15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
How many solutions does the pair \(x + 2y = 5\), \(3x + 6y = 12\) have?
Why: a₁/a₂ = b₁/b₂ = 1/3 but c₁/c₂ = 5/12, so the lines are parallel.
\(\dfrac{a_1}{a_2} = \dfrac13\), \(\dfrac{b_1}{b_2} = \dfrac26 = \dfrac13\), \(\dfrac{c_1}{c_2} = \dfrac{5}{12} \ne \dfrac13\). The lines are parallel, so there is no solution.
The solution of \(2x + y = 11\) and \(x - y = 1\) is
Why: Adding the equations gives 3x = 12.
Adding: \(3x = 12 \Rightarrow x = 4\). Then \(y = x - 1 = 3\). Check: \(2(4) + 3 = 11\).
For which value of \(k\) do \(3x - y = 2\) and \(6x + ky = 4\) have infinitely many solutions?
Why: Need 3/6 = −1/k = 2/4, so −1/k = 1/2.
\(\dfrac36 = \dfrac{-1}{k} = \dfrac24\). From \(\dfrac{-1}{k} = \dfrac12\), \(k = -2\).
The lines \(4x + my = 7\) and \(2x + 3y = 5\) intersect at exactly one point when
Why: A unique solution needs a₁/a₂ ≠ b₁/b₂, i.e. 2 ≠ m/3.
\(\dfrac{a_1}{a_2} = \dfrac42 = 2\). Unique solution when \(\dfrac{m}{3} \ne 2\), i.e. \(m \ne 6\). (At \(m = 6\) the lines are parallel: \(\tfrac{c_1}{c_2} = \tfrac75 \ne 2\).)
The graph of \(2x - 5y = 10\) meets the \(x\)-axis at
Why: On the x-axis y = 0, so 2x = 10.
Put \(y = 0\): \(2x = 10 \Rightarrow x = 5\). The point is \((5, 0)\).
If \(x + y = 7\) and \(2x - 3y = -1\), then \(x - y\) equals
Why: Substitute x = 7 − y: 14 − 5y = −1 gives y = 3, x = 4.
\(x = 7 - y \Rightarrow 2(7 - y) - 3y = -1 \Rightarrow 14 - 5y = -1 \Rightarrow y = 3\), \(x = 4\). So \(x - y = 1\).
The lines represented by \(3x - 4y = 8\) and \(9x - 12y = 24\) are
Why: All three ratios are equal: 3/9 = −4/−12 = 8/24 = 1/3.
\(\dfrac39 = \dfrac{-4}{-12} = \dfrac{8}{24} = \dfrac13\). The second equation is \(3\) times the first, so the lines coincide.
Which statement is true for the pair \(5x + 2y = 9\) and \(2x - y = 0\)?
Why: 5/2 ≠ 2/(−1), so the lines intersect: one solution (1, 2).
\(\dfrac{a_1}{a_2} = \dfrac52\), \(\dfrac{b_1}{b_2} = -2\); these differ, so there is exactly one solution: \(y = 2x\), \(9x = 9\), \(x = 1\), \(y = 2\).
The sum of two numbers is \(50\) and their difference is \(14\). The larger number is
Why: Add x + y = 50 and x − y = 14: 2x = 64.
\(x + y = 50\), \(x - y = 14\). Adding, \(2x = 64\), so \(x = 32\) (and \(y = 18\)).
\(3\) pens and \(2\) notebooks cost ₹\(110\), while \(2\) pens and \(3\) notebooks cost ₹\(115\). The cost of one notebook is
Why: Add: 5(p + n) = 225; subtract: n − p = 5.
\(3p + 2n = 110\), \(2p + 3n = 115\). Adding: \(p + n = 45\); subtracting: \(n - p = 5\). So \(n = 25\), \(p = 20\).
For which value of \(a\) do \(ax + 4y = 8\) and \(5x + 10y = 20\) have infinitely many solutions?
Why: b₁/b₂ = c₁/c₂ = 2/5, so a/5 must also be 2/5.
\(\dfrac{4}{10} = \dfrac{8}{20} = \dfrac25\), so \(\dfrac{a}{5} = \dfrac25 \Rightarrow a = 2\).
The lines \(y = 2x\) and \(y = -x + 6\) meet at the point
Why: 2x = −x + 6 gives x = 2, then y = 4.
\(2x = -x + 6 \Rightarrow 3x = 6 \Rightarrow x = 2\), \(y = 4\). The point is \((2, 4)\).
The lines \(x - y = 1\) and \(x + y = 5\) and the \(y\)-axis enclose a triangle. Its area (in square units) is
Why: Vertices (0, −1), (0, 5) and (3, 2): base 6 on the y-axis, height 3.
The lines meet at \((3, 2)\) and cut the \(y\)-axis at \((0, -1)\) and \((0, 5)\). Area \(= \dfrac12 \times 6 \times 3 = 9\).
Riya is \(4\) years older than twice her brother's age, and the sum of their ages is \(25\) years. Riya's age is
Why: r = 2b + 4 and r + b = 25 give 3b + 4 = 25.
\(r = 2b + 4\), \(r + b = 25 \Rightarrow 3b + 4 = 25 \Rightarrow b = 7\), so \(r = 18\).
The pair \(2x - 3y = 4\) and \(6x - 9y = c\) is inconsistent when
Why: a₁/a₂ = b₁/b₂ = 1/3, so no solution exactly when 4/c ≠ 1/3.
\(\dfrac26 = \dfrac{-3}{-9} = \dfrac13\). The pair has no solution when \(\dfrac{4}{c} \ne \dfrac13\), i.e. \(c \ne 12\); at \(c = 12\) the lines coincide.
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
On Monday a group of friends bought \(4\) samosas and \(3\) glasses of juice from the school canteen for ₹\(130\). On Tuesday another group bought \(2\) samosas and \(5\) glasses of juice for ₹\(170\). Let one samosa cost ₹\(x\) and one glass of juice cost ₹\(y\).
A residents' welfare association is fencing a rectangular park. The fence around it is \(84\) m long, and the length of the park is \(6\) m more than twice its width. Let the length be \(l\) m and the width be \(w\) m.
This free set is separate from the chapter's question bank. On the Pair of Linear Equations in Two Variables chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 43 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.