General Instructions
- The question paper has 21 questions divided into six sections, carrying a total of 40 marks.
- Section A: Questions 1–8, one mark each (8 marks). Section B: Questions 9–10, one mark each (2 marks).
- Section C: Questions 11–14, two marks each (8 marks). Section D: Questions 15–18, three marks each (12 marks).
- Section E: Questions 19–20, four marks each (8 marks). Section F: Question 21, two marks (2 marks).
- In Questions 19 and 20, attempt either part (a) or part (b). All other questions are compulsory.
- Write the full solution, not just the final answer, wherever marks are awarded for working.
Section A – Multiple Choice Questions
8 questions × 1 mark = 8 marks
Each question below has four options (A, B, C, D). Choose the one correct answer.
If \(x = 1\) and \(y = 2\) is a solution of the equations \(2x - 3y + a = 0\) and \(2x + 3y - b = 0\), then \(a\) and \(b\) satisfy CBSE PYQ · 2025
For which value of \(p\) does the pair of equations \(px + 3y = p - 3\) and \(12x + py = p\) have infinitely many solutions? CBSE PYQ · 2025
The value of \(k\) for which the system of equations \(3x - y + 8 = 0\) and \(6x - ky + 16 = 0\) has infinitely many solutions is CBSE PYQ · 2024
Which of the following types of straight lines will be represented by the system of equations \(3x + 4y = 5\) and \(6x + 8y = 7\)? CBSE PYQ · 2024
The pair of linear equations whose graphs are two distinct parallel straight lines has
The graphs of a pair of linear equations intersect at the single point \((3, -2)\). How many solutions does this pair of equations have?
If \(x + y = 7\) and \(x - y = 1\), then the point \((x, y)\) is
The pair of linear equations \(kx + 2y = 5\) and \(3x + 6y = 10\) will have a unique solution when
Section B – Assertion and Reason
2 questions × 1 mark = 2 marks
In each of the following questions, an Assertion (A) and a Reason (R) are given. Choose the correct option from (a) to (d).
- (a) Both A and R are true, and R is the correct explanation of A.
- (b) Both A and R are true, but R is not the correct explanation of A.
- (c) A is true, but R is false.
- (d) A is false, but R is true.
Assertion (A): The pair of linear equations \(px + 3y + 59 = 0\) and \(2x + 6y + 118 = 0\) has infinitely many solutions when \(p = 1\).
Reason (R): The pair of equations has a unique solution when \(p \neq 1\). CBSE PYQ · 2025
Assertion (A): The pair of linear equations \(2x + 5y = 11\) and \(4x + 10y = 22\) has infinitely many solutions.
Reason (R): The graphs of these two equations are coincident straight lines.
Section C – Short Answer Questions
4 questions × 2 marks = 8 marks
Solve the following system of linear equations and verify your answer: CBSE PYQ · 2024
\[7x - 2y = 5 \text{ and } 8x + 7y = 15\]
Solve the following pair of linear equations by the substitution method:
\[3x - 2y = 5 \text{ and } 2x + 3y = 12\]
For what value of \(k\) will the pair of equations \(kx + 2y = 3\) and \(6x + 4y = 11\) have no solution?
Two pens and three erasers cost ₹46, while three pens and two erasers cost ₹54. Form a pair of linear equations and find the cost of one pen and one eraser.
Section D – Application-Based Questions
4 questions × 3 marks = 12 marks
These questions test application of concepts. Show your working clearly.
Three years ago, Rashmi was thrice as old as Nazma. Ten years later, Rashmi will be twice as old as Nazma. How old are Rashmi and Nazma now? CBSE PYQ · 2024
Solve the following pair of linear equations graphically: \[x - y = 1 \text{ and } x + y = 5\]
(a) Draw the two lines and write the solution of the pair. [2]
(b) Find the points where the two lines cut the y-axis and the area of the triangle formed by the two lines and the y-axis. [1]
A two-digit number is such that the product of its digits is 12. When 36 is added to the number, its digits interchange their places.
(a) Taking the digits as \(x\) and \(y\), form the pair of linear equations. [1]
(b) Solve the equations to find the number. [2]
The perimeter of a rectangle is 36 cm and its length is 3 cm more than twice its breadth. Taking the breadth as \(x\) cm and the length as \(y\) cm:
(a) Form the pair of linear equations in \(x\) and \(y\). [1]
(b) Find the length and the breadth of the rectangle. [1]
(c) Find the area of the rectangle. [1]
Section E – Long Answer Questions
2 questions × 4 marks = 8 marks
Attempt either part (a) or part (b).
(a) Solve the following system of equations graphically and verify the solution algebraically: \[x + y = 5 \text{ and } 3x - y = 3\] [4]
(b) A fraction becomes \(\frac{4}{5}\) when 1 is added to its numerator, and becomes \(\frac{1}{2}\) when 1 is added to its denominator. Find the fraction. [4]
Attempt either part (a) or part (b). CBSE PYQ · 2024
(a) Using the graphical method, solve the following system of equations: \[3x + y + 4 = 0 \text{ and } 3x - y + 2 = 0\] [4]
(b) Tara scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks deducted for each wrong answer, then Tara would have scored 50 marks. Assuming that Tara attempted all questions, find the total number of questions in the test. [4]
Section F – Case-Based Question
1 question × 2 marks = 2 marks
Essel World is one of India's largest amusement parks, offering a diverse range of thrilling rides, water attractions and entertainment options for visitors of all ages. The park is known for its iconic "Water Kingdom" section, making it a popular destination for family outings and fun-filled adventure. The ticket charges for the park are ₹150 per child and ₹250 per adult.
On a day, the cashier of the park found that 300 tickets were sold and an amount of ₹55,000 was collected.
(a) If the number of children who visited is \(x\) and the number of adults is \(y\), write the given situation as a pair of linear equations. [1]
(b) How many children visited the amusement park that day? CBSE PYQ · 2024 [1]
Check your work: Section A (8) + Section B (2) + Section C (8) + Section D (12) + Section E (8) + Section F (2) = 40 marks. Attempt the paper fully before revealing the answers below.
Answer Key
Section A – Multiple Choice Questions
| Question | Answer | Marks | Source |
|---|---|---|---|
| 1 | B | 1 | CBSE PYQ (2025) |
| 2 | B | 1 | CBSE PYQ (2025) |
| 3 | B | 1 | CBSE PYQ (2024) |
| 4 | A | 1 | CBSE PYQ (2024) |
| 5 | C | 1 | Original |
| 6 | B | 1 | Original |
| 7 | B | 1 | Original |
| 8 | C | 1 | Original |
Section B – Assertion and Reason
| Question | Answer | Marks | Source |
|---|---|---|---|
| 9 | (b) | 1 | CBSE PYQ (2025) |
| 10 | (a) | 1 | Original |
Section C – Short Answer Questions
| Question | Answer | Marks | Source |
|---|---|---|---|
| 11 | \(x = 1,\ y = 1\) | 2 | CBSE PYQ (2024) |
| 12 | \(x = 3,\ y = 2\) | 2 | Original |
| 13 | \(k = 3\) | 2 | Original |
| 14 | Pen ₹14, eraser ₹6 | 2 | Original |
Section D – Application-Based Questions
| Question | Answer | Marks | Source |
|---|---|---|---|
| 15 | Rashmi is 42, Nazma is 16 | 3 | CBSE PYQ (2024) |
| 16 | (a) \((3, 2)\) (b) \((0, -1)\), \((0, 5)\); area \(9\) sq units | 3 | Original |
| 17 | (a) \(xy = 12\), \(x - y = -4\) (b) number is 26 | 3 | Original |
| 18 | (a) \(y = 2x + 3\), \(x + y = 18\) (b) length 13, breadth 5 (c) 65 cm² | 3 | Original |
Section E – Long Answer Questions
| Question | Answer | Marks | Source |
|---|---|---|---|
| 19 | (a) \((2, 3)\) OR (b) \(\frac{3}{5}\) | 4 | Original |
| 20 | (a) \((-1, -1)\) OR (b) 20 questions (15 right, 5 wrong) | 4 | CBSE PYQ (2024) |
Section F – Case-Based Question
| Question | Answer | Marks | Source |
|---|---|---|---|
| 21 | (a) \(x + y = 300\), \(150x + 250y = 55000\) (b) 200 children | 2 | CBSE PYQ (2024) |
Detailed Solutions
Source: CBSE Board Paper 2025 (Mathematics Standard, 30/1/1)
Concept Used: Substituting a given solution into a linear equation to find unknown constants.
Put \(x = 1\), \(y = 2\) into both equations:
\(2(1) - 3(2) + a = 0 \Rightarrow 2 - 6 + a = 0 \Rightarrow a = 4\).
\(2(1) + 3(2) - b = 0 \Rightarrow 2 + 6 - b = 0 \Rightarrow b = 8\).
Now \(2a = 2 \times 4 = 8 = b\), so \(2a = b\).
Source: CBSE Board Paper 2025 (Mathematics Standard, 30/4/1)
Concept Used: Comparing coefficients of \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\) for infinitely many solutions \(\left(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\right)\).
Here \(a_1 = p,\ b_1 = 3,\ c_1 = p - 3\) and \(a_2 = 12,\ b_2 = p,\ c_2 = p\).
\(\frac{p}{12} = \frac{3}{p} \Rightarrow p^2 = 36 \Rightarrow p = \pm 6\).
Now check which value also satisfies the ratio of the constant terms. The two equations in standard form are \(px + 3y - (p - 3) = 0\) and \(12x + py - p = 0\), so \(\frac{c_1}{c_2} = \frac{p - 3}{p}\).
For \(p = 6\): \(\frac{p - 3}{p} = \frac{3}{6} = \frac{1}{2} = \frac{6}{12}\) ✓ — infinitely many solutions.
For \(p = -6\): \(\frac{p - 3}{p} = \frac{-9}{-6} = \frac{3}{2} \neq \frac{-6}{12} = -\frac{1}{2}\) ✗ — the pair has no solution.
Hence only \(p = 6\) works.
Source: CBSE Board Paper 2024 (Mathematics Standard, 30/3/1)
Concept Used: Condition for infinitely many solutions: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
Comparison: \(\frac{3}{6} = \frac{1}{2}\), \(\frac{-1}{-k} = \frac{1}{k}\), and \(\frac{8}{16} = \frac{1}{2}\).
For infinitely many solutions all three ratios are equal:
\(\frac{1}{2} = \frac{1}{k} = \frac{1}{2} \Rightarrow k = 2\).
Source: CBSE Board Paper 2024 (Mathematics Standard, 30/5/1)
Concept Used: Condition for parallel (inconsistent) lines: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
\(\frac{3}{6} = \frac{1}{2}\), \(\frac{4}{8} = \frac{1}{2}\) but \(\frac{5}{7} \neq \frac{1}{2}\).
The ratio of the \(x\)-coefficients equals the ratio of the \(y\)-coefficients, but the ratio of the constant terms is different. Hence the lines are parallel and the system has no solution.
Source: Original question
Concept Used: Geometry of a pair of linear equations — parallel distinct lines never meet.
A solution of a pair of linear equations is the point where the two lines meet. Two distinct parallel lines never meet, so the pair has no common point — that is, no solution. Such a system is called inconsistent.
Source: Original question
Concept Used: The solution of a pair of linear equations is the intersection point of its graphs.
Two lines that meet at exactly one point represent a consistent pair with a unique solution — the coordinates of that single intersection point, \((3, -2)\).
Source: Original question
Concept Used: Solving a pair by elimination (adding the equations).
Add the two equations: \((x + y) + (x - y) = 7 + 1 \Rightarrow 2x = 8 \Rightarrow x = 4\).
Then \(4 + y = 7 \Rightarrow y = 3\). The point is \((4, 3)\).
Source: Original question
Concept Used: Condition for a unique solution: \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\).
\(\frac{a_1}{a_2} = \frac{k}{3}\) and \(\frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3}\).
A unique solution requires \(\frac{k}{3} \neq \frac{1}{3} \Rightarrow k \neq 1\).
Source: CBSE Board Paper 2025 (Mathematics Standard, 30/2/1)
Concept Used: Consistency of a pair by comparing ratios of coefficients.
For \(px + 3y + 59 = 0\) and \(2x + 6y + 118 = 0\):
\(\frac{a_1}{a_2} = \frac{p}{2}\), \(\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{59}{118} = \frac{1}{2}\).
Assertion: Infinitely many solutions when all three ratios are equal, i.e. \(\frac{p}{2} = \frac{1}{2} \Rightarrow p = 1\). The Assertion is true.
Reason: When \(p \neq 1\), \(\frac{p}{2} \neq \frac{1}{2}\) so the pair has a unique solution. The Reason is also true.
However, the Reason only states when the pair is not consistent with infinitely many solutions; it does not explain why \(p = 1\) produces infinitely many solutions (which requires comparing all three ratios). Hence R is not the correct explanation of A.
Source: Original question
Concept Used: Coincident lines represent a dependent pair with infinitely many solutions.
Dividing \(4x + 10y = 22\) by 2 gives \(2x + 5y = 11\), which is exactly the first equation. So both equations represent the very same straight line (coincident lines). Every point of this line is a common solution, so the pair has infinitely many solutions. The Reason correctly explains the Assertion.
Source: CBSE Board Paper 2024 (Mathematics Standard, 30/1/1)
Concept Used: Elimination method for a pair of linear equations, followed by verification.
Make the coefficients of \(y\) equal in magnitude:
Multiply the first equation by 7: \(49x - 14y = 35\).
Multiply the second equation by 2: \(16x + 14y = 30\).
Adding: \(65x = 65 \Rightarrow x = 1\).
Substitute \(x = 1\) in \(7x - 2y = 5\):
\(7(1) - 2y = 5 \Rightarrow 2y = 2 \Rightarrow y = 1\).
Verification: In \(8x + 7y = 15\): \(8(1) + 7(1) = 15\). ✓
Source: Original question
Concept Used: Substitution method — expressing one variable in terms of the other.
From \(3x - 2y = 5\): \(x = \frac{5 + 2y}{3}\).
Substitute into \(2x + 3y = 12\):
\(2\left(\frac{5 + 2y}{3}\right) + 3y = 12\)
Multiplying by 3: \(10 + 4y + 9y = 36 \Rightarrow 13y = 26 \Rightarrow y = 2\).
Then \(x = \frac{5 + 2(2)}{3} = \frac{9}{3} = 3\).
Check: \(3(3) - 2(2) = 5\). ✓
Source: Original question
Concept Used: Condition for no solution: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
For \(kx + 2y = 3\) and \(6x + 4y = 11\):
\(\frac{a_1}{a_2} = \frac{k}{6}\), \(\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{3}{11}\).
No solution requires \(\frac{k}{6} = \frac{1}{2} \Rightarrow k = 3\).
Then \(\frac{c_1}{c_2} = \frac{3}{11} \neq \frac{1}{2}\), so the pair is inconsistent (parallel lines, no solution).
Source: Original question
Concept Used: Forming equations from a real-life (cost) situation and solving them.
Let the cost of one pen be ₹\(p\) and one eraser be ₹\(e\).
\(2p + 3e = 46\) …(1)
\(3p + 2e = 54\) …(2)
Adding (1) and (2): \(5p + 5e = 100 \Rightarrow p + e = 20\). …(3)
From (3), \(p = 20 - e\). Putting this in (1):
\(2(20 - e) + 3e = 46 \Rightarrow 40 + e = 46 \Rightarrow e = 6\).
Hence \(p = 20 - 6 = 14\).
Check: \(2(14) + 3(6) = 46\) and \(3(14) + 2(6) = 54\). ✓
Source: CBSE Board Paper 2024 (Mathematics Standard, 30/1/1)
Concept Used: Age problems — expressing age conditions as linear equations.
Let Rashmi's present age be \(R\) years and Nazma's be \(N\) years.
Three years ago: \(R - 3 = 3(N - 3) \Rightarrow R = 3N - 6\). …(1)
Ten years later: \(R + 10 = 2(N + 10) \Rightarrow R = 2N + 10\). …(2)
From (1) and (2): \(3N - 6 = 2N + 10 \Rightarrow N = 16\).
Then \(R = 2(16) + 10 = 42\).
Check: 3 years ago: 39 and 13 (thrice) ✓ 10 years later: 52 and 26 (twice) ✓
Source: Original question
Concept Used: Solving a pair graphically and computing an area bounded by the lines.
(a) Plotting points for each line:
\(x - y = 1\): \((1, 0)\), \((0, -1)\), \((3, 2)\).
\(x + y = 5\): \((5, 0)\), \((0, 5)\), \((3, 2)\).
Both lines pass through \((3, 2)\), and they are not parallel, so the unique solution is \((3, 2)\).
Check: \(3 - 2 = 1\) and \(3 + 2 = 5\). ✓
(b) On the y-axis, \(x = 0\). The cuts are \((0, -1)\) for \(x - y = 1\) and \((0, 5)\) for \(x + y = 5\).
The vertical base of the triangle, along the y-axis, has length \(5 - (-1) = 6\) units.
The third vertex is \((3, 2)\), whose perpendicular distance from the y-axis is 3 units.
Area \(= \frac{1}{2} \times 6 \times 3 = 9\) square units.
Source: Original question
Concept Used: Two-digit-number problems — expressing digits as variables.
(a) Let the tens digit be \(x\) and the units digit be \(y\). The number is \(10x + y\).
Product condition: \(xy = 12\). …(1)
Interchange condition: \(10x + y + 36 = 10y + x \Rightarrow 9x + 36 = 9y \Rightarrow x - y = -4\). …(2)
(b) From (2), \(y = x + 4\). Putting this in (1):
\(x(x + 4) = 12 \Rightarrow x^2 + 4x - 12 = 0 \Rightarrow (x + 6)(x - 2) = 0 \Rightarrow x = 2\) (rejecting \(x = -6\) as a digit).
So \(y = 6\), and the number is \(10(2) + 6 = 26\).
Check: Product \(2 \times 6 = 12\) and \(26 + 36 = 62\) (digits interchanged). ✓
Source: Original question
Concept Used: Perimeter and area of a rectangle from a linear-equations setup.
(a) Length \(y = 2\)(breadth) \(+ 3 \Rightarrow y = 2x + 3\). …(1)
Perimeter \(2(x + y) = 36 \Rightarrow x + y = 18\). …(2)
(b) Substitute (1) in (2): \(x + (2x + 3) = 18 \Rightarrow 3x = 15 \Rightarrow x = 5\).
Then \(y = 2(5) + 3 = 13\). Breadth is 5 cm and length is 13 cm.
(c) Area \(= 13 \times 5 = 65\) cm².
Check: Perimeter \(= 2(13 + 5) = 36\) cm. ✓
Source: Original question
Concept Used: Graphical solution followed by algebraic verification.
Plotting points:
\(x + y = 5\): \((5, 0)\), \((0, 5)\), \((2, 3)\).
\(3x - y = 3\): \((1, 0)\), \((0, -3)\), \((2, 3)\).
The two lines meet at \((2, 3)\), so the solution is \((2, 3)\).
Algebraic verification: Add the equations \(x + y = 5\) and \(3x - y = 3\):
\(4x = 8 \Rightarrow x = 2\), then \(y = 5 - 2 = 3\).
Check: \(3(2) - 3 = 3\). ✓
Source: Original question
Concept Used: Fraction problems — setting up equations from given conditions.
Let the fraction be \(\frac{n}{d}\).
Adding 1 to the numerator gives \(\frac{4}{5}\): \(\frac{n + 1}{d} = \frac{4}{5} \Rightarrow 5n + 5 = 4d\). …(1)
Adding 1 to the denominator gives \(\frac{1}{2}\): \(\frac{n}{d + 1} = \frac{1}{2} \Rightarrow 2n = d + 1\). …(2)
From (2), \(d = 2n - 1\). Substitute in (1):
\(5n + 5 = 4(2n - 1) \Rightarrow 5n + 5 = 8n - 4 \Rightarrow 9 = 3n \Rightarrow n = 3\).
Then \(d = 2(3) - 1 = 5\).
Check: \(\frac{3 + 1}{5} = \frac{4}{5}\) and \(\frac{3}{5 + 1} = \frac{1}{2}\). ✓
Source: CBSE Board Paper 2024 (Mathematics Standard, 30/2/1)
Concept Used: Graphical solution of a pair of linear equations.
Plotting points:
\(3x + y + 4 = 0\) i.e. \(y = -3x - 4\): \((-2, 2)\), \((-1, -1)\), \((0, -4)\).
\(3x - y + 2 = 0\) i.e. \(y = 3x + 2\): \((-2, -4)\), \((-1, -1)\), \((0, 2)\).
The two lines intersect at \((-1, -1)\), so the solution is \((-1, -1)\).
Check: \(3(-1) + (-1) + 4 = 0\) and \(3(-1) - (-1) + 2 = 0\). ✓
Source: CBSE Board Paper 2024 (Mathematics Standard, 30/2/1)
Concept Used: Test-marks problem — forming equations with right and wrong answers.
Let the number of right answers be \(x\), wrong answers be \(y\), so total questions \(= x + y\).
First scheme: \(3x - y = 40\). …(1)
Second scheme: \(4x - 2y = 50\). …(2)
From (1), \(y = 3x - 40\). Substitute in (2):
\(4x - 2(3x - 40) = 50 \Rightarrow 4x - 6x + 80 = 50 \Rightarrow 2x = 30 \Rightarrow x = 15\).
Then \(y = 3(15) - 40 = 5\).
Total questions \(= x + y = 15 + 5 = 20\).
Check: \(3(15) - 5 = 40\) and \(4(15) - 2(5) = 50\). ✓
Source: CBSE Board Paper 2024 (Mathematics Standard, 30/5/1) — Case Study (scene text lightly abridged)
Concept Used: Interpreting a real-life situation as a pair of linear equations.
(a) With \(x\) children and \(y\) adults:
Total tickets: \(x + y = 300\). …(1)
Total collection: \(150x + 250y = 55000\). …(2)
(b) From (1), \(x = 300 - y\). Substitute in (2):
\(150(300 - y) + 250y = 55000 \Rightarrow 45000 + 100y = 55000 \Rightarrow y = 100\).
Then \(x = 300 - 100 = 200\).
Check: \(200 + 100 = 300\) tickets and \(150(200) + 250(100) = 30000 + 25000 = 55000\). ✓