M.C.Q

MCQ 11 Mark

Let $f ( x )$ be a polynomial such that $f\left(-\frac{1}{2}\right)=0$, then a factor of $f ( x )$ is

A
$X+1$
B
$2 x+1$
C
$X -1$
D
$X -1$

Answer

(b) $2 x+1$
Explanation: Let $f ( x )$ be a polynomial such that $f\left(-\frac{1}{2}\right)=0$
i.e., $x+\frac{1}{2}=0$ is a factor.
On rearranging $x+\frac{1}{2}=0$ can be written as $(2 x +1)=0$
Thus, $(2 x+1)$ is a factor of $f(x)$.

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MCQ 21 Mark

The graph of x + y = 6 intersect coordinate axes at

A
Both $(0,6)$ and $(6,0)$
B
$(6,0)$
C
$(0,6)$
D
$(2,3)$

Answer

(a) Both (0, 6) and (6, 0)
Explanation: Both (0, 6) and (6, 0)

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MCQ 31 Mark

The co-ordinates of two points A and B are (4, 3) and (4, -5) respectively. The co-ordinates of the point at which the line segment AB meets the x-axis are

A
$(-5,0)$
B
$(3,0)$
C
$(0,4)$
D
$(4,0)$

Answer

(d) (4, 0)
Explanation: Since it will meet at x-axis and we know that any point on x-axis have ordinate equals to zero. So point will be (4, 0).

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MCQ 41 Mark

In the given figure, equilateral $\triangle A B C$ is inscribed in a circle and $\ce{ABCD}$ is a quadrilateral, as shown. Then, $\angle BDC =$ ?

✓
$120^{\circ}$
B
$60^{\circ}$
C
$150^{\circ}$
D
$90^{\circ}$

Answer

Correct option: A.

$120^{\circ}$

$\triangle ABC$ is an equilateral triangle so $\angle BAC =60^{\circ}$
In cyclic quadrilateral $\ce{ABCD},$ we have:
$\angle BDC +\angle BAC =180^{\circ}$
$\Rightarrow \angle BDC +60^{\circ}=180^{\circ}$
$\therefore \angle BDC =\left(180^{\circ}-60^{\circ}\right)$
$=120^{\circ}$

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MCQ 51 Mark

If $\frac{3^{5 x} \times 81^2 \times 6561}{2^{2 x}}=3^7$, then $x=$

A
$-\frac{1}{3}$
B
$3$
✓
$-3$
D
$\frac{1}{3}$

Answer

Correct option: C.

$-3$

$\frac{3^{5 x} \times 81^2 \times 6561}{3^{2 x}}=3^7$
$\Rightarrow 3^{5 x} \times 81^2 \times 6561$
$=3^7 \times 3^{2 x}$

$\Rightarrow 3^{5 x } \times\left(3^4\right)^2 \times 3^8=3^7 \times 3^{2 x }$
$\Rightarrow 3^{5 x +8+8}=3^{7+2 x }$
$\Rightarrow 5 x +16=7+2 x$
$\Rightarrow 5 x -2 x =7-16$
$\Rightarrow 3 x =-9$
$\Rightarrow x =-3$

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MCQ 61 Mark

In the given figure, the value of $x$ which makes $\ce{POQ}$ a straight line is:

A
$40^{\circ}$
B
$30^{\circ}$
C
$35^{\circ}$
✓
$25^{\circ}$

Answer

Correct option: D.

$25^{\circ}$

We know that he measure of a straight angle is $180^{\circ}$
$\left(2 x+30^{\circ}\right)+4 x=180^{\circ}$
$2 x+30^{\circ}+4 x=180^{\circ}$
$6 x=180^{\circ}-30^{\circ}$
$6 x=150^{\circ}$
$x=\frac{150^{\circ}}{6}$
$=25^{\circ}$

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MCQ 71 Mark

The graph of the linear equation y = x passes through the point

A
$\left(\frac{3}{2}, \frac{-3}{2}\right)$
B
$\left(0, \frac{3}{2}\right)$
C
$\left(\frac{-1}{2}, \frac{1}{2}\right)$
D
$(1,1)$

Answer

(d) $(1,1)$
Explanation: $y = x , \Rightarrow$ both the coordinates are the same. Hence $(1,1)$ is correct option.

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MCQ 81 Mark

The number $x =1.242424.........$. can be expressed in the form $x =\frac{p}{q}$, where $p$ and $q$ are positive integers having no common factors. Then $p+q$ equals

A
$41$
✓
$74$
C
$53$
D
$72$

Answer

Correct option: B.

$74$

We have, $x = 1.2424.....(i)$
$\Rightarrow 100 x=124.2424 ........(ii)$
Subtracting $(i)$ from $(ii),$ we get
$100x - x = 123$
$\Rightarrow 99 x=123 $
$\Rightarrow x=\frac{123}{99}=\frac{41}{33}$
$\therefore p=41, q=33 $ and $p+q=41+33=74$

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MCQ 91 Mark

In which of the following figures are the diagonals equal?

A
Rhombus
B
Rectangle
C
Parallelogram
D
Trapezium

Answer

(b) Rectangle
Explanation: Rectangle is the correct answer. As we know that from all the quadrilaterals given in other options, diagonals of a rectangle are equal.

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MCQ 101 Mark

If $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, then $x^2+x y+y^2=$

A
$102$
B
$101$
✓
$99$
D
$98$

Answer

Correct option: C.

$99$

Given $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$,
Consider,
$x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3})^2+(\sqrt{2})^2}$
$=\frac{(\sqrt{3})^2+(\sqrt{2})^2-2(\sqrt{3})(\sqrt{2})}{3-2}$
$=\frac{3+2-2 \sqrt{6}}{1}$
$=5-2 \sqrt{6}$
Hence $x=5-2 \sqrt{6}$
$\Rightarrow x^2=(5-2 \sqrt{6})^2$
$=25+24-20 \sqrt{6}$
$=49-20 \sqrt{6}$
i.e. $x^2=49-20 \sqrt{6}........(i)$
Again consider
$ y =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}$
$=\frac{(\sqrt{3})^2+(\sqrt{2})^2+2(\sqrt{3})(\sqrt{2})}{3-2}$
$=\frac{3+2+2 \sqrt{6}}{1}$
$=5+2 \sqrt{6}$
Hence $y=5+2 \sqrt{6}$
$\Rightarrow y^2=(5+2 \sqrt{6})^2$
$=25+24+20 \sqrt{6}$
$=49+20 \sqrt{6}$
i.e. $y ^2=49+20 \sqrt{6}-\ldots (ii)$
Then $x^2+x y+y^2$
$=49-20 \sqrt{6}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+49+20 \sqrt{6}[$ from $(i)$ nd $(ii)]$
$=98+1$
$=99$

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MCQ 111 Mark

Which of the following is a true statement?

A
$5 x^3$ is a monomial
B
$x^2+5 x-3$ is a linear polynomial
C
$x +1$ is a monomial
D
$x^2+4 x-1$ is a binomial

Answer

(a) $5 x^3$ is a monomial
Explanation: $5 x ^3$ is a monomial as it contains only one term.

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MCQ 121 Mark

Express$ ' x\ '$ in terms of $'y\ '$ in the equation $2 x-3 y-5=0$.

A
$x=\frac{3 y-5}{2}$
✓
$x=\frac{3 y+5}{2}$
C
$x=\frac{5-3 y}{2}$
D
$x=\frac{3+5 y}{2}$

Answer

Correct option: B.

$x=\frac{3 y+5}{2}$

$2 x-3 y-5=0$
$2 x=3 y+5$
$x=\frac{3 y+5}{2}$

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MCQ 131 Mark

In the adjoining figure, $AB = BC$ and $\angle ABD =\angle CBD$, then another angle which measures $30^{\circ}$ is

A
$\angle BCA$
B
$\angle BCD$
C
$\angle BDA$
D
$\angle BAD$

Answer

(c) $\angle BDA$
Explanation: In triangle ABD and CBD
$AB = BC$ and $\angle ABD =\angle CBD$ (Given)
BD (Common)
Therefore In triangle ABD and CBD are congruent by SAS criteria.
Therefore, $\angle BDA =30^{\circ}$ (by CPCT)

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MCQ 141 Mark

The value of $\frac{2}{\sqrt{5}-\sqrt{3}}$ is

A
$\frac{1}{\sqrt{5}-\sqrt{3}}$
B
$\sqrt{5}-\sqrt{3}$
✓
$\sqrt{5}+\sqrt{3}$
D
$\frac{1}{\sqrt{5}+\sqrt{3}}$

Answer

Correct option: C.

$\sqrt{5}+\sqrt{3}$

$\frac{2}{\sqrt{5}-\sqrt{3}}$
multiplying $nu$ nominator and denominator by
$\sqrt{5}+\sqrt{3} \text {, we get }$
$\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$
$=\frac{2(\sqrt{5}+\sqrt{3})}{5-3}$
$=\sqrt{5}+\sqrt{3}$

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MCQ 151 Mark

$E$ and $F$ are the mid$-$points of the sides $A B$ and $A C$ of a $\triangle A B C$. If $A B=6 \ cm, B C=5 \ cm$ and $A C=6 \ cm$, Then $EF$ is equal to

A
$4 \ cm$
B
$3 \ cm$
✓
$2.5 \ cm$
D
$2 \ cm$

Answer

Correct option: C.

$2.5 \ cm$

since $E$ and $F$ are the mid points of sides $AB$ and $AC$ respectively.
according to mid point theorem of triangle;
according to mid point theorem of triangle;
$ EF =\frac{1}{2} \times B C$
$EF =\frac{1}{2} \times 5$

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MCQ 161 Mark

In the figure, O is the centre of the circle. If $\angle O P Q=25^{\circ}$ and $\angle O R Q=20^{\circ}$, then the measures of $\angle P O R$ and $\angle P Q R$ are respectively :

A
$150^{\circ}, 30^{\circ}$
B
$120^{\circ}, 60^{\circ}$
C
$90^{\circ}, 45^{\circ}$
D
$60^{\circ}, 30^{\circ}$

Answer

(c) $90^{\circ}, 45^{\circ}$
Explanation: Here, given
$OP = OQ$ and $OR = OQ$ (Radius of circle)
So, $\{$angles opposite to equal sides are also equal $\}$
Hence,
$PQR=25^{\circ}+20^{\circ}=45^{\circ}$
and $PQR =2 PQR =245^{\circ}=90^{\circ}$
$\{$Angle subtended by same sides on centre is double the angle at opposite vertex$\}$

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MCQ 171 Mark

Each equal side of an isosceles triangle is $13 \ cm$ and its base is $24 \ cm$ Area of the triangle is :

A
$40 \sqrt{3} \ cm^2$
B
$25 \sqrt{3} \ cm^2$
✓
$60 \ cm^2$
D
$50 \sqrt{3} \ cm^2$

Answer

Correct option: C.

$60 \ cm^2$

$s =\frac{13+13+24}{2}=25 \ cm$
$\text { Area of triangle }=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{25(25-13)(25-13)(25-24)}$
$=\sqrt{25 \times 12 \times 12 \times 1}$
$=60 sq . \ cm $

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MCQ 181 Mark

Abscissa of a point is negative in

A
quadrant IV only
B
quadrant II and III
C
quadrant I and IV
D
quadrant I only

Answer

(b) quadrant II and III
Explanation: The abscissa (x-axis) is -ve in 2nd and 3rd quadrant only because, Sign of point in 2nd quadrant is $(-,+)$, and in 3rd quadrant, it is $(-,-)$.

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