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MATHS M.C.Q. [1 Marks Each]

Lines and Angles · Gujarat Board (GSEB) STD 7 (GSEB - English Medium). 70 questions.

Questions · Page 1 of 2

M.C.Q. [1 Marks Each]

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50 questions · timed · auto-graded

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,
$\text{x}+\frac{1}{3}(180^\circ-\text{x})=90^\circ$
$\Rightarrow 3\text{x}+180^\circ-\text{x}=270^\circ$
$\Rightarrow 2\text{x}=90^\circ$
$\Rightarrow \text{x}=45^\circ$
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 $AOB$ and $COD$ 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 $EF$ 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 $AOB$ and $COD$ 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 $=180^{\circ}-\mathrm{x}$
Then,
$x=2\left(180^{\circ}-\mathrm{x}\right)$
$\Rightarrow x=360^{\circ}-2 x$
$\Rightarrow 3 x=360^{\circ}$
$\Rightarrow x=120^{\circ}$
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 $P$ and $Q$ 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 $AOC$ is a straight line, then $x =​​​​​​$
  • A
    $42^\circ$
  • ✓
    $52^\circ$
  • C
    $142^\circ$
  • D
    $38^\circ$
Answer
Correct option: B.
$52^\circ$
$\angle \text{AOD}+\angle \text{DOB}+\angle \text{BOC}=180^\circ  [ \because$ $AOC$ 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 $AB, CD$ and $EF$ 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]
$\angle \text{AOC}+\angle \text{COF}+\angle \text{BOF}=180^\circ [AOB$ 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 $AB$ is parallel to $CD,$ 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.
$75^\circ$
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.
$32$
$\angle \text{AOC}+\angle \text{COD}+\angle \text{BOD}=180^\circ [AOB$ 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. $AOB$ is a straight line and the ray $OC$ 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. $AOB$ 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 [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 $AB$ is parallel to $CO$ and $EF$ is a transversal, then $x =$
  • A
    $19$
  • B
    $29$
  • C
    $39$
  • ✓
    None of these
Answer
Correct option: D.
None of these
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
The 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,
$\text{x}+\frac{1}{2}(90^\circ-\text{x})=75^\circ$
$\Rightarrow2\text{x}+90^\circ-\text{x}=150^\circ$
$\Rightarrow \text{x}=150-90^\circ$
$\Rightarrow \text{x}=60^\circ$
Hence, the correct answer is option $(c).$
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MCQ 311 Mark
In Fig. if $AB, CD$ and $EF$ are straight lines, then $x + y + z =$
  • A
    $180$
  • ✓
    $203$
  • C
    $213$
  • D
    $134$
Answer
Correct option: B.
$203$
$\angle \text{BAE}+\angle \text{BAD}+\angle \text{BAF}=180^\circ [EAF$ 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, $\angle \text{CAF}+\angle \text{CAF}=180^\circ[\because EAF$ 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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MCQ 361 Mark
In Fig. if $A C \| D F$ and $A B \| C E$, then
Image
  • ✓
    x =145 x y = 223
  • B
    x = 223 y = 145
  • C
    x = 135 y = 233
  • D
    x = 233 y = 135
Answer
Correct option: A.
x =145 x y = 223
A
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MCQ 421 Mark
In Fig. if AOB and COD are straight lines, then
Image
  • ✓
    X=29, Y=100
  • B
    x = 110, y = 29
  • C
    X=29, Y=130
  • D
    x =39 ,y = 110
Answer
Correct option: A.
X=29, Y=100
A
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MCQ 441 Mark
In Fig. $A B \mid C D, \angle O A B=150^{\circ}$ and $\angle O C D=120^{\circ}$. Then, $\angle A O C=$

Image
  • A
    $80^{\circ}$
  • ✓
    $90^{\circ}$
  • C
    $70^{\circ}$
  • D
    $100^{\circ}$
Answer
Correct option: B.
$90^{\circ}$
B
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MCQ 461 Mark
In Fig.  $P Q \mid R S$ and $\angle P A B=60^{\circ}$ and $\angle A C S=100^{\circ}$. Then, $\angle B A C=$
Image
  • A
    $40^{\circ}$
  • B
    $60^{\circ}$
  • ✓
    $80^{\circ}$
  • D
    $50^{\circ}$
Answer
Correct option: C.
$80^{\circ}$
C
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MCQ 471 Mark
In Fig. $A B\|C D\| E F, \angle A B G=110^{\circ}, \angle G C D=100^{\circ}$ and $\angle B G C=x^{\circ}$. The value of $x$ is
Image
  • A
    35
  • ✓
    50
  • C
    30
  • D
    40
Answer
Correct option: B.
50
B
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MCQ 491 Mark
In Fig., $A B \| C D$ and $E F$ is a transversal intersecting $A B$ and $C D$ at $P$ and $Q$ respectively. The measure of $\angle D Q P$ is
Image
  • A
    65
  • ✓
    25
  • C
    115
  • D
    105
Answer
Correct option: B.
25
B
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MCQ 501 Mark
In Fig. AB ||CD and EF is a transversal intersecting AB and CD at P and Q respectively. The measure of DPQ is
Image
  • ✓
    $100^{\circ}$
  • B
    $80^{\circ}$
  • C
    $110^{\circ}$
  • D
    $70^{\circ}$
Answer
Correct option: A.
$100^{\circ}$
A
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