MCQ 11 Mark
x = 2 y = - 1 is a solution of the linear equation
- ✓x + 2y = 0
- Bx + 2y = 4
- C2x + y = 0
- D2x + y = 5
Answer
View full question & answer→Correct option: A.
x + 2y = 0
a
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MCQ 21 Mark
The work done by a body on application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. If the constant force is 3 units, y is the work done and x is the distance travelled, then the linear equation in two variables to express the above statement is
- Ax = 3y
- By = 3x
- Cy = x + 3
- Dx = y + 3
Answer
View full question & answer→B. y = 3x
$\begin{array}{l}\text { It is given that: Work done }=\text { Constant force } \times \text { Distance } \\ \therefore \quad y=3 x \text {, which is the required linear equation. }\end{array}$
$\begin{array}{l}\text { It is given that: Work done }=\text { Constant force } \times \text { Distance } \\ \therefore \quad y=3 x \text {, which is the required linear equation. }\end{array}$
MCQ 31 Mark
The solution of the linear equation x + 2y = 8 which represents a point on y-axis, is
- A(0,4)
- B(8,0)
- C(4,2)
- D(2, 3)
Answer
View full question & answer→A. (0, 4)
Let $(0, a)$ be a point on $y$-axis representing a solution of the equation. Then, $x=0$ and $y=a$ is a solution of equation $x+2 y=8$.
$\therefore$ $0+2 a=8 \Rightarrow a=4
$
Hence, $(0,4)$ is the required point.
Let $(0, a)$ be a point on $y$-axis representing a solution of the equation. Then, $x=0$ and $y=a$ is a solution of equation $x+2 y=8$.
$\therefore$ $0+2 a=8 \Rightarrow a=4
$
Hence, $(0,4)$ is the required point.
MCQ 41 Mark
The solution of the linear equation x + 2y = 8 which represents a point on x-axis, is
- A(4,0)
- B(0,4)
- C(8,0)
- D(4,2)
Answer
View full question & answer→C. (8, 0)
Let $(a, 0)$ be a point on $x$-axis representing a solution of the equation $x+2 y=8$. Then, $x=a$ and $y=0$ is a solution of $x+2 y=8$.
$
\therefore \quad a+2 \times 0=8 \Rightarrow a=8
$
Hence, $(8,0)$ is the point representing a solution of the given equation.
Let $(a, 0)$ be a point on $x$-axis representing a solution of the equation $x+2 y=8$. Then, $x=a$ and $y=0$ is a solution of $x+2 y=8$.
$
\therefore \quad a+2 \times 0=8 \Rightarrow a=8
$
Hence, $(8,0)$ is the point representing a solution of the given equation.
MCQ 51 Mark
The positive solutums of the equation $a x+b y+c=0$ always lie in the
- A$l^2$ quadrant
- B$2^{\text {nt }}$ quadrant
- C$3^{\text {te }}$ quadrant
- D$4^{\text {te }}$ quadrant
Answer
View full question & answer→A. $l^2$ quadrant
The points representing positive solutions of $a x+b y+c=0$ have both the coordinates poritive So, they lie in the $1^{st}$ quadrant.
The points representing positive solutions of $a x+b y+c=0$ have both the coordinates poritive So, they lie in the $1^{st}$ quadrant.
MCQ 61 Mark
The point on the graph of the linear equation $2 x+5 y=19$, whose ordinate is $1 \frac{1}{2}$ times its abscissa, is
- A$(2,3)$
- B$(3,2)$
- C$(-2,-3)$
- D$(-3,-2)$
Answer
View full question & answer→A. $(2,3)$
Let $P(x, y)$ be the required point. It is given that the ordinate is $1 \frac{1}{2}$ times the abscissa.
$
\therefore \quad y=\frac{3}{2} x
$
Putting $y=\frac{3}{2} x$ in $2 x+5 y=19$, we obtain
$
2 x+\frac{15}{2} x=19 \Rightarrow \frac{19}{2} x=19 \Rightarrow x=2
$
Putting $x=2$ in (i), we obtain $y=3$. Hence the required point is $(2,3)$.
Let $P(x, y)$ be the required point. It is given that the ordinate is $1 \frac{1}{2}$ times the abscissa.
$
\therefore \quad y=\frac{3}{2} x
$
Putting $y=\frac{3}{2} x$ in $2 x+5 y=19$, we obtain
$
2 x+\frac{15}{2} x=19 \Rightarrow \frac{19}{2} x=19 \Rightarrow x=2
$
Putting $x=2$ in (i), we obtain $y=3$. Hence the required point is $(2,3)$.
MCQ 71 Mark
The point on the graph of the equation $3 x-2 y-12=0$ whose $y$-coordinate is $3 / 4$ times the $x$-coordinate, is
- A$(8,6)$
- B$(8,-6)$
- C$(-8,-6)$
- D$(-6,-8)$
Answer
View full question & answer→C. $(-8,-6)$
Let the required point be $(x, y)$. It is given that the $y$-coordinate is $3 / 4$ times the $x$-coordinate. Therefore, $y=\frac{3}{4} x$.
Putting $y=\frac{3}{4} x$ in $3 x-2 y+12=0$, we obtain$
3 x-2\left(\frac{3}{4} x\right)+12=0=3 x-\frac{3}{2} x+12=0=\frac{3}{2} x=-12 \Rightarrow x=-8
$
'utting $x=-8$ in $y=\frac{3}{4} x$, we get $y=-6$. Hence, the required point is $(-8,-6)$.
Let the required point be $(x, y)$. It is given that the $y$-coordinate is $3 / 4$ times the $x$-coordinate. Therefore, $y=\frac{3}{4} x$.
Putting $y=\frac{3}{4} x$ in $3 x-2 y+12=0$, we obtain$
3 x-2\left(\frac{3}{4} x\right)+12=0=3 x-\frac{3}{2} x+12=0=\frac{3}{2} x=-12 \Rightarrow x=-8
$
'utting $x=-8$ in $y=\frac{3}{4} x$, we get $y=-6$. Hence, the required point is $(-8,-6)$.
MCQ 81 Mark
The point of the form (a, a) always lies on
- Ax-axis
- By-axis
- ✓on the line y = x
- Don the line x + y = 0
Answer
View full question & answer→Correct option: C.
on the line y = x
c
MCQ 91 Mark
The linear equation 2x - 5y = 7 has
- Aa unique solution
- Btwo solutions
- ✓infinitely many solutions
- Dno solutions
Answer
View full question & answer→Correct option: C.
infinitely many solutions
c
MCQ 101 Mark
The graph of y = 6 is a line
- ✓parallel to x-axis at a distance 6 units from the origin.
- Bparallel to y-axis at a distance 6 units from the origin.
- Cmaking an intercept 6 on the x-axis.
- Dmaking an intercept 6 on both the axes.
Answer
View full question & answer→Correct option: A.
parallel to x-axis at a distance 6 units from the origin.
a
MCQ 111 Mark
The graph of the linear equation y = x passes through the point
- A$\left(\frac{3}{2},-\frac{3}{2}\right)$
- B$\left(0, \frac{3}{2}\right)$
- ✓(1,1)
- D$\left(-\frac{1}{2}, \frac{1}{2}\right)$
Answer
View full question & answer→Correct option: C.
(1,1)
c
MCQ 121 Mark
The graph of the linear equation 4x - 3y - 12 = 0 cuts x-axis at point
- A(3,0)
- B(-3,0)
- C(4,0)
- D(-4, 0)
Answer
View full question & answer→A. (3, 0)
Let the graph of the equation $4 x-3 y-12=0$ meet $x$-axis at $(a, 0)$. Then, $x=a$ and $y=0$, is a solution of the equation.
$\therefore$ $4 \times a-3 \times 0-12=0$ $\Rightarrow a=3$
Hence, the required point is $(3,0)$.
Let the graph of the equation $4 x-3 y-12=0$ meet $x$-axis at $(a, 0)$. Then, $x=a$ and $y=0$, is a solution of the equation.
$\therefore$ $4 \times a-3 \times 0-12=0$ $\Rightarrow a=3$
Hence, the required point is $(3,0)$.
MCQ 131 Mark
The graph of the linear equation 2x - y = 4 cuts x-axis at
- ✓(2,0)
- B(-2,0)
- C(0,-4)
- D(0,4)
Answer
View full question & answer→Correct option: A.
(2,0)
a
MCQ 141 Mark
The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point
- A(2,0)
- B(0,3)
- C(3,0)
- ✓(0,2)
Answer
View full question & answer→Correct option: D.
(0,2)
d
MCQ 151 Mark
The graph of the equation 2x + 3y = 6 is a line which meets the x and y axes respectively at the points
- ✓(3,0) and (0, 2)
- B(2,0) and (0,3)
- C(0, 3) and (2,0)
- D(3, 2) and (2, 3)
Answer
View full question & answer→Correct option: A.
(3,0) and (0, 2)
a
MCQ 161 Mark
The equation y = - 4 in two variables x and y, can be written as
- A$1 \cdot x+1 \cdot y=-4$
- B$1 \cdot x+0 y=-4$
- C$0 x+1 \cdot y+4=0$
- D$0 x+1 \cdot y=4$
Answer
View full question & answer→C. $0 x+1 \cdot y+4=0$
Equation $y=-4$ can also be written as $0 \cdot x+1 \cdot y=-4$ or, $0 \cdot x+1 \cdot y+4=0$.
Equation $y=-4$ can also be written as $0 \cdot x+1 \cdot y=-4$ or, $0 \cdot x+1 \cdot y+4=0$.
MCQ 171 Mark
The equation x = 7 in two variables x and y, can be written as
- A$1 \cdot x+1 \cdot y=7$
- B$1 \cdot x+0 y=7$
- C$0 x+1 \cdot y=7$
- D$0 x+0 y=7$
Answer
View full question & answer→B. $1 \cdot x+0 y=7$
Equation $x=7$ can be written as $1 \cdot x+0 y=7$.
Equation $x=7$ can be written as $1 \cdot x+0 y=7$.
MCQ 181 Mark
The equation x - 2 = 0 on number line is represented by
- Aa line
- ✓a point
- Cinfinitely many lines
- Dtwo lines
Answer
View full question & answer→Correct option: B.
a point
b
MCQ 191 Mark
The equation of the x-axis is of the form
- Ax = 0
- ✓y = 0
- Cx + y = 0
- Dx = y
Answer
View full question & answer→Correct option: B.
y = 0
b
MCQ 201 Mark
The equation 2x + 5y = 7 has a unique solution, if x y are
- ✓natural numbers
- Bpositive real numbers
- Creal numbers
- Drational numbers
Answer
View full question & answer→Correct option: A.
natural numbers
a
MCQ 211 Mark
The distance between the graphs of the equations y = - 1 and y = 3 is
- A2
- ✓4
- C3
- D1
Answer
View full question & answer→Correct option: B.
4
b
MCQ 221 Mark
The distance between the graph of the equations x = -3 and x = 2 is
- A1
- B2
- C3
- ✓5
Answer
View full question & answer→Correct option: D.
5
d
MCQ 231 Mark
The Autorikshaw fare in a city is charged @ ₹ 10 for the first kilometer and @ ₹4 per kilometer for subsequent distance covered. The linear equation to express the statement is
- Ay = 4x + 10
- By = 4x + 6
- Cy + 4x = 10
- Dy + 4x = 6
Answer
View full question & answer→B. y = 4x + 6
Let the total distance covered be x km and the fare charged be ₹y. Then, for the first km, fare charged is ₹10 and for the remaining (x - 1) km fare charged is ₹ 4(x - 1) .
$
\therefore \quad y=4(x-1)+10 \Rightarrow y=4 x+6
$
The required equation is $y=4 x+6$.
Let the total distance covered be x km and the fare charged be ₹y. Then, for the first km, fare charged is ₹10 and for the remaining (x - 1) km fare charged is ₹ 4(x - 1) .
$
\therefore \quad y=4(x-1)+10 \Rightarrow y=4 x+6
$
The required equation is $y=4 x+6$.
Question 241 Mark
x = 2, y = -1 is a solution of the linear equation:
- x + 2y = 0
- x + 2y = 4
- 2x + y = 0
- 2x + y = 5
Answer
View full question & answer→- x + 2y = 0
Solution:
Substituting x = 2 and y = -1 in the following equations:
L.H.S. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 = R.H.S.
L.H.S. = x + 2y = 2 + 2(-1) = 2 - 2 = 0 ≠ 4 ≠ R.H.S.
L.H.S. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 ≠ 0 ≠ R.H.S.
L.H.S. = 2x + y = 2(2) + (-1) = 4 - 1 = 3 ≠ 5 ≠ R.H.S.
Hence, correct option is (a).
Question 251 Mark
The graph of the linear equation 2x - y = 4 cuts x-axis at:
- (2, 0)
- (-2, 0)
- (0, -4)
- (0, 4)
Answer
View full question & answer→- (2, 0)
Solution:
On x-axis, the y-co-ordinate is always 0.
So, 2x - y = 4 will cut the x-axis where y = 0
i.e. 2x = 4
i.e. x = 2
Thus, 2x - y = 4 will cut the x-axis at (2, 0).
Hence, correct option is (a).
Question 261 Mark
The equation x - 2 = 0 on number line is represented by:
- A line.
- A point.
- Infinitely many lines.
- Two lines.
Answer
View full question & answer→- A point.
Solution:
The equation x - 2 = 0 is represented by a point on the number line.
Therefore, the correct answer is (b).
Question 271 Mark
The distance between the graphs of the equations y = -1 and y = 3 is:
- 2
- 4
- 3
- 1
Answer
View full question & answer→- 4
Solution:
The distance between given two graphs
= 3 - (-1)
= 3 + 1
= 4
Hence, correct option is (b).
Question 281 Mark
The distance between the graph of the equations x = -3 and x = 2 is:
- 1
- 2
- 3
- 5
Answer
View full question & answer→- 5
Solution:
The distance between the graph of the equations x = -3 and x = 2
= 2 - (-3)
= 2 + 3
= 5
Hence, correct option is (d).
Question 291 Mark
lf the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length:
- 4 units.
- 3 units.
- 5 units.
- None of these.
Answer
View full question & answer→- 5 units.
Solution:
4x + 3y = 12
At x = 0, 3y = 12 ⇒ y = 4 units
At y = 0, 4x = 12 ⇒ x = 3 units
The triangle formed is $\triangle\text{AOB},$ where
OB = 4 units
OA = 3 units
Hypotenuse $=\text{AB}=\sqrt{\text{OB}^2+\text{OA}^2}=\sqrt{16+9}=5\text{ units}$
Hence, correct option is (c).
Question 301 Mark
If (a, 4) lies on the graph of 3x + y = 10, then the value of a is:
- 3
- 1
- 2
- 4
Answer
View full question & answer→- 2
Solution:
3x + y = 10
If (a, 4) lies on its graph, then it must satisfy the equation.
Thus, we have
3(a) + 4 = 10
i.e. 3a = 6
i.e. a = 2
Hence, correct option is (c).
Question 311 Mark
If (4, 19) is a solution of the equation y = ax + 3, then a =
- 3
- 4
- 5
- 6
Answer
View full question & answer→- 4
Solution:
y = ax + 3
If (4, 19) is its solution, then it must satisfy the equation.
Thus, we have
19 = a × 4 + 3
i.e. 4a = 16
i.e. a = 4
Hence, correct option is (b).
Question 321 Mark
If (2k - 1, k) is a solution of the equation 10x - 9y = 12, then k =
- 1
- 2
- 3
- 4
Answer
View full question & answer→- 2
Solution:
If (2k - 1, k) is solution of equation 10x - 9y = 12, then it must satisfy this equation.
Thus, we have
10(2k - 1) - 9k = 12
20k - 10 - 9k = 12
11k = 22
k = 2
Hence, correct option is (b).
Question 331 Mark
How many linear equations are satisfied by x = 2 and y = -3?
- Only one.
- Two.
- Three.
- Infinitely many.
Answer
View full question & answer→- Infinitely many.
Solution:
From Point (2, -3) there are infinitely many lines passing in every-direction.
So (2, -3) is satisfied with infinite linear equations.
Hence, correct option is (d).
MCQ 341 Mark
If x = k + 1 y = 2k - 1 is a solution of the equation 3x - 2y + 7 = 0 then k =
- A10
- B6
- C4
- D12
Answer
View full question & answer→D. 12
Given that $x=k+1, y=2 k-1$ is a solution of $3 x-2 y+7=0$.$
\therefore \quad 3(k+1)-2(2 k-1)+7=0 \Rightarrow-k+12=0 \Rightarrow k=12 .
$
Given that $x=k+1, y=2 k-1$ is a solution of $3 x-2 y+7=0$.$
\therefore \quad 3(k+1)-2(2 k-1)+7=0 \Rightarrow-k+12=0 \Rightarrow k=12 .
$
MCQ 351 Mark
If $x=1$ and $y=6$ is a solution of the equation $8 x-a y+a^2=0$, then $a=$
- A$-2-4$
- B2,4
- C$-2,4$
- D$2,-4$
Answer
View full question & answer→B. 2, 4
Given that $x=1$ and $y=6$ is a solution of the equation $8 x-a y+a^2=0$.$
8 \times 1-6 \times a+a^2=0 \Rightarrow a^2-6 a+8=0 \Rightarrow(a-2)(a-4)=0 \Rightarrow a=2, 4$
Given that $x=1$ and $y=6$ is a solution of the equation $8 x-a y+a^2=0$.$
8 \times 1-6 \times a+a^2=0 \Rightarrow a^2-6 a+8=0 \Rightarrow(a-2)(a-4)=0 \Rightarrow a=2, 4$
MCQ 361 Mark
If $x=1$ and $y=1$ is a solution of both the equations $2 x-3 a y+a^2=0$ and $4 x-5 a y+a^2=0$, then $a=$
- A1
- B4
- C2
- D-1
Answer
View full question & answer→A. 1
Given that $x=1$ and $y=1$ is a solution of both the equations$
\begin{array}{ll}
\therefore & 2-3 a+a^2=0 \text { and } 4-5 a+a^2=0 \\
\Rightarrow & a^2-3 a+2=0 \text { and } a^2-5 a+4=0 \\
\Rightarrow & (a-1)(a-2) \text { and }(a-1)(a-4)=0 \Rightarrow a=1,2 \text { and } a=1,4 \Rightarrow a=1
\end{array}
$
Given that $x=1$ and $y=1$ is a solution of both the equations$
\begin{array}{ll}
\therefore & 2-3 a+a^2=0 \text { and } 4-5 a+a^2=0 \\
\Rightarrow & a^2-3 a+2=0 \text { and } a^2-5 a+4=0 \\
\Rightarrow & (a-1)(a-2) \text { and }(a-1)(a-4)=0 \Rightarrow a=1,2 \text { and } a=1,4 \Rightarrow a=1
\end{array}
$
MCQ 371 Mark
If we multiply or divide both sides of a linear equation with non-zero number, then the solution of the linear equation
- Achanges
- ✓remains some
- Cchanges in case of division only
- Dchanges in case of multiplication only.
Answer
View full question & answer→Correct option: B.
remains some
b
MCQ 381 Mark
If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length
- A4 units
- B3 units
- ✓5 units
- Dnone of these
Answer
View full question & answer→Correct option: C.
5 units
c
MCQ 391 Mark
If a linear equation has solutions (-2, 2) (0, 0) and (2,2), then it is of the form
- ✓y = - x
- By = x
- Cy = 2x
- Dx = 2y
Answer
View full question & answer→Correct option: A.
y = - x
a
MCQ 401 Mark
If (a, 4) lies on the graph of 3x + y = 10 then the value of a is
- A3
- B1
- ✓2
- D4
Answer
View full question & answer→Correct option: C.
2
c
MCQ 411 Mark
If (4, 19) is a solution of the equation y = ax + 3 then a =
- A3
- ✓4
- C5
- D6
Answer
View full question & answer→Correct option: B.
4
b
MCQ 421 Mark
If (2k - 1, k) is a solution of the equation 10x - 9y = 12 then k =
- A1
- ✓2
- C3
- D4
Answer
View full question & answer→Correct option: B.
2
b
MCQ 431 Mark
If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is
- ✓4
- B6
- C5
- D2
Answer
View full question & answer→Correct option: A.
4
a
MCQ 441 Mark
How many linear equations in x and y can be satisfied by x = 1 and y = 2?
- Aonly one
- Btwo
- Cthree
- ✓infinitely many
Answer
View full question & answer→Correct option: D.
infinitely many
d
MCQ 451 Mark
How many linear equations are satisfied by x = 2 and y = - 3 ?
- Aonly one
- Btwo
- Cthree
- ✓infinitely many
Answer
View full question & answer→Correct option: D.
infinitely many
d
MCQ 461 Mark
Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form
- ✓$\left(-\frac{9}{2}, a\right)$
- B$\left(b,-\frac{9}{2}\right)$
- C$\left(0,-\frac{9}{2}\right)$
- D$(-9,0)$
Answer
View full question & answer→Correct option: A.
$\left(-\frac{9}{2}, a\right)$
a
MCQ 471 Mark
Any solution of linear equation 0x - 2y + 11 = 0 in two variables is of the form
- A$(m, 11 / 2)$
- B$(m,-11 / 2)$
- C$(m, 11)$
- D$(m,-11)$
Answer
View full question & answer→A. $(m, 11 / 2)$
We find that $x=m$ and $y=\frac{11}{2}$ satisfies the equation $0 x-2 y+11=0$.
Hence, ( $m, 11 / 2$ ) represents a solution.
We find that $x=m$ and $y=\frac{11}{2}$ satisfies the equation $0 x-2 y+11=0$.
Hence, ( $m, 11 / 2$ ) represents a solution.
MCQ 481 Mark
Any point on the x-axis is of the form
- A(x, y)
- B(0, y)
- ✓(x, 0)
- D(x, x)
Answer
View full question & answer→Correct option: C.
(x, 0)
c
MCQ 491 Mark
Any point on the line y = x is of the form
- A(a, - a)
- B(0, a)
- ✓(a, a)
- D(a, 0)
Answer
View full question & answer→Correct option: C.
(a, a)
c
MCQ 501 Mark
Any point on the line y = - x is of the form
- ✓(a, - a)
- B(- a, - a)
- C(a, a)
- D(a, 0)
Answer
View full question & answer→Correct option: A.
(a, - a)
a