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
| Page 5 of 8 pages
Answer: Option D. -> 22
:
D
If a=−1,b=4 is the solutionof equation −2a=−5b+k, then they will satisfy the given equation.
Putting values of a&b in the given equation.
So,
−2(−1)=−5(4)+k2=−20+k⇒k=20+2⇒k=22.
Therefore, the value of k is 22.
:
D
If a=−1,b=4 is the solutionof equation −2a=−5b+k, then they will satisfy the given equation.
Putting values of a&b in the given equation.
So,
−2(−1)=−5(4)+k2=−20+k⇒k=20+2⇒k=22.
Therefore, the value of k is 22.
Answer: Option C. -> a line passing through the origin
:
C
Given equation: y=mx
Substituting x=0,
y=m×0=0
So, if x = 0, then y = 0.
i.e, (0,0) is a point on the line.
Therefore, the line passes through the origin.
:
C
Given equation: y=mx
Substituting x=0,
y=m×0=0
So, if x = 0, then y = 0.
i.e, (0,0) is a point on the line.
Therefore, the line passes through the origin.
Answer: Option C. -> y=2x+3
:
C
The coordinates of the points of the graph (0,3) & (-1.5,0) satisfy only the equation y=2x+3:
(0, 3):
3 = 2 x 1 + 3
(-1.5, 0):
0 = 2 × (-1.5) + 3
:
C
The coordinates of the points of the graph (0,3) & (-1.5,0) satisfy only the equation y=2x+3:
(0, 3):
3 = 2 x 1 + 3
(-1.5, 0):
0 = 2 × (-1.5) + 3
Answer: Option A. -> True
:
A
Putting x = 3 , y = -1 in the given equation,
k x 3 - 2 x k x -1 + 15 = 0
⇒ 3k + 2k + 15 = 0
⇒ k = -3
:
A
Putting x = 3 , y = -1 in the given equation,
k x 3 - 2 x k x -1 + 15 = 0
⇒ 3k + 2k + 15 = 0
⇒ k = -3
:
Putting x = 1, y = 2 in the given equation,
3 x 1 + 2 x 2 = k
⇒ k = 7
Answer: Option A. -> x=1, y=3
:
A
Substitutingeach set of the values for x and y in the equation 2x+3y=11, we have
(i)x=1,y=3LHS=(2×1)+(3×3)=11=RHS(ii)x=0,y=5LHS=(2×0)+(3×5)=15≠RHS(iii)x=−2,y=−5LHS=(2×−2)+(3×−5)=−19≠RHS(iv)x=9,y=1LHS=(2×9)+(3×1)=21≠RHS
∴x=1,y=3 is solution to the given equation.
:
A
Substitutingeach set of the values for x and y in the equation 2x+3y=11, we have
(i)x=1,y=3LHS=(2×1)+(3×3)=11=RHS(ii)x=0,y=5LHS=(2×0)+(3×5)=15≠RHS(iii)x=−2,y=−5LHS=(2×−2)+(3×−5)=−19≠RHS(iv)x=9,y=1LHS=(2×9)+(3×1)=21≠RHS
∴x=1,y=3 is solution to the given equation.
Answer: Option D. -> 2x−3y+4=0
:
D
The standard form of a linear equation in 2 variables x and y is given by ax + by + c = 0.
It is given that a = 2, b = -3 and c = 4.
Therefore, the linear equation is 2x−3y+4=0
:
D
The standard form of a linear equation in 2 variables x and y is given by ax + by + c = 0.
It is given that a = 2, b = -3 and c = 4.
Therefore, the linear equation is 2x−3y+4=0
Answer: Option A. -> True
:
A
When x=1,
2+5y=7
⇒ y = 1
When x=2,
4+5y=7
⇒y=35
When x=3,
6+5y=7
⇒y=15
When x=4,
8+5y=7
⇒y=−15
So, in natural numbers, when x>1, the value of y is <1 and progressivelydecreases, and then becomes negative at x = 4.
When y=2,
2x+10=7
⇒y=−32
When y=3,
2x+15=7
⇒y=−4
In natural numbers, when y>1, x<0 and progressively decreases.
Hence, in natural numbers, there is only one pair i.e., (1,1) which satisfy the given equation but in real numbers and rational numbers there are many pairs to satisfy the given linear equation.
:
A
When x=1,
2+5y=7
⇒ y = 1
When x=2,
4+5y=7
⇒y=35
When x=3,
6+5y=7
⇒y=15
When x=4,
8+5y=7
⇒y=−15
So, in natural numbers, when x>1, the value of y is <1 and progressivelydecreases, and then becomes negative at x = 4.
When y=2,
2x+10=7
⇒y=−32
When y=3,
2x+15=7
⇒y=−4
In natural numbers, when y>1, x<0 and progressively decreases.
Hence, in natural numbers, there is only one pair i.e., (1,1) which satisfy the given equation but in real numbers and rational numbers there are many pairs to satisfy the given linear equation.
Answer: Option A. -> -4, 1, 0
:
A
y = 4x can be rewritten as -4x +y = 0.
On comparing4x - y = 0 with
ax + by + c= 0, we get
a = -4
b = 1
c = 0
∴ 4, -1, 0 are the values of a, b and c respectively.
:
A
y = 4x can be rewritten as -4x +y = 0.
On comparing4x - y = 0 with
ax + by + c= 0, we get
a = -4
b = 1
c = 0
∴ 4, -1, 0 are the values of a, b and c respectively.