Chapter 3 Pair of Linear Equations in Two Variables from CBSE Class 10 Maths textbook explains all about the pair of linear equations in two variables. Two linear equations in two variables also called the pair of linear equations in two variables can be represented graphically and algebraically. Algebraically, students can use the Substitution method, Elimination method or Crossmultiplication method to solve the pair of linear equations in two variables.
From this chapter, we have compiled some 20 MCQs, which have been categorized topicwise in the PDF link given below:
List of SubTopics Covered
Some of the main subtopics covered in this chapter of the CBSE Class 10 Maths have been listed below. Students also find these CBSE Class 10 Maths Objective Questions very useful. Students can practice the MCQs from the chapter given topic wise and master the concepts from the topics of this chapter.Â Here are the list of topics given:
3.1 Algebraic Solution (4 MCQs Listed From This Topic)
3.2 All about Lines (4 MCQs From The Topic)
3.3 Basics Revisited (4 MCQs From This Topic)
3.4 Graphical Solution (4 MCQs Listed From The Topic)
3.5 Solving Linear Equations (4 MCQs From The Mentioned Topic)
Download Free CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Objective Questions PDF
Algebraic Solution

 Half the perimeter of a rectangular room is 46 m, and its length is 6 m more than its breadth. WhatÂ is the length and breadth of the room?
 2m, 20m
 2m, 3m
 56m, 40m
 26m, 20m
 Half the perimeter of a rectangular room is 46 m, and its length is 6 m more than its breadth. WhatÂ is the length and breadth of the room?
Answer: (D) 26m, 20m
Solution: LetÂ lÂ andÂ bÂ be the length and breadth of the room. Then, the perimeter of the room =Â 2(l+b)Â metres
From question,Â l=6+b… (1)
Ã—2(l+b) =46âŸ¹l+b=46… (2)
Using Substitution method:
Substituting the value ofÂ lÂ from (1) in (2), we get
6+b+b=46
â‡’Â 6+2b=46âŸ¹2b=40
âŸ¹b=20Â m.
Thus,Â l=26Â m

 Solve the following pair of equations:
2x+y=7
3x+2y=12
Choose the correct answer from the given options.
 (3,2)
 (1,0)
 (3,2)
 (2,3)
 Solve the following pair of equations:
Answer: (D) (2, 3)
Solution: We have,
2x+y=7Â Â … (1)
3x+2y=12… (2)
Multiply equation (1)Â by 2, we get:
2(2x+y) =2(7)
â‡’4x+2y=14… (3)
Subtracting (2) from (3) we get,
x=2
Substituting the value ofÂ xÂ in (1) we get,
2(2) +y=7âŸ¹y=3
Thus, the solution for the given pair of linear equations is (2, 3).

 Solve
 Y= 51/19
 X= 51/19
 Y= 94/57
 X= 117/54
 Solve
Answer: (A) Y= 51/19
Solution: (3x/2) – (5y/3) =2
LCM of 2 and 3 is 6. Multiply by 6 on both sides
9xâˆ’10y=âˆ’12Â ———— (1)
(X/2) + (y/2) = 13/6
LCM is 6. Multiply by 6 on both sides
3x+3y=13Â ————–(2)
Multiply equation (2) by 3 to eliminate x; so we get,
9x+9y=39……………(3)
Subtract (3) from (1) we have
âˆ’19y=âˆ’51â‡’y= 51/19
Substitute this in one of the equation and we get

 Given:Â 3xâ€“5y=4; 9x=2y+7
Solve above equations by Elimination method and find the value of x.


 X= 9/13
 Y= 5/13
 X= (5) / 13
 Y= 9/13

Answer: (D) X= 9/13
Solution: Given:
3xâ€“5y=4……(1)
9x=2y+7
9xâˆ’2y=7…..(2)
Multiply equation (1) by 3
â‡’9xâ€“15y=12…….(3)
Subtracting (2) from (3) we get,
âˆ’13y=5
Y= – ( (5) / 13)
Substituting the value ofÂ yÂ in (2)
All about Lines

 Choose the pair of equations which satisfy the point (1,1)
 4xâ€“y=3,4x+y=3
 4x+y=3,3x+2y=1
 2x+3y=5,2x+3y=âˆ’1
 2x+y=3,2xâ€“y=1]
 Choose the pair of equations which satisfy the point (1,1)
Answer: (B) 4x+y=3, 3x+2y=1
Solution: For a pair of equationsÂ to satisfyÂ a point, the point should be the unique solution of them.
Solve the pair equationsÂ 4x+y=3,3x+2y=1
letÂ 4x+y=3…..(1)
andÂ 3x+2y=1Â …..(2)
y=3âˆ’4xÂ Â [ From (1)]
Substituting value if y in (2)
3x+2y=1
3x+2(3âˆ’4x)=1
3x+6âˆ’8x=1
âˆ’5x=âˆ’5â‡’x=1
Substituting x = 1 in (1),
4(1)+y=3â‡’y=âˆ’1
â‡’Â (1,1) is the solution of pair of equation.
âˆ´Â Pair of equations which satisfy the point (1,1)
Note: – We can also substitute the valueÂ (1,1) in the given equations and check if it satisfies the pair of equations or not. In this case it only satisfies the pair of equationÂ 4x+y=3, 3x+2y=1Â and hence (1, 1) is the unique solution of the equation.

 54 is divided into twoÂ parts such that sum of 10 times the first part and 22 times the second part is 780. What is the bigger part?
 34
 32
 30
 24
 54 is divided into twoÂ parts such that sum of 10 times the first part and 22 times the second part is 780. What is the bigger part?
Answer: (A) 34
Solution: Let the 2 parts of 54 be x and y
x+y = 54…. (i)
And 10x + 22y = 780 ——————– (ii)
Multiply (i) by 10, we get
Substituting y = 20 in x + y = 54, we have x + 20 = 54; x = 34
Hence, x = 34 and y = 20.

 What are the values ofÂ a,Â bÂ andÂ cÂ for the equationÂ y=0.5x+âˆš7Â when written in the standard form:Â ax+by+c=0?
 0.5, 1, âˆš7
 0.5, 1, – âˆš7
 0.5, 1, âˆš7
 0.5, 1, âˆš7
 What are the values ofÂ a,Â bÂ andÂ cÂ for the equationÂ y=0.5x+âˆš7Â when written in the standard form:Â ax+by+c=0?
Answer: (C) 0.5, 1, âˆš7
Solution: Y= 0.5X + âˆš7
â‡’0.5xâˆ’y+ âˆš7= 0
The general form of an equation isÂ ax+by+c=0.
Here, on comparing, we get
a=0.5, b=âˆ’1Â andÂ c= âˆš7

 Which of the following pair of linear equations has infinite solutions?
Answer: (D) x+2y=7;Â 3x+6y=21
Solution: If two equations are consistent and overlapping, then they will have infinite solutions. Option A consists of two equations where the second equation can be reduced to an equation which is same as the first equation.
x+2y=7…. (i)
3x+6y=21….. (ii)
Dividing equation (ii) by 3, we get
x+2y=7Â which is the same as equation (i).
The equations coincide and will have an infinite solution.
Alternate Method:
Let the two equations be
Basics Revisited

 Which of these points lie on the line 7x+8y=61
 (3,4)
 (2,5)
 (3,7)
 (3,5)
 Which of these points lie on the line 7x+8y=61
Answer: (D) (3, 5)
Solution: Substituting the value of x = 3 and y = 4,
7x+8y=61Â =Â 7(3) +8(4) =53.
Substituting the value of x = 2 and y = 5,
7x+8y=61Â =Â 7(2) +8(5) =54.
Substituting the value of x = 3 and y = 7,
7x+8y=61Â =Â 7(âˆ’3) +8(7) =35.
Substituting the value of x = 3 and y = 5,
7x+8y=61Â =Â 7(3) +8(5) =61
Hence, (3, 5) lies on the given line

 Which of these following equations haveÂ x=âˆ’3, y=2Â as solutions?
 3xâˆ’2y=0
 3x+2y=0
 2x+3y=0
 2xâˆ’3y=0
 Which of these following equations haveÂ x=âˆ’3, y=2Â as solutions?
Answer: (C) 2x+3y=0
Solution: Substituting the values in LHS,
L.H.S=2x+3y
L.H.S=2(âˆ’3) +3(2)
L.H.S=0=R.H.S
HenceÂ x=âˆ’3, y=2Â is the solution of the equationÂ 2x+3y=0

 If y =Â 1/2 (3x+7) is rewritten in the formÂ ax+by+c=0,Â what are the values of a, b and c?
 Â½, 7/2, 3/2
 7,2,3
 2, 3, 7
 3,2,7
 If y =Â 1/2 (3x+7) is rewritten in the formÂ ax+by+c=0,Â what are the values of a, b and c?
Answer: (D) 3, 2,7
Solution: The given equation is:
y =Â 1/2 (3x+7)
Simplifying the equation we get:
2yâˆ’3xâˆ’7=0
â‡’âˆ’3x+2yâˆ’7=0Â Â (1)
Thus, the value of a, b and c is 3,Â 2Â and 7 respectively.
The equation can also be written as,
3xâˆ’2y+7=0
Thus, the value of a, b and c is +3, 2Â and +7 respectively.
The option 3, 2 and 7 is correctÂ [Since +3, 2Â and +7Â is not an option]

 xâˆ’y=0Â is a line:
 Passing through origin
 Passing through (1,1)
  to y axis
  to x axis
 xâˆ’y=0Â is a line:
Answer: (A) Passing through origin
Solution: xâˆ’y=0, is a line passing through the origin as point (0, 0) satisfies the given equation
Graphical Solution
Statement 1: This is theÂ condition for inconsistent equations
Statement 2: There exists infinitely many solutions
Statement 3: The equations satisfying the condition are parallel
Which of the aboveÂ statements are true?


 S_{1} only
 S_{1} and S_{2}
 S_{1} and S_{3}
 S_{2} only

Answer: (C) S_{1} and S_{3}
The condition is for inconsistent pair of equations which are parallel and have no solution.
âˆ´Â Statement 1 and 3 are correct
Choose the correct statement.


 For k3(m/5), a unique solution exists
 For k=3 (m/5), infinitely many solutions exists
 For k=Â 5( m/3), a unique solution exists
 Fork=Â 5(m/3) , infinitely many solutions exist

Answer: For k=Â 5 (m/3), infinitely many solutions exist

 The figure showsÂ the graphical representation of a pair of linear equations. On the basis of graph,Â the pair of linear equation gives _______________solutions.


 Four
 Only one
 Infinite
 Zero

Answer: (B) Only one
Solution: If the graph ofÂ linear equations represented by the lines intersects at a point, thisÂ point gives the unique solution. Here the lines meet at the point (1,1) which is the unique solution of the given pair of linear equations.

 For what value ofÂ k, the pair of linear equationsÂ 3x+ky =9Â andÂ 6x+4y=18Â has infinitely many solutions?
 5
 6
 1
 2
 For what value ofÂ k, the pair of linear equationsÂ 3x+ky =9Â andÂ 6x+4y=18Â has infinitely many solutions?
Answer: (D) 2
Solution: Given equations gives infinitely many solutions if,
The given linear equations are:
3x+ky=9; 6x+4y=18.
â‡’ 3/6 = k/4 = (9)/ (18)
â‡’ Â½ = k/4
â‡’ k = 2
Solving Linear Equations

 Solve the following pair of linear equation


 9,8
 4,9
 3,2
 Â½ , 1/3

Answer: (B) 4, 9
Solution: The pair of equations is not linear. We will substitute 1/x as u^{2} and 1/y as v^{2} then we will get the equation as
2u+3v=2
4uâˆ’9v=âˆ’1
We will use method of elimination to solve the equation.
Multiply the first equation by 3, we get
6u+9v=6
4uâˆ’9v=âˆ’1
Adding the above two equations
10u=5
u= Â½
SubstitutingÂ uÂ in equationÂ 4uâˆ’9v=âˆ’1Â we getÂ v= 1/3

 Solve the following pair of equation
(7x2y) / xy=5
(8x+6y) /xy =15


 None of these
 2, not defined
 5/2 , not defined
 (2) /5 , not defined

Answer: (D) (2) /5 , not defined
Solution: First separate the variables
Similarly we can do separation of variables for second equation
Now we see that the equation is not linear.
So we will substitute 1/x = u and 1/y = v
The pair of equation can be written as
2u+7v=5
6u+8v=15
We can solve the pair of equation by method of elimination.
6u+21v=15————————— (1)
6u+8v = 15————————— (2)
Subtracting (2) from (1), we getÂ 13v = 0; v = 0
Substituting v = 0 in 2u + 7v = 5
we getÂ u= – (5/2)
1/x= u and 1/y = v
âˆ´x = âˆ’ (2/5) and y = not defined


 x = 2,Â y = 2
 x =Â 7,Â y =Â 8
 x =Â 0,Â y =Â 8
 x =Â 1,Â y =Â 7

Answer: (D) x =Â 1,Â y =Â 7

 Solve the following pair of equations:
(1/x) + (3/y) =1
(6/x) 12/y=2


 X= 5/3 , Y= 15/2
 x=4,y=9
 x=3,y=11
 X= Â¾ , Y= 7/3

Answer: (A) X= 5/3 , Y= 15/2
Solution: Let 1/x= a and 1/y = b
(AsÂ xâ‰ 0, yâ‰ 0)
Then, the given equations become
a+3b=1Â â€¦ (1)
6aâˆ’12b=2Â â€¦ (2)
Multiplying equation (1) by 4, we getÂ 4a+12b=4Â â€¦ (3)
On adding equation (2) and equation (3), we getÂ 10a=6
â‡’a= 3/5
Putting a= 3/5 in equation (1), we get
(3/5) + 3b =1
â‡’b= 2/15
Hence,Â x= 5/3 and y= 15/2
The students appearing for the board examinations will find it helpful in scoring well, as according to the latest modified exam pattern, the question paper will contain more objective type questions.
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