MCQ - Maths STD 7 Questions - Vidyadip

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35 questions · auto-graded multiple-choice test.

MCQ 11 Mark

The sum of an angle and one third of its supplementary angle is $90^{\circ}$. The measure of the angle is:

A
$135^{\circ}$
B
$120^{\circ}$
C
$60^{\circ}$
✓
$45^{\circ}$

Answer

Correct option: D.

$45^{\circ}$

Let the required angle be $x$
Now, supplementary of the required angle $=180^{\circ}- x$
Then,
$x+\frac{1}{3}\left(180^{\circ}-x\right)=90^{\circ}$
$\Rightarrow 3 x+180^{\circ}-x=270^{\circ}$
$\Rightarrow 2 x=90^{\circ}$
$\Rightarrow x=45^{\circ}$
Hence, the correct answer is option $(d).$

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MCQ 21 Mark

In Fig. the value of $x$ is:

A
$75$
✓
$65$
C
$45$
D
$55$

Answer

Correct option: B.

$65$

$\angle \text{AOC}$ and $\angle \text{BOC}=180^\circ [\because$ Linear pair angles$]$
$\Rightarrow 44^\circ+(2\text{x}+6)^\circ=180^\circ$
$\Rightarrow (2\text{x+6})^\circ=136^\circ$
$\Rightarrow 2\text{x}+6=136$
$\Rightarrow 2\text{x}=130$
$\Rightarrow \text{x}=65$
Hence, the correct answer is option $(b).$

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MCQ 31 Mark

In Fig. if $\text{AOB}$ and $\text{COD}$ are straight lines, then:

A
$x = 29, y = 100$
B
$x = 110, y = 29$
✓
$x = 29, y = 110$
D
$x = 39, y = 110$

Answer

Correct option: C.

$x = 29, y = 110$

$\angle \text{AOD}+\angle \text{BOD}=180^\circ [$Linear pair angles$]$
$\Rightarrow \text{y}^\circ+70^\circ=180^\circ$
$\Rightarrow \text{y}^\circ=110^\circ$
$\Rightarrow \text{y}=110$
Now, $\angle \text{AOC}= \angle \text{BOD}=70^\circ [$Vertically opposite angles$]$
Now, $\angle \text{AOC}+\angle \text{COE}+\angle \text{EOB}+\angle \text{BOD}+\angle \text{AOD}=360^\circ [$Complete angle$]$
$\Rightarrow 70^\circ+28^\circ+(3\text{x}-5)^\circ+70^\circ+110^\circ=360^\circ$
$\Rightarrow (3\text{x})^\circ+273^\circ=360^\circ$
$\Rightarrow3\text{x}=87$
$\Rightarrow \text{x}=29$
Hence, the correct answer is option $(c).$

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MCQ 41 Mark

If angles of $a$ linear pair are equal, then the measure of each angle is:

A
$30^\circ$
B
$45^\circ$
C
$60^\circ$
✓
$90^\circ$

Answer

Correct option: D.

$90^\circ$

Let the required angle be $x$
Now, Sum of linear pair angles $= 180^\circ $
$\Rightarrow x + x = 180^\circ $
$\Rightarrow 2x = 180^\circ $
$\Rightarrow x = 90^\circ $
Hence, the correct answer is option $(d).$

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MCQ 51 Mark

In Fig. $AB \| CD$ and $\text{EF}$ is a transversal. The value of $y - x$ is:

A
$30$
✓
$35$
C
$95$
D
$25$

Answer

Correct option: B.

$35$

Since, $AB \| CD$
$\therefore \angle \text{BPQ}=\angle \text{DQF} [$Corresponding angles$]$
$\Rightarrow (5\text{x}-20)^\circ=(3\text{x}+40)^\circ$
$\Rightarrow 5\text{x}-20=3\text{x}+40$
$\Rightarrow 2\text{x}=60$
$\Rightarrow \text{x}=30$
$\therefore \text{BPQ}=(5\times 30-20)^\circ=130^\circ$
Now, $\angle \text{APE}=\angle \text{BPQ} [$Vertically opposite angles$]$
$\Rightarrow 2\text{y}^\circ=130^\circ$
$\Rightarrow \text{y}=65$
$\therefore \text{y}-\text{x}=65-30$
$=35$
Hence, the correct answer is option $(b).$

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MCQ 61 Mark

In Fig. the value of $x$ is:

✓
$22$
B
$20$
C
$21$
D
$24$

Answer

Correct option: A.

$22$

$(8x - 41)^\circ + (3x)^\circ + (3x + 10)^\circ + (4x - 5)^\circ = 360^\circ $
$\Rightarrow 8x - 41 + 3x + 3x + 10 + 4x - 5 = 360$
$\Rightarrow 18x - 36 = 360$
$\Rightarrow 18x = 396$
$\Rightarrow x = 22$
Hence, the correct answer is option $(a).$

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MCQ 71 Mark

Two supplementary angles are in the ratio $3 : 2.$ The smaller angle measures:

A
$108^\circ$
B
$81^\circ$
✓
$72^\circ$
D
$68^\circ$

Answer

Correct option: C.

$72^\circ$

Let the angles be $3x$ and $2x$
Now, $3x + 2x = 180^\circ $
$\Rightarrow 5x = 180^\circ $
$\Rightarrow x = 36^\circ $.
$\therefore$ Smaller angle $= 2x = 2 \times 36^\circ = 72^\circ $
Hence, the correct answer is option $(c).$

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MCQ 81 Mark

In Fig. if $\text{AOB}$ and $\text{COD}$ are straight lines. Then, $x + y =$

A
$120$
✓
$140$
C
$100$
D
$160$

Answer

Correct option: B.

$140$

$\angle \text{AOD}+\angle \text{BOD}=180^\circ [$Linear pair angles$]$
$\Rightarrow (7\text{x}-20)^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 7\text{x}-20+3\text{x}=180$
$\Rightarrow 10\text{x}=200$
$\Rightarrow \text{x}=20$
$\therefore \angle \text{AOD}=(7\times 20-20)^\circ=120^\circ$
Now, $\angle \text{AOD}=\angle \text{BOC}=120^\circ [$Vertically opposite angles$]$
$\therefore \text{y}=120$
Now, $x + y = 20 + 120$
$= 140$
Hence, the correct answer is option $(b).$

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MCQ 91 Mark

An angle is double of its supplement. The measure of the angle is:

A
$60^{\circ}$
✓
$120^{\circ}$
C
$40^{\circ}$
D
$80^{\circ}$

Answer

Correct option: B.

$120^{\circ}$

Let the required angle be $x$
Now, supplementary of the required angle $=180^{\circ}- x$
Then,
$x=2\left(180^{\circ}-x\right)$
$\Rightarrow x=360^{\circ}-2 x$
$\Rightarrow 3 x=360^{\circ}$
$\Rightarrow x=120^{\circ}$
Hence, the correct answer is option $(b).$

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MCQ 101 Mark

In Fig. $AB \| CD$ and $EF$ is a transversal intersecting $AB$ and $CO$ at Pand $Q$ respectively. The measure of $\angle \text{DPQ}$ is:

A
$100^\circ$
✓
$80^\circ$
C
$110^\circ$
D
$70^\circ$

Answer

Correct option: B.

$80^\circ$

$\angle \text{BQF}=\angle \text{AQP}=(4\text{x})^\circ [$Vertically opposite angles$]$
Since, $AB \| CD$
$\therefore \angle \text{AQP}+ \angle \text{CPQ}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow (4\text{x})^\circ+(5\text{x})^\circ=180^\circ$
$\Rightarrow 9\text{x}=180$
$\Rightarrow \text{x}=20$
$\therefore \angle \text{BQF}=(4\times20)^\circ=80^\circ$
Now, $\angle \text{BQF}=\angle \text{DPQ}=80^\circ [$Corresponding angles$]$
Hence, the correct answer is option $(b).$

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MCQ 111 Mark

In Fig. if $\text{AOC}$ is a straight line, then $x =$

A
$42^\circ$
✓
$52^\circ$
C
$142^\circ$
D
$38^\circ$

Answer

Correct option: B.

$52^\circ$

$\angle \text{AOD}+\angle \text{DOB}+\angle \text{BOC}=180^\circ [\because \text{AOC}$ is a straight line$]$
$\Rightarrow 38^\circ+\text{x}+90^\circ=180^\circ$
$\Rightarrow \text{x}+128^\circ=180^\circ$
$\Rightarrow \text{x}=52^\circ$
Hence, the correct answer is option $(b).$

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MCQ 121 Mark

In fig. if $\text{AB, CD}$ and $\text{EF}$ are straight lines, then $x =$

A
$5$
B
$10$
✓
$20$
D
$30$

Answer

Correct option: C.

$20$

Let all the lines intersect at $O$.

$\angle \text{COF}=\angle \text{DOE}=4\text{x}^\circ [$Vertically opposite angles$]$
$\angle \text{AOC}+\angle \text{COF}+\angle \text{BOF}=180^\circ [\text{AOB}$ is a straight line$]$
$\Rightarrow 2\text{x}^\circ+4\text{x}^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 9\text{x}^\circ=180^\circ$
$\Rightarrow9\text{x}=180$
$\Rightarrow \text{x}=20$
Hence, the correct answer is option $(c).$

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MCQ 131 Mark

In Fig. if $\text{AB}$ is parallel to $\text{CD}$, then the value of $\angle \text{BPE}$ is:

A
$106^\circ$
B
$76^\circ$
✓
$74^\circ$
D
$84^\circ$

Answer

Correct option: C.

$74^\circ$

Since, $AB \| CD$
$\therefore \angle \text{BPQ}= \angle \text{PQC} [$Alternate interior angles$]$
$\Rightarrow(3\text{x}+34)^\circ=(5\text{x}-14)^\circ$
$\Rightarrow 3\text{x}+34=5\text{x}-14$
$\Rightarrow 48=2\text{x}$
$\Rightarrow \text{x}=24$
$\therefore \angle\text{BPQ}=(3\times 24+34)^\circ=106^\circ$
$\angle \text{BPQ}+\angle \text{BPE}=180^\circ [EF$ is a straight line$]$
$\Rightarrow 106^\circ+\angle \text{BPE}=180^\circ$
$\Rightarrow \angle \text{BPE}=74^\circ$
Hence, the correct answer is option $(c).$

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MCQ 141 Mark

In Fig. if $AB \| CO$ then $x =$

✓
$154$
B
$139$
C
$144$
D
$164$

Answer

Correct option: A.

$154$

Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
Since, $PQ \| AB$
$\therefore \angle \text{AME}+\angle \text{QEM}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow 139^\circ+\angle \text{QEM}=180^\circ$
$\Rightarrow \angle \text{QEM}=41^\circ$
Now, $\angle \text{QEM}+\angle \text{DEQ}=\angle \text{MED}$
$\Rightarrow 41^\circ+\angle \text{DEQ}=67^\circ$
$\Rightarrow \angle \text{DEQ}=26^\circ$
Now, $\angle \text{PED}+\angle \text{DEQ}=180^\circ [$Linear Pair angles$]$
$\Rightarrow \angle \text{PED}+26^\circ=180^\circ$
$\Rightarrow \angle \text{PED}=154^\circ$
Since, $PQ \| AB$
$\therefore \text{x}^\circ=\angle \text{PED} [$Corresponding angles$]$
$\Rightarrow \text{x}^\circ-154^\circ$
$\Rightarrow \text{x}=154$
Hence, the correct answer is option $(a).$

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MCQ 151 Mark

In Fig. if $AB \| CD$. The value of $x$ is:

A
$122$
B
$238$
C
$58$
✓
$119$

Answer

Correct option: D.

$119$

Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
Since, $PQ \| CD$
$\therefore \angle \text{EFC}=\angle \text{FEQ}=37^\circ$ [Alternate angles]
Now, $\angle \text{AEQ}+\angle \text{FEQ}=\angle \text{AEF}$
$\Rightarrow \angle \text{AEQ}+37^\circ=95^\circ$
$\Rightarrow \angle \text{AEQ}=58^\circ$
Since, $PQ \| AB$
$\therefore \angle \text{EAB}+\angle \text{AEQ}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow \angle\text{EAB}+58^\circ=180^\circ$
$\Rightarrow \angle\text{EAB}=122^\circ$
$\angle\text{EAB}+\text{Reflex}\angle\text{EAB}=360^\circ$ [Complete angle]
$\therefore 122^\circ+(2\text{x})^\circ=360^\circ$
$\Rightarrow 2\text{x}=238$
$\Rightarrow \text{x}=119$
Hence, the correct answer is option $(d).$

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MCQ 161 Mark

An angle is thrice its supplement. The measure of the angle is:

A
$120^\circ$
B
$105^\circ$
✓
$135^\circ$
D
$150^\circ$

Answer

Correct option: C.

$135^\circ$

Let the required angle be $x$
Then,
$x = 3(180^\circ - x)$
$\Rightarrow x = 540^\circ - 3x$
$\Rightarrow 4x = 540^\circ $
$\Rightarrow x = 135^\circ $
Hence, the correct answer is option $(c).$

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MCQ 171 Mark

In Fig. $AB \| CO, \angle \text{OAB}=150^\circ$ and $\angle \text{OCO}=120^\circ.$ Then, $\angle \text{AOC}=$

A
$80^\circ$
✓
$90^\circ$
C
$70^\circ$
D
$100^\circ$

Answer

Correct option: B.

$90^\circ$

Construction: Draw a line $OE$ from the point $O$ parallel to $AB$ and $CD$

Since, $AB \| OE$
$\therefore\angle \text{BAO}+\angle \text{AOE}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow 150^\circ+\angle \text{AOE}=180^\circ$
$\Rightarrow \angle \text{AOE}=30^\circ$
Again, $CD \| OE$
$\therefore \angle \text{DCO}+\angle \text{COE}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow 120^\circ+\angle \text{COE}=180^\circ$
$\Rightarrow \angle \text{COE}=60^\circ$
Now, $\angle \text{AOC}=\angle \text{AOE}+\angle \text{COE}$
$=30^\circ+60^\circ$
$=90^\circ$
Hence, the correct answer is option $(b).$

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MCQ 181 Mark

The measure of an angle which is its own complement is:

A
$30^\circ$
B
$60^\circ$
C
$90^\circ$
✓
$45^\circ$

Answer

Correct option: D.

$45^\circ$

Let the required angle be $x$
Now, complementary of the required angle $= 90^\circ - x$
Then,
$x = 90^\circ - x$
$\Rightarrow x = 90^\circ - x$
$\Rightarrow 2x = 90^\circ $
$\Rightarrow x = 45^\circ $
Hence, the correct answer is option $(d).$

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MCQ 191 Mark

In Fig. PR is a straight line and $\angle \text{PQS}:\angle \text{SQR}=7:5$ The measure of $\angle \text{SQR}$ is:

A
$60^{\circ}$
B
$62 \frac{1^{\circ}}{2}$
C
$67 \frac{1^{\circ}}{2}$
✓
$75^{\circ}$

Answer

Correct option: D.

$75^{\circ}$

d. $75^{\circ}

Solution:

Let the measures of the angle $\angle \text{PQS}$ and $\angle \text{SQR}$ be 7x and 5x

Now, $\angle \text{PQS}+\angle \text{SQR}=180^\circ$[Linear pair angles]

$\Rightarrow 7\text{x}+5\text{x}=180^\circ$

$\Rightarrow 12\text{x}=180^\circ$

$\Rightarrow \text{x}=15^\circ$

$\therefore \angle \text{SQR}=5\text{x}=5\times 15^\circ$

$=75^\circ$

Hence, the correct answer is option (d).

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MCQ 201 Mark

In Fig. if $AB \| CD$, then $x =$

A
$32$
✓
$42$
C
$52$
D
$31$

Answer

Correct option: B.

$42$

Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
$\angle \text{CDP}+\text{Reflex}\angle \text{CDP}=360^\circ [$Complete angle$]$
$\therefore \text{CDP}+249^\circ=360^\circ$
$\Rightarrow \angle \text{CDP}=111^\circ$
Since,$ PQ \| AB$
$\therefore \angle \text{BAP}=\angle \text{APQ} [$Alternate angles$]$
$\Rightarrow \angle \text{BAP}=28^\circ$
Now, $\angle \text{APQ}+\angle \text{QPD}=\angle \text{APD}$
$\Rightarrow 28^\circ+\angle \text{QPD}=(2\text{x}+13)^\circ$
$\Rightarrow \angle \text{QPD}=(2\text{x}+13)^\circ-28^\circ$
Since, $PQ \| CD$
$\therefore \angle \text{QPD}+\angle \text{CDP}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow (2\text{x}+13)^\circ-286\circ+111^\circ=180^\circ$
$\Rightarrow 2\text{x}+13-28+111=180$
$\Rightarrow 2\text{x}=84$
$\Rightarrow \text{x}=42$
Hence, the correct answer is option $(b).$

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MCQ 211 Mark

In Fig. AOB is a straight line such that $\angle \text{AOC}=(3\text{x}+10)^\circ. \angle \text{COD}=50^\circ$ and $\angle \text{BOD}=(\text{x}-8)^\circ.$ The value of $x$ is:

✓
$32$
B
$36$
C
$42$
D
$52$

Answer

Correct option: A.

$32$

$\angle \text{AOC}+\angle \text{COD}+\angle \text{BOD}=180^\circ [\text{AOB}$ is a straight line$]$
$\Rightarrow (3\text{x}+10)^\circ+50^\circ+(\text{x}-8)^\circ=180^\circ$
$\Rightarrow 3\text{x}+10+50+\text{x}-8=180$
$\Rightarrow 4\text{x}+52=180$
$\Rightarrow 4\text{x}=128$
$\Rightarrow \text{x}=32$
Hence, the correct answer is option $(a).$

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MCQ 221 Mark

In Fig. if $AB \| CD$ then the value of $x$ is:

A
$87$
B
$93$
✓
$147$
D
$141$

Answer

Correct option: C.

$147$

Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
$\angle \text{FCD}+\text{Reflex}\angle \text{FCD}=360^\circ ($Complere angle$)$
$\Rightarrow \angle \text{FCD}+273^\circ=360^\circ$
$\Rightarrow \angle \text{FCD}=87^\circ$
Since, $PQ \| CD$
$\therefore \angle \text{QFC}+\angle \text{FCD}=180^\circ ($Angles on the same side of a transversal line are supplementary$)$
$\Rightarrow \angle \text{QFC}+87^\circ=180^\circ$
$\Rightarrow \angle \text{QFC}=93^\circ$
Now, $\angle \text{ABF}=\angle \text{BFQ} ($Corresponding angles$)$
$=\angle \text{BFC}+\angle \text{QFC}$
$=54^\circ+93^\circ$
$=147^\circ$
$\therefore \text{x}^\circ=147^\circ$
$\Rightarrow \text{x}=147$
Hence, the correct answer is option $(c).$

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MCQ 231 Mark

In Fig. $\text{AOB}$ is a straight line and the ray $\text{OC}$ stands on it. The value of $x$ is:

A
$16$
✓
$26$
C
$36$
D
$46$

Answer

Correct option: B.

$26$

$\angle\text{AOC}+\angle\text{BOC}=180^\circ [\because$ Linear pair angles$]$
$\Rightarrow (2\text{x}+15)^\circ+(3\text{x}+35)^\circ=180^\circ$
$\Rightarrow (5\text{x}+50)^\circ=180^\circ$
$\Rightarrow 5\text{x}+50=180$
$\Rightarrow 5\text{x}=130$
$\Rightarrow \text{x}=26$
Hence, the correct answer is option $(b).$

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MCQ 241 Mark

In Fig. $AB \| CO$ and $EF$ is a transversal intersecting $AB$ and $CD$ at $P$ and $Q$ respective. The measure of $\angle \text{OOP}$ is:

A
$65$
B
$25$
✓
$115$
D
$105$

Answer

Correct option: C.

$115$

$\angle \text{BPE}=\angle \text{APQ}=(5\text{x}-10)^\circ [$Vertically opposite angles$]$
Since, $AB \| CD$
$\therefore \angle \text{APQ}+ \angle \text{CQP}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow(5\text{x}-10)^\circ+(3\text{x}-10)^\circ=180^\circ$
$\Rightarrow 8\text{x}-20=180$
$\Rightarrow 8\text{x}=200$
$\Rightarrow \text{x}=25$
$\therefore \angle \text{BPE}=(5\times 25-10)^\circ=115^\circ$
Now, $\angle \text{BPE}=\angle \text{DQP}=115^\circ [$Corresponding angles$]$
Hence, the correct answer is option $(c).$

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MCQ 251 Mark

In Fig. $\text{AOB}$ is a straight line and $4x = 5y.$ The value of $x$ is:

✓
$100$
B
$105$
C
$110$
D
$115$

Answer

Correct option: A.

$100$

$\angle \text{AOC}+\angle\text{BOC}=180^\circ [\because$ Linear pair angles$]$
$\Rightarrow \text{y}^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{y}+\text{x}=180$
$\Rightarrow \frac{4\text{x}}{5}+\text{x}=180$ $\big[\because 4\text{x}=5\text{y}\Rightarrow \text{y}=\frac{4\text{x}}{5}\big]$
$\Rightarrow 4\text{x}+5\text{x}=180\times 5$
$\Rightarrow 9\text{x}=180\times 5$
$\Rightarrow \text{x}=100$
Hence, the correct answer is option $(a).$

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MCQ 261 Mark

$\angle \text{A}$ is an obtuse angle. The measure of $\angle \text{A}$ and twice its supplementary differ by $30^\circ $. Then $\angle \text{A}$ can be:

A
$150^\circ $
✓
$110^\circ $
C
$140^\circ $
D
$120^\circ $

Answer

Correct option: B.

$110^\circ $

Supplementary of $\angle \text{A}=180^\circ-\angle \text{A}$
Now,
$\angle \text{A}+30^\circ=2(180^\circ-\angle \text{A})$
$\Rightarrow \angle \text{A}+30^\circ=360^\circ-2\angle \text{A}$
$\Rightarrow 3\angle \text{A}=360^\circ-30^\circ$
$\Rightarrow 3\angle \text{A}=330^\circ$
$\Rightarrow \angle \text{A}=110^\circ$
Hence, the correct answer is option $(b).$

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MCQ 271 Mark

In Fig. if $AB \| CD$ then the value of $x$ is:

✓
$34$
B
$124$
C
$24$
D
$158$

Answer

Correct option: A.

$34$

Construction: Draw a line $PQ$ parallel to $AB$ which is also parallel to $CD$
$\angle \text{QFC}+\angle \text{ECD}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow \angle \text{QEC}+56^\circ=180^\circ$
$\Rightarrow \angle \text{QEC}=124^\circ$
Now, $\angle \text{BEQ}+\angle \text{QEC}=\angle \text{BEC}$
$\Rightarrow \angle \text{BEQ}+124^\circ=158^\circ$
$\Rightarrow\angle \text{BEQ}=34^\circ$
Now, $\angle \text{ABE}=\angle \text{BEQ}=34^\circ [$Corresponding angles$]$
$\therefore \text{x}^\circ=34^\circ$
$\Rightarrow \text{x}=34$
Hence, the correct answer is option $(a).$

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MCQ 281 Mark

In fig. if $\angle \text{AOC}$ is a straight line, then the value of $x$ is:

A
$15$
✓
$18$
C
$20$
D
$16$

Answer

Correct option: B.

$18$

$\angle\text{AOD}+\angle\text{DOB}+\angle \text{BOC}=180^\circ [\text{AOC}$ is a straight line$]$
$\Rightarrow 2\text{x}^\circ+90^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 5\text{x}^\circ+90^\circ=180^\circ$
$\Rightarrow 5\text{x}=90$
$\Rightarrow \text{x}=18$
Hence, the correct answer is option $(b).$

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MCQ 291 Mark

In Fig. if $\text{AB}$ is parallel to $\text{CO}$ and $\text{EF}$ is a transversal, then $x =$

A
$19$
✓
$29$
C
$39$
D
$49$

Answer

Correct option: B.

$29$

Let the line $EF$ intersect $AB$ and $CD$ at $P$ and $Q$ respectively.

Since, $AB \| CD$
$\therefore \angle \text{BPQ}+ \angle \text{PQD}=180^\circ ($Angles on the same side of a transversal line are supplementary$)$
$\Rightarrow (7\text{x}-12)^\circ+(4\text{x}+17)^\circ=180^\circ$
$\Rightarrow 7\text{x}-12+4\text{x}+17=180$
$\Rightarrow 11\text{x}+5=180$
$\Rightarrow 11\text{x}=175$
$\Rightarrow \text{x}=15.90$
Disclaimer: No option is correct.

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MCQ 301 Mark

he sum of an angle and half of its complementary angle is $75^{\circ}$. The measure of the angle is:

A
$40^{\circ}$
B
$50^{\circ}$
✓
$60^{\circ}$
D
$80^{\circ}$

Answer

Correct option: C.

$60^{\circ}$

Let the required angle be $x$
Now, complementnary of the required angle $=90^{\circ}- x$
Then,
$x+\frac{1}{2}\left(90^{\circ}-x\right)=75^{\circ}$
$\Rightarrow 2 x+90^{\circ}-x=150^{\circ}$
$\Rightarrow x=150-90^{\circ}$
$\Rightarrow x=60^{\circ}$
Hence, the correct answer is option $(c).$

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MCQ 311 Mark

In Fig. if $\text{AB, CD}$ and $\text{EF}$ are straight lines, then $x + y + z =$

A
$180$
✓
$203$
C
$213$
D
$134$

Answer

Correct option: B.

$203$

$\angle \text{BAE}+\angle \text{BAD}+\angle \text{BAF}=180^\circ [\text{EAF}$ is a straight line$]$
$\Rightarrow 3\text{x}^\circ+49^\circ+62^\circ=180^\circ$
$\Rightarrow 3\text{x}^\circ+111^\circ=180^\circ$
$\Rightarrow 3\text{x}^\circ=69^\circ$
$\Rightarrow 3\text{x}=69$
$\Rightarrow \text{x}=23$
Now, $\angle \text{CAF}+\angle \text{CAF}=180^\circ [\because \text{EAF}$ is a straight line$]$
$\Rightarrow \text{z}^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{z}+\text{y}=180$
Now, $x + y + z = 23 + 180$
$= 203$
Hence, the correct answer is option$(b)$.

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MCQ 321 Mark

In Fig. $AB \| CD \| EF, \angle \text{ABG}=110^\circ,\angle \text{GCO}=100^\circ$ and $\angle \text{BGC}=\text{x}^\circ.$ The value of $x$ is:

A
$35$
B
$50$
✓
$30$
D
$40$

Answer

Correct option: C.

$30$

Since, $AB \| EG$
$\therefore \angle \text{ABG}+\angle \text{EGB}=180^\circ ($Angles on the same side of a transversal line are supplementary$)$
$\Rightarrow 110^\circ+\angle \text{EGB}=180^\circ$
$\Rightarrow \angle \text{EGB}=70^\circ$
Again, $CD \| GF$
$\therefore \angle \text{DCG}+\angle \text{FGC}=180^\circ ($Angles on the same side of a transversal line are supplementary$)$
$\Rightarrow 100^\circ+\angle \text{FGC}=180^\circ$
$\Rightarrow \angle \text{FGC}=80^\circ$
Now, $\angle \text{EGB}+\angle \text{BGC}+\angle \text{FGC}=180^\circ$
$\Rightarrow 70^\circ+\text{x}^\circ+80^\circ=180^\circ$
$\Rightarrow 150^\circ+ \text{x}^\circ=180^\circ$
$\Rightarrow \text{x}^\circ=30^\circ$
$\Rightarrow \text{x}=30$
Hence, the correct answer is option $(c).$

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MCQ 331 Mark

In Fig. if $AC \| OF$ and $AB \| CE$, then

✓
$x = 145, y = 223$
B
$x = 135, y = 233$
C
$x = 223, y = 145$
D
$x = 233, y = 135$

Answer

Correct option: A.

$x = 145, y = 223$

Construction: Produce $FD$ towards $D$ to the point $M$
$\angle \text{DCA}+\text{Reflex}\angle \text{DCA}=360^\circ [$Complete angle$]$
$\therefore \angle \text{DCA}+(\text{y}+15)^\circ=360^\circ$
$\Rightarrow \angle \text{DCA}=345^\circ-\text{y}^\circ$
Now,
$\angle \text{MDC}=\angle \text{EDF}=58^\circ [$Vertically Opposite angles$]$
Since, MF || AC
$\therefore \angle \text{MDC}+\angle \text{QPD}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow 58^\circ+345^\circ-\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}=223$
$\therefore \angle\text{DCA}=345^\circ-223^\circ=122^\circ$
Again, $\angle \text{BAC}+\text{Reflex}\angle \text{BAC}=360^\circ [$Complete angle$]$
$\therefore \angle \text{BAC}+(2\text{x}+12)^\circ=360^\circ$
$\Rightarrow \angle \text{DCA}=348^\circ-(2\text{x})^\circ$
Since, $AB \| CD$
$\therefore \angle \text{DCA}+\angle \text{DCA}=180^\circ [$Angles on the same side of a transversal line are supplementary$]$
$\Rightarrow 348^\circ-(2\text{x})^\circ+122^\circ=180^\circ$
$\Rightarrow (2\text{x})^\circ=290^\circ$
$\Rightarrow\text{x}=145$
Hence, the correct answer is option $(a).$

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MCQ 341 Mark

Two complemntary angles are in the ratio $2 : 3$. The measure of the larger angle is:

A
$60^\circ$
✓
$54^\circ$
C
$66^\circ$
D
$48^\circ$

Answer

Correct option: B.

$54^\circ$

Let the angles be $2x$ and $3x$
Now, $2x + 3x = 90^\circ $
$\Rightarrow 5x = 90^\circ $
$\Rightarrow x = 18^\circ $
$\therefore$ Larger angle $= 3x = 3 \times 18^\circ = 54^\circ $
Hence, the correct answer is option $(b).$

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MCQ 351 Mark

In Fig. $PO \| RS$ and $\angle \text{PAB}=60^\circ$ and $\angle \text{ACS}=100^\circ.$ Then, $\angle \text{BAC}=$

✓
$40$
B
$60$
C
$80$
D
$50$

Answer

Correct option: A.

$40$

Since, $PQ \| RS$
$\therefore \angle \text{PAC}= \angle \text{ACS}=100^\circ [$Corresponding angles$]$
Now, $\angle \text{PAC}=100^\circ$
$\Rightarrow \angle \text{PAB}+\angle \text{BAC}=100^\circ$
$\Rightarrow 60^\circ+\angle \text{BAC}=100^\circ$
$\Rightarrow \angle \text{BAC}=40^\circ$
Hence, the correct answer is option $(a).$

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