Alphabet series............[competitive. exam prep.... topic 22]

Alphabet series tricks....

Introduction :Almost every test on reasoning contains questions on alphabetical series. In such a question, if it consists of a single series of alphabets/combination, the alphabets/combinations are arranged in a particular manner and each alphabet/combination is related to the  earlier and the following alphabets in a particular way. The examinees is supposed to decode the logic involved in the sequence and then fill in the space containing the question mark with a suitable choice out of those given. But before we proceed to discuss the various types of questions related to alphabetical series, we will talk of some basic facts which are essential to an understanding of these types of questions.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

I. THE ALPHABET: The normal English alphabet contains 26 letters in all, as shown above

(Usually, questions on alphabet are accompanied by this normal alphabet). From A to M, the alphabet completes its first half, while the other half starts from N and ends at Z.

A-M – 1-13 (First Alphabetical Half)

N-Z –14-26 (Second Alphabetical Half)

II. EJOTY: For purpose of convenience, it is helpful to remember this simple formula called EJOTY, with the help of which you can easily find the position of any letter without much effort. But for practical purposes, you need to learn by heart the positions of different letters in the alphabet.

E   J    O    T    Y

5     10   15   20  25

Now, for instance, we wish to find the position of, say, the 17th letter from the left side. You already know that the 15th letter from the left side is O, therefore, the only thing you have to do is to find a letter which is two positions ahead of O, which is Q (The Answer). Using this simple formula, you can quickly find the position of any letter from the left side without much brain-rattling. Remembering the positions of different alphabets is basic to solving any question on alphabetical series. One of the best ways to achieve it is to practice EJOTY. Simply write down the full names of any 200 people you can imagine and do as follows:

For example, let’s say the name of the person imagined is ZUBINA. Now from EJOTY, we know that Z stands for 26, U stands for 21, B stands for 2, I stands for 9, N stands for 14 and A stands for 1. Now add up all these positions (26+21+2+9+14+1). What you get on addition does not have any significance, but it can be a very good way to try to make out and remember the individual positions of letters in the alphabet.

III. FINDING POSITIONS: Much more commonly, you get questions in the tests which provide you alphabetical positions from the right side. Since we are used to counting from the left side i.e. A, B, C… and not Z, Y, X…, the formula we discussed earlier will be applicable with a bit of modification.  But before we proceed to discuss it, it is essential to remember one simple mathematical fact.

*Let’s say there is a row of 7 boys in which a boy is standing 3rd from left. We want to know his position from the right side.

 I       I       I        I        I       I       I

                        1st     2nd    3rd    4th    5th    6th    7th  

You can see for yourself that the boy who was 3rd from the left is placed 5th from the right side.

The sum of both the positions is 8 (3+5), while the total number of boys is 7. This happens because we are counting a single boy twice in the calculation process. If we had subtracted 3 from 7 (as some of us might do), we would have got 4, which is obviously not the correct position from the right side. An important conclusion emerges from this discussion. If we are dealing with an alphabet and we have been given the position of any letter from either side, we will add 1 to the total no. of letters and then subtract its position from one side to get its position from the other side. For example, let’s find the position from the right of a letter which is the 9th from the left side.

    A   B    C   D   E    F    G    H    I    J   K     L  M    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    14    15    16    17   18   19   20  21  22    23   24   25   26

    26  25    24  23  22  21   20   19  18 17  16    15  14    13    12     11    10    9    8    7    6     5     4    3     2    1

As you can see for yourself, the 9th letter from the left side, I, comes out to be the 18th from the right side. Their sum (9+18=27) is again one more than the total number of alphabets i.e. 26. We can do this operation easily by adding one to the total number of letters (26+1=27) and then subtracting 9 from it. It gives you the letter position 18th from the right, which you can verify yourself from the above alphabet. The same procedure will be applicable if we are given an initial right position and are supposed to find it from the left side. Take for example, a letter which is placed 11th from the right side. If we want to locate its position from the left side, we will add 1 to total no. of letters and then subtract the right position from it to get its position from the left side. 27 – 11 gives you 16. Using EJOTY, you can easily conclude that the letter is P (16th from left, 11th from right)

The same logic is applicable if we are dealing with a situation in which the position of an item from the top is given to us and we want to find it from the bottom side or vice-versa.

IV. Still another type of question concerns finding the midpoint between two letters in the alphabet. For instance, let’s talk of a case which requires us to find the mid-point between the 11th letter and the 17th letter from the left side.

 A   B    C   D   E    F    G    H   I     J   K      L   M    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    14    15    16    17   18   19   20  21  22    23   24   25   26

26    25  24  23  22  21   20    19 18   17  16     15   14   13    12     11    10    9      8   7     6     5     4    3   2    1

You can see that there are five letters between these two positions i.e.  L, M, N, O and P. Obviously, the midpoint of 5 items is the third item from either side, whether counted from the left or the right. It comes out to be N, which is the correct answer. But frankly speaking, so much labour is not exactly required in solving such questions. Let’s let the cat out of the bag. In such questions, if the positions are given from the same side (i.e. either both are from left or both are from right), simply add up the two positions, get their average and you have the answer. In this case, the two positions are 11 and 17 from left. Adding them and averaging them gives you 14. Recollect the EJOTY formula and you immediately come up with the letter which is 14th from the left side (preceding O). The same procedure will be applicable if you are given a case in which both the positions are counted from the right side. Remember that the answer you get will be from the same sides which you have been given. Let’s make this thing clearer by taking a practical example.

*Consider a case in which we have to find the mid-point between the 13th and the 19th letter from the right side. Adding the two positions gives us 32, the average of which is 16. So we get the mid-point, which is 16th from the right side (the same as the sides given in the question). Now we have to convert this position into a position from the left.

Applying the logic discussed earlier, we subtract 16 from 27 and get 11th  from the left, which is obviously K. You can verify this answer by looking up the above alphabet. In fact, for such questions, one should have so much practice that one does not need to look up the alphabet, which proves to be time-consuming.

A   B    C   D   E    F    G    H     I      J    K     L    M    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    14    15    16    17   18   19   20  21  22    23   24   25   26

26 25     24  23  22  21    20  19    18   17  16    15   14    13    12   11     10    9      8    7    6     5     4    3     2    1

Now let us consider the third case in which we have to find the mid-point between two alphabetical positions, one of which is given from the left and the other from the right.

*Take, for instance, a case in which we have to find the mid-point between the 6th position from the left side and the 11th position from the right side. The first thing we have to do is to convert the right position into a left position to make the data comparable in nature. Doing so gives us 16. Now add up 16 and 6 (because now both are from the same side), average them, apply EJOTY and you get the correct answer.

V. REVERSING: Many questions concerning reversing of the alphabet are a part of reasoning tests.

Alphabet series tricks :

Introduction :

A  B  C  D  E

1  2  3  4  5

F  G  H  I

6  7  8  9

J  K  L  M  N

10  11  12  13  14

O  P  Q  R

15  16  17  18

S  T  U  V

19  20  21  22

W  X  Y  Z

23  24  25  26

Just mug up all the numbers related with alphabet like when i say K you must quikly count it as 11. So you must byheard every number associated with the alphabet .So I will explain you with below example of alphabet series

AD::GJ::MP:: ?

So now we have to solve the question part. So if you have mug up the numbers you can replace the alphabets as

AD :: GJ :: MP :: ?

1,4 :: 7,10 :: 13,16 ::

So now we have to only play with numbers like there is difference between 4 & 7 i.e is 3 again if you see 10 & 13 its 3.So now we know that we have to add 3 in 16 so its 19 and 19 is “S”.

AD :: GJ :: MP :: S

1,4 :: 7,10 :: 13,16 :: 19

Now we need to find second number again if you see between 1 & 4 the gap is 3 . So add in 19 so its 19 + 3= 22 and 22 is “V”.So now answer is “SV”

AD :: GJ :: MP :: SV

1,4 :: 7,10 :: 13,16 :: 19,22

TYPES ALPHABET SERIES COMPLETION IN REASONING ABILITY

TYPE-1 :- Alphabet series

a) Increasing by a definite number

e.g i) IJKL? ( each letter increases by 1)

ii) AGMSY? ( each letter increases by 6 place to its right position)

b) Decreasing by a definite number

e.g. i) ZXVTRP ? ( each letter decreases by 2 places to its left )

c) Increasing successively

e.g. DEGJNS? ( +1,+2,+3,+4,+5)

d) Decreasing successively

e.g.

i) ZYWTP ( -1,-2,-3,-4 ..)

ii) ZTOKHFE ( -6,-5,-4,-3,-2,-1)

e) Decreasing and Increasing by a constant value.

e.g. i) DFCEBDACZ (+2,-3,+2,-3,…)

TYPE-II :- ALPHANUMERIC SERIES

EX-1: Z1A, X2D,V6G,T21J,R88M, P445P,?

First letter: ZXVTRP (-2,-2,-2,…..)

Second letter: ADGJMP ( +3, +3,+3,…)

Series of numerals: 1,2,6,21,88,445 ( x1+1, x2+2, x3+3…)

So next term is N2676S.

EX.2:- 2Z5,7Y7,14X9,23W11,34V13,?

First numeral- 2,7,14,23,34 (+5,+7,+9,+11..)

Second letter- ZYXWV ( decreases by 1 each time)

Third numeral- 5,7,9,11,13 ( increases by 2 each time)

EX-3 :- W-144 , U-121, S-100, Q-81,?

First letter- decreases by 2 each time

Second numeral- square of 12,11,10,9,8..

Type-III :- Continuous patterns series

Ex-1 : ab_ _ baa_ _ ab_

options i) aaaaa ii) aabaa iii) caabab iv) baabb

solution: our answer is ii) . Here series aba is repeated

Ex-2 :ab_aa_bbb_aaa_bbba

options i) abba ii) baab iii) aabb iv) abab

Solution- our answer is ii) . The series is abb/aaabbb/aaaabbbb/a. Thus the letter are repeated twice , then thrice , then four times and so on .

Ex.3 – _bc_ca_aba_c_ca

Options i)abcbb ii)bbbcc iii)bacba iv)abbcc

Solutions- our answer is i) . The series is abc/bca/cab/abc/bca. Thus the letter change in cyclic order .

Ex.4- _c_bd_cbcda_a_db_a

Options i) adabcd ii) bdbcba iii) cdbbca iv)daabbc

Solutions- our answer is i). The series is acdb/dacb/cdab/acdb/da. Each group of four letters contains the letters of the previous group in the order – third , first , second and fourth.

Ex.5:- a_bb_baa_bbb_aa_

Options i) aabba ii) bbaab iii)abaaa iv)baabb

Solutions:- our answer is iii). The series is aabbbb/aaabbb/aaaa. At each step , the number of a’s increases by one while the number of b’s decrease by one.

Ex.6- _aba_cabc_dcba_bab _a

Options i) abdca ii) bcadc iii) abcdb iv) cbdaa

Solutions- Our answer is i) . The series is aababcabcd/dcbacbabaa. The letters equidistant from the beginning and the end of the series is same .

Ex.7- mnonopqopqrs_ _ _ _ _

Options- i) mnopq ii)oqrst iii)pqrst iv) qrstu

Solutions- our answer is iii) . The series is mno/nopq/opqrs

1. In the following alphabet series , one term missing as shown by question mark (?). Choose missing term from options.

U, O, I, ?, A

Options:

(A.) E (B.) C (C.) S (D.) G

Answer: A . E

Justification:

The series consists of vowels A, E, I, O, U written in a reverse order.

2. In the following alphabet series , one term missing as shown by question mark . Choose missing term from options.

A, C, F, H, ?, M

Options:

(A.) L (B.) K (C.) J (D.) I

Answer: B . K

Justification:

The letters are alternately moved two and three steps forward to obtain the successive terms.

3. In following alphabet series , one term missing as shown by question mark . Choose missing term from options.

W, V, T, S, Q, P, N, M, ?, ?

Options:

(A.) I, J (B.) J, I (C.) J, K (D.) K, J

Answer: D . K, J

Justification:

The letters are alternately moved one and two steps backward to obtain the successive terms.

4. In the following alphabet series , one term missing as shown by question mark . Choose missing term from options.Z, L, X, J, V, H, T, F, ?, ?

Options:

(A.) R, D (B.) R, E (C.) S, E (D.) Q, D

Answer: A . R, D

Justification:

The given sequence consists of two series — Z, X, V, T, ? and L, J, H, F, ?, both consisting of alternate letters in a reverse order.

5. In following alphabet series , one term missing as shown by question mark . Choose missing term from options.

AZ, GT, MN, ?, YB

Options:

(A.) KF (B.) RX (C.) SH (D.) TS

Answer: C . SH

Justification:

The first letter of each term is moved six steps forward while the second letter is moved six steps backward to obtain the corresponding letter of the next term.

6. In following alphabet series , one term missing as shown by question mark . Choose missing term from options.

AZ, CX, FU, ?

Options:

(A.) IR (B.) IV (C.) JQ (D.) KP

Answer: C . JQ

Justification:

The first letter of the first, second, third, …. terms are respectively moved two, three, four, … steps forward to obtain the first letter of the successive term.

The second letter of the first, second, third…. terms are respectively moved two, three, four, … steps backward to obtain the second letter of the successive terms.