Maths Questions for CTET Exam : 6th september 2018_00.1
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Maths Questions for CTET Exam : 6th september 2018

Maths Questions for CTET Exam : 6th september 2018_40.1

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 ,KVS,STET Exam.

Q1. If a=

Maths Questions for CTET Exam : 6th september 2018_50.1

, then the value of a is

(a) 2001
(b) 999
(c) 1001
(d) 1002

Q2. The value of

Maths Questions for CTET Exam : 6th september 2018_60.1

where a = 2, b = 3 and c = 0, is
(a) 2/5
(b) 1
(c) 0
(d) -1

Q3. In the standard form, the number 7345000 is expressed as   

Maths Questions for CTET Exam : 6th september 2018_70.1

 the value of k + n is
(a) 13.345
(b) 11.345
(c) 12.345
(d) 7.345

Q4. LCM of 22, 54, 135 and 198 is
(a) 2² × 3³ × 5 × 11
(b) 2 × 3³ × 5 × 11
(c) 2² × 3³ × 5 × 11
(d) 2³ × 3³ × 5 × 11

Q5. The LCM of two prime number is 
(a) sum of numbers
(b) difference of numbers
(c) quotient of numbers
(d) product of numbers

Maths Questions for CTET Exam : 6th september 2018_80.1

Q7. The LCM of two prime number x and y, (x > y), is 209. The value of 2y – x is
(a) 2
(b) –2
(c) –3
(d) 3

Q8. In 1999, the population of a country was 30.3 million. The number which is the same as 30.3 million is
(a) 303000000
(b) 30300000
(c) 3030000
(d) 3030000000

Q9. If a³ = 1 + 7, b³ = 1 + 26 and c³ = 1 + 63, where a, b and c are different positive integers, then a + b+ c is equal to
(a) a²
(b) a³
(c) b²
(d) b³

Q10. Which of the following is the least 3-digit number using the digits 1, 3 and 9?
(a) 139
(b) 931
(c) 319
(d) 391

Solutions
Maths Questions for CTET Exam : 6th september 2018_90.1

S3. Ans.(a)
Sol. 7345000 can be written in the standard form as
7.345 × 10⁶
Therefore, k = 7.345 and n = 6
Hence, k + n = 7.345 + 6
= 13.345

S4. Ans.(b)
Sol. We have,
22 = 2 × 11
54 = 2 × 3 × 3 × 3
135 = 3 × 3 × 3 × 5
198 = 2 × 3 × 3 × 11
Now, multiply each factor the highest number of times it occurs in either number. If the same factor occurs more than once in both numbers, multiply the factor that occurs the highest number of times.
Therefore, LCM of 22, 54, 135 and 198 = 2 × 3 × 3 × 3 × 5 × 11
= 2 × 3³ × 5 × 11

S5. Ans.(d)
Sol. The LCM of two prime numbers is their product.

Maths Questions for CTET Exam : 6th september 2018_100.1

S7. Ans.(d)
Sol. We know that the LCM of two prime numbers is the product of those numbers.
209 is the product of prime numbers 11 and 19.
Therefore, 209 is the LCM of 11 and 19.
We get x = 19 and y = 11
Now, 2y – x = 2 × 11 – 19
= 22 – 19 = 3

S8. Ans.(b)
Sol. 1 million = 1000000
∴ 30.3 million = 30.3 × 1000000
= 30300000

S9. Ans.(c)
Sol. We have,
a³ = 1 + 7
⇒ a³ = 8
⇒ a = 2
b³ = 1 + 26
⇒ b³ = 27
⇒ b = 3
c³ = 1 + 63
⇒ c³ = 64
⇒ c = 4
∴ a + b + c = 2 + 3 + 4
= 9 = 3² = b²

S10. Ans.(a)
Sol. Here, 1 < 3 < 9
Therefore, the least 3-digit number is 139.

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