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# Maths Questions for DSSSB Exam :22nd June 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 & amp; STET Exams.

Q1. What is the value of 0.5% in decimals?
(a) 0.0005
(b) 0.005
(c) 0.05
(d) 0.5

Q2. The multiplication of the age of Rajeev, before 5 yr and after 9 yr is 15. What is the present age of Rajeev?
(a) 6 yr
(b) 7 yr
(c) 5 yr
(d) 8 yr

(a) 80°
(b) 100°
(c) 120°
(d) 140°

Q4. In a ∆ABC, if 3∠A = 4∠B = 6∠C, then the value of ∠A is equal to
(a) 60°
(b) 80°
(c) 30°
(d) 40°

(a) 30°
(b) 40°
(c) 60°
(d) 80°

Q6. If the radius of the circle is increased 3 times, then how many times the new circumference increase from its original size?
(a) 3
(b) 1/3
(c) 9
(d) None of these

Q7. A room is 12 m long, 9 m wide and 8 m high. What is maximum length of a rod that can be kept in the room?
(a) 17 m
(b) 16 m
(c) 15 m
(d) 14 m

Directions (8-9): The students (in Lakh) who appeared and ratio of pass and fail in different cities are given in the following table. Read carefully and answer following questions.

Q8. How many applicants pass in City E?
(a) 13000
(b) 1110000
(c) 113000
(d) 111000

Q9. Which city has highest number of failures in exam?
(a) F
(b) C
(c) B
(d) D

Q10. Histogram is used to represent
(a) Non-continuous grouped frequency distribution
(b) Ungrouped frequency distribution
(c) Continuous grouped frequency distribution
(d) All of the above

Solution

S2. Ans.(a)
Sol. Let the present age of Rajeev = x
∴ (x – 5)(x + 9) = 15
⇒ x² + 9x – 5x – 45 = 15
⇒ x² + 4x – 60 = 0
⇒ x² + 10x – 6x – 60 = 0
⇒ x (x + 10) – 6x (x – 10) = 0
⇒ (x – 6) (x + 10) = 0
∴ x = 6

S3. Ans.(c)
Sol. Let a = 2x,  b = x
We know that,
2x + x = 180°
⇒ 3x = 180° ⇒ x = 60°
∴ a = 2x
= 2 × 60° = 120°

S5. Ans.(d)

Sol. ∵ ∠ACB = 40°
∴ ∠AOB = 2 × ∠ACB = 2 × 40° = 80°

S6. Ans.(a)
Sol. If radius will be increased by 3 times, then circumference will also increased by 3 times.

S8. Ans.(d)
Sol. Number of applicants pass in City E
=3/5 × 1.85 = 111000

S10. Ans.(c)