Series Test

SERIES TEST

Introduction

A series is a sequence of letters or numbers obtained by some particular predefined rule(s) and applying that predefined rule(s) it is possible to find out the next term of the series.

Types of series:

  1. Letter series
  2. Number series
  3. Letter-Number series

Letter Series (with repeating letters)

In this type of questions group of letters, usually given in small letters, are repeated in a systematic way and thus a series is established. Some cases are presented below:

(i)       A sequence of letters in series may be formed by simply repeating the same group of letters, skipping one letter in turn, in cyclic order.

                                      _b_bca_a_ab_

1) abbca                          

2) babac

3) accbc                          

4) abbac

          If the letters of group (3) are placed in the blanks, respectively, we get a b c b c a b a b c. In cyclic order from a we have abc, from b we get bca and from c, cab.

(ii)      A series of letters may be formed on the basis of some rhythm, e.g.

                                      b_ba_b_b_a_b

1) aabba                          

2) abaab

3) abbab                          

4) bbabb

          If we put the letters of group (4) in the blanks, we get the series: abbabbabbabb- which produces a sort or rhythm.

(iii)     A sequence of letters may be formed by repeating the same group of letters, skipping one and repeating another, For example,

                                      ab_c_c_a_ab_a_cc

1) bbcca                          

2) aabcba

3) cbacbb                        

4) bacbbc

          If we put the letters of choice (3) in the gaps of the series, we obtain abccbcaacabbabcc. Here in the group of letters abcc, c is repeated. Next, a is dropped from abc. We start from b and the last letter in the cycle is repeated.

(iv)     The letter a, b, c, may be arranged in cyclic order to form a group and then repeated to form a series. For example:-

                                      a_c_b_ab_a_ca_c.

1) abaccb                        

2) accbab

3) aabbcc                        

4) baccbb

          If we put the letters of the fourth choice, ie baccbb, in the gaps then we get the series, a b c a b c a b c a b c a b c. Here a b c is repeated several times to form a series. Hence, the answer is (4).

(v)      A sequence may be formed by repeating some letters in a group. Consider

                                      _b_a_a_ba_b_

1) babaab                        

2) aabaab

3) aabaaa                        

4) abbaab

          When the letters of (3) are placed, respectively, in each gap, we get the series in which the letters aba are repeated as aba aba aba aba aba. Therefore, answer is (3).

Letter Series (for non-repeating letters)

          In another type of questions, a series of letters is given, usually in bold letters where the letters do not repeat, rather there is a definite rule on the basis of which successive letters are formed. For example, each next letter may be formed by skipping one letter or may be formed by skipping two letters or it may be formed by going two letters back.

          Though there is no definite rule or principle in completion of order of alphabets or letters in a series, yet each question carries a definite pattern or sequence. The candidates should try to find out that pattern and apply it in the functioning of a series. However, certain hints given below can be kept in mind.

          Keep in mind the order of letters with their respective numbers i.e., A-1 to Z-26 and vice versa Z-1 to A-26 as given below in the chart:

1

2

3

4

5

6

7

8

9

10

11

12

13

A

B

C

D

E

F

G

H

I

J

K

L

M

14

15

16

17

18

19

20

21

22

23

24

25

26

N

O

P

Q

R

S

T

U

V

W

X

Y

Z

 

1

2

3

4

5

6

7

8

9

10

11

12

13

Z

Y

X

W

V

U

T

S

R

Q

P

O

N

14

15

16

17

18

19

20

21

22

23

24

25

26

M

L

K

J

I

H

G

F

E

D

C

B

A

 

Note:

When a group of questions of the ‘letter series’ type are given, then write down the alphabet and number the letters as above. A few examples will show how this is helpful. Write the alphabet down on a piece of paper. Number it as shown above, draw the lines, and keep it in front of you.

Example:       A C F J O ?

Answer:         U

Look at letters A, C, F, J and O on the numbered alphabet. A (skip 1) C (skip 2) F (skip 3) J (skip 4) O.Therefore the next ‘skip’ should be 5 letters .Hence the answer is ‘U’.

Example:       A D H M S ?

Answer:         Z

The skipping pattern is 2, 3, 4, 5 and 6 letters.

Example:       D F I M R ?

Answer:         X

The skipping pattern is 1, 2, 3, 4 and 5 letters.

Example:       U, O, I, ?, A

Answer:         E.

The pattern consists of vowels a, e, i, o, u in reverse order.

Example:       Y, W, U, S, Q, ?, ?

Answer:         M, O.

The series consists of alternate letter in reverse order

Series having Group of Letters

Like the non-repeating series of letters in this type of series, the position of various letters plays an important role.

Direction: In each of the following questions various terms of a letter series are given with one term missing as shown by (?). Choose the missing term out of the given alternatives.

Example:       ak, fp, ?, pz, ue, zj

1) ku                     

2) jt            

3) ju

4) kv                     

5) uk

Answer:       1;

If we write the position numbers of each letter group we will get the following series:

(1, 11), (6, 16), ?, (16, 26), (21, 5), (26, 10).

We see that the first numbers of every pair form a definite series:

1, 6, ?, 16, 21, 26; where each successive number is obtained by adding 5 to previous number. So the first number in the unknown pair is 11. Consequently the second number of the pair would be 11 + 10 = 21. Hence we have (11, 21) which is equivalent to (k, u).

Short cut method: Look at fp and pz; pz and zj. We see that any letters coming at the second place comes at the first place after a gap as one word. Since ak has k as the second letter and since the gap between ak and ? is of one word, k must come to be the first letter here. Now look at the ? and ue. Since u has come at the first place, it must have been at the second place in the ?. Hence correct answer is ku.

Number Series

In mathematics you must have read various types of number series. For example, arithmetic series, geometric series, arithmetic-geometric series etc.

 

Tricks for solving number series are:

  1. Each number may be the multiple of the other.
  2. In each number something may be added or subtracted to get the second number and so on.
  3. The numbers may be divided or multiplied to get the next number.
  4. The square root or cube root of the numbers may be taken to get the next number.
  5. The number may be squared or cubed to get the next number.
  6. Sometimes odd numbers are followed or preceded by even numbers and vice versa.
  7. Sometimes one number is squared, the other multiplied thrice, fourth multiplied four times etc. and then something is added or subtracted.

Two different cases of number series are:

Case I: Completing the given series

Example:     Find the next number in the series: 1, 3, 6, 10, 15, __

Answer:         Looking at the series we find that the series is an "INCREASING" sequence.

So we try to find what is being added. Here the series is

1, 1+2, 1+2+3, 1+2+3+4

1+2 =3, 3+3 = 6, 6+4 = 10, 10+5 = 15

So the answer is 15+ 6 = 21.

Example:     Find out the missing number.

                             7, __, 15, 23, 38, 61

Answer:       In the given series 15 + 23 = 38, 23 + 38 = 61.

Therefore the missing number is 8 so that 7 + 8 = 15 and 8 + 15 = 23.

Case II: Finding the wrong term in the given series

Example:     Find the wrong number in the series.

                             3, 8, 15, 24, 34, 48, 63

Answer:       Here the difference between consecutive terms of the series is 5, 7, 9, 11, and 13 respectively. So the wrong number is 34.

Letter-Number Series

Letter-number series is the combinations of two series, letter series and number series.

Example:      Choose the term which will continue the following series:

                             P3C, R5F, T81, V12L, ?

1) Y17O        

2) X17M       

3) X17O       

4) X16O

Answer:       3; clearly, the first letters of the terms are alternate. The sequence followed by the number is +2, +3, +4, … . The last letter of each term is the steps ahead of the last letter of the preceding term. Thus, the next term would be X17O. Hence, the answer is (3).