The reasoning section of almost every competitive examination contains questions on Alphabet Reasoning.In this topic, questions are asked relating to positions of English alphabets. This topic is very important from exam point of view. So you need to memorize the positions of the alphabets so that you can handle each and every question based on this topic. Let us discuss various types of alphabet reasoning:

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

From A to M, the alphabet completes the first 13 letters that is the first half, while the second half starts from 14

(First Alphabetical Half) A to M ⇒ 1 to 13

(Second Alphabetical Half) N to Z ⇒ 14 to 26

E | J | O | T | Y |

5 | 10 | 15 | 20 | 25 |

For example, we are asked to find the 24^{th} letter from the left side of the alphabet. We already know that the 25^{th} letter from the left-hand side is Y, now we need to find letter before Y and that is X. By using this simple method, we can easily find out the position of any letter in the alphabet. Memorizing the positions & sequence of letters is a basic to solve any questions of this type, so you should try to memorize these positions. For this particular reason, you should practice EJOTY. Write down the names of ten of your friends & do as given below:

Let us take an example of name RAVI. With the help of EJOTY, we know that the letter R stands for number 18, A stand for 1, V stands for 22, I stands for 9. Adding all these numbers, we get (18+1+22+9). This is a very good way to remember the positions of all the letters in the alphabet.

Let us take an example of name RAVI. With the help of EJOTY, we know that the letter R stands for number 18, A stand for 1, V stands for 22, I stands for 9. Adding all these numbers, we get (18+1+22+9). This is a very good way to remember the positions of all the letters in the alphabet.

Must Read Series Alphabet Articles

- Reasoning Concepts & Tricks
- Reasoning Practice Questions Level 01

Suppose there is a row of 8 persons in which a person is standing 5th from left. Let's find out his/her position from the right side.

I | I | I | I | I | I | I | 1 |

1^{st} | 2^{nd} | 3^{rd} | 4^{th} | 5^{th} | 6^{th} | 7^{th} | 8^{th} |

You can see that the person who was standing fifth from the left hand side is placed fourth from the right side. Suppose there are ‘n’ persons, the order can be horizontal (from left to right) or vertical (from top to bottom) and if the position of a person is given which is rth from left then his position from right will be (n+1-r). Likewise this concept can also be used if their sitting order is vertical. Sum of positions from left and from right will always be (n+1). As given in the above mentioned example there are 8 persons, position of 5th person from left will be 4th (8+1-5)from right. Sum of both the positions is (4 + 5) = 9. If we are dealing with letters & we are given the position of any letter from either side, we need to add 1 more to the total number of letters & then subtract the position from left side to get its position from the right side.

For instance, let us find the position from right side of a letter, which is the 10^{th} from left side.

A | B | C | D | E | F | G | H | I | J | K | L | M | N |

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |

O | P | Q | R | S | T | U | V | W | X | Y | Z | from LHS |

15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | from RHS |

As you can see, the 10^{th} letter from the left hand side of the alphabet is J, which is 17^{th} letter (26+1-10) from the right hand side of the alphabet. We did this operation by adding 1 to the total number of letters (26 + 1 = 27) and then subtracting 10 from it. This same logic is to be applied if we have given an initial right position and we need to find the position from the left side, or we have given initial position from top and we have to calculate position from bottom and vice-versa

A | B | C | D | E | F | G | H | I | J | K | L | M |

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |

→ |

N | O | P | Q | R | S | T | U | V | W | X | Y | Z |

14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 |

→ |

You can see that there are 7 letters between these 2 positions i.e. M, N, O, P, Q, R and S. We know that the midpoint of 7 items is the 4^{th} item from either of the sides, whether counted from the right hand side of the alphabet or the left hand side of the alphabet. It comes out to be P, which is the correct answer in this particular case. But so much effort is not needed in solving such questions. In these type of questions, if the two different positions are given from the same side (i.e. either both are from the left hand side of the alphabet or both are from the right hand side of the alphabet), we can simply add up the 2 different positions from the same side, get their simple average and thus, the correct answer. In this particular case, the two positions are 12 and 20 from the left hand side of the alphabet. Adding and then averaging them, we get 16. Recollecting the EJOTY formula that we discussed earlier, we come up with the letter which is 16th from left side (after O) is P. The same process can be applied if we are given a case in which both the positions are counted from the right hand side of the alphabet. Remember, the answer we get will be from the same from both sides. Let’s make this clearer by discussing an illustration.

Consider a scenario in which we have to find the mid-point between the 11^{th} and the 17^{th} letter from the right hand side of the alphabet. Adding the two positions, we get total as 28. The average of these two numbers is therefore 14. So, the mid-point is 14th from the right hand side of the alphabet (the same as the sides given in the question). Now we will convert this position into a position from the left hand side of the alphabet. Applying the logic which we had applied earlier, we will subtract 14 from 27 and get the answer- 13^{th} from the left hand side of the alphabet, which is M. You can verify this answer by looking up the above alphabet.

These types of series consist of small letters which follow a specific pattern or series. Some spaces are left blank in between the series given. We have to fill in the blanks from given options to make a pattern. This can be clearly explained by this illustration:

1. ppppp

2. ppqpp

3. ppqpq

4. qppqq

Sol: In order to solve these kinds of series, we should fill the given blanks by taking each option one by one & see where it forms a logical pattern. When you try to fill the first option, it becomes pqppqpppppqp. It does not result into any particular logical pattern. If you fill the second option you get pqp/pqp/pqp/pqp. It has been separated by the symbol “/” for your better understanding. Now, this becomes a pattern of writing pqp again & continously. So, the second option becomes the correct answer to this given question.