MCQ(1M)

MCQ 11 Mark

In figure, if lines $l$ and $m$ are parallel, then $x =$

A
$20^\circ$
✓
$45^\circ$
C
$65^\circ$
D
$85^\circ$

Answer

Correct option: B.

$45^\circ$

From figure,
$\angle\text{ABD}=\angle\text{CDF} [$Correspondence angles$]$
$\Rightarrow\ \angle\text{CDF}=65^\circ$
Now,
$\angle\text{FDE}=180^\circ-\angle\text{CDF}=180^\circ-65^\circ$
$\Rightarrow\ \angle\text{FDE}=115^\circ$
In $\triangle\text{EDF},$
$\angle\text{FDE}+\angle\text{DEF}+\angle\text{EFD}=180^\circ$
$\Rightarrow\ 115^\circ+\text{x}+20^\circ=180^\circ [$Sum of all interior angles of a $\triangle$ as $180^\circ ]$
$\Rightarrow\ \text{x}=180^\circ-20^\circ-115^\circ=45^\circ$

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MCQ 21 Mark

In figure, if $ l_1 || l_2,$ what isb the value of $y$ ?

A
$100$
B
$120$
✓
$135$
D
$150$

Answer

Correct option: C.

$135$

Let angle supplement of $3 x^{\circ}$ be $Z^{\circ}$
$\Rightarrow z^{\circ}=180^{\circ}-3 x^{\circ}$
$\Rightarrow \angle AHF+\angle FHB=180^{\circ}$
$\Rightarrow z^{\circ}+3 x^{\circ}=180^{\circ}$
$\Rightarrow z^{\circ}=180^{\circ}-3 x^{\circ}$
Now,
$x^{\circ}+y^{\circ}=180^{\circ}$
Also, $x^{\circ}=z^{\circ} \ [$Correspondence angles$]$
$\Rightarrow x^{\circ}=180^{\circ}-3 x^{\circ}$
$\Rightarrow 4 x^{\circ}=180^{\circ}$
$\Rightarrow x^{\circ}=45^{\circ}$
$x^{\circ}+y^{\circ}=180^{\circ}$
$\Rightarrow y^{\circ}=180^{\circ}-x^{\circ}=180^{\circ}-45^{\circ}=135^{\circ}$

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MCQ 31 Mark

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $2 : 3,$ then the measure of the larger angle is:

A
$54^\circ$
B
$120^\circ$
✓
$108^\circ$
D
$136^\circ$

Answer

Correct option: C.

$108^\circ$

Let $AB$ and $CD$ are two parallel lines and $PQ$ is transverce to it.
According to question,
$\frac{\angle\text{BRS}}{\angle\text{DSR}}=\frac{2}{3}$
$\Rightarrow\ \angle\text{BRS}=\frac{2}{3}\angle\text{DSR}\dots(1)$
Now,
$\angle\text{CSR}=\angle\text{BRS} \ [$Alternate angles$]$
$\Rightarrow\ \angle\text{CSR}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \angle\text{BRS}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \frac{2}{3}\angle\text{DSR}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \angle\text{DSR}=\frac{180\times3}{5}=108^\circ$
$\Rightarrow\ \angle\text{BRS}=\frac{2}{3}\times108^\circ=72^\circ$
Thus,
$\angle\text{DSR}=108^\circ$ and $\angle\text{BRS}=72^\circ$
$\Rightarrow$ Larger angle is $\angle\text{DSR}.$

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MCQ 41 Mark

In figure, $\text{AOB}$ is a straight line. If $\angle\text{AOC}+\angle\text{BOD}=85^\circ,$ then $\angle\text{COD}=$

A
$85 ^{\circ}$
B
$90 ^{\circ}$
✓
$95 ^{\circ}$
D
$100 ^{\circ}$

Answer

Correct option: C.

$95 ^{\circ}$

From figure, we can see

$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$

But,

$\angle\text{AOC}+\angle\text{BOD}=85^\circ$ [Given]

$\Rightarrow\ 85^\circ+\angle\text{COD}=180^\circ$

$\Rightarrow\ \angle\text{COD}=95^\circ$

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MCQ 51 Mark

In figure, if $l_1 || l_2$ and $l_3 || l_4$, what is $y$ in terms of $x$?

A
$90+\text{x}$
B
$90+2\text{x}$
✓
$90-\frac{\text{x}}{2}$
D
$90-2\text{x}$

Answer

Correct option: C.

$90-\frac{\text{x}}{2}$

From figure,
$\angle\text{EPR}=\angle\text{PQS} \ [$Correspondence angles are equal$]$
$\Rightarrow\ \angle\text{PQS}=\text{x}^\circ$
Also $, \angle\text{PQS}=\angle\text{RSD} \ [$Correspondence angles are equal$]$
$\Rightarrow\ \angle\text{RSD}=\text{x}^\circ$
Now $, \angle\text{RSD}+\text{y}^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ+2\text{y}^\circ=180^\circ$
$\Rightarrow\ \text{y}^\circ=\frac{180^\circ-\text{x}^\circ}{2}$
$\Rightarrow\ \text{y}^\circ=90^\circ-\frac{\text{x}^\circ}{2}$

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MCQ 61 Mark

Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is:

A
$45^\circ$
B
$30^\circ$
✓
$36^\circ$
D
None of these

Answer

Correct option: C.

$36^\circ$

Let one angle be $\theta$
Then, its complementary $=90-\theta$
According to question,
$2\theta=3(90-\theta)$
$=5\theta=270$
$\theta=54^\circ$
Then, $90-\theta^\circ=36^\circ$
Hence, the smaller angle is $36^\circ .$

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MCQ 71 Mark

In figure, if $l_1|| l_2$, what is $x + y$ in terms of $w$ and $z$?

✓
$180 - w + z$
B
$180 + w - z$
C
$180 - w - z$
D
$180 + w + z$

Answer

Correct option: A.

$180 - w + z$

Let angle supplement of $w^{\circ}$ be $a^{\circ}$.
$\Rightarrow a^{\circ}=180^{\circ}-w^{\circ}$
Now $, a^{\circ}=x^\circ \ [$Alternate opposite angles$]$
$\Rightarrow x^{\circ}=180-w^{\circ} . . .(1)$
Now $, y^{\circ}=z^{\circ} . .. (2) \ [$Alternate angles$]$
Adding $(1)$ and $(2),$ we get
$x^{\circ}+y^{\circ}=180^{\circ}-w^{\circ}+z^{\circ}$

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MCQ 81 Mark

One angle is equal to three times its supplement. The measure of the angle is:

A
$130^\circ$
✓
$135^\circ$
C
$90^\circ$
D
$120^\circ$

Answer

Correct option: B.

$135^\circ$

Let the required angle be ${\theta}.$
Then, measure of its supplement $180^\circ-\theta$
According to question, we have
$\theta=3(180-\theta)$
$\Rightarrow\ \theta=540^\circ-3\theta$
$\Rightarrow\ 4\theta=540^\circ $
$\Rightarrow\ \theta=135^\circ$

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MCQ 91 Mark

In figure, if $l_1 || l_{2,}$ what is the value of $x$?

A
$90^\circ$
✓
$85^\circ$
C
$75^\circ$
D
$70^\circ$

Answer

Correct option: B.

$85^\circ$

From figure,
$\angle\text{ERC}=\angle\text{RPA}\ [$Corresponding angles are equal$]$
$\Rightarrow \angle\text{ERC}=37^\circ=\angle\text{RPA}$
Also $, \angle\text{RPA}=\angle\text{BPF} \ [$Opposite angles$]$
$\Rightarrow \angle\text{RPA}=37^\circ=\angle\text{BPF}$
Now $, \angle\text{QPB}+\angle\text{BPF}+\angle\text{FPG}=180^\circ$
$\Rightarrow\ \text{x}^\circ+37^\circ+58^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ=85^\circ$

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MCQ 101 Mark

In figure, if $l \| m,$ then $x =$

✓
$105^\circ$
B
$65^\circ$
C
$40^\circ$
D
$25^\circ$

Answer

Correct option: A.

$105^\circ$

From figure,
$\angle\text{AGE}=\angle\text{FGB}\ [$Opposite angles$]$
$\Rightarrow\ \angle\text{FGB}=65^\circ$
Also,
$\angle\text{FGB}=\angle\text{HJI}\ [$Corresponding angle$]$
$\Rightarrow\ \angle\text{HJI}=65^\circ$
Now, in $\angle\text{HJI},$
$\angle\text{HJI}+\angle\text{JIH}+\angle\text{IHJ}=180^\circ$
$\Rightarrow\ 65^\circ+40^\circ+\angle\text{IHJ}=180^\circ$
$\Rightarrow\ \angle\text{IHJ}=180^\circ-65^\circ-40^\circ=75^\circ$
Now,
$\text{x}=180^\circ-\angle\text{IHJ}=180^\circ-75^\circ$
$=105^\circ$

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MCQ 111 Mark

Two lines $AB $ and $CD $ intersect at $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ,$ then $\angle\text{AOC}=$

A
$70^\circ$
B
$80^\circ$
✓
$90^\circ$
D
$180^\circ$

Answer

Correct option: C.

$90^\circ$

$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ \ [$Given$]$
From figure,
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}+\angle\text{DOA}=360^\circ$
$\Rightarrow\ 270^\circ+\angle\text{DOA}=360^\circ$
$\Rightarrow\ \angle\text{DOA}=360^\circ-270^\circ=90^\circ$
Now,
$\angle\text{DOA}+\angle\text{AOC}=180^\circ$
$\Rightarrow\ \angle\text{AOC}=180^\circ-90^\circ=90^\circ$

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MCQ 121 Mark

In figure, if $AB \| CD,$ then the value of $x$ is:

A
$20^\circ$
✓
$30^\circ$
C
$45^\circ$
D
$60^\circ$

Answer

Correct option: B.

$30^\circ$

From figure,
$\angle\text{DPQ}+\angle\text{x}^\circ=180^\circ\dots(1) \ [$linear pair$]$
Also,
$\angle\text{DPQ}=\angle\text{AQP} \ [$Interior opposite angles$]$
$\Rightarrow\ \angle\text{DPQ}=120^\circ+\text{x}$
From $(1),$
$120^\circ+\text{x}+\text{x}=180^\circ$
$\Rightarrow\ 2\text{x}=60^\circ$
$\Rightarrow\text{x}=30^\circ$

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MCQ 131 Mark

In figure, if $l \| m,$ what is the value of $x$?

✓
$60$
B
$50$
C
$45$
D
$30$

Answer

Correct option: A.

$60$

$3y^\circ = 2y^\circ + 25^\circ\ [$Alternate angles$]$
$\Rightarrow y^\circ = 25^\circ$
Now,
$x^\circ + 15^\circ = 2y^\circ + 25^\circ\ [$Opposite angles$]$
$\Rightarrow x = 2y^\circ + 25^\circ - 15^\circ$
$\Rightarrow x = 2y^\circ + 10^\circ$
$\Rightarrow x = 2 \times 25^\circ + 10^\circ$
$\Rightarrow x = 60^\circ$

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