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Practice Test: Number Series- 1


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10 Questions MCQ Test CSAT Preparation for UPSC CSE | Practice Test: Number Series- 1

Practice Test: Number Series- 1 for Banking Exams 2022 is part of CSAT Preparation for UPSC CSE preparation. The Practice Test: Number Series- 1 questions and answers have been prepared according to the Banking Exams exam syllabus.The Practice Test: Number Series- 1 MCQs are made for Banking Exams 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Number Series- 1 below.
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Practice Test: Number Series- 1 - Question 1

17, 19, 23, 29, 31, 37, _____

Detailed Solution for Practice Test: Number Series- 1 - Question 1

The given numbers are consecutive prime numbers in increasing order starting with 17.
Hence, the next number in the series is 41.

Practice Test: Number Series- 1 - Question 2

225, 196, 169, _____, 121, 100, 81

Detailed Solution for Practice Test: Number Series- 1 - Question 2

The given numbers are squares of consecutive natural numbers in decreasing order starting with 15,
i.e., the numbers 225, 196, 169, ____, 100, 81 can be written as 152, 142, 132 , ____, 112 , 102 , 92.

Hence, the missing number is 122 = 144.

Practice Test: Number Series- 1 - Question 3

54, 66, 82, 102, 126, _____

Detailed Solution for Practice Test: Number Series- 1 - Question 3

1st term = 54
2nd term = 54 + 12 = 66
3rd term = 66 + 16=82
4th term = 82 + 20 =102
5th term = 102 + 24 = 126...

The difference is increasing by 4, starting with 12.
So, the next difference is 24 + 4 = 28.

Hence, the next number is 126 + 28 = 154.

Practice Test: Number Series- 1 - Question 4

97, 83, 73, 67, 59, _____

Detailed Solution for Practice Test: Number Series- 1 - Question 4

List of Prime Numbers from 1 to 100:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,...

The given numbers are alternate prime numbers in decreasing order, starting with 97.

Hence, the next number in the series is 47.

Practice Test: Number Series- 1 - Question 5

8, 16, 48, 96, 288, 576, _____

Detailed Solution for Practice Test: Number Series- 1 - Question 5

1st term = 8
2nd term = 8 x 2 = 16
3rd term = 16 x 3 = 48
4th term = 48 x 2 = 96
5th term = 96 x 3 = 288
6th term = 288 x 2 = 576...

⇒ For every term at even position, the previous term is multiplied by 2.
⇒ For every term at odd position, the previous term is multiplied by 3.

So the missing term is: 576 × 3 = 1728

Practice Test: Number Series- 1 - Question 6

75, 291, 416, 480, 507, ____

Detailed Solution for Practice Test: Number Series- 1 - Question 6

1st term = 75
2nd term = 75 + 216 = 75 + 63 = 291
3rd term = 291 + 125 = 291 + 53 = 416
4th term = 416 + 64 = 416 + 43 = 480
5th term = 480 + 27 = 480 + 33 = 507...

The differences are cubes of consecutive natural numbers in decreasing order from 6.
So, the missing term = 507 + 8 = 507 + 23 = 515

Practice Test: Number Series- 1 - Question 7

225, 224, _____, 222, 221

Detailed Solution for Practice Test: Number Series- 1 - Question 7

The given numbers are consecutive natural numbers in decreasing order starting with 225. Hence, the missing number is 223.

Practice Test: Number Series- 1 - Question 8

5, 13, 41, 85, 257, _____

Detailed Solution for Practice Test: Number Series- 1 - Question 8

1st term = 5
2nd term = 5 × 2 + 3 = 13
3rd term = 13 × 3 + 2 = 41
4th term = 41 × 2 + 3 = 85 
5th term = 85 × 3 + 2 =257...

⇒ For every term at even position, the previous term is multiplied by 2 and 3 is added.
⇒ For every term at odd position, the previous term is multiplied by 3 and 2 is added.

So the missing term is: 257 × 2 + 3 = 517

Practice Test: Number Series- 1 - Question 9

123 , 129, 141, _________, 159, 165

Detailed Solution for Practice Test: Number Series- 1 - Question 9

The common difference between consecutive terms is 6, 12, 6, 12 & so on. So, the term next to 141 is 141 + 6 = 147

Practice Test: Number Series- 1 - Question 10

0, 5, 22, 57, 116, _________ 

Detailed Solution for Practice Test: Number Series- 1 - Question 10

The 1st to 5th terms are 0, 5, 22, 57, 116.
These terms are 1 – 1, 8 – 3, 27 – 5, 64 – 7, 125 – 9.
i.e. 13 – 1, 23 – 3, 33 – 5, 43 – 7, 53 – 9

So, the nth term of the series: n3 – (2n – 1).

So, the 6th term of the series is 63 – (2 × 6 –1) = 216 – 11 = 205.

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