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CBSE Class 9  >  Class 9 Test  >  Mathematics Olympiad  >  Math Olympiad Test: Lines and Angles- 2 - Class 9 MCQ

Math Olympiad Test: Lines and Angles- 2 - Free MCQ with solutions


MCQ Practice Test & Solutions: Math Olympiad Test: Lines and Angles- 2 (15 Questions)

You can prepare effectively for Class 9 Mathematics Olympiad for Class 9 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Math Olympiad Test: Lines and Angles- 2". These 15 questions have been designed by the experts with the latest curriculum of Class 9 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 15 minutes
  • - Number of Questions: 15

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Math Olympiad Test: Lines and Angles- 2 - Question 1
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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 larger angle is:

Detailed Solution: Question 1

Let the measure of angle be x.
∴ Its complement = (90° - x)
∴  2x = 3(90° - x)
⇒  5x = 3 × 90°
⇒ x = 54°
and  (90° - x ) = 36°
∴ Measure of larger angle = 54°

Math Olympiad Test: Lines and Angles- 2 - Question 2
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In the figure, OA and OB are opposite rays ∠AOC + ∠BOD = 63°. The measure of angle ∠COD is;

Detailed Solution: Question 2

∵ OA and OB are opposite rays.
∴ ∠AOB is a straight angle.
⇒ ∠AOB = 180°
⇒ (∠AOC + ∠BOD) + ∠COD = 180°
⇒ 63° + ∠COD = 180°
⇒ ∠COD = 117°

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Math Olympiad Test: Lines and Angles- 2 - Question 3
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Determine the value of x from the given figure;

Detailed Solution: Question 3

∠AOB = 180°
⇒ ∠AOP + ∠POB = 180°
⇒ ∠AOP + ∠POQ + ∠BOQ = 180°
⇒ 90° + 3x + x = 180°
⇒  4x = 90° 
⇒ x = 90°/4 = 22.5°

Math Olympiad Test: Lines and Angles- 2 - Question 4
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b = a + 20°, then a =

Detailed Solution: Question 4

∵ Sum of angles around a point = 360°
∠AOB + ∠BOC = 360°
⇒ 90° + a + b = 360°
⇒ a + b = 270° ...(i)  and 
b = a + 20° ...(ii)
Using (ii) in (i)
a + (a + 20°) = 270°
⇒ 2a = 250°
⇒ a = 125°

Math Olympiad Test: Lines and Angles- 2 - Question 5
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The value of m is

Detailed Solution: Question 5

∠m = ∠x  [Vertically opposite ∠s]
∵ ∠AOB = 180°
⇒ ∠BOF + ∠COF + ∠AOC = 180°
⇒ ∠BOF + ∠DOE +∠AOC = 180°
⇒ x° + 2x° + 3x° = 180°
⇒  6x° = 180°
⇒ x = 30° 
∴ m = 30°

Math Olympiad Test: Lines and Angles- 2 - Question 6
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Find x from the figure.

Detailed Solution: Question 6

Here ∠COQ = ∠POD   [vertically opposite ∠s]
∵ ∠AOB = 180°   (AOB is a straight line)
⇒ ∠POA + ∠POD + ∠BOD = 180°
⇒ 2x° + 3x° + 20° + 3x° = 180°
⇒ 8x = 160°
⇒ x = 20°

Math Olympiad Test: Lines and Angles- 2 - Question 7
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x = 3y = 6/7z, then, find the value of y.

Detailed Solution: Question 7

x = 3y, z = 21/6 y = 7/2y.
∴ x + y + z = 180°
⇒ 3y + y + 7/2 y = 180° 
⇒ 4y + 7/2 y = 180°
⇒ 15y = 180° × 2
⇒ y = 24°

Math Olympiad Test: Lines and Angles- 2 - Question 8
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Find x from the adjoining figure.

Detailed Solution: Question 8

∠AOD = ∠BOC [vertically opposite ∠s]
⇒ y = 60° + x   ...(i) and,
∵ ∠DOC = 180°
⇒ 60° + x + 100° = 180°
⇒ x = 20°

Math Olympiad Test: Lines and Angles- 2 - Question 9
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In the figure, l ∥ m, Find k.

Detailed Solution: Question 9

Since l ∥ m, the corresponding angles are supplementary.

So, 105° + 5x = 180°

5x = 180° - 105° = 75°

x = 75° ÷ 5 = 15°

Therefore, the value of x is 15°.

Math Olympiad Test: Lines and Angles- 2 - Question 10
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AB || CD, and ∠RQB = 115°, and ∠PRQ = 30°. The measure of ∠APC is:

Detailed Solution: Question 10

Here ∠RQB + ∠RQP =180°   (∵ AB is a straight line)
⇒ ∠RQP = 180° - 115° = 65°
Now ∠PRQ = 30°
∵ ∠PRQ, ∠RQP and ∠APQ are the ∠S of D
∴ ∠PRQ + ∠RQP + ∠APQ = 180°
⇒ ∠APQ = 180°- 65°- 30° = 85°
∠APC = ∠APQ [Vertically opposite ∠S]
∴ ∠APC = 85°

Math Olympiad Test: Lines and Angles- 2 - Question 11
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If x : y = 2 : 3, then the value of y is equal to:

Detailed Solution: Question 11

∵  X : Y = 2 : 3 

∴ Let the angles x and y be 2k and 3k respectively
∴ 2k + 3k = 180° 
[Sum of ∠S in the interior of transversal]
⇒ 5k = 180° ⇒ k = 36°
∴ y = 3k = 3 × 36° = 108°

Math Olympiad Test: Lines and Angles- 2 - Question 12
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AB || CD || EF and GH || KL. The measure of ∠HKL is

Detailed Solution: Question 12

The figure below is an extended version of the question figure, where we have added extra points and lines to make the angle calculation easier.

Extending GH to M, we have,

∠CHG = ∠APH = 60° [Corresponding ∠S]
∠APH = ∠MPD = 60°  [Vertically opposite ∠S]
∠APH = ∠MPD, then,
∠MPD + ∠LKP = 180° [Sum of interior ∠S]
⇒ ∠LKP = 180° - 60° = 120°, also
∠KHD = ∠PKH = 25° (Alternate ∠S)
∴ ∠HKL = ∠LKP + ∠PKH
= 120° + 25° = 145° 

Math Olympiad Test: Lines and Angles- 2 - Question 13
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Find x from the given figure (CP || DQ):

Detailed Solution: Question 13

Construct a line l || AB. ∠a = 25° [Alternate angles] ∠c = 105° ∴ ∠b = 105° [Vertically opposite angles] ∴ x = a + b = 25° + 105° = 130°

Math Olympiad Test: Lines and Angles- 2 - Question 14
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AB || CD and PQ || RS, then x - y =

Detailed Solution: Question 14

∠DD1S = ∠B1D1C1 = 2x   [Vertically opposite ∠S]
∠B1D1C1 + ∠A1B1D1 = 180° [Interior ∠S]
⇒  2x + 2y = 180°
⇒ x + y = 90° ...(i)
Also,
∠AA1C1 + ∠A1B1D1  [Corresponding ∠S]
⇒ 30° = 2y
⇒ y = 15° ...(ii)
Using (ii) in (i), we get
x = 90° - y = 90° - 15° = 75° 

Math Olympiad Test: Lines and Angles- 2 - Question 15
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The value of y, if AB || PQ is

Detailed Solution: Question 15

 From the figure,

(2y + y + y) + 35° = 180° (Interior ∠S)
⇒ 4y = 180° - 35°
⇒ 4y = 145°
⇒ y = 145° ÷ 4
⇒ y = 36.25°

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About this Class 9 Practice Test

This test contains 15 MCQs on Math Olympiad Test: Lines and Angles- 2 with detailed solutions for each question. It is part of the Mathematics Olympiad for Class 9 course on EduRev, designed to align with the current Class 9 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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