10th Grade > Mathematics
LINEAR EQUATIONS MCQs
Linear Equations In One Variable, Linear Equations In Two Variables, Pair Of Linear Equations In Two Variables (8th, 9th And 10th Grade)
Total Questions : 74
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Answer: Option A. -> 34
:
A
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
10 x + 10 y = 540 ...(iii)
Subtracting (ii) from (iii)
- 12y = - 240
y = 20
Subsituting y = 20 in x + y = 54 ,
⟹ x + 20 = 54
⟹x = 34
Hence, x = 34 and y = 20
:
A
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
10 x + 10 y = 540 ...(iii)
Subtracting (ii) from (iii)
- 12y = - 240
y = 20
Subsituting y = 20 in x + y = 54 ,
⟹ x + 20 = 54
⟹x = 34
Hence, x = 34 and y = 20
Answer: Option B. -> consistent
:
B
A pair of linear equations in two variables which has a solution, is said to beconsistent.
Here the lines representing the given linear equations are meeting only at point (-2, 3.5) giving a unique solution. Therefore, it isa consistent pair of linear equations.
Now, a pair of linear equations in two variables is said to be dependent when the lines representing them are coinciding i.e. will have many solutions.
Here, the lineshave only one solution (-2, 3.5). Therefore, they aren't dependent.
:
B
A pair of linear equations in two variables which has a solution, is said to beconsistent.
Here the lines representing the given linear equations are meeting only at point (-2, 3.5) giving a unique solution. Therefore, it isa consistent pair of linear equations.
Now, a pair of linear equations in two variables is said to be dependent when the lines representing them are coinciding i.e. will have many solutions.
Here, the lineshave only one solution (-2, 3.5). Therefore, they aren't dependent.
Answer: Option D. -> 2
:
D
Given equations gives infinitely many solutions if,
a1a2=b1b2=c1c2
The given linear equations are:
3x+ky=9;6x+4y=18.
⇒a1=3,b1=k,c1=−9 and a2=6,b2=4,c2=−18
⇒36=k4=−9−18
⇒12=k4
⇒k=2
:
D
Given equations gives infinitely many solutions if,
a1a2=b1b2=c1c2
The given linear equations are:
3x+ky=9;6x+4y=18.
⇒a1=3,b1=k,c1=−9 and a2=6,b2=4,c2=−18
⇒36=k4=−9−18
⇒12=k4
⇒k=2
Answer: Option A. -> (2,3)
:
A
We have,
2x+y=7 ...(1)
3x+2y=12...(2)
Multiply equation (1)by 2, we get:
2(2x+y)=2(7)
⇒4x+2y=14...(3)
Subtracting (2) from (3) we get ,
x=2
Substituting the value of x in (1) we get,
2(2)+y=7⟹y=3
Thus, the solution for the given pair of linear equations is (2,3).
:
A
We have,
2x+y=7 ...(1)
3x+2y=12...(2)
Multiply equation (1)by 2, we get:
2(2x+y)=2(7)
⇒4x+2y=14...(3)
Subtracting (2) from (3) we get ,
x=2
Substituting the value of x in (1) we get,
2(2)+y=7⟹y=3
Thus, the solution for the given pair of linear equations is (2,3).
Answer: Option A. -> x - 3y = 0
:
A
Given, the price of 1 kg of oranges isx and that of 1 kg of apples isy.
According to the given condition, the price of the 1 kg oranges is three times the price of the 1 kg apple.
So, x = 3y.
Therefore, x - 3y = 0
:
A
Given, the price of 1 kg of oranges isx and that of 1 kg of apples isy.
According to the given condition, the price of the 1 kg oranges is three times the price of the 1 kg apple.
So, x = 3y.
Therefore, x - 3y = 0
Answer: Option D. -> y = 2x - 4
:
D
By looking at the graph we can easily say that the line passes through the points (2,0) and (0,-4).
We can identify which line passes through these points by substituting the points in the equation of the line.
Plugging x = 2, y = 0 in equation y = 2x - 4, we get
0 = 2(2) - 4 = 0
Plugging x = 0, y = -4 in equation y = 2x - 4, we get
-4 = 2(0) - 0 = -4
Theequation y = 2x - 4 is satisfied by both the points. So, given graph belongs to y = 2x - 4.
:
D
By looking at the graph we can easily say that the line passes through the points (2,0) and (0,-4).
We can identify which line passes through these points by substituting the points in the equation of the line.
Plugging x = 2, y = 0 in equation y = 2x - 4, we get
0 = 2(2) - 4 = 0
Plugging x = 0, y = -4 in equation y = 2x - 4, we get
-4 = 2(0) - 0 = -4
Theequation y = 2x - 4 is satisfied by both the points. So, given graph belongs to y = 2x - 4.
Answer: Option B. -> False
:
B
A linear equation in two variables is of the form ax + by + c = 0, where a and b are both non-zero. If any one of a or b becomes 0, it would reduce to a linear equation in one variable. So, the equation x = -5 cannot be expressed as a linear equation in 2 variables.
:
B
A linear equation in two variables is of the form ax + by + c = 0, where a and b are both non-zero. If any one of a or b becomes 0, it would reduce to a linear equation in one variable. So, the equation x = -5 cannot be expressed as a linear equation in 2 variables.
Answer: Option B. -> The geometrical representation is a straight line
:
B
The degree (the highest power on any variable) of the given equation is 1. Hence, its geometrical representationis a straight line.
:
B
The degree (the highest power on any variable) of the given equation is 1. Hence, its geometrical representationis a straight line.
Answer: Option A. -> X axis
:
A
We can observe that all the points have zero as their y coordinates.
(1,0), (-4,0) and (5,0)
y = 0 is the equation of X - axis.
Hence, all the points lie on the X axis.
:
A
We can observe that all the points have zero as their y coordinates.
(1,0), (-4,0) and (5,0)
y = 0 is the equation of X - axis.
Hence, all the points lie on the X axis.
Answer: Option C. -> x+y=6
:
C
Putting x=4 and y=2 in each of the given equations, we get the following:
2x+3y=12:
LHS =2×4+3×2=14≠ RHS
4x+y=19:
LHS =4×4+2=18≠ RHS
x+y=6:
LHS =4+2=6= RHS
3x+2y=15:
LHS =3×4+2×2=16≠ RHS
Hence, x=4,y=2 isa solution of the equation x+y=6.
:
C
Putting x=4 and y=2 in each of the given equations, we get the following:
2x+3y=12:
LHS =2×4+3×2=14≠ RHS
4x+y=19:
LHS =4×4+2=18≠ RHS
x+y=6:
LHS =4+2=6= RHS
3x+2y=15:
LHS =3×4+2×2=16≠ RHS
Hence, x=4,y=2 isa solution of the equation x+y=6.