MCQ Questions for Class 11 Engineering Maths Straight Lines Quiz 1

Area of a triangle formed by the points A(5, 2), B(4, 7) and C(7, -4) is _____.

0%
2 sq. units
0%
$2\sqrt2$ sq. units
0%
4 sq. units
0%
$4\sqrt2$ sq. units

Explanation

Area of a triangle ABC = $\frac { 1 }{ 2 } \left|[ x_{ 1 }\left( y_{ 2 }-y_{ 3 } \right) +x_{ 2 }\left( y_{ 3 }-y_{ 1 } \right) +x_{ 3 }\left( y_{ 1 }-y_{ 2 } \right) \right]|$

Substituting the given coordinates, we have

= $\frac { 1 }{ 2 } \left[ 5 \left(7+4) \right) -\left( 4+2\right) +7\left( 2-7) \right) \right]$

$=\frac { 1 }{ 2 } \left( 5 ( 11) +4\times( -6) +7\times( -5)\right)$

$=\frac { 1 }{ 2 } \left[(55) - ( 24) - ( 35) \right]$

$=\frac { 1 }{ 2 } \times |-4| = 2$ square units.

$A(0, 0), B(7, 2), C(7, 7)$ and

$D(2, 7)$ are the vertices of a quadrilateral. The respective slopes of diagonals

$AC$ and

$BD$ are

0%
$-1, 1$
0%
$1, -1$
0%
$0, -1$
0%
$1, 0$

Explanation

The slope of a line passing through the point

$( { x }_{ 1 },{ y }_{ 1 })$ and

$( { x }_{ 2 },{ y }_{ 2 })$ is given by

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$
Given :

$A(0 , 0), B (7, 2), C (7, 7)$ and

$D (2, 7)$ are the vertices of quadrilateral

$AC$ and

$BD$ are diagonals of a quadrilateral

Then, slope of diagonal

$AC$ is

$m_{AC} = \dfrac{7 - 0}{7 - 0}$

$\implies m_{AC} = \dfrac{7}{7}$

$\therefore\ m_{AC} = 1$
Similarly, slope of

$BD$ is

$m_{BD} = \dfrac{7 - 2}{2 - 7}$

$\implies m_{BD} = \dfrac{5}{-5}$

$\therefore\ m_{BD} = -1$ .

Hence, the slopes of diagonals

$AC$ and

$BD$ are

$1$ and

$-1$ respectively.

Let

$\mathrm{P}(\mathrm{x}_{1},\mathrm{y}_{1})\mathrm{b}\mathrm{e}$ any point on the cartesian plane then match the following lists:

LIST - I	LIST - II
$\mathrm{A})$ The distance from $\mathrm{P}$ to X-axis	1) $0$
$\mathrm{B})$ The distance from $\mathrm{P}$ to Y-axis	2) $\|\mathrm{y}_{1}\|$
$\mathrm{C})$ The distance from $\mathrm{P}$ to origin is	3) $\sqrt{x_{1}^{2}+y_{1}^{2}}$
	4) $\|x_{1}\|$

0%
$A - 4, B - 2, C - 1$
0%
$A - 2, B - 4, C - 3$
0%
$A - 4, B - 2, C - 3$
0%
$A - 2, B - 4, C - 1$

Explanation

$(A)=\ |y_{1}|\rightarrow 2$

$(B)= |x_{1}| \rightarrow 4$

$(C)= \sqrt{x_{1}^{2}+y_{1}^{2}}\rightarrow 3$

If A and B are the points

$(-6,7)$ and

$(-1,-5)$ respectively, then the distance 2AB is equal to

0%
$28$
0%
$26$
0%
$169$
0%
$238$

Explanation

The distance between two points can be found by the equation

$=\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-{ y }_{ 1 })^{ 2 } }$

here point

$A(-6,7)$ and $$B(-1,-5)$$

$\therefore$ AB

$=\sqrt { (-1+6)^{ 2 }+(-5-7)^{ 2 } } =\sqrt { 169 } =13$

$\therefore$

$2AB = 2 \times 13 = 26$

The vertices of a triangle are

$(-2,0) ,(2,3)$ and

$(1, -3)$ , then the type of the triangle is

0%
scalene
0%
equilateral
0%
isosceles
0%
Right triangle

Explanation

Consider give the vertex of a triangle are $(-2,0),(2,3)$ and $(1,-3)$ .

Let, a triangle ABC, which have vertex $A(-2,0),B(2,3)$ and $C(1,-3)$ .

Now, sides of triangle AB,BC and CA are,

$AB=\sqrt{{{\left( 2+2 \right)}^{2}}+{{\left( 3-0 \right)}^{2}}}=5unit$

$BC=\sqrt{{{\left( 1-2 \right)}^{2}}+{{\left( -3-3 \right)}^{2}}}=\sqrt{37}unit$

And $CA=\sqrt{{{\left( 1+2 \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}=3\sqrt{2}unit$

Hence, $AB\ne BC\ne CA$

So, required triangle is scalene triangle.

Hence,this is the answer,

The vertices P, Q, R, and S of a parallelogram are at (3,-5), (-5,-4), (7,10) and (15,9) respectively The length of the diagonal PR is

0%
$\displaystyle \sqrt{241}$
0%
$\displaystyle \sqrt{124}$
0%
$\displaystyle \sqrt{200}$
0%
$\displaystyle \sqrt{180}$

Explanation

PR =

$\displaystyle \sqrt{\left ( 3-7 \right )^{2}+\left ( -5-10 \right )^{2}}=\sqrt{16+225}=\sqrt{241}$

The distance between (a,-b) and (-a, -b) is given by

0%
$\displaystyle \sqrt{a^{2}+b^{2}}$
0%
2a
0%
2b
0%
$\displaystyle 2\sqrt{a^{2}+b^{2}}$

Explanation

d =

$\displaystyle \sqrt{\left ( a+a \right )^{2}+\left ( -b+b \right )^{2}}=2a$

The area of a triangle whose vertices are

$(a, c+a), (a, c)$ and

$(-a, c-a)$ are

0%
$\displaystyle a^{2}$
0%
$\displaystyle b^{2}$
0%
$\displaystyle c^{2}$
0%
$\displaystyle a^{2}+c^{2}$

Explanation

Area of a triangle with vertices

$({ x }_{ 1 },{ y }_{ 1 })$ ;

$({ x }_{ 2 },{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is

$\left| \frac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$
Hence, area of the triangle with given vertices is

$\left| \frac { { a }(c-c+a)+a(c-a-c-a)-a(c+a-c) }{ 2 } \right| = \left| \frac { { a }^{ 2 }-2{ a }^{ 2 }-{ a }^{ 2 } }{ 2 } \right| = { a }^{ 2 }$

Equation of the line y = 0 represents

0%
x-axis
0%
y-axis
0%
both x-axis and y-axis
0%
neither x-axis nor y-axis

Explanation

It is clear that

$y$ coordinate is constant i.e.

$0$ for given equation and only

$x$ coordinate is variable

Hence it represent the

$x$ -axis.

The points

$(-2, -1), (1, 0),(4, 3),$ and

$(1, 2)$ are the vertices

0%
Parallelogram
0%
Rectangle
0%
Square
0%
Both $(1)$ & $(2)$

Explanation

$To\quad identify-\\ whether\quad ABCD\quad is\quad a\quad (i)\quad parallelogram,\quad (ii)\quad rectangle,\quad (iii)\quad square,\quad \\ (iv)\quad scalene\quad one.\\ Solution-\\ (i)\quad We\quad know\quad that\quad in\quad a\quad parallelogrm\quad the\quad diagonals\quad bisect\quad each\quad \\ other\quad i.e\quad the\quad mid\quad points\quad of\quad AC\quad \& \quad BD\quad are\quad same.\\ Let\quad the\quad mid\quad point\quad of\quad AC\quad be\quad P(x,y).\\ Applying\quad mid\quad point\quad theorem,\\ P(x,y)=\left( \dfrac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\dfrac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } \right) =\left( \dfrac { -2+4 }{ 2 } ,\dfrac { -1+3 }{ 2 } \right) =(1,1).\\ Again\quad \\ let\quad the\quad mid\quad point\quad of\quad BD\quad be\quad Q(p,q).\\ Applying\quad mid\quad point\quad theorem,\\ Q(p,q)=\left( \dfrac { { x }_{ 2 }+{ x }_{ 4 } }{ 2 } ,\dfrac { { y }_{ 2 }+{ y }_{ 4 } }{ 2 } \right) =\left( \dfrac { 1+1 }{ 2 } ,\dfrac { 0+1 }{ 2 } \right) =(1,1).\\ \therefore \quad P(x,y)=Q(p,q)\\ \Longrightarrow the\quad mid\quad points\quad of\quad AC\quad \& \quad BD\quad are\quad same.\\ So\quad ABCD\quad is\quad a\quad parallelogram.\\ (ii)\quad If\quad the\quad diagonals\quad of\quad the\quad parallelogram\quad ABCD\quad are\quad equal\quad \\ then\quad it\quad is\quad a\quad rectangle.\\ i.e\quad AC=BD.\\ Applying\quad distance\quad formula,\\ AC=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 3 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 3 } \right) }^{ 2 } } =\sqrt { { \left( -2-4 \right) }^{ 2 }+{ \left( -1-3 \right) }^{ 2 } } units=2\sqrt { 13 } units.\\ BD=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 4 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-y_{ 4 } \right) }^{ 2 } } =\sqrt { { \left( 1-1 \right) }^{ 2 }+{ \left( 0-2 \right) }^{ 2 } } units=2units.\\ \therefore \quad AC\neq BD.\quad \\ So\quad ABCD\quad is\quad not\quad a\quad rectagle.\\ Consequently,\quad it\quad is\quad not\quad a\quad square.\\ Ans-\quad Option\quad A.\\$

The line joining

$(-1,0)$ and

$(-2, \displaystyle -\sqrt{3} )$ makes with the

$x$ -axis an angle equal to

0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 45^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 75^{\circ}$

Explanation

Let the straight line joining (-1, 0) and

$\displaystyle \left ( -\sqrt{2},-\sqrt{3} \right )$ makes with the x-axis an
angle equal to

$\displaystyle \theta$

$\displaystyle \therefore \tan \theta =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{\left ( -\sqrt{3} \right )-0}{-2-\left ( -1 \right )}=\sqrt{3}=\tan 60^{\circ}$

If the distance between the origin and (x,3)is 5 then x=

0%
5
0%
-5
0%
$\displaystyle \pm 4$
0%
$\displaystyle \pm 3$

Explanation

$\displaystyle \left ( x-0 \right )^{2}+\left ( 3-0 \right )^{2}=5^{2}$
or

$\displaystyle x^{2}=16$ or

$\displaystyle x=\pm 4$

If P(1,4), Q(9,-2), and R(5,1) are collinear then

0%
P lies between Q and R
0%
Q lies between P and R
0%
R lies between P and Q
0%
none of these

Explanation

PQ =

$\displaystyle \sqrt{\left ( 1-9 \right )^{2}+\left ( 4+2 \right )^{2}}=10$
QR =

$\displaystyle \sqrt{\left ( 9-5 \right )^{2}+\left ( -2-1 \right )^{2}}=5$
PR =

$\displaystyle \sqrt{\left ( 1-5 \right )^{2}+\left ( 4-1 \right )^{2}}=5$

$\displaystyle \therefore PQ=PR+RQ$
Therefore R lies between P and Q

What is the acute angle between the lines y = 3x + 2 and y = 4x + 9?

0%
4.4 $\displaystyle ^{\circ}$
0%
28.3 $\displaystyle ^{\circ}$
0%
5.2 $\displaystyle ^{\circ}$
0%
18.6 $\displaystyle ^{\circ}$
0%
none of these

Explanation

$y=3x+2$

$m_1=3$

$=>tan\theta_1=3$

$=>\theta_1=tan^{-1}(3)$

$=>\theta_1=71.57^{0}$

Again,

$y=4x+9$

$=>m_2=4$

$=>tan\theta_2=4$

$=>\theta_2=tan^{1}(4)$

$=>\theta_2=75.96^{0}$

$\therefore$ Angle between the lines

$=\theta_2-\theta_1$

$=75.96-71.56$

$=4.4^{0}$

If the area of the triangle formed by the points

$(-2,3), (4,-5)$ and

$(-3,y)$ is 10 square units then

$y =$

0%
$\dfrac{19}{3}$
0%
$\dfrac{21}{3}$
0%
$\dfrac{23}{3}$
0%
$\dfrac{25}{3}$

Explanation

Area of triangle having vertices

$(x_1,y_1), (x_2,y_2)$ and

$(x_3,y_3)$ is given by

$= \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ]$

$\displaystyle \text{Area} =\frac{1}{2}\left [ -2\left ( -5-y \right )+4\left ( y-3 \right )-3\left ( 3+5 \right ) \right ]$

$\displaystyle =\frac{1}{2}\left ( 10+2y+4y-12-24 \right )$

$\displaystyle =\frac{1}{2}\left ( 6y-26 \right )=3y-13$

$\displaystyle 3y-13=10\ \mathrm{sq. unit}$

$\displaystyle y=\frac{23}{3}$

Find the inclination of the line passing through (-5, 3) and (10, 7)

0%
14.73
0%
14.93
0%
14.83
0%
14.63
0%
none of these

Explanation

Slope of the line passing through

$(-5,3)$ and

$(10,7)$ is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$=>m=\dfrac{7-3}{10-(-5)}$

$=>m=\dfrac{4}{15}$

Now,

$tan \theta=\dfrac{4}{15}$

$=>\theta=tan^{-1}(\dfrac{4}{15})$

$=>\theta=14.93^{0}$

$\therefore$ inclination of the line =

$14.93^{0}$

If the point (P,0) is equidistant from (3,2) and (5,3) then P=

0%
5
0%
5.1
0%
5.25
0%
5.5

Explanation

$\displaystyle \left ( p-3 \right )^{2}+\left ( 0-2 \right )^{2}=\left ( p-5 \right )^{2}+\left ( 0-3 \right )^{2}$
or

$\displaystyle p^{2}-6p+9+4=p^{2}-10p+25+9$
or

$\displaystyle p=\frac{21}{4}=5\cdot 25$

Find the angle between the lines 3x + 2y = 6 and x + y = 6

0%
12 $\displaystyle ^{\circ}$ 20'
0%
11 $\displaystyle ^{\circ}$ 19'
0%
14 $\displaystyle ^{\circ}$ 25'
0%
13 $\displaystyle ^{\circ}$ 06'

Explanation

Line

$3x+2y=6$

$=>2y=-3x+6$

$=>y=\frac{-3}{2}x+\frac{6}{2}$

$=>m'=\frac{-3}{2}x$

$\therefore tan\theta_1=\frac{-3}{2}$

$=>\theta_1=tan^{-1}(\frac{-3}{2})$

$=>\theta_1=-56.31$

Line is

$x+y=6$

$=>y=x+6$

$=>m_2=-1$

$=>tan\theta_2=-1$

$=>\theta_2=tan^{-1}(-1)$

$=>\theta_2=-45$

Thus, ale between them

$=(-45-56.31)$

$=11.31^0$

If the area of the triangle formed by

$(-2,5), (x,-3)$ and

$(3,2)$ is

$14$ square units, then

$x=$ ____.

0%
1
0%
2
0%
-2
0%
-1

Explanation

Area of triangle having vertices

$(x_1,y_1), (x_2,y_2)$ and

$(x_3,y_3)$ is given by

$= \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ]$

$\therefore \displaystyle Area=\frac{1}{2}\left [ -2\left ( -3-2 \right )+x\left ( 2-5 \right )+3\left ( 5+3 \right ) \right ]$

$\displaystyle =\frac{1}{2}\left [ 10-3x+24 \right ]=\frac{34-3x}{2}$

$\displaystyle \therefore \frac{34-3x}{2}=14$ or x = 2

If the inclination of a line is

$45^\circ$ , then the slope of the line is ?

0%
$0$
0%
$-1$
0%
$1$
0%
$2$

Explanation

The inclination of line is

$45^0$

Hence slope of line

$m=\tan 45^0=1$

The distance between the points (sin x, cos x) and (cos x -sin x) is

0%
$1$
0%
$\displaystyle \sqrt{2}$
0%
$2 sin x cos x$
0%
$4 sin x cos x$
0%
none of these

Explanation

Distance between

$(sinx,cosx)$ and

$(cosx,-sinx)$ is

$r=\sqrt{(sinx-cox)^2+(cosx-(-sinx)^2)}$

$=>r=\sqrt{(sin^2x+cos^2x-2sinxcosx)+(cox^2x+sin^2x-2sinxcosx)}$

$=>r=\sqrt{1+1}$

$=>r=\sqrt{2}$

Find the distance from the point (2, 3) to the line 3x + 4y + 9 = 0

0%
$5$
0%
$5.4$
0%
$5.8$
0%
$6.2$
0%
none of these

Explanation

Distance from point

$(2,3)$ to the line

$3x+4y+9=0$

$=>r=\dfrac{|3(2)+4(3)+9|}{\sqrt {3^2+4^2}}$

$=>r=\dfrac{|16+12+9|}{\sqrt {9+16}}$

$=>r=\dfrac{27}{5}$

$=>r=5.4$

How far is the line 3x - 4y + 15 = 0 from the origin?

0%
$1$
0%
$2$
0%
$3$
0%
$4$
0%
Answer Required

Explanation

Distance of

$O(0,0)$ from the line

$3x-14y+15=0$ is

$=>r=\dfrac{|3(0)-4(0)+15|}{\sqrt {3^3+4^2}}$

$=>r=\dfrac{15}{\sqrt {25}}$

$=>r=\dfrac{15}{5}$

$=>r=3$

The point

$(0,\,5)$ lies on the line

$x+y=5$ or not

0%
True
0%
False
0%
Cant say
0%
None

Explanation

Given the line

$x+y=5$

when

$x=0 ,y =5 \Rightarrow(0,5)$

$\therefore$ The point (0,5) lies on the line

$x+y=5$

What is the slope of the line 3x + 2y + 1 = 0?

0%
3/2
0%
2/3
0%
-3/2
0%
-2/3

Explanation

The given equation of the line is

$3x+2y+1=0$

$=>3x+2y=-1$

$=>2y=-3x-1$

$=>y=\frac{-3}{2}x-\frac{-1}{2}$

Here

$m=\frac{-3}{2}$

Thus, slope of the line is

$\frac{-3}{2}$

Which of the following is perpendicular to the line x/3 + y/4 = 1?

0%
x - 4y - 8 = 0
0%
4x - 3y - 6 = 0
0%
3x - 4y - 5 = 0
0%
4x + 3y - 11 = 0

Explanation

Line is

$\frac{x}{3}+\frac{y}{4}=1$

$=>4x+3y=12$

$=>3y=-4x+12$

$=>y=\frac{-4}{3}+4$

Here

$m=\frac{-4}{3}$

$\therefore$ Slope of line perpendicular to given line is

$\frac{3}{4}$

Option A

$x-4y-8=0$

$=>y=\frac{-x}{4}-\frac{8}{4}$ (not perpendicular)

Option B

$4x-3y-6=0$

$=>y=\frac{4x}{3}-\frac{6}{3}$ (not perpendicular)

Option C

$3x-4y-11=0$

$=>y=\frac{3}{4}x-\frac{5}{4}$ (perpendicular)

Option D

$4x+3y-11=0$

$=>y=\frac{-4}{3}x+\frac{11}{4}$ (not perpendicular)

Find the distance from the point (5, -3) to the line 7x - 4y - 28 = 0

0%
$2.62$
0%
$2.36$
0%
$2.48$
0%
$2.54$
0%
none of these

Explanation

Distance of point

$(5,-3)$ from the line

$7x-4y-28=0$ is

$=>r=\dfrac{|7(5)-4(-3)-28|}{\sqrt {7^2+4^2}}$

$=>r=\dfrac{|35+12-281|}{\sqrt {49+16}}$

$=>r=\dfrac{19}{\sqrt 65}$

$=>r=\dfrac{19}{8.06}$

$=>r=2.36$

Given three vertices of a triangle whose coordinates are A (1, 1), B (3, -3) and (5, -3). Find the area of the triangle.

0%
3 square units
0%
4 square units
0%
5 square units
0%
6 square units

Explanation

We know that the area of the triangle whose vertices are

$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$ and

$(x_{3},y_{3})$ is

$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$

The three vertices of the triangle are

$A(1,1), B(3,-3), C(5,-3)$

Area of triangle

$=\dfrac{|1(-3-(-3)+3(-3-1)+5(1-(-3))|}{2}$

$=\dfrac{|1(0)+3(-4)+4(4)|}{2}$

$=\dfrac{|-12+20|}{2}$

$=\dfrac{8}{2}$

$=4$ square units.

The area of triangle whose vertices are

$A (-3, -1), B(5, 3)$ and

$C(2, -8)$ is ____

$\text{ sq. units}$ .

0%
$34\text{ sq. units}$
0%
$36\text{ sq. units}$
0%
$38\text{ sq. units}$
0%
$32\text{ sq. units}$

Explanation

We know that the area of the triangle whose vertices are

$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$ and

$(x_{3},y_{3})$ is

$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$

The given vertices of the triangle are

$A(-3,-1), B(5,3)$ and

$C(2,-8)$ .

So, by using the above formula,

$\begin{aligned}{}\text{Area of the triangle} &= \frac{1}{2} {[ - 3(3 - ( - 8)) + 5( - 8 - ( - 1)) + 2( - 1 - 3)]}\\ &= \frac{1}{2}{[ - 33 - 35 - 8]}\\ &= \frac{{[ - 76]}}{2}\\& = \frac{{76}}{2}\quad\quad\quad\quad\dots[\text{Area can never be negative so, we ignore negative sign}]\\& = 38\text{ sq. units}\end{aligned}$

So, the area of the triangle is equal to

$38\text{ sq. units}$ .

$P$ is the point

$(-5,3)$ and

$Q$ is the point

$(-5,m)$ . If the length of the straight line

$PQ$ is

$8$ units, then the possible value of

$m$ is:

0%
$-5$ or $5$
0%
$-5$ or $11$
0%
$-5$ or $-11$
0%
$5$ or $11$

Explanation

$\sqrt{(-5-(-5))^{2}+(m-3)^{2}}=8$

$\Rightarrow (m-3)^{2}=8^{2}$

$\Rightarrow m-3=\pm 8$

$\Rightarrow m=-5+11$

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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