 \implies x + y = 900 \quad \text{— (3)}\]Subtracting (1) from (2):
\[15x – 15y = 1500 \implies x – y = 100 \quad \text{— (4)}\]Adding (3) and (4):
\[2x = 1000 \implies x = 500\]Substitute \(x = 500\) into (3): \(y = 400\).
Answer: Cost price of table \(= \mathbf{\text{Rs } 500}\), Cost price of chair \(= \mathbf{\text{Rs } 400}\).
Let the cost of a pen be \(\text{Rs } x\) and that of a pencil box be \(\text{Rs } y\):
\[4x + 4y = 100 \implies x + y = 25 \quad \text{— (1)}\] \[3x = y + 15 \implies 3x – y = 15 \quad \text{— (2)}\]Adding (1) and (2):
\[(x + y) + (3x – y) = 25 + 15 \implies 4x = 40 \implies x = 10\]Substituting \(x = 10\) into (1):
\[10 + y = 25 \implies y = 15\]Answer: Cost of a pen \(= \mathbf{\text{Rs } 10}\), Cost of a pencil box \(= \mathbf{\text{Rs } 15}\).
Let Salim’s present age be \(x\) years and his daughter’s age be \(y\) years.
Two years ago: \((x – 2) = 3(y – 2) \implies x – 2 = 3y – 6 \implies x – 3y = -4 \quad \text{— (1)}\)
Six years later: \((x + 6) = 2(y + 6) + 4 \implies x + 6 = 2y + 12 + 4 \implies x – 2y = 10 \quad \text{— (2)}\)
Subtracting (1) from (2):
\[(x – 2y) – (x – 3y) = 10 – (-4) \implies y = 14\]Substitute \(y = 14\) into (2):
\[x – 2(14) = 10 \implies x – 28 = 10 \implies x = 38\]Answer: Salim’s age \(= \mathbf{38 \text{ years}}\), Daughter’s age \(= \mathbf{14 \text{ years}}\).
Let the two numbers be \(x\) and \(y\):
\[\frac{x}{y} = \frac{5}{6} \implies 6x – 5y = 0 \quad \text{— (1)}\]Subtracting \(8\) from each:
\[\frac{x – 8}{y – 8} = \frac{4}{5} \implies 5(x – 8) = 4(y – 8) \implies 5x – 40 = 4y – 32 \implies 5x – 4y = 8 \quad \text{— (2)}\]Multiply (1) by 4 and (2) by 5:
\[24x – 20y = 0 \quad \text{— (3)}\] \[25x – 20y = 40 \quad \text{— (4)}\]Subtracting (3) from (4):
\[x = 40\]Substitute \(x = 40\) into (1):
\[6(40) – 5y = 0 \implies 5y = 240 \implies y = 48\]Answer: The numbers are \(\mathbf{40}\) and \(\mathbf{48}\).
Let fixed charge for the first two days be \(\text{Rs } x\) and additional charge per subsequent day be \(\text{Rs } y\).
For Latika (\(6\) days total \(= 2\) fixed days \(+ 4\) extra days):
\[x + 4y = 22 \quad \text{— (1)}\]For Anand (\(4\) days total \(= 2\) fixed days \(+ 2\) extra days):
\[x + 2y = 16 \quad \text{— (2)}\]Subtracting (2) from (1):
\[2y = 6 \implies y = 3\]Substitute \(y = 3\) into (2):
\[x + 2(3) = 16 \implies x + 6 = 16 \implies x = 10\]Answer: Fixed charge \(= \mathbf{\text{Rs } 10}\), Daily charge thereafter \(= \mathbf{\text{Rs } 3 \text{ per day}}\).
Let the number of correct answers be \(x\) and incorrect answers be \(y\):
\[x + y = 120 \quad \text{— (1)}\] \[1(x) – \frac{1}{2}y = 90 \implies 2x – y = 180 \quad \text{— (2)}\]Adding (1) and (2):
\[(x + y) + (2x – y) = 120 + 180 \implies 3x = 300 \implies x = 100\]Substitute \(x = 100\) into (1):
\[100 + y = 120 \implies y = 20\]Answer: She answered \(\mathbf{100}\) questions correctly.
