CBSE Class 10 · Maths 
Since BCDE forms a rectangle, $ \text{CD} = \text{BE} = 7 $: $$x + y = 7 \quad \text{…(1)}$$ Using the perimeter condition: $$AB + BC + CD + DE + AE = 27$$ $$5 + (x-y) + (x+y) + (x-y) + 5 = 27$$ $$3x – y + 10 = 27 \Rightarrow 3x – y = 17 \quad \text{…(2)}$$ From (1): $ y = 7 – x $. Substituting in (2): $$3x – (7-x) = 17 \Rightarrow 4x = 24 \Rightarrow x = 6,\quad y = 1$$ Final Answer: $ x = 6,\ y = 1 $.
CBSE Class 10 Maths Pair of Linear Equations in Two Variables Previous Year Questions
Help your child master CBSE Class 10 Maths Pair of Linear Equations in Two Variables with this carefully curated collection of previous year questions sourced from real CBSE board papers spanning 2015–2025. Every question — from MCQs on consistency conditions to graphical and word-problem long answers — comes with a detailed step-by-step solution prepared by Angle Belearn’s verified CBSE Maths specialists.
CBSE Class 10 Maths Pair of Linear Equations in Two Variables — Questions with Solutions
Multiple Choice Questions (1 Mark Each)
The pair of linear equations $ 2x = 5y + 6 $ and $ 15y = 6x – 18 $ represents two lines which are?
Solution
Ans. Option (C) is correct.
Explanation: Rewriting both equations in standard form: $ 2x – 5y – 6 = 0 $ and $ -6x + 15y + 18 = 0 $.
$$\frac{a_1}{a_2} = \frac{2}{-6} = -\frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-5}{15} = -\frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-6}{18} = -\frac{1}{3}$$ Since $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $, the given pair of equations represents coincident lines.
Explanation: Rewriting both equations in standard form: $ 2x – 5y – 6 = 0 $ and $ -6x + 15y + 18 = 0 $.
$$\frac{a_1}{a_2} = \frac{2}{-6} = -\frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-5}{15} = -\frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-6}{18} = -\frac{1}{3}$$ Since $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $, the given pair of equations represents coincident lines.
The point of intersection of the line represented by $ 3x – y = 3 $ and the y-axis is given by?
Solution
Ans. Option (A) is correct.
Explanation: On the y-axis, $ x = 0 $. Substituting into $ 3x – y = 3 $: $$3(0) – y = 3 \Rightarrow y = -3$$ Thus, the point of intersection is $ (0, -3) $.
Explanation: On the y-axis, $ x = 0 $. Substituting into $ 3x – y = 3 $: $$3(0) – y = 3 \Rightarrow y = -3$$ Thus, the point of intersection is $ (0, -3) $.
3 chairs and 1 table cost ₹900; whereas 5 chairs and 3 tables cost ₹2,100. If the cost of 1 chair is ₹x and the cost of 1 table is ₹y, then the situation can be represented algebraically as?
Solution
Ans. Option (C) is correct.
Explanation: Cost of 1 chair = $ x $, cost of 1 table = $ y $.
3 chairs + 1 table = ₹900 gives: $ 3x + y = 900 $
5 chairs + 3 tables = ₹2100 gives: $ 5x + 3y = 2100 $
Explanation: Cost of 1 chair = $ x $, cost of 1 table = $ y $.
3 chairs + 1 table = ₹900 gives: $ 3x + y = 900 $
5 chairs + 3 tables = ₹2100 gives: $ 5x + 3y = 2100 $
The lines representing the given pair of linear equations are non-intersecting. Which of the following statements is true?
Solution
Ans. Option (B) is correct.
Explanation: Non-intersecting lines are parallel to each other — they have no solution (inconsistent system). The condition for parallel lines is: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$
Explanation: Non-intersecting lines are parallel to each other — they have no solution (inconsistent system). The condition for parallel lines is: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$
Graphically, the pair of equations $6x – 3y + 10 = 0$ and $2x – y + 9 = 0$ represents two lines which are?
Solution
Ans. Option (D) is correct.
Explanation: From the given equations: $$\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}$$ Since $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} $, the lines are parallel.
Explanation: From the given equations: $$\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}$$ Since $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} $, the lines are parallel.
If $ x = 1 $ and $ y = 2 $ is a solution of the pair of linear equations $ 2x + 3y + a = 0 $ and $ 2x + 3y + b = 0 $, then:
Solution
Ans. Option (B) is correct.
Explanation: Substituting $ x = 1,\ y = 2 $ in the first equation: $$2(1) + 3(2) + a = 0 \Rightarrow 8 + a = 0 \Rightarrow a = -8$$ Substituting in the second equation: $$2(1) + 3(2) + b = 0 \Rightarrow 8 + b = 0 \Rightarrow b = -8$$ Now, $ 2a = 2(-8) = -16 $ and $ b = -8 $, giving $ 2a = b $. Hence Option (B).
Explanation: Substituting $ x = 1,\ y = 2 $ in the first equation: $$2(1) + 3(2) + a = 0 \Rightarrow 8 + a = 0 \Rightarrow a = -8$$ Substituting in the second equation: $$2(1) + 3(2) + b = 0 \Rightarrow 8 + b = 0 \Rightarrow b = -8$$ Now, $ 2a = 2(-8) = -16 $ and $ b = -8 $, giving $ 2a = b $. Hence Option (B).
Two lines are given to be parallel. The equation of one of the lines is $3x – 2y = 5$. The equation of the second line can be:
Solution
Ans. Option (C) is correct.
Explanation: For two parallel lines $ a_1x + b_1y + c_1 = 0 $ and $ a_2x + b_2y + c_2 = 0 $: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$ Checking Option (C): $ \dfrac{3}{-12} = \dfrac{-2}{8} = -\dfrac{1}{4} $, but $ \dfrac{5}{7} \neq -\dfrac{1}{4} $. ✓ Condition satisfied, so the lines are parallel.
Explanation: For two parallel lines $ a_1x + b_1y + c_1 = 0 $ and $ a_2x + b_2y + c_2 = 0 $: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$ Checking Option (C): $ \dfrac{3}{-12} = \dfrac{-2}{8} = -\dfrac{1}{4} $, but $ \dfrac{5}{7} \neq -\dfrac{1}{4} $. ✓ Condition satisfied, so the lines are parallel.
What is the value of $ k $ such that the following pair of equations have infinitely many solutions?
$$x – 2y = 3 \quad \text{and} \quad -3x + ky = -9$$
Solution
Ans. Option (D) is correct.
Explanation: Rewriting: $ x – 2y – 3 = 0 $ and $ -3x + ky + 9 = 0 $.
$ a_1=1,\ b_1=-2,\ c_1=-3 $ and $ a_2=-3,\ b_2=k,\ c_2=9 $.
For infinitely many solutions: $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $ $$\frac{1}{-3} = \frac{-2}{k} = \frac{-3}{9}$$ From $ \dfrac{-2}{k} = \dfrac{-3}{9} \Rightarrow 3k = 18 \Rightarrow k = 6 $.
Explanation: Rewriting: $ x – 2y – 3 = 0 $ and $ -3x + ky + 9 = 0 $.
$ a_1=1,\ b_1=-2,\ c_1=-3 $ and $ a_2=-3,\ b_2=k,\ c_2=9 $.
For infinitely many solutions: $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $ $$\frac{1}{-3} = \frac{-2}{k} = \frac{-3}{9}$$ From $ \dfrac{-2}{k} = \dfrac{-3}{9} \Rightarrow 3k = 18 \Rightarrow k = 6 $.
If the system of equations $3x + y = 1$ and $(2k-1)x + (k-1)y = 2k+1$ is inconsistent, then $ k = $?
Solution
Ans. Option (D) is correct.
Concept: For an inconsistent (no solution) system: $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} $.
Here $ a_1=3,\ b_1=1,\ c_1=-1 $ and $ a_2=2k-1,\ b_2=k-1,\ c_2=-(2k+1) $.
Setting $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} $: $$\frac{3}{2k-1} = \frac{1}{k-1} \Rightarrow 3(k-1) = 2k-1 \Rightarrow 3k-3 = 2k-1 \Rightarrow k = 2$$
Concept: For an inconsistent (no solution) system: $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} $.
Here $ a_1=3,\ b_1=1,\ c_1=-1 $ and $ a_2=2k-1,\ b_2=k-1,\ c_2=-(2k+1) $.
Setting $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} $: $$\frac{3}{2k-1} = \frac{1}{k-1} \Rightarrow 3(k-1) = 2k-1 \Rightarrow 3k-3 = 2k-1 \Rightarrow k = 2$$
The values of $ x $ and $ y $ satisfying the two equations $32x + 33y = 34$ and $33x + 32y = 31$ respectively are:
Solution
Ans. Option (A) is correct.
Explanation: Subtracting equation (ii) from equation (i): $$(32x + 33y) – (33x + 32y) = 34 – 31 \Rightarrow -x + y = 3 \Rightarrow y = x + 3$$ Substituting $ y = x + 3 $ into equation (i): $$32x + 33(x+3) = 34 \Rightarrow 65x = -65 \Rightarrow x = -1$$ Then $ y = -1 + 3 = 2 $. Hence $ x = -1,\ y = 2 $.
Explanation: Subtracting equation (ii) from equation (i): $$(32x + 33y) – (33x + 32y) = 34 – 31 \Rightarrow -x + y = 3 \Rightarrow y = x + 3$$ Substituting $ y = x + 3 $ into equation (i): $$32x + 33(x+3) = 34 \Rightarrow 65x = -65 \Rightarrow x = -1$$ Then $ y = -1 + 3 = 2 $. Hence $ x = -1,\ y = 2 $.
In a triangle $ \triangle ABC $, $ \angle A = x^\circ $, $ \angle B = (3x – 2)^\circ $, $ \angle C = y^\circ $. Also, $ \angle C – \angle B = 9^\circ $. The sum of the greatest and the smallest angles of this triangle is:
Solution
Ans. Option (A) is correct.
Explanation: Angle sum of a triangle = 180°: $$x + (3x-2) + y = 180 \Rightarrow 4x + y = 182 \quad \text{…(i)}$$ From $ \angle C – \angle B = 9 $: $$y – (3x-2) = 9 \Rightarrow y – 3x = 7 \quad \text{…(ii)}$$ Subtracting (ii) from (i): $ 7x = 175 \Rightarrow x = 25^\circ $.
From (ii): $ y = 7 + 75 = 82^\circ $. And $ \angle B = 3(25)-2 = 73^\circ $.
Sum of greatest (82°) and smallest (25°) = 107°.
Explanation: Angle sum of a triangle = 180°: $$x + (3x-2) + y = 180 \Rightarrow 4x + y = 182 \quad \text{…(i)}$$ From $ \angle C – \angle B = 9 $: $$y – (3x-2) = 9 \Rightarrow y – 3x = 7 \quad \text{…(ii)}$$ Subtracting (ii) from (i): $ 7x = 175 \Rightarrow x = 25^\circ $.
From (ii): $ y = 7 + 75 = 82^\circ $. And $ \angle B = 3(25)-2 = 73^\circ $.
Sum of greatest (82°) and smallest (25°) = 107°.
If a pair of linear equations given by $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has a unique solution, then which of the following is true?
Solution
Ans. Option (B) is correct.
Explanation: For a unique solution, the two lines must intersect at exactly one point. The condition is: $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$ This is equivalent to $ a_1b_2 \neq a_2b_1 $, which is Option (B).
Explanation: For a unique solution, the two lines must intersect at exactly one point. The condition is: $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$ This is equivalent to $ a_1b_2 \neq a_2b_1 $, which is Option (B).
Short Answer Questions — 1 Mark
For what value of $k$, the pair of linear equations $3x + y = 3$ and $6x + ky = 8$ does not have a solution?
Answer
Given: $ 3x + y – 3 = 0 $ …(i) and $ 6x + ky – 8 = 0 $ …(ii)
Comparing with standard form: $ a_1=3,\ b_1=1,\ c_1=-3 $ and $ a_2=6,\ b_2=k,\ c_2=-8 $.
For no solution (inconsistent): $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{3}{6} = \frac{1}{k} \neq \frac{-3}{-8}$$ From $ \dfrac{3}{6} = \dfrac{1}{k} \Rightarrow \dfrac{1}{2} = \dfrac{1}{k} \Rightarrow k = 2 $.
Check: $ \dfrac{1}{2} \neq \dfrac{3}{8} $ ✓
Hence, the value of $k$ is 2.
Comparing with standard form: $ a_1=3,\ b_1=1,\ c_1=-3 $ and $ a_2=6,\ b_2=k,\ c_2=-8 $.
For no solution (inconsistent): $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{3}{6} = \frac{1}{k} \neq \frac{-3}{-8}$$ From $ \dfrac{3}{6} = \dfrac{1}{k} \Rightarrow \dfrac{1}{2} = \dfrac{1}{k} \Rightarrow k = 2 $.
Check: $ \dfrac{1}{2} \neq \dfrac{3}{8} $ ✓
Hence, the value of $k$ is 2.
Find the value of $k$ for which the system of linear equations $x + 2y = 3$ and $5x + ky + 7 = 0$ is inconsistent.
Answer
Rewriting: $ x + 2y – 3 = 0 $ and $ 5x + ky + 7 = 0 $.
$ a_1=1,\ b_1=2,\ c_1=-3 $ and $ a_2=5,\ b_2=k,\ c_2=7 $.
For inconsistency: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{1}{5} = \frac{2}{k} \neq \frac{-3}{7}$$ From $ \dfrac{1}{5} = \dfrac{2}{k} \Rightarrow k = 10 $.
Check: $ \dfrac{1}{5} \neq \dfrac{-3}{7} $ ✓
Hence, the value of $k$ is 10.
$ a_1=1,\ b_1=2,\ c_1=-3 $ and $ a_2=5,\ b_2=k,\ c_2=7 $.
For inconsistency: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{1}{5} = \frac{2}{k} \neq \frac{-3}{7}$$ From $ \dfrac{1}{5} = \dfrac{2}{k} \Rightarrow k = 10 $.
Check: $ \dfrac{1}{5} \neq \dfrac{-3}{7} $ ✓
Hence, the value of $k$ is 10.
For which value(s) of $p$, will the lines represented by the following pair of linear equations be parallel?
$3x – y – 5 = 0$ and $6x – 2y – p = 0$
$3x – y – 5 = 0$ and $6x – 2y – p = 0$
Answer
$ a_1=3,\ b_1=-1,\ c_1=-5 $ and $ a_2=6,\ b_2=-2,\ c_2=-p $.
For parallel lines: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{3}{6} = \frac{-1}{-2} \neq \frac{-5}{-p}$$ $$\frac{1}{2} = \frac{1}{2} \neq \frac{5}{p} \Rightarrow p \neq 10$$ Hence, $p$ can be any real number except 10.
For parallel lines: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{3}{6} = \frac{-1}{-2} \neq \frac{-5}{-p}$$ $$\frac{1}{2} = \frac{1}{2} \neq \frac{5}{p} \Rightarrow p \neq 10$$ Hence, $p$ can be any real number except 10.
Short Answer Questions — 2 Marks
Find $c$ if the system of equations $cx + 3y + (3 – c) = 0$ and $12x + cy – c = 0$ has infinitely many solutions.
Answer
$ a_1=c,\ b_1=3,\ c_1=3-c $ and $ a_2=12,\ b_2=c,\ c_2=-c $.
For infinitely many solutions: $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $
Equating $ \dfrac{a_1}{a_2} = \dfrac{c_1}{c_2} $: $$\frac{c}{12} = \frac{3-c}{-c} \Rightarrow -c^2 = 12(3-c) \Rightarrow -c^2 = 36 – 12c$$ $$c^2 – 12c + 36 = 0 \Rightarrow (c-6)^2 = 0 \Rightarrow c = 6$$ Verifying with $ \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $: $ \dfrac{3}{6} = \dfrac{-3}{-6} = \dfrac{1}{2} $ ✓
Hence, the value of $c$ is 6.
For infinitely many solutions: $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $
Equating $ \dfrac{a_1}{a_2} = \dfrac{c_1}{c_2} $: $$\frac{c}{12} = \frac{3-c}{-c} \Rightarrow -c^2 = 12(3-c) \Rightarrow -c^2 = 36 – 12c$$ $$c^2 – 12c + 36 = 0 \Rightarrow (c-6)^2 = 0 \Rightarrow c = 6$$ Verifying with $ \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $: $ \dfrac{3}{6} = \dfrac{-3}{-6} = \dfrac{1}{2} $ ✓
Hence, the value of $c$ is 6.
For what value of $k$, the following pair of linear equations have infinitely many solutions?
$2x + 3y = 7$ and $(k + 1)x + (2k – 1)y = 4k + 1$
$2x + 3y = 7$ and $(k + 1)x + (2k – 1)y = 4k + 1$
Answer
Rewriting: $ 2x + 3y – 7 = 0 $ and $ (k+1)x + (2k-1)y – (4k+1) = 0 $.
$ a_1=2,\ b_1=3,\ c_1=-7 $ and $ a_2=k+1,\ b_2=2k-1,\ c_2=-(4k+1) $.
For infinitely many solutions: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \Rightarrow \frac{2}{k+1} = \frac{3}{2k-1}$$ Cross-multiplying: $$2(2k-1) = 3(k+1) \Rightarrow 4k – 2 = 3k + 3 \Rightarrow k = 5$$ Hence, for $k = 5$, the pair of equations will have infinitely many solutions.
$ a_1=2,\ b_1=3,\ c_1=-7 $ and $ a_2=k+1,\ b_2=2k-1,\ c_2=-(4k+1) $.
For infinitely many solutions: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \Rightarrow \frac{2}{k+1} = \frac{3}{2k-1}$$ Cross-multiplying: $$2(2k-1) = 3(k+1) \Rightarrow 4k – 2 = 3k + 3 \Rightarrow k = 5$$ Hence, for $k = 5$, the pair of equations will have infinitely many solutions.
Find the value(s) of $k$ so that the pair of equations $x + 2y = 5$ and $3x + ky + 15 = 0$ has a unique solution.
Answer
Rewriting: $ x + 2y – 5 = 0 $ and $ 3x + ky + 15 = 0 $.
$ a_1=1,\ a_2=3,\ b_1=2,\ b_2=k $.
For a unique solution: $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow \frac{1}{3} \neq \frac{2}{k} \Rightarrow k \neq 6$$ Hence, for all values of $k$ except $k = 6$, the given pair of equations has a unique solution.
$ a_1=1,\ a_2=3,\ b_1=2,\ b_2=k $.
For a unique solution: $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow \frac{1}{3} \neq \frac{2}{k} \Rightarrow k \neq 6$$ Hence, for all values of $k$ except $k = 6$, the given pair of equations has a unique solution.
Long Answer Questions — 3 Marks
Determine graphically the co-ordinates of the vertices of a triangle, the equations of whose sides are given by $2y – x = 8$, $5y – x = 14$ and $y – 2x = 1$.
Answer
Setting up tables of values:
Line 1: $ 2y – x = 8 \Rightarrow x = 2y – 8 $
Line 2: $ 5y – x = 14 \Rightarrow x = 5y – 14 $
Line 3: $ y – 2x = 1 \Rightarrow y = 1 + 2x $
Plotting and finding intersections graphically:
The co-ordinates of the vertices of the triangle are: $A(1,\ 3)$, $B(2,\ 5)$, $C(-4,\ 2)$.
Line 1: $ 2y – x = 8 \Rightarrow x = 2y – 8 $
| $ y $ | 0 | 4 | 5 |
|---|---|---|---|
| $ x = 2y-8 $ | −8 | 0 | 2 |
Line 2: $ 5y – x = 14 \Rightarrow x = 5y – 14 $
| $ y $ | 3 | 4 | 2 |
|---|---|---|---|
| $ x = 5y-14 $ | 1 | 6 | −4 |
Line 3: $ y – 2x = 1 \Rightarrow y = 1 + 2x $
| $ x $ | 0 | 1 | 2 |
|---|---|---|---|
| $ y = 1+2x $ | 1 | 3 | 5 |
Plotting and finding intersections graphically:
The co-ordinates of the vertices of the triangle are: $A(1,\ 3)$, $B(2,\ 5)$, $C(-4,\ 2)$.
Three lines $x + 3y = 6$, $2x – 3y = 12$ and $x = 0$ are enclosing a beautiful triangular park. Find the points of intersection of the lines graphically.
Answer
Line 1: $ x + 3y = 6 \Rightarrow y = \dfrac{6-x}{3} $
Line 2: $ 2x – 3y = 12 \Rightarrow y = \dfrac{2x-12}{3} $
Line 3: $ x = 0 $ is the y-axis.
The points of intersection (vertices of the triangle) are: $A(6,\ 0)$, $B(0,\ 2)$, $C(0,\ -4)$.
| $ x $ | 6 | 0 | 3 |
|---|---|---|---|
| $ y $ | 0 | 2 | 1 |
Line 2: $ 2x – 3y = 12 \Rightarrow y = \dfrac{2x-12}{3} $
| $ x $ | 6 | 0 | 3 |
|---|---|---|---|
| $ y $ | 0 | −4 | −2 |
Line 3: $ x = 0 $ is the y-axis.
The points of intersection (vertices of the triangle) are: $A(6,\ 0)$, $B(0,\ 2)$, $C(0,\ -4)$.
The incomes of two persons A and B are in the ratio 8 : 7 and the ratio of their expenditures is 19 : 16. If their savings are ₹2550 per month each, find their monthly incomes.
Answer
Let income of A = $ 8x $ and income of B = $ 7x $.
Let expenditure of A = $ 19y $ and expenditure of B = $ 16y $.
Using Saving = Income − Expenditure and savings = ₹2550 each: $$8x – 19y = 2550 \quad \text{…(i)}$$ $$7x – 16y = 2550 \quad \text{…(ii)}$$ Multiplying (i) by 7 and (ii) by 8, then subtracting: $$56x – 133y = 17850$$ $$56x – 128y = 20400$$ $$\Rightarrow -5y = -2550 \Rightarrow y = 510$$ Substituting $ y = 510 $ in (i): $ 8x = 2550 + 9690 = 12240 \Rightarrow x = 1530 $.
Income of A = $ 8 \times 1530 = \textbf{₹12,240} $ per month.
Income of B = $ 7 \times 1530 = \textbf{₹10,710} $ per month.
Let expenditure of A = $ 19y $ and expenditure of B = $ 16y $.
Using Saving = Income − Expenditure and savings = ₹2550 each: $$8x – 19y = 2550 \quad \text{…(i)}$$ $$7x – 16y = 2550 \quad \text{…(ii)}$$ Multiplying (i) by 7 and (ii) by 8, then subtracting: $$56x – 133y = 17850$$ $$56x – 128y = 20400$$ $$\Rightarrow -5y = -2550 \Rightarrow y = 510$$ Substituting $ y = 510 $ in (i): $ 8x = 2550 + 9690 = 12240 \Rightarrow x = 1530 $.
Income of A = $ 8 \times 1530 = \textbf{₹12,240} $ per month.
Income of B = $ 7 \times 1530 = \textbf{₹10,710} $ per month.
(6, 0) and (0, 2) are two of the points of intersection of two lines represented by a pair of linear equations.
(i) How many points of intersection does the pair of linear equations have in total? Justify your answer.
(ii) Find the equation that represents one of the lines of the above pair.
(i) How many points of intersection does the pair of linear equations have in total? Justify your answer.
(ii) Find the equation that represents one of the lines of the above pair.
Answer
(i) The pair has infinitely many points of intersection.
Since more than one point of intersection is given, the lines must be coincident (overlapping). Every point on one line is also on the other, so there are infinitely many common points.
(ii) Using the general form $ ax + by = c $ and substituting the two given points:
Point $ (6, 0) $: $ 6a = c \Rightarrow a = \dfrac{c}{6} $
Point $ (0, 2) $: $ 2b = c \Rightarrow b = \dfrac{c}{2} $
Substituting and dividing by $ c $: $$\frac{x}{6} + \frac{y}{2} = 1 \Rightarrow x + 3y = 6$$ One equation of the pair is $ x + 3y = 6 $.
Since more than one point of intersection is given, the lines must be coincident (overlapping). Every point on one line is also on the other, so there are infinitely many common points.
(ii) Using the general form $ ax + by = c $ and substituting the two given points:
Point $ (6, 0) $: $ 6a = c \Rightarrow a = \dfrac{c}{6} $
Point $ (0, 2) $: $ 2b = c \Rightarrow b = \dfrac{c}{2} $
Substituting and dividing by $ c $: $$\frac{x}{6} + \frac{y}{2} = 1 \Rightarrow x + 3y = 6$$ One equation of the pair is $ x + 3y = 6 $.
If 16 is subtracted from twice the greater of two positive numbers, the result is half the other number. If 1 is subtracted from half the greater number, the result is still half the other number. Find the two numbers.
Answer
Let the two numbers be $ x $ and $ y $ with $ x > y $.
From the first condition: $ 2x – 16 = \dfrac{y}{2} \Rightarrow 4x – y = 32 $ …(i)
From the second condition: $ \dfrac{x}{2} – 1 = \dfrac{y}{2} \Rightarrow x – y = 2 $ …(ii)
From (ii): $ x = y + 2 $. Substituting in (i): $$4(y+2) – y = 32 \Rightarrow 3y = 24 \Rightarrow y = 8$$ Then $ x = 10 $.
The two numbers are 10 and 8.
From the first condition: $ 2x – 16 = \dfrac{y}{2} \Rightarrow 4x – y = 32 $ …(i)
From the second condition: $ \dfrac{x}{2} – 1 = \dfrac{y}{2} \Rightarrow x – y = 2 $ …(ii)
From (ii): $ x = y + 2 $. Substituting in (i): $$4(y+2) – y = 32 \Rightarrow 3y = 24 \Rightarrow y = 8$$ Then $ x = 10 $.
The two numbers are 10 and 8.
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Answer
Let the ten’s digit = $ x $ and unit’s digit = $ y $. Original number = $ 10x + y $, reversed = $ 10y + x $.
From sum condition: $ (10x+y) + (10y+x) = 66 \Rightarrow 11(x+y) = 66 \Rightarrow x + y = 6 $ …(i)
Digits differ by 2: either $ x – y = 2 $ …(ii) or $ y – x = 2 $ …(iii).
Case 1: Solving (i) and (ii): $ x = 4,\ y = 2 $. Number = 42.
Case 2: Solving (i) and (iii): $ y = 4,\ x = 2 $. Number = 24.
There are two such numbers: 42 and 24.
From sum condition: $ (10x+y) + (10y+x) = 66 \Rightarrow 11(x+y) = 66 \Rightarrow x + y = 6 $ …(i)
Digits differ by 2: either $ x – y = 2 $ …(ii) or $ y – x = 2 $ …(iii).
Case 1: Solving (i) and (ii): $ x = 4,\ y = 2 $. Number = 42.
Case 2: Solving (i) and (iii): $ y = 4,\ x = 2 $. Number = 24.
There are two such numbers: 42 and 24.
Solve:
$$\frac{2}{x} + \frac{3}{y} = 2, \quad \frac{4}{x} – \frac{9}{y} = -1, \quad x,\ y > 0$$
Answer
Let $ m = \dfrac{1}{x} $ and $ n = \dfrac{1}{y} $. The equations become:
$$2m + 3n = 2 \quad \text{…(1)}$$
$$4m – 9n = -1 \quad \text{…(2)}$$
Multiply (1) by $ -2 $: $ -4m – 6n = -4 $ …(3). Adding (2) and (3):
$$-15n = -5 \Rightarrow n = \frac{1}{3}$$
Substituting in (1): $ 2m + 1 = 2 \Rightarrow m = \dfrac{1}{2} $.
Since $ m = \dfrac{1}{x} = \dfrac{1}{2} \Rightarrow x = 2 $ and $ n = \dfrac{1}{y} = \dfrac{1}{3} \Rightarrow y = 3 $.
Final Answer: $ x = 2,\ y = 3 $.
Since $ m = \dfrac{1}{x} = \dfrac{1}{2} \Rightarrow x = 2 $ and $ n = \dfrac{1}{y} = \dfrac{1}{3} \Rightarrow y = 3 $.
Final Answer: $ x = 2,\ y = 3 $.
In the figure, ABCDE is a pentagon with BE ∥ CD and BC ∥ DE. BC is perpendicular to CD. AE = AB = 5 cm, BE = 7 cm, BC = x – y and CD = x + y. If the perimeter of ABCDE is 27 cm, find the value of x and y, given that $x, y \neq 0$.
Answer
Since BCDE forms a rectangle, $ \text{CD} = \text{BE} = 7 $: $$x + y = 7 \quad \text{…(1)}$$ Using the perimeter condition: $$AB + BC + CD + DE + AE = 27$$ $$5 + (x-y) + (x+y) + (x-y) + 5 = 27$$ $$3x – y + 10 = 27 \Rightarrow 3x – y = 17 \quad \text{…(2)}$$ From (1): $ y = 7 – x $. Substituting in (2): $$3x – (7-x) = 17 \Rightarrow 4x = 24 \Rightarrow x = 6,\quad y = 1$$ Final Answer: $ x = 6,\ y = 1 $.
The present age of a father is three years more than three times the age of his son. Three years hence, the father’s age will be 10 years more than twice the age of the son. Determine their present ages.
Answer
Let the present age of the son = $ x $ years and the father’s age = $ 3x + 3 $ years.
Three years hence: Father’s age = $ 3x + 6 $, Son’s age = $ x + 3 $.
According to the condition: $$3x + 6 = 10 + 2(x + 3) \Rightarrow 3x + 6 = 2x + 16 \Rightarrow x = 10$$ Father’s present age: $ 3(10) + 3 = 33 $ years.
Son’s present age = 10 years; Father’s present age = 33 years.
Three years hence: Father’s age = $ 3x + 6 $, Son’s age = $ x + 3 $.
According to the condition: $$3x + 6 = 10 + 2(x + 3) \Rightarrow 3x + 6 = 2x + 16 \Rightarrow x = 10$$ Father’s present age: $ 3(10) + 3 = 33 $ years.
Son’s present age = 10 years; Father’s present age = 33 years.
Long Answer Questions — 5 Marks
Represent the following pair of linear equations graphically and hence comment on the condition of consistency of this pair.
$$x – 5y = 6 \quad \text{and} \quad 2x – 10y = 12$$
Answer
Line 1: $ x – 5y = 6 \Rightarrow y = \dfrac{x-6}{5} $
Line 2: $ 2x – 10y = 12 \Rightarrow y = \dfrac{x-6}{5} $
Observation: Both equations represent the same line — the graphs coincide completely.
Condition of Consistency: The system has infinitely many solutions. The pair of linear equations is consistent and dependent.
| $ x $ | 6 | 1 | −4 |
|---|---|---|---|
| $ y $ | 0 | −1 | −2 |
Line 2: $ 2x – 10y = 12 \Rightarrow y = \dfrac{x-6}{5} $
| $ x $ | 6 | 1 | −4 |
|---|---|---|---|
| $ y $ | 0 | −1 | −2 |
Observation: Both equations represent the same line — the graphs coincide completely.
Condition of Consistency: The system has infinitely many solutions. The pair of linear equations is consistent and dependent.
For what values of $m$ and $n$, the following system of linear equations has infinitely many solutions?
$$3x + 4y = 12 \quad \text{and} \quad (m+n)x + 2(m-n)y = 5m – 1$$
Answer
For infinitely many solutions: $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $, i.e.,
$$\frac{3}{m+n} = \frac{4}{2(m-n)} = \frac{12}{5m-1}$$ Step 1: Equating first two ratios:
$$\frac{3}{m+n} = \frac{2}{m-n} \Rightarrow 3(m-n) = 2(m+n) \Rightarrow m – 5n = 0 \quad \text{…(1)}$$ Step 2: Equating second and third ratios:
$$\frac{2}{m-n} = \frac{12}{5m-1} \Rightarrow 2(5m-1) = 12(m-n) \Rightarrow -2m + 12n = 2 \Rightarrow m – 6n = -1 \quad \text{…(2)}$$ Step 3: Solving (1) and (2): From (1), $ m = 5n $. Substituting: $ 5n – 6n = -1 \Rightarrow n = 1 \Rightarrow m = 5 $.
The values are $ m = 5 $ and $ n = 1 $.
The values are $ m = 5 $ and $ n = 1 $.
For Uttarakhand flood victims, two Sections A and B of Class X contributed ₹1,500. If the contribution of X-A was ₹100 less than that of X-B, find graphically the amounts contributed by both the sections.
Answer
Let the amounts contributed by X-A = ₹$ x $ and X-B = ₹$ y $.
Total: $ x + y = 1500 $ …(i) X-A was ₹100 less: $ y – x = 100 $ …(ii)
From (i): $ y = 1500 – x $
From (ii): $ y = x + 100 $
The lines intersect at $ (700, 800) $.
Section X-A contributed ₹700 and Section X-B contributed ₹800.
Total: $ x + y = 1500 $ …(i) X-A was ₹100 less: $ y – x = 100 $ …(ii)
From (i): $ y = 1500 – x $
| $ x $ | 0 | 700 | 1500 |
|---|---|---|---|
| $ y $ | 1500 | 800 | 0 |
From (ii): $ y = x + 100 $
| $ x $ | 0 | 700 | 500 |
|---|---|---|---|
| $ y $ | 100 | 800 | 600 |
The lines intersect at $ (700, 800) $.
Section X-A contributed ₹700 and Section X-B contributed ₹800.
If 3 chairs and 1 table cost ₹1500 and 6 chairs and 1 table cost ₹2400, form linear equations to represent this situation.
Answer
Let the cost of 1 chair = ₹$ x $ and the cost of 1 table = ₹$ y $.
From the first condition (3 chairs + 1 table = ₹1500): $$3x + y = 1500$$ From the second condition (6 chairs + 1 table = ₹2400): $$6x + y = 2400$$ The required pair of linear equations is $ 3x + y = 1500 $ and $ 6x + y = 2400 $.
From the first condition (3 chairs + 1 table = ₹1500): $$3x + y = 1500$$ From the second condition (6 chairs + 1 table = ₹2400): $$6x + y = 2400$$ The required pair of linear equations is $ 3x + y = 1500 $ and $ 6x + y = 2400 $.
Aruna has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹75, then find the number of ₹1 and ₹2 coins.
Answer
Let the number of ₹1 coins = $ x $ and ₹2 coins = $ y $.
Total coins: $ x + y = 50 $ …(i)
Total amount: $ x + 2y = 75 $ …(ii)
Subtracting (i) from (ii): $ y = 25 $. Substituting back: $ x = 25 $.
Number of ₹1 coins = 25 and Number of ₹2 coins = 25.
Total coins: $ x + y = 50 $ …(i)
Total amount: $ x + 2y = 75 $ …(ii)
Subtracting (i) from (ii): $ y = 25 $. Substituting back: $ x = 25 $.
Number of ₹1 coins = 25 and Number of ₹2 coins = 25.
Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?
Answer
Let Sumit’s present age = $ x $ and son’s age = $ y $.
From condition 1: $ x = 3y $ …(i)
After five years: $ x + 5 = \dfrac{5}{2}(y + 5) \Rightarrow 2x + 10 = 5y + 25 \Rightarrow 2x = 5y + 15 $ …(ii)
Substituting (i) in (ii): $ 6y = 5y + 15 \Rightarrow y = 15 $. Then $ x = 45 $.
Sumit’s present age is 45 years.
From condition 1: $ x = 3y $ …(i)
After five years: $ x + 5 = \dfrac{5}{2}(y + 5) \Rightarrow 2x + 10 = 5y + 25 \Rightarrow 2x = 5y + 15 $ …(ii)
Substituting (i) in (ii): $ 6y = 5y + 15 \Rightarrow y = 15 $. Then $ x = 45 $.
Sumit’s present age is 45 years.
Solve the following simultaneous equations for $x$ and $y$:
$$3x – 5y = 4 \quad \text{…(i)}$$
$$9x – 2y = 7 \quad \text{…(ii)}$$
Answer
Multiply equation (i) by 3: $ 9x – 15y = 12 $ …(iii)
Subtract (ii) from (iii): $$(9x – 15y) – (9x – 2y) = 12 – 7 \Rightarrow -13y = 5 \Rightarrow y = -\frac{5}{13}$$ Substituting in (i): $ 3x = 4 + 5\left(-\dfrac{5}{13}\right) = 4 – \dfrac{25}{13} = \dfrac{27}{13} \Rightarrow x = \dfrac{9}{13} $.
Final Answer: $ x = \dfrac{9}{13},\ y = -\dfrac{5}{13} $.
Subtract (ii) from (iii): $$(9x – 15y) – (9x – 2y) = 12 – 7 \Rightarrow -13y = 5 \Rightarrow y = -\frac{5}{13}$$ Substituting in (i): $ 3x = 4 + 5\left(-\dfrac{5}{13}\right) = 4 – \dfrac{25}{13} = \dfrac{27}{13} \Rightarrow x = \dfrac{9}{13} $.
Final Answer: $ x = \dfrac{9}{13},\ y = -\dfrac{5}{13} $.
Given below is a pair of linear equations: $2x – my = 9$ and $4x – ny = 9$.
Find at least one pair of possible values of $m$ and $n$, if they exist, for which the above pair has:
(i) a unique solution (ii) infinitely many solutions (iii) no solution
Find at least one pair of possible values of $m$ and $n$, if they exist, for which the above pair has:
(i) a unique solution (ii) infinitely many solutions (iii) no solution
Answer
$ a_1=2,\ a_2=4,\ b_1=m,\ b_2=n,\ c_1=c_2=9 $. Note $ \dfrac{a_1}{a_2} = \dfrac{1}{2} $ and $ \dfrac{c_1}{c_2} = 1 $.
(i) Unique Solution — need $ \dfrac{m}{n} \neq \dfrac{1}{2} $.
Take $ m = 2,\ n = 6 $: $ \dfrac{2}{6} = \dfrac{1}{3} \neq \dfrac{1}{2} $ ✓
Equations become $ 2x – 2y = 9 $ and $ 4x – 6y = 9 $. One valid pair: $ m = 2,\ n = 6 $.
(ii) Infinitely Many Solutions — need $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $, i.e., $ \dfrac{1}{2} = \dfrac{m}{n} = 1 $.
But $ \dfrac{1}{2} \neq 1 $, so this condition is impossible.
No such values of $m$ and $n$ exist for infinitely many solutions.
(iii) No Solution — need $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} $, i.e., $ \dfrac{m}{n} = \dfrac{1}{2} $ and $ \dfrac{c_1}{c_2} = 1 \neq \dfrac{1}{2} $ ✓
Take $ m = 3,\ n = 6 $: $ \dfrac{3}{6} = \dfrac{1}{2} $ and $ \dfrac{c_1}{c_2} = 1 \neq \dfrac{1}{2} $ ✓
Equations become $ 2x – 3y = 9 $ and $ 4x – 6y = 9 $. One valid pair: $ m = 3,\ n = 6 $.
(i) Unique Solution — need $ \dfrac{m}{n} \neq \dfrac{1}{2} $.
Take $ m = 2,\ n = 6 $: $ \dfrac{2}{6} = \dfrac{1}{3} \neq \dfrac{1}{2} $ ✓
Equations become $ 2x – 2y = 9 $ and $ 4x – 6y = 9 $. One valid pair: $ m = 2,\ n = 6 $.
(ii) Infinitely Many Solutions — need $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} $, i.e., $ \dfrac{1}{2} = \dfrac{m}{n} = 1 $.
But $ \dfrac{1}{2} \neq 1 $, so this condition is impossible.
No such values of $m$ and $n$ exist for infinitely many solutions.
(iii) No Solution — need $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} $, i.e., $ \dfrac{m}{n} = \dfrac{1}{2} $ and $ \dfrac{c_1}{c_2} = 1 \neq \dfrac{1}{2} $ ✓
Take $ m = 3,\ n = 6 $: $ \dfrac{3}{6} = \dfrac{1}{2} $ and $ \dfrac{c_1}{c_2} = 1 \neq \dfrac{1}{2} $ ✓
Equations become $ 2x – 3y = 9 $ and $ 4x – 6y = 9 $. One valid pair: $ m = 3,\ n = 6 $.
Additional Multiple Choice Questions (2025 Board)
If x = 1 and y = 2 is a solution of the pair of linear equations $2x – 3y + a = 0$ and $2x + 3y – b = 0$, then:
Solution
Ans. Option (B) is correct.
Explanation: Substituting $ x=1,\ y=2 $ in the first equation: $$2(1) – 3(2) + a = 0 \Rightarrow 2 – 6 + a = 0 \Rightarrow a = 4$$ Substituting in the second equation: $$2(1) + 3(2) – b = 0 \Rightarrow 8 – b = 0 \Rightarrow b = 8$$ Now, $ 2a = 2(4) = 8 = b $. Hence $ 2a = b $, which is Option (B).
Explanation: Substituting $ x=1,\ y=2 $ in the first equation: $$2(1) – 3(2) + a = 0 \Rightarrow 2 – 6 + a = 0 \Rightarrow a = 4$$ Substituting in the second equation: $$2(1) + 3(2) – b = 0 \Rightarrow 8 – b = 0 \Rightarrow b = 8$$ Now, $ 2a = 2(4) = 8 = b $. Hence $ 2a = b $, which is Option (B).
The value of $k$ for which the pair of linear equations $5x + 2y – 7 = 0$ and $2x + ky + 1 = 0$ does not have a solution is:
Solution
Ans. Option (B) is correct.
Explanation: For no solution (parallel lines), $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} $.
$$\frac{5}{2} = \frac{2}{k} \Rightarrow 5k = 4 \Rightarrow k = \frac{4}{5}$$ Check: $ \dfrac{-7}{1} \neq \dfrac{5}{2} $ ✓. Hence $ k = \dfrac{4}{5} $, which is Option (B).
Explanation: For no solution (parallel lines), $ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} $.
$$\frac{5}{2} = \frac{2}{k} \Rightarrow 5k = 4 \Rightarrow k = \frac{4}{5}$$ Check: $ \dfrac{-7}{1} \neq \dfrac{5}{2} $ ✓. Hence $ k = \dfrac{4}{5} $, which is Option (B).
Explore More CBSE Class 10 Maths Chapters
- CBSE Class 10 Maths Real Numbers Questions →
- CBSE Class 10 Maths Polynomials Questions →
- CBSE Class 10 Maths Quadratic Equations Questions →
- CBSE Class 10 Maths Arithmetic Progressions Questions →
- CBSE Class 10 Maths Triangles Questions →
- CBSE Class 10 Maths Coordinate Geometry Questions →
- CBSE Class 10 Maths Introduction to Trigonometry Questions →
- CBSE Class 10 Maths Some Applications of Trigonometry Questions →
- CBSE Class 10 Maths Circles Questions →
- CBSE Class 10 Maths Areas Related to Circles Questions →
- CBSE Class 10 Maths Surface Areas and Volumes Questions →
- CBSE Class 10 Maths Statistics Questions →
- CBSE Class 10 Maths Probability Questions →
Frequently Asked Questions
What does the Pair of Linear Equations chapter cover in CBSE Class 10 Maths? ⌄
This chapter covers graphical and algebraic methods for solving a pair of linear equations in two variables — including the substitution, elimination, and cross-multiplication methods. Students also learn to interpret the consistency of a system (unique solution, infinitely many solutions, or no solution) using the ratios of coefficients, and apply these concepts to real-life word problems.
How many marks does this chapter carry in the CBSE Class 10 board exam? ⌄
Pair of Linear Equations is part of the Algebra unit, which together with other algebra chapters carries approximately 20 marks in the CBSE Class 10 Mathematics board paper. This chapter contributes across question types — 1-mark MCQs on consistency conditions, 2-mark direct solving questions, 3-mark word problems, and 5-mark graphical or multi-part questions.
What are the most important topics students should focus on in this chapter? ⌄
The most frequently tested topics in board exams are: the conditions for consistency using coefficient ratios ($ \frac{a_1}{a_2},\ \frac{b_1}{b_2},\ \frac{c_1}{c_2} $), finding the value of an unknown constant for a given type of solution, solving word problems (age, coins, income, geometry), and the graphical method of finding the vertices of a triangle formed by three lines.
What common mistakes do students make when solving questions from this chapter? ⌄
A very common error is confusing the three consistency conditions — students often mix up when to use $ = = $ (coincident) versus $ = \neq $ (parallel). Another frequent mistake is errors in sign while applying the elimination method, especially when multiplying equations before subtracting. In word problems, students sometimes set up the equations incorrectly by swapping the roles of the two variables. Practising a range of board questions helps your child avoid these under exam pressure.
How does Angle Belearn help students score well in this chapter? ⌄
Angle Belearn’s CBSE specialists have curated this question bank from real board papers spanning a decade, ensuring your child practises exactly the kind of questions the board asks. Each solution is written step-by-step with clear reasoning — mirroring the structured working that earns full marks in board evaluation. Regular practice with these verified questions builds both the speed and the accuracy your child needs to walk into the exam with confidence.
