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# Maths Questions for DSSSB Exam : 11 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. The LCM of 20, 24, 30 and 32 would be
(a) 398
(b) 410
(c) 450
(d) 480

Q2. The HCF of 12, 15 and 27 will be
(a) 2
(b) 3
(c) 4
(d) 6

Q3. The HCF of 20, 24 and 48 will be
(a) 4
(b) 6
(c) 5
(d) 7

Q4. Find the HCF of 13, 78 and 117 by division method.
(a) 7
(b) 9
(c) 11
(d) 13

Q5. Find the HCF of 2 × 3³ × 5² × 7² and 3 × 5² × 7 × 11².
(a) 3 × 5 × 7
(b) 2 × 3 × 5 × 7 × 11
(c) 3 × 5² × 7
(d) 3³ × 5² × 7²

Q6. What is the HCF of 2a²b² and (ab)⁷ – 4a²b⁹?
(a) ab
(b) a²b³
(c) a²b²
(d) a³b²

Q7. The LCM of two numbers is 48. The numbers are in the ratio of 2:3. Find the sum of the numbers.
(a) 28
(b) 32
(c) 40
(d) 64

Q8. The HCF and LCM of two numbers are 8 and 48 respectively. If one of the number is 24, then the other number is
(a) 16
(b) 24
(c) 36
(d) 48

Q9. The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is
(a) 46
(b) 48
(c) 50
(d) 56

Q10. The largest number which divides 25, 73 and 97 to leave the same remainder in each case, is
(a) 6
(b) 21
(c) 23
(d) 24

Solutions

S5. Ans.(c)
Sol. Required HCF = 3 × 5² × 7

S6. Ans.(c)
Sol. We have  2a²b²
and (ab)⁷ – 4a²b⁹ = a²b⁷ (a⁵ – 4b²)
∴ Required HCF = a²b²

S7. Ans.(c)
Sol. Let the numbers are 2x and 3x, respectively.
Also,
48 = 2 × 2 × 2 × 2 × 3
= 2⁴ × 3 = 2 × 3 × 2³
∴ x = 8
Hence, the sum of numbers = 16 + 24 = 40

S8. Ans.(a)
Sol. We know that,
HCF × LCM = First number × Second number
⇒ 8 × 48 = 24 × Second number
⇒ Second number
=(8×48)/24
= 8 × 2 = 16

S9. Ans.(c)
Sol. LCM of 4, 6, 8, 12 and 16 = 48
∴ Required number = 48 + 2 = 50

S10. Ans.(d)
Sol. Let x be the remainder.
Then, (25 – x), (73 – x) and (97 – x) will be exactly divisible by the required number.
∴ Required number
= HCF of {(73 – x) – (25 – x)}, {(97 – x) – (73 – x)} and {(97 – x) – (25 – x)}
= HCF of (73 – 25), (97 – 73) and (97 – 25)
= HCF of 48, 24 and 72
= 24