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# Maths Questions for DSSSB Exam : 31st July 2018 (Solutions)

Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB Exam & STET Exam.
Q1. If the three angles of a quadrilateral are 120°, 45° and 100°. Find the value of the fourth angle.
(a) 90°
(b) 50°
(c) 75°
(d) 95°
Q2. What is the sum of the measures of angles of a quadrilateral?
(a) 90°
(b) 360°
(c) 180°
(d) 95°
Q3. If two supplementary angles differ by 16°, then the value of the larger angle is
(a) 84°
(b) 96°
(c) 106°
(d) 98°
Q4. A polyhedron has 6 faces and 8 vertices. The number of its edges is
(a) 10
(b) 12
(c) 14
(d) 16
Q5. The sum of all interior angles of a convex polygon having 5 sides is
(a) 540°
(b) 450°
(c) 720°
(d) 900°
Q6. Angles of an isosceles triangle can be
(a) 30°, 60°, 90°
(b) 70°, 70°, 50°
(c) 50°, 50°, 60°
(d) 54°, 72°, 54°
Q7. Find the measure of each exterior angle of a regular pentagon.
(a) 54°
(b) 60°
(c) 72°
(d) 90°

Q9. All angles of a regular polygon are of
(a) equal measure
(b) equal length
(c) 60°
(d) 90°
Q10. The sum of all interior angles of a polygon with n sides is given by
(a) (n + 2) × 180°
(b) (n – 2) × 180°
(c) 180° ÷ (n + 2)
(d) 180° ÷ (n – 2)
Solutions
S1. Ans.(d)
Sol. Let the fourth angle be x.
∴ 120° + 45° + 100° + x = 360°
(Sum of angles of a quadrilateral is 360°)
⇒ 265° + x = 360°
⇒ x = 360° – 265° = 95°
S2. Ans.(b)
Sol. The sum of the measures of angles of a quadrilateral is 360°
S3. Ans.(d)
Sol. Let the larger angle be x.
∴ x – (180° – x) = 16°
⇒ x + x – 180° = 16°
⇒ 2x = 196°
⇒ x = 98°
∴ The larger angle is equal to 98°.
S4. Ans.(b)
Sol. For any polyhedron that does not intersect itself,
F + V – E = 2
Where,
F is the number of faces, V is the number of vertices and E is the number of edges.
Putting the given values in F + V – E = 2 we get
6 + 8 – E = 2
⇒ E = 6 + 8 – 2 = 12
S5. Ans.(a)
Sol. Sum of interior angles of a n-sided polygon = (n – 2) 180°
⇒ Sum of interior angles of a 5-sided polygon = (5 – 2) 180°
= 540°
S6. Ans.(d)
Sol. In an isosceles triangle, two angles are equal to each other and the sum of three angles of the triangle is 180°.
Only option (d) satisfies this condition, as 54° + 72° + 54° = 180°
S7. Ans.(c)
Sol. A pentagon has 5 sides,
n = 5
Each exterior angle (regular polygon)
= (360°)/n
= (360°)/5 = 72°
S8. Ans.(b)
Sol. ∠4 and ∠5 are two angles of one side of a straight line.
∴ ∠4 + 150° = 180°
⇒ ∠4 + 150° = 180°
⇒ ∠4 = 30°
∴ ∠2 = ∠3 = ∠4 = 30°
Now,
∠1, ∠2 and ∠3 are three angles of one side of a straight line.
∠1 + ∠2 + ∠3 = 180°
⇒ ∠1 + 30° + 30° = 180°
∴ ∠1 = 120°

S9. Ans.(a)
Sol. All angles of a regular polygon are of equal measure.

S10. Ans.(b)
Sol. Sum of all interior angles of a polygon with n sides = (n – 2) × 180°.