--- title: "Pair of Linear Equations in Two Variables: CBSE Class 10 Maths previous year questions" url: https://www.swavid.com/cbse/class-10/maths/pyq/pair-of-linear-equations-in-two-variables --- # Pair of Linear Equations in Two Variables: CBSE Class 10 Maths previous year questions 23 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions ### Question 5 *1 mark · MCQ* If a pair of linear equations in two variables is represented by two coincident lines, then the pair of equations has : - a unique solution - two solutions - no solution - an infinite number of solutions **Solution** 1. For a pair of linear equations represented by coincident lines, every point on the line is a common solution. 2. Therefore, coincident lines have infinitely many solutions. **Answer:** (d) an infinite number of solutions > Common mistake: Confusing coincident lines with parallel lines which have no solution. ### Question 32 *5 marks · Long answer* Determine graphically, the coordinates of vertices of a triangle whose equations are $2x - 3y + 6 = 0$; $2x + 3y - 18 = 0$ and $x = 0$. Also, find the area of this triangle. **Solution** 1. For the equation $2x - 3y + 6 = 0$, when $x = 0$, $y = 2$; when $x = -3$, $y = 0$; when $x = 3$, $y = 4$. 2. For the equation $2x + 3y - 18 = 0$, when $x = 0$, $y = 6$; when $x = 3$, $y = 4$; when $x = 9$, $y = 0$. 3. For the equation $x = 0$, it represents the y-axis. 4. Plot the lines on a graph paper and find the intersection points which form the vertices of the triangle. 5. The three lines intersect at $(0, 2)$, $(3, 4)$, and $(0, 6)$, which are the vertices of the triangle. 6. The base of the triangle along the y-axis is from $y = 2$ to $y = 6$, so base = $6 - 2 = 4\text{ units}$. 7. The corresponding height is the x-coordinate of the vertex $(3, 4)$, which is height = $3\text{ units}$. 8. Area of the triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6\text{ square units}$. **Answer:** Vertices are $(0, 2)$, $(3, 4)$, and $(0, 6)$; Area = $6\text{ sq. units}$. > Common mistake: Wrong calculation of base and height from the plotted graph. ## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions ### Question 20 *1 mark · Assertion and reason* Assertion (A): The value of p for which the system of equations $4x+py+8=0$ and $2x+2y+2=0$ is consistent is 4. Reason (R): The system of equations $a_{1}x+b_{1}y=c_{1}$ and $a_{2}x+b_{2}y=c_{2}$ is consistent with infinitely many solutions, if $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$. - Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). - Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). - Assertion (A) is true, but Reason (R) is false. - Assertion (A) is false, but Reason (R) is true. **Solution** 1. A system of linear equations is consistent if it has at least one solution (unique or infinitely many). 2. For the given equations, $\frac{a_{1}}{a_{2}} = \frac{4}{2} = 2$ and $\frac{b_{1}}{b_{2}} = \frac{p}{2}$. For unique solution, $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$, so $p \neq 4$. 3. When $p = 4$, the system has infinitely many solutions and is consistent, making Assertion (A) true. 4. However, Reason (R) only defines the condition for infinitely many solutions, whereas consistency also includes a unique solution. **Answer:** Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). > Common mistake: Assuming consistent only means infinitely many solutions, ignoring the unique solution case. ### Question 21 *2 marks · Very short answer* Solve the following system of equations for x and y : $\frac{x}{2}+\frac{2y}{3}=-1$ and $x-\frac{y}{3}=3$ **Solution** 1. Write the given equations as $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$. 2. Multiply the first equation by 6 to get $3x + 4y = -6$ (equation 1) and the second equation by 3 to get $3x - y = 9$ (equation 2). 3. Subtract equation 2 from equation 1: $(3x + 4y) - (3x - y) = -6 - 9$, which gives $5y = -15$, so $y = -3$. 4. Substitute $y = -3$ in $3x - y = 9$: $3x - (-3) = 9 \implies 3x + 3 = 9 \implies 3x = 6 \implies x = 2$. **Answer:** $x = 2$, $y = -3$ > Common mistake: Sign errors while subtracting equations or multiplying by denominators. ### Question 28 (a) *3 marks · Short answer* Solve the following system of equations graphically : $x+3y=6$; $2x-3y=12$ **Solution** 1. For the first equation $x + 3y = 6$, express $x$ as $x = 6 - 3y$. When $y = 0$, $x = 6$; when $y = 2$, $x = 0$; when $y = 1$, $x = 3$. 2. For the second equation $2x - 3y = 12$, express $x$ as $x = \frac{12 + 3y}{2}$. When $y = 0$, $x = 6$; when $y = -2$, $x = 3$; when $y = 2$, $x = 9$. 3. Plot both lines on the graph paper; the two lines intersect at the point $(6, 0)$ where both equations are satisfied. 4. The solution of the given system of equations is $x = 6$ and $y = 0$. **Answer:** $x = 6, y = 0$ > Common mistake: Interchanging the $x$ and $y$ coordinates or plotting points incorrectly from the table of values. ### Question 28 (OR) *3 marks · Short answer* x and y are complementary angles such that $x:y=1:2$. Express the given information as a system of linear equations in two variables and hence solve it. **Solution** 1. Since $x$ and $y$ are complementary angles, their sum is $90^\circ$, so $x + y = 90$. 2. Given the ratio $x : y = 1 : 2$, we can write $2x = y$, which gives $2x - y = 0$. 3. Adding the two equations: $(x + y) + (2x - y) = 90 + 0$, which gives $3x = 90$, so $x = 30$. 4. Substitute $x = 30$ into $x + y = 90$ to get $30 + y = 90$, so $y = 60$. 5. The values of the angles are $x = 30^\circ$ and $y = 60^\circ$. **Answer:** $x = 30^\circ$, $y = 60^\circ$ > Common mistake: Forgetting to define the equations based on complementary angle properties. ## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions ### Question 2 *1 mark · MCQ* 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 : - $a = 2b$ - $2a = b$ - $a + 2b = 0$ - $2a + b = 0$ **Solution** 1. Substitute $x = 1$ and $y = 2$ in the first equation $2x - 3y + a = 0$ to get $2(1) - 3(2) + a = 0$, which implies $2 - 6 + a = 0$ or $a = 4$. 2. Substitute $x = 1$ and $y = 2$ in the second equation $2x + 3y - b = 0$ to get $2(1) + 3(2) - b = 0$, which implies $2 + 6 - b = 0$ or $b = 8$. 3. Relating $a$ and $b$, we find $b = 2a$ or $2a = b$. **Answer:** (b) $2a = b$ > Common mistake: Confusing the relation between $a$ and $b$ by writing $a = 2b$ instead of $2a = b$. ### Question 32 *5 marks · Long answer* Vijay invested certain amounts of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. He received ₹ 1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received ₹ 20 more as annual interest. How much money did he invest in each scheme? **Solution** 1. Let the amount invested in scheme A be ₹ $x$ and in scheme B be ₹ $y$. 2. According to the first condition, interest from scheme A at 8% plus interest from scheme B at 9% is ₹ 1,860, so $\frac{8x}{100} + \frac{9y}{100} = 1860$, which simplifies to $8x + 9y = 186000$. 3. According to the second condition, if amounts are interchanged, the interest is ₹ 20 more, so $\frac{9x}{100} + \frac{8y}{100} = 1880$, which simplifies to $9x + 8y = 188000$. 4. Adding the two equations, $17x + 17y = 374000$, giving $x + y = 22000$, or $y = 22000 - x$. 5. Subtracting the two equations, $x - y = -2000$, or $y - x = 2000$. 6. Solving the system of linear equations, $x + (x + 2000) = 22000$, which gives $2x = 20000$, so $x = 10000$. 7. Substituting $x = 10000$, we get $y = 12000$. 8. Vijay invested ₹ 10,000 in scheme A and ₹ 12,000 in scheme B. **Answer:** ₹ 10,000 in scheme A and ₹ 12,000 in scheme B > Common mistake: Mixing up the percentages or the increased interest amount in the second condition. ## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions ### Question 17 *1 mark · MCQ* The value of $k$ for which the pair of linear equations $5x + 2y - 7 = 0$ and $2x + ky + 1 = 0$ don't have a solution, is : - $5$ - $\frac{4}{5}$ - $\frac{5}{4}$ - $\frac{5}{2}$ **Solution** 1. For a pair of linear equations not to have a solution, the lines must be parallel, which gives the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. 2. Here $a_1 = 5, b_1 = 2, c_1 = -7$ and $a_2 = 2, b_2 = k, c_2 = 1$. 3. Therefore, $\frac{5}{2} = \frac{2}{k} \implies 5k = 4 \implies k = \frac{4}{5}$. **Answer:** (b) $\frac{4}{5}$ > Common mistake: Using the condition for infinitely many solutions instead of no solution. ### Question 22 *2 marks · Short answer* Solve the following pair of linear equations for $x$ and $y$ algebraically : $x + 2y = 9$ and $y - 2x = 2$ **Solution** 1. The given equations are $x + 2y = 9$ (1) and $-2x + y = 2$ (2). 2. Multiply equation (1) by 2: $2x + 4y = 18$ (3). 3. Add equation (2) and equation (3): $(-2x + y) + (2x + 4y) = 2 + 18$, which simplifies to $5y = 20$. 4. Solving for $y$ gives $y = 4$. 5. Substitute $y = 4$ into equation (1): $x + 2(4) = 9$, which gives $x = 1$. 6. The solution is $x = 1$ and $y = 4$. **Answer:** $x = 1, y = 4$ > Common mistake: Sign errors while adding or subtracting equations during elimination. ### Question 22 (OR) *2 marks · Short answer* Check whether the point $(-4, 3)$ lies on both the lines represented by the linear equations $x + y + 1 = 0$ and $x - y = 1$. **Solution** 1. Substitute $x = -4$ and $y = 3$ into the first equation $x + y + 1 = 0$. 2. LHS = $-4 + 3 + 1 = 0$, which equals RHS. Thus, the point lies on the first line. 3. Substitute $x = -4$ and $y = 3$ into the second equation $x - y = 1$. 4. LHS = $-4 - 3 = -7$, which is not equal to RHS ($1$). Thus, the point does not lie on the second line. 5. Since the point does not satisfy both equations, it does not lie on both lines. **Answer:** No, the point does not lie on both lines. > Common mistake: Checking only one equation and concluding for both. ### Question 27 *3 marks · Short answer* The greater of two supplementary angles exceeds the smaller by $18^\circ$. Find measures of these two angles. **Solution** 1. Let the greater angle be $x^\circ$ and the smaller angle be $y^\circ$. 2. Since the angles are supplementary, $x + y = 180^\circ$. 3. According to the question, the greater angle exceeds the smaller by $18^\circ$, so $x - y = 18^\circ$. 4. Add the two equations: $(x + y) + (x - y) = 180^\circ + 18^\circ$, which gives $2x = 198^\circ$, so $x = 99^\circ$. 5. Substitute $x = 99^\circ$ into $x + y = 180^\circ$, giving $99^\circ + y = 180^\circ$, so $y = 81^\circ$. 6. The measures of the two angles are $99^\circ$ and $81^\circ$. **Answer:** $99^\circ$ and $81^\circ$ > Common mistake: Mixing up supplementary angles with complementary angles. ## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions ### Question 18 *1 mark · MCQ* In the given figure, graphs of two linear equations are shown. The pair of these linear equations is : - consistent with unique solution. - consistent with infinitely many solutions. - inconsistent. - inconsistent but can be made consistent by extending these lines. **Solution** 1. The figure shows two lines intersecting at a single point on the Cartesian plane. 2. Intersecting lines represent a pair of linear equations that have a unique solution, hence they are consistent. **Answer:** (a) consistent with unique solution. > Common mistake: Confusing intersecting lines with parallel lines (inconsistent). ### Question 21 *2 marks · Very short answer* Solve the following system of linear equations $7x - 2y = 5$ and $8x + 7y = 15$ and verify your answer. **Solution** 1. Multiply the first equation by $7$ and the second equation by $2$ to equate coefficients of $y$: $49x - 14y = 35$ and $16x + 14y = 30$. 2. Add the two new equations: $(49x + 16x) = 35 + 30$, which gives $65x = 65$, so $x = 1$. 3. Substitute $x = 1$ in $7x - 2y = 5$: $7(1) - 2y = 5$, giving $2y = 2$, so $y = 1$. 4. Verification: For $7(1) - 2(1) = 5$ and $8(1) + 7(1) = 15$, both equations are satisfied. **Answer:** $x = 1$, $y = 1$ > Common mistake: Arithmetic errors while multiplying equations for elimination method. ### Question 29 *3 marks · Short answer* 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 ? **Solution** 1. Let the present age of Rashmi be $x$ years and the present age of Nazma be $y$ years. 2. Three years ago, Rashmi's age was $(x - 3)$ and Nazma's age was $(y - 3)$. 3. According to the first condition, $x - 3 = 3(y - 3)$, which simplifies to $x - 3 = 3y - 9$, or $x - 3y = -6$ (Equation 1). 4. Ten years later, Rashmi's age will be $(x + 10)$ and Nazma's age will be $(y + 10)$. 5. According to the second condition, $x + 10 = 2(y + 10)$, which simplifies to $x + 10 = 2y + 20$, or $x - 2y = 10$ (Equation 2). 6. Subtracting Equation 1 from Equation 2: $(x - 2y) - (x - 3y) = 10 - (-6)$, which gives $y = 16$. 7. Substitute $y = 16$ in Equation 2: $x - 2(16) = 10 \implies x - 32 = 10 \implies x = 42$. **Answer:** Rashmi's age is 42 years and Nazma's age is 16 years. > Common mistake: Forgetting to add or subtract years for both persons in age word problems. ## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions ### Question 16 *1 mark · MCQ* The larger of two supplementary angles exceeds the smaller by $18$ degrees. What is the measure of larger angle ? - $81^{\circ}$ - $99^{\circ}$ - $36^{\circ}$ - $54^{\circ}$ **Solution** 1. Let the larger angle be $x$ and the smaller angle be $y$. 2. We have $x + y = 180^\circ$ and $x - y = 18^\circ$. 3. Adding the two equations gives $2x = 198^\circ$, so $x = 99^\circ$. **Answer:** (b) $99^{\circ}$ > Common mistake: Finding the smaller angle ($81^\circ$) instead of the larger angle. ### Question 20 *1 mark · Assertion and reason* Assertion (A) : The system of linear equations $3x + 5y - 4 = 0$ and $15x + 25y - 25 = 0$ is inconsistent. Reason (R) : The pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is inconsistent if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. - Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). - Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). - Assertion (A) is true, but Reason (R) is false. - Assertion (A) is false, but Reason (R) is true. **Solution** 1. For the given equations, $a_1 = 3, b_1 = 5, c_1 = -4$ and $a_2 = 15, b_2 = 25, c_2 = -25$. 2. Checking ratios: $\frac{a_1}{a_2} = \frac{3}{15} = \frac{1}{5}$, $\frac{b_1}{b_2} = \frac{5}{25} = \frac{1}{5}$, and $\frac{c_1}{c_2} = \frac{-4}{-25} = \frac{4}{25}$. 3. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the system is inconsistent, making Assertion (A) true. 4. Reason (R) states the correct condition for inconsistency, so Reason (R) is true and is the correct explanation for Assertion (A). **Answer:** Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). > Common mistake: Incorrectly simplifying the constant ratio or mixing up consistency conditions. ### Question 24 *2 marks · Short answer* Solve for $x$ and $y$ : $x + y = 6$, $2x - 3y = 4$. **Solution** 1. Write the given equations as $x + y = 6$ -- (1) and $2x - 3y = 4$ -- (2). 2. Multiply equation (1) by 3 to get $3x + 3y = 18$ -- (3). 3. Add equation (2) and equation (3) to eliminate $y$, giving $(2x - 3y) + (3x + 3y) = 4 + 18$, which simplifies to $5x = 22$, so $x = \frac{22}{5}$. 4. Substitute $x = \frac{22}{5}$ into equation (1): $\frac{22}{5} + y = 6$, giving $y = 6 - \frac{22}{5} = \frac{8}{5}$. **Answer:** $x = \frac{22}{5}, y = \frac{8}{5}$ > Common mistake: Making calculation errors while substituting fractions or failing to multiply the entire equation during elimination. ### Question 24 (OR) *2 marks · Short answer* Find out whether the following pair of linear equations are consistent or inconsistent : $5x - 3y = 11$, $-10x + 6y = 22$ **Solution** 1. Write the given equations in standard form and identify the coefficients: $a_1 = 5, b_1 = -3, c_1 = -11$ and $a_2 = -10, b_2 = 6, c_2 = -22$. 2. Find the ratios of the coefficients: $\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{-11}{-22} = \frac{1}{2}$. 3. Compare the ratios: since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel. 4. Conclude that the pair of linear equations is inconsistent. **Answer:** Inconsistent > Common mistake: Not writing the equations with constants on the same side before comparing $c_1/c_2$. ## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions ### Question 2 *1 mark · MCQ* The value of $k$ for which the pair of equations $kx = y + 2$ and $6x = 2y + 3$ has infinitely many solutions, - is $k = 3$ - does not exist - is $k = -3$ - is $k = 4$ **Solution** 1. Express the given equations in standard form: $kx - y - 2 = 0$ and $6x - 2y - 3 = 0$. 2. For a pair of linear equations to have infinitely many solutions, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. 3. Substitute the coefficients: $\frac{k}{6} = \frac{-1}{-2} = \frac{-2}{-3}$, which gives $\frac{k}{6} = \frac{1}{2}$ and $\frac{1}{2} = \frac{2}{3}$ (false). 4. Since $\frac{1}{2} \neq \frac{2}{3}$, the ratios of constants are not equal to the ratios of coefficients, so no such value of $k$ exists. **Answer:** (b) does not exist > Common mistake: Only comparing $a_1/a_2 = b_1/b_2$ and forgetting to check the constant term ratio $c_1/c_2$. ### Question 21 *2 marks · Short answer* Solve the pair of equations $x = 3$ and $y = -4$ graphically. **Solution** 1. The given equations are $x = 3$ and $y = -4$. 2. The graph of $x = 3$ is a straight line parallel to the y-axis at a distance of $3\text{ units}$ to the right of it. 3. The graph of $y = -4$ is a straight line parallel to the x-axis at a distance of $4\text{ units}$ below it. 4. The two lines intersect at the unique point $(3, -4)$, which gives the solution of the pair of linear equations. **Answer:** x = 3, y = -4 (or point of intersection is (3, -4)) > Common mistake: Drawing the line $x = 3$ parallel to the x-axis instead of the y-axis. ### Question 21 (OR) *2 marks · Short answer* Using graphical method, find whether following system of linear equations is consistent or not: $x = 0$ and $y = -7$ **Solution** 1. The given equations are $x = 0$ (which is the y-axis) and $y = -7$ (a line parallel to the x-axis). 2. Plotting both lines on a graph, the line $x = 0$ and the line $y = -7$ intersect at the point $(0, -7)$. 3. Since the pair of linear equations has a unique solution, the system of equations is consistent. **Answer:** Consistent > Common mistake: Confusing $x = 0$ with the x-axis instead of the y-axis. ### Question 26 *3 marks · Short answer* Half of the difference between two numbers is 2. The sum of the greater number and twice the smaller number is 13. Find the numbers. **Solution** 1. Let the greater number be $x$ and the smaller number be $y$. 2. According to the first condition, half of their difference is 2, so $\frac{x - y}{2} = 2$, which gives $x - y = 4$. 3. According to the second condition, the sum of the greater number and twice the smaller number is 13, so $x + 2y = 13$. 4. Subtracting the first equation from the second equation gives $(x + 2y) - (x - y) = 13 - 4$, which simplifies to $3y = 9$, so $y = 3$. 5. Substitute $y = 3$ into $x - y = 4$ to get $x - 3 = 4$, which gives $x = 7$. 6. The numbers are 7 and 3. **Answer:** The greater number is 7 and the smaller number is 3. > Common mistake: Mixing up the greater and smaller numbers while forming the linear equations. ## Related pages - [Pair of Linear Equations in Two Variables: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions) - [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student