Missing number series questions and answers, Missing Number Series Questions for SBI Clerk ~ Simplifying Reasoning, maths ncert class 12 solutions, logical reasoning questions with answers

Ten most important Missing number series questions and answers, Number Series Questions for SBI Clerk ,SBI PO with solution , for SSC GL , SSC CHSL , RRB NTPC and other competitive Exams have been discussed in this post. These types of series questions are asked in many other competitive exams like SI, CPO and various entrance exams.

Problem #1

This problem is a series problem where we have to find the value of question mark using all its previous terms by analysing the trend of all its terms whether they are in increasing order or decreasing order or in any other format.

Formula :-

Every term is the product of its two preceding terms.

1st number = 3
2nd number = 3

3rd number = (1st number )×( 2nd number ) = 3 × 3 = 9

4th number = ( 2nd number )×( 3rd number ) = 3 × 9 = 27

5th number = ( 3rd number )×( 4th number ) = 9 × 27 = 243

6th number = ( 4th number )×( 5th number ) = 243 × 27 = 6561

Option (4)6561 is correct option

Problem #2

This problem is a series problem where we have to find the value of question mark using all its previous terms means by analysing the trend of all its terms whether they are in increasing order or decreasing order. This reasoning problem can be solved by two method . we shall discuss both these methods one by one.

Method #1

35 -15 = 20 (Here difference of 1st and 2nd term is 20)

63 - 35 = 28 (Here difference of 2nd and 3rd term is 20 + 8 = 28, i.e. 8 more than previous difference)

99 - 63 = 36 (Here difference of 3rd and 4th term is 28 + 8 = 36, i.e. 8 more than previous difference)

143 - 99 = 44 (Here difference of 4th and 5th term is 36 + 8 = 44, i.e. 8 more than previous difference)

? - 143 = 52 (Here difference of 5th and 6th term must be 44 + 8 = 52, i.e. 8 more than previous difference)

⇒ ? = 52 + 143 = 195

Option (2)195 is correct option

Method #2

All these terms are one less than perfect squares of even numbers.

4² - 1 = 16 -1 = 15

6² - 1 = 36 -1 = 35

8² - 1 = 64 -1 = 63

10² - 1 = 100 -1 = 99

12² - 1 = 144 -1 = 143

14² - 1 = 196 -1 = 195

Option (2)195 is correct option

Problem #3

This reasoning problem is a series problem where we have to find the value of question mark using all its previous terms means by analysing the trend of all its terms whether they are in increasing order or decreasing order. The value of question mark can be found using the difference of two of its previous terms. And after analysing the difference so obtained we can find the value of question mark.

13 - 6 = 7

27 - 13 = 14

55 - 27 = 28

111 - 55 = 56

? - 111 = 112

⇒ ? = 112 + 111

⇒ ? = 223

Option (3)223 is correct option

Problem #4

1st number = 8

2nd number = (1st number × 1 ) + 1

( 8 × 1 ) + 1 = 8 + 1 = 9

3rd number = ( 2nd number × 2 ) + 2

( 9 × 2 ) + 2 = 18 + 2 = 20

4th number = ( 3rd number × 3 ) + 3

( 20 × 3 ) + 3 = 60 + 3 = 63

5th number = ( 4th number × 4 ) + 4

( 63 × 4 ) + 4 = 252 + 4 = 256

6th number = ( 5th number × 5 ) + 5

( 256 × 5 ) + 5 = 1280 + 5 = 1285

Option (2)1285 is correct option.

Problem #5

These numbers are in binary number system. So 1st of all convert these numbers to decimal number system.

11(binary number system) = 3 (decimal number system) .

100(binary number system) = 4 (decimal number system)

111(binary number system) = 5 (decimal number system)

Similarly

1000(binary number system) = 6 (decimal number system)

Option (2)1000 is correct option

Problem #6

(9 × 1) - 2 = 9 - 2 = 7 ( 2nd Number )
(7 × 2) - 3 = 14 - 3 = 11 ( 3rd Number )
(11 × 3) - 4 = 33 - 4 = 29 ( 4th Number )
(29 × 4) - 5 = 116 - 5 = 111 ( 5th Number )
(111 × 5) - 6 = 555 - 6 = 549 ( 6th Number ) = value of question mark

Option (4)549 is correct option

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Problem #7

57 - 56 = 1= 1²

56- 55 = 1 = 1³

55 - 51 = 4 = 2²

51 - 43 = 8 = 2³

43 - 34 = 9 = 3²

Similarly ? - 34 = 4² = 16

⇒? - 34 = 16

⇒? = 16 + 34

⇒? = 50

Option (2)50 is correct option

Problem #8

336 - 224 = 112

224 - 168 = 56

168 - 140 = 28

140 - 126 = 14

Similarly continuing in the same pattern

126 - ? = 7

⇒-? = 7 - 126

⇒? = -7 + 126

⇒? = 119

Option (3)119 is correct option.

Problem #9

Formula :- Sum of both digits of each numbers are same except one.

Sum of both the digits of 30 ⇒ 3 + 0 = 3

Sum of both the digits of 27 ⇒ 2 + 7 = 9

Sum of both the digits of 36 ⇒ 3 + 6 = 9

Sum of both the digits of 45 ⇒ 4 + 5 = 9

Sum of both the digits of 72 ⇒ 7 + 2 = 9

All the numbers except 30 have have it's digits sum equal to 9. Only the number 30 have it's digits sum equal to 3 . So 30 is the odd number and it must be out.

Option (2)30 is correct option

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Problem #10

Separating these given numbers to two series by picking odd number and even number position.

1st series

5 , 7 , 10 , 14 ,

Difference 2 , 3 , 4 , 5

2nd series

6 , 8 , 11 , ?

Difference 2, 3, 4

Hence ? = 15

Option (3)15 is correct option.

Your Queries

(By Jai Singh Nayak)

Missing number in series
1, 6, 26, _____ , 426

Formula

Next Term = (4 × Previous Term ) + 2.
1st Term = 1
2nd Term = (4 × 1st Term ) + 2
2nd Term = (4 × 1 ) + 2
= 4 + 2
= 6

3rd Term = (4 × 2nd Term ) + 2
3rd Term = (4 × 6 ) + 2
= 24 + 2
= 26
4th Term = (4 × 3rd Term ) + 2
4th Term = (4 × 26 ) + 2
= 104 + 2
= 106
5th Term = (4 × 4th Term ) + 2
5th Term = (4 × 106 ) + 2
= 424 + 2
= 426
6th Term = (4 × 5th Term ) + 2
6th Term = (4 × 426 ) + 2
= 1704 + 2
= 1706
7th Term = (4 × 6th Term ) + 2
7th Term = (4 × 1706 ) + 2
= 6824 + 2
= 6826.

By MD Kaif

Complete the series

1, 3, 4, 8, 15 , 27, ?

Formula

To find any term after 3rd Term = Sum of its previous three terms .

4th term = 1st term + 2nd term + 3rd term = 1 + 3 + 4 = 8

5th term = 2nd term + 3rd term + 4th term = 3 + 4 + 8 = 15

6th term = 3rd term + 4th term + 5th term = 4 + 8 + 15 = 27

7th term = 4th term + 5th term + 6th term = 8 + 15 + 27 = 50 ( the value of question mark ).

By Shavnam Sharma

Complete the series

343, 729, 1331, 2197, ?

Formula

Any Term = (n)³ ,where n is odd number greater than equal to 7

Any Term = (2n+1)³ ,where n is odd number greater than equal to 3.

That means every term has been expressed as the cube of some odd numbers.

1st Term = 343 = 7³

2nd Term = 729 = 9³

3rd Term = 1331 = 11³

4th Term = 2197 = 13³

5th Term = 15³ = 3375

So the value of question mark = 3375

Similarly if you want to find out the value of next terms then it will be

6th Term = 17³ = 4913

7th Term = 19³ = 6859

8th Term = 21³ = 9261

9th Term = 23³ = 12167

10th Term = 25³ = 15625

Ten most important Missing number series questions and answers, Number Series Questions for SBI Clerk ,SBI PO with solution for SSC GL , SSC CHSL , RRB NTPC and other competitive Exams were discussed in this post. comment your valuable suggestions regarding this post and for further improvement.