IBM Number Series Questions | IBM Recruitments Questions

IBM Number Series Questions | Number Series Questions for IBM Recruitments

The IBM Written Exam consists of two sections: Quantitative Ability and Reasoning Skills.  Find the complete  IBM Recruitment Process, here.  Below you will find IBM Number Series Questions based on previous IBM Number Series Questions asked. Read on to discover the topics of the questions commonly asked in IBM Number Series Questions for freshers. 

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IBM Number Series

  • A total of 18 questions to be solved in 38 minutes.
  • IBM Test Pattern of Number Series is designed to be highly logical and tricky.
  • Every question will consist of different difficulty and logic.
  • Do not to spent much time on trickier questions, try to solve more number of easier questions first.

To know more about IBM Aptitude Syllabus, click here.

Types of Number Series

Perfect Square
A series based on perfect squares is mostly based on perfect squares of numbers in a specific order and usually one of the numbers is missing in this type of series.

Example:512, 729, 1000,?
Sol:83, 93, 103, 113
 

Arithmetic Series
This is a series in which the following term is obtained by adding / subtracting a constant number from its previous term

Example: 1, 2, 3, 4, ?
Ans . 5
 

Geometric Series
It is based on either descending or ascending order of numbers and each successive number is obtained by dividing or multiplying the previous number by a specific number.

Example: 4, 36, 324, 2916?
Sol: 4 x 9 = 36, 36 x 9 = 324, 324 x 9 = 2916, 2916 x 9 = 26244.
 

Two-Step Arithmetic Series
In a two-step arithmetic series, the differences in consecutive numbers themselves form an arithmetic series.

Example: 1, 3, 6, 10, 15…..
Sol:3 – 1 = 2, 6 – 3 = 3, 10 – 6 = 4, 15 – 10 = 5….
Now, we get an arithmetic sequence 2, 3, 4, 5

Arithmetic-Geometric Series
As the name suggests, the Arithmetic-Geometric series is formed of a particular combination of Arithmetic and Geometric series. An important property of the Arithmetic-Geometric series is that the differences between the consecutive terms are expressed in a geometric sequence.

Example:1, 4, 8, 11, 22, 25, ?
Sol :Series Type +3 , X2 ( i.e Arithmetic and Geometric Mixing)
1 + 3 = 4, 4 X 2 = 8, 8 + 3 = 11, 11 X 2 = 22, 22 + 3 = 25, 25 X 2 = 50

Twin/Alternate Series 
As the name of the series indicates, this type of series can consist of two series combined into a single series. Alternate terms in this series may constitute an independent series in itself.

Example: 3, 4, 8, 10, 13, 16 ? ?
Sol: As we can see, there are two series formed
Series 1: 3, 8, 13 with a common difference of 5
Series 2: 4, 10, 16 with a common difference of 6
So, the next two terms of the series should be 18 & 22 respectively.

IBM Number Series Questions

Try IBM’s free Number Series Questions. The questions will be based on the logic building, general aptitude and mind puzzles. IBM Number Series Questions should be attempted laying emphasis on getting shortlisted by IBM

(1) 1, 3, 4, 5, 13 , 2 , 3 , 4 , 22, 1, 2, 3, ? 
Solution – > 1 + 3 + 4 + 5 = 13, 13 + 2 + 3 + 4 = 22, 22 + 1 + 2 + 3 = 28 

(2) 48 , 24 , 35 , 7, 16, 8, 75, 15, 80, ? 
Solution – > 48/2 = 24    35/5 = 7 
16/2 = 8    75/5 = 15 
80/2 = 40 

(3) 5, 10, 9, 3, 6, 5, 4, 8, 7, 7, ? 
Solution – > 5 x 2 = 10   10 – 1 = 9 
3 x 2 = 6     6 – 1 = 5 
4 x 2 = 8 8 – 1 = 7 
7 x 2 = 14 

(4) 7, 4, 6, 3, 4.5, 1.5, 2.25 , – 0.75, ? 
Solution – > 7 – 3 = 4; 4 x (3/2) = 6 ; 
6 – 3 = 3;  3 x (3/2) = 4.5; 
4.5 – 3 = 1.5;   1.5 x (3/2) = 2.25; 
2.25 – 3 = – 0.75; – 0.75 x (3/2) = – 1.125 

(5) 1/4, 1/4, 1/2, 3/2, 6, ? 
Solution – > 1/4 x 1 = 1/4, 1/4 x 2 = 1/2, 1/2 x 3 = 3/2, 3/2 x 4 = 6, 6 x 5 = 30 Ans 

(6) 49, 7, 98, 16, 4, 48, 9, 3, 36, 25, 5, ? 
Solution – > 7^2, 7, 49 x 2 
4^2, 4, 16 x 3 
3^2, 3, 9 x 4 
5^2, 5, 25 x 5 = 125 

(7) 507, 169, 248, 62, 36, 12, 168, 42, 168, ? 
Solution – > 169 x 3, 169, 62 x 4, 62, 
12 x 3, 12 42 x 4, 42 
56 x 3 = 168 So 56 is Ans 

(8) 1, 3, 1, 2.5, 5, 1, 4, 5, 4, 8, 7, ? 
Solution – > 1 + 3 + 1 = 5 /2 = 2.5 
5 + 1 + 4 = 10 /2 = 5 
4 + 8 + 7 = 19 /2 = 9.5 

(9) 13, 24, 36, 23, 34, 56, ? 
Solution – > 13, 23, 33 24, 34 36, 56 

(10) 7, 0, 1, 8, 5, 12, 9, 26, 3, 23, 2, ? 
Solution – > 28 

(11) 2, 3, 10, 15, 26, ? 
Solution – >1 x 1 + 1 = 2, 2 x 2 – 1 = 3, 3 x 3 + 1 = 10, 4 x 4 – 1 = 15, 5 x 5 + 1 = 26, 6 x 6 – 1 = 35 Ans 

(12) 0.03, 0.0018, 0.00108, 0.00648, 0.3888, ? 
Solution – >Multiply by 0.06, 0.6, 6, 60, 600 
Ans 233.28 

(13) 4, 11, 66, 74, 370, 379, ? 
Solution – > 4, 11, 11 x 6, 74, 74 x 5, 379, 379 x 4 = 1516 Ans 

(14) 7, 9, 11, 6, 11, 8, 5, 13, 5, 4, 15, ? 
Solution – > 3 series 7, 6, 5, 4 
9, 11, 13, 1511, 8, 5, 2(Ans) 

(15) 2, 1, 2, 3, 2, 9, 9, 0, 1, 1, 9, ? 
Solution – >2^1 = 2 3^2 = 9 9^0 = 1 1^9 = 1 

(16) 2, 7, 36, 4, 14, 225, 6, 21, ? 
Solution – > 2 + 7 = 9 (9 – 3)^2 = 36 
4 + 14 = 18 (18 – 3)^2 = 225 
6 + 21 = 27 (27 – 3)^2 = 576 

17) 7, 4, 6, 3, 4.5, 1.5, 2.25, – 0.75, ? 

Sol. – 1.125 
7 – 4 = 3 
4 x 3/2 = 6 
6 – 3 = 3 
3 x 3/2 = 4.5 
4.5 – 1.5 = 3 
1.5 x 3/2 = 2.25 
2.25 – ( – 0.75) = 3 
– 0.75 x 3/2 = – 1.125 

18) 11, 15, 23, , 29, 71, ? 

Sol. one solution may be like : 
11 x 2 + 1 = 23 
15 x 2 – 1 = 29 
23 x 3 + 2 = 71 
29 x 3 – 2 = 85 (ans)

19) 22, 99, 34, 21, 10, 31, 18, 104, 28, 13 ?

Sol. There should be 101 in place of 10 in the given series. 
Hence the series is like 22, 99, 34, 21, 101, 31, 18, 104, 28, 13, ? 
On dividing into three sub series: we will get three types of series which are described below, 
(22, 21, 18, 13), (99, 101, 104), (34, 31, 28) 
22 – 1 = 21, 99 + 2 = 101, 34 – 3 = 34 
21 – 3 = 18, 101 + 3 = 104, 31 – 3 = 28 
18 – 5 = 13, 104 + 4 = 108(Ans)

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