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CBSE Important Questions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables is part of the CBSE Class 10 Maths syllabus. It explains how two linear equations in two variables can be solved together, both by drawing their graphs and by algebraic methods such as substitution, elimination and cross multiplication. Students also learn when a pair of equations has one solution, no solution or infinitely many solutions.

These CBSE important questions for Class 10 Maths Chapter 3 cover reading the number of solutions from a graph, consistent and inconsistent systems, the condition for a unique solution, solving equations by the elimination method, the shortcut for equations of the form ax + by = c and bx + ay = d, parallel lines, and word problems on boats and streams, work and time, and speed and distance. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Examine the pair of linear equations represented by the lines in the given graph. Determine the number of solutions for this pair of equations and state whether the system is consistent or inconsistent. Give a reason.

Graph showing two parallel lines representing a pair of linear equations

Answer: The two lines in the graph are parallel, so they never meet.
Since the lines do not intersect at any point, the pair of equations has no solution.
A system that has no solution is called an inconsistent system.
(For such a pair, a₁/a₂ = b₁/b₂ ≠ c₁/c₂.)

Q2. Solve: 217x + 131y = 913 and 131x + 217y = 827.

Answer: When the coefficients are interchanged like this, we add and subtract the equations.
Adding both equations:
348x + 348y = 1740
Dividing by 348: x + y = 5 ...(iii)
Subtracting the second from the first:
86x − 86y = 86
Dividing by 86: x − y = 1 ...(iv)
Adding (iii) and (iv): 2x = 6, so x = 3
Putting x = 3 in (iii): 3 + y = 5, so y = 2
Therefore x = 3 and y = 2.

Q3. For what value of u does the system 3x + 5y = 0 and ux + 10y = 0 have a unique solution?

Answer: A pair of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has a unique solution when a₁/a₂ ≠ b₁/b₂.
Here a₁ = 3, b₁ = 5, a₂ = u, b₂ = 10.
For a unique solution: 3/u ≠ 5/10
3/u ≠ 1/2
So u ≠ 6.
Therefore the system has a unique solution for every value of u except 6.

Q4. Solve the system: 3x + 2y = 7 and 2x + 3y = 5.

Answer: Use the elimination method.
Multiply the first equation by 3 and the second by 2:
9x + 6y = 21 ...(i)
4x + 6y = 10 ...(ii)
Subtracting (ii) from (i): 5x = 11, so x = 11/5
Putting this in 3x + 2y = 7:
3(11/5) + 2y = 7
33/5 + 2y = 7
2y = 7 − 33/5 = 2/5
y = 1/5

Q5. The solution of 47x + 31y = 63 and 31x + 47y = 15 is
(a) x = −2, y = 1
(b) x = 2, y = −1
(c) x = 2, y = 1
(d) x = 3, y = 4

Answer: (b) x = 2, y = −1
Adding both equations: 78x + 78y = 78, so x + y = 1
Subtracting the second from the first: 16x − 16y = 48, so x − y = 3
Adding these two: 2x = 4, so x = 2.
Then 2 + y = 1, so y = −1.

Q6. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Answer: Let the speed of Ritu in still water be x km/h and the speed of the stream be y km/h.
Speed downstream = (x + y) km/h and speed upstream = (x − y) km/h.
Downstream: 2(x + y) = 20, so x + y = 10 ...(1)
Upstream: 2(x − y) = 4, so x − y = 2 ...(2)
Adding (1) and (2): 2x = 12, so x = 6
Putting x = 6 in (1): y = 4
So Ritu's speed in still water is 6 km/h and the speed of the current is 4 km/h.

Q7. Two women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and by 1 man alone to finish the work.

Answer: Let 1 woman finish the work in x days and 1 man in y days.
So, 1 woman's one day work = 1/x and 1 man's one day work = 1/y.
From the first condition: 2/x + 5/y = 1/4
From the second condition: 3/x + 6/y = 1/3
Let 1/x = p and 1/y = q. The equations become:
2p + 5q = 1/4 ...(1)
3p + 6q = 1/3 ...(2)
Multiplying (1) by 3 and (2) by 2:
6p + 15q = 3/4 and 6p + 12q = 2/3
Subtracting: 3q = 3/4 − 2/3 = 1/12, so q = 1/36
Putting q in (1): 2p + 5/36 = 1/4, so 2p = 1/9, giving p = 1/18
Therefore x = 18 and y = 36.
So 1 woman alone takes 18 days and 1 man alone takes 36 days.

Q8. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the rest by bus. If she travels 100 km by train and the rest by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer: Let the speed of the train be x km/h and that of the bus be y km/h.
First case: 60/x + 240/y = 4
Second case: 100/x + 200/y = 4 + 10/60 = 25/6
Let 1/x = p and 1/y = q:
60p + 240q = 4 ...(1)
100p + 200q = 25/6 ...(2)
Solving these two equations gives p = 1/60 and q = 1/80.
So x = 60 and y = 80.
The speed of the train is 60 km/h and the speed of the bus is 80 km/h.

Q9. Write the conditions for a pair of linear equations to have (i) a unique solution, (ii) no solution and (iii) infinitely many solutions.

Answer: For the pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
(i) Unique solution (lines intersect at one point, consistent): a₁/a₂ ≠ b₁/b₂
(ii) No solution (lines are parallel, inconsistent): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
(iii) Infinitely many solutions (lines coincide, dependent and consistent): a₁/a₂ = b₁/b₂ = c₁/c₂

Q10. Solve the following pair of equations by the substitution method: x + y = 14 and x − y = 4.

Answer: From the first equation, x = 14 − y.
Putting this value in the second equation:
(14 − y) − y = 4
14 − 2y = 4
2y = 10
y = 5
Putting y = 5 back in x = 14 − y:
x = 9
So the solution is x = 9 and y = 5.

Q11. The sum of the digits of a two-digit number is 9. If 27 is added to the number, the digits get reversed. Find the number.

Answer: Let the digit at the tens place be x and at the units place be y.
The number = 10x + y, and the reversed number = 10y + x.
From the first condition: x + y = 9 ...(1)
From the second condition: (10x + y) + 27 = 10y + x
9x − 9y = −27
x − y = −3 ...(2)
Adding (1) and (2): 2x = 6, so x = 3
Then y = 6.
So the number is 10(3) + 6 = 36. (Check: 36 + 27 = 63, digits reversed ✓)

Q12. Five years ago, a father's age was seven times his son's age. Five years from now, the father will be three times as old as his son. Find their present ages.

Answer: Let the present age of the father be x years and that of the son be y years.
Five years ago: x − 5 = 7(y − 5), which gives x − 7y = −30 ...(1)
Five years later: x + 5 = 3(y + 5), which gives x − 3y = 10 ...(2)
Subtracting (1) from (2): 4y = 40, so y = 10
Putting y = 10 in (2): x − 30 = 10, so x = 40
So the father is 40 years old and the son is 10 years old.

Related Study Material on Extramarks

Q.1

$\begin{array}{l} Solve: \\ \frac{1}{2 x} \frac{1}{y} = 1 \\ \frac{1}{x} + \frac{1}{2 y} = 8 \end{array}$

Marks:6
Ans

$\begin{array}{l} Let \frac{1}{x} = u, \frac{1}{y} = v \\ then the given equations becomes \\ \frac{u}{2} v = 1 . \dots (i) \\ u + \frac{v}{2} = 8 . \dots (ii) \\ Multiplyby2 \\ u-2v=-2 . \dots (iii) \\ 2u+v=16 . \dots (iv) \\ Multiply (iii) by2andsubtractingEquation (iv) \\ 2 u 4 v = 4 \\ 2 u + v = 16 \\ \bar{5 v = 20} \\ v = \frac{20}{5} = 4 \\ Puttingtheequationinequation (iii) \\ u 2 (4) = 2 \\ u = 2 + 8 = + 6 \\ u = \frac{1}{x}, x = \frac{1}{u}, x = \frac{1}{6} \\ v = \frac{1}{y}, y = \frac{1}{v} y = \frac{1}{4} \end{array}$

Q.2 A boat goes 30 km upstream and 44 km downstream is 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of boat in still water.

Marks:6
Ans

Let the speed of the boat in still water be x km/hr
And the speed of the stream be y km/hr
Speed of the boat upstream = (x – y)km/hr
Speed of boat downstream = (x + y)km/hr
According to the given conditions

$\begin{array}{l} \frac{30}{x y} + \frac{44}{x + y} = 10 . .. (1) \\ \frac{40}{x y} + \frac{55}{x + y} = 13 . .. (2) \\ Let \frac{1}{x y} = u, \frac{1}{x + y} = v \\ Therefore, \\ 30u+44v=10 . .. (3) \\ 40 u + 55 v = 13 . .. (4) \\ Onsolvingtheseequations,weget \\ u= \frac{1}{5}, v = \frac{1}{11} \\ Since \frac{1}{x y} = u = \frac{1}{5}, \frac{1}{x + y} = v = \frac{1}{11} \\ x y = 5 \\ x + y = 11 \\ x = 8, y = 3 \\ The speed of boat in still water is 8 km/hr and speed of stream is 3 km/hr. \end{array}$

Q.3 Find the values of m and n for which the system of linear equations has infinite number of solutions.
2x + 3y = 7, 2mx + (m+n)y = 28

Marks:3
Ans

$\begin{array}{l} The given pair of equations are \\ 2x+3y=7 \\ 2mx+ (m + n) y = 28 \\ For infinite number of solutions we have, \\ \frac{a_{1}}{a_{2}} = \frac{b_{2}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ \frac{2}{2 m} = \frac{3}{m + n} = \frac{7}{284} \\ \frac{1}{m} = \frac{3}{m + n} = \frac{1}{4} \\ From first and last ratio we have, \\ m=4 \\ From last two ratios we get, \\ m + n = 12 \\ 4 + n = 12 [m = 4] \\ n = 8 \\ Hence, the given system of equations will have \\ infinite number of solutions if m=4andn=8. \end{array}$

Q.4 Find the values of k for which the system of linear equations has unique solution.
2x + 3y = 7, kx + 9y = 28.

Marks:2
Ans

$\begin{array}{l} The given pair of equations are \\ 2 x + 3 y = 7 \\ kx + 9 y = 28 \\ For unique solutions we have, \\ \frac{a_{1}}{a_{2}} \frac{b_{1}}{b_{2}} \\ \frac{2}{k} \frac{3}{9} \\ \frac{2}{k} \frac{1}{3} \\ k 6 \\ Hence, the given system of equations will have \\ unique solutions if the value of k is other than 6. \end{array}$

CBSE Important Questions Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

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