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Test: Lines & Angles - 3 - GMAT MCQ


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10 Questions MCQ Test Quantitative for GMAT - Test: Lines & Angles - 3

Test: Lines & Angles - 3 for GMAT 2024 is part of Quantitative for GMAT preparation. The Test: Lines & Angles - 3 questions and answers have been prepared according to the GMAT exam syllabus.The Test: Lines & Angles - 3 MCQs are made for GMAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Lines & Angles - 3 below.
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Test: Lines & Angles - 3 - Question 1

An angle which is greater than 180° but less than 360° is called :

Detailed Solution for Test: Lines & Angles - 3 - Question 1

An angle which is greater than 180° but less than 360° is called a reflex angle.

Test: Lines & Angles - 3 - Question 2

The complement of 72° 40’ is :

Detailed Solution for Test: Lines & Angles - 3 - Question 2

Complement of 72° 40’ is 90° ‒ (72° 40’)
= (89° 60’) ‒ (72° 40’) {since 1° = 60’}
= 17° 20’

Test: Lines & Angles - 3 - Question 3

The supplement of 154° 30’ is :

Detailed Solution for Test: Lines & Angles - 3 - Question 3

Supplement of 154° 30’ is 180° ‒ (154° 30’)

= (179° 30’) ‒ (154° 30’) {since 1° = 60’}

= 25° 30’.

Test: Lines & Angles - 3 - Question 4

Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC is:

Detailed Solution for Test: Lines & Angles - 3 - Question 4

As given ∠BOD = 63° 
Since COD is a straight line, we have:

∠BOC + ∠BOD = 180°. So, ∠BOC = (180° ‒ 63°) = 117°.

Test: Lines & Angles - 3 - Question 5

The straight lines AD and BC intersect one another at the point O. If ∠AOB + ∠BOD + ∠DOC = 274°, then ∠AOC is :

Detailed Solution for Test: Lines & Angles - 3 - Question 5

As we know that the sum of all the angles around a point is 360°.

(∠AOB + ∠BOD + ∠DOC) + ∠AOC = 360°
∴ 274° + ∠AOC = 360° or ∠AOC = 86°.
Hence, option A is correct.

Test: Lines & Angles - 3 - Question 6

In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 85°, then ∠COD is:

Detailed Solution for Test: Lines & Angles - 3 - Question 6

Clearly,
∠AOC + ∠COD + ∠BOD = 180°
∴ 85° + ∠COD = 180°. So, ∠COD = (180° ‒ 85°) = 95°.
Hence, option C is correct.

Test: Lines & Angles - 3 - Question 7

In the given figure, if AOB is a straight line, then the value of x is:

Detailed Solution for Test: Lines & Angles - 3 - Question 7

As, (x + 30°) + 45° + (x + 15°) = 180°
⇒ x = 45°.

Test: Lines & Angles - 3 - Question 8

In the given figure, the value of x, that would make POQ a straight line, is :

Detailed Solution for Test: Lines & Angles - 3 - Question 8

POQ will be a straight line,
If 80° + 66° + x = 180°, i.e. x = 34°.

Test: Lines & Angles - 3 - Question 9

If two angles are complementary of each other, then each angle is :

Detailed Solution for Test: Lines & Angles - 3 - Question 9

If two angles are complementary, then clearly each angle is less than 90° and is therefore an acute angle.

Test: Lines & Angles - 3 - Question 10

In the given figure, if AB || CD, then ∠FXE is equal to:

Detailed Solution for Test: Lines & Angles - 3 - Question 10

As per the given figure,
∠BFE = ∠CEF = 110° (alt. ∠s).
So, ∠XFE = ∠BFE ‒ ∠BFX = (110° ‒ 50°) = 60°.
And on straight line CD,
110° + ∠FEX + 30° = 180° ⇒ ∠FEX = 40°.
Now, ∠XFE + ∠FEX + ∠FXE = 180° ⇒ 60° + 40° + ∠FXE = 180°.
∴ ∠FXE = 80°.
Hence, option D is correct.

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