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Test: Lines & Angles - 2 - GMAT MCQ


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10 Questions MCQ Test Quantitative for GMAT - Test: Lines & Angles - 2

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Test: Lines & Angles - 2 - Question 1

In the given figure AB || CD, ∠A = 128°, ∠E = 144°. Then, ∠FCD is equal to :

Detailed Solution for Test: Lines & Angles - 2 - Question 1

As per the given figure,
Through E draw EE’ || AB || CD.
Then, ∠AEE’ = 180° ‒ ∠BAE = (180° ‒ 128°) = 52°.
(Interior angles on the same side of the transversal are supplementary.)
Now, ∠E’EC = (144° ‒ 52°) = 92°.
∠FCD = ∠E’EC = 92° (Corr. ∠s).
Hence, option D is correct.

Test: Lines & Angles - 2 - Question 2

In the trapezium PQRS, QR || PS, ∠Q = 90°, PQ = QR and ∠PRS = 20°. If ∠TSR = θ, then the value of θ is:

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

In the given figure, 
PQ = QR and ∠PQR = 90° ⇒ ∠QPR = ∠QRP = 45°.
∴ ∠QRS = (45° + 20°) = 65°.
∴ θ = ∠QRS = 65° (alt. ∠s)

Hence, option C is correct.

Test: Lines & Angles - 2 - Question 3

In the adjoining figure, ∠ABC = 100°, ∠EDC = 120° and AB || DE. Then, ∠BCD is equal to:

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

In the given figure,
Produce AB to meet CD at F.
∠BFD = ∠EDF = 120° (alt. ∠s)
∠BFC = (180° ‒ 120°) = 60°.
∠CBF = (180° ‒ 100°) = 80°.
∴ ∠BCF = 180° ‒ (60° + 80°) = 40°.
Hence, option C is correct.

Test: Lines & Angles - 2 - Question 4

In the given figure, AB || CD, ∠ABO = 40° and ∠CDO = 30°. If ∠DOB = x°, then the value of x is:

Detailed Solution for Test: Lines & Angles - 2 - Question 4

In the given figure,

Through O draw EOF parallel to AB & so to CD.
∴ ∠BOF = ∠ABO = 40° (alt. ∠s)
Similarly, ∠FOD = ∠CDO = 30° (alt. ∠s)
∴ ∠BOD = (40° + 30°) = 70°.
So, x = 70°.

Hence, option C is correct.

Test: Lines & Angles - 2 - Question 5

In the given figure, AB || CD, m∠ABF = 45° and m∠CFC = 110°. Then, m∠FDC is:

Detailed Solution for Test: Lines & Angles - 2 - Question 5

As in the given figure,
∠FCD = ∠FBA = 45° (alt. ∠s)
∴ ∠FDC = 180° ‒ (110° + 45°) = 25°.
Hence, option A is correct.

Test: Lines & Angles - 2 - Question 6

In the given figure, line CE is drawn parallel to DB. If ∠BAD = 110°, ∠ABD = 30°, ∠ADC = 75° and ∠BCD = 60°, then the value of x° is:

Detailed Solution for Test: Lines & Angles - 2 - Question 6

As in the given figure,
∠ADB = 180° ‒ (110° + 30°) = 40°.
So, ∠BDC = (75° ‒ 40°) = 35°.

∴ ∠DBC = 180° ‒ (60° + 35°) = 85°.

∴ ∠BCE = ∠DBC = 85° (alt. ∠s).

So, x = 85°.
Hence, option C is correct.

Test: Lines & Angles - 2 - Question 7

If two supplementary angles differ by 44°, then one of the angle is:

Detailed Solution for Test: Lines & Angles - 2 - Question 7

Let the two angles are x and y. Therefore, as per the given information,
x ‒ y = 44° and 
x + y = 180°
[ As the total of supplementary angles is 180°]
On solving these two linear equations we get,
2x = 224, 
x = 112°. 
Therefore the other angle y = 180° ‒ 112° = 68°
Hence, option D is correct.

Test: Lines & Angles - 2 - Question 8

Consider the following statements

If two straight lines intersect, then
I. vertically opposite angles are equal.
II. vertically opposite angles are supplementary.
III. adjacent angles are complementary.

Which of the statements given above is/are correct?

Detailed Solution for Test: Lines & Angles - 2 - Question 8

Here, AB and CD are two lines.

If two straight lines intersect, then opposite vertically angles are equal.
Hence, option B is correct.

Test: Lines & Angles - 2 - Question 9


In the figure given above LOM is a straight line. What is the value of x°?

Detailed Solution for Test: Lines & Angles - 2 - Question 9

From the given figure,
∠LOQ + ∠QOP + ∠POM = 180°
(straight line)
∴   (x° + 20°) + 50° + (x° – 10°) = 180°
⇒   2x° + 60° = 180° ⇒ 2x° = 120°
∴   x° = 60°
Hence, optuion B is correct.

Test: Lines & Angles - 2 - Question 10

In the figure given above, EC is parallel to AB, ∠ECD = 70° and ∠BDO = 20°. What is the value of ∠OBD?

Detailed Solution for Test: Lines & Angles - 2 - Question 10

Given that, EC || AB
∴    ∠ECO + ∠AOC = 180°

⇒   ∠AOC = 180° – 70° = 110°

∴    ∠BOD = ∠AOC = 110°

(alternate angle)

Now, in ΔOBD

∠BOD + ∠ODB + ∠DBO = 180°

∴   110° + 20° + x° = 180° ⇒ x° = 50°.

Hence, option D is correct.

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