Types of Questions in Letter Series - CSAT Preparation - UPSC

Type - 1

To find out the position of the letter according to the given condition.

Note: If we are moving to the left hand side of any letter that means positions are decreasing and if we are moving towards right hand side of the series that means Positions of the letters are increasing.

If the first half and second half of the alphabet series are Arranged in reverse order, then find out the answers to the given questions:

Q.1. Which letter is 5^th to the right of the 12^th letter from the left-hand side of the series?
Ans. 5^th to the right of 12^th letter from left means 12+5 = 17^th letter of the series.
Because series is in reverse order for that the trick is to:

• Add the first and last position of the group in which this position falls.
• Deduct the given position from the sum we will get the position of the same letter in the real series.

Hence, the letter 17^th falls in the 2^nd half which starts from position 14^th and ends on 26^th.
∴ 14 + 26 = 40, 40 - 17 = 23
which means the letter which is in 17^th position in the given series is 23^rd in the real series. Hence, the letter is “W”.

Q.2. Which letter is 5^th to the left of the 12^th letter from the left-hand side on the series?
Ans. 5^th to the left of 12^th letter from left means, 12 – 5 = 7^th letter of the serries
Which falls in the first half of the series which starts from position 1^st and ends on 13th.
∴ 1 + 13 = 14, 14 – 7 = 7^th letter i.e.“g” in the real series.
(Position is same even in reversed series, because letter 7 is in the middle of the series, and its position will not be Changed even after letter get reversed)

Q.3. Which letter is the 5^th to the left of the 12^th letter from the right-hand side of the series?
Ans. 5^th to the left of the 12^th letter from the right.
In this case first of all we will convert the 12^th letter from right to its position from left. (27 – 12 = 15)
⇒ We have to find the 5^th letter to the left of the 15^th letter from the left.
Hence,  15 – 5 = 10^th letter, that falls in first half of the series.
∴ 1 + 13 = 14,   14 – 10 = 4^th letter i.e. “d”.

Q.4. Which letter is 5^th to the right of the 12^th letter from the right-hand side of the series?
Ans. 5^th to the right of the 12^th letter from the right end.
► 12^th from the right end means 27-12 = 15^th from left.
⇒ 15 + 5 = 20^th letter, which falls in the second half of the series.
∴ 14 + 26 = 40,  40 – 20 = 20^th letter i.e. “t”.
(because position 20 is in the middle of the second half of the series).

Type - 2 

To find out the next letter of the series.

Find The Letter In Place Of Question Mark (?) In The Series Given Below:
Q.1. B, F, ?, N, R
(a) G
(b) K
(c) J
(d) L
(e) None of The Above
Ans. (c) 

Q.2. P, ?, J, G, D, A
(a) Q
(b) N
(c) K
(d) M
(e) None of The Above
Ans. (d)

Q.3. ?, E, G, J, N
(a) A   
(b) B
(c) D
(d) Z
(e) None of The Above
Ans. (c)

Q.4. X, F, Y, G, ?, H
(a) Z
(b) A
(c) B
(d) Y
(e) None of The Above
Ans. (a)

Q.5. B, B, A, D, ?, F    
(a) B
(b) A
(c) Z
(d) C
(e) None of The Above 
Ans. (c)

Type - 3 

Fill in the blank series

Example 1.   a _ c a a _ b c c _ a a b bb _ c c
(a) bbca
(b) abac
(c) bbac
(d) bacb
Ans. (b)
Explanation. We can solve questions of these types by following steps:

• Check out the total number of positions in the statement which is 18 here.
• Now,18 can be divided by 2, 3, 6, 9 if the pattern here is of groups of equal size.
• We can’t divide them into the group of 2 or 3 letters here because of repeated letters in the last of the series.
• If we divide them into a group of 6 or 9 letters each then also we are not getting a particular pattern here.
• Hence, groups here are of different sizes.
• Considering this part _ a a b bb _ c c, we can see that letter “b” comes thrice here, which means the letter “a” and “c” also could be thrice and if we fill this with a and c respectively, we will get that – a aa b bb c cc .
• That means, in the previous group letters are repeated twice- a a _ b c c. if we fill it with “b”.
• So, groups here are formed by all the three letters first one time then twice, and at last thrice.
• By looking at the options we will get the option (c) – bbac is our answer.

Example 2.  a _ b b a a _ b a a _ b a a b _ a a b
(a) abab
(b) baba
(c) abbb
(d) babb
Ans. (c)
Explanation.
We can solve this by following steps:

• If we look at the total no. of positions that are 19.
• Because, 19 is a prime number that means we can’t divide this, so we can use either 18 or 16 letters to divide into groups.
• If we are taking 18 letters, we can divide them into groups of 3, 6, and 9 letters from which we will get optimum numbers and size of the group if we divide by 6. But, it will also not lead to a particular pattern.
• If we are taking 16, we can divide it into groups of 2, 4, and 8 letters of which we will get the optimum number and size of the group if we divide it by 4. therefore,-a _ b b a a _ b a a _ b a a b _ a a b,
• Now, we will get same group of same size, therefore pattern is - a a b b a a b b a a b b a a b b a a b.
• Thus, we will get with the help of option (c) – abbb.

Example 3. a _ b c a a _ _ aa b _ a a _ c a a _ c
(a) accbbb
(b) acbcbb
(c) cbcbba
(d) abccbb
Ans. (d)
Explanation. 
We can solve this by following steps:

• Total no. of positions here are 20 which can be divided by 2, 4, 5 and 10.
• To make an optimum group of optimum size either we can divide it by 4 or by 5.
• So, if we make groups of 4 letters each: a _ b c a a _ _ aa b _ a a _ c a a _ c,
  We will get identical groups.
• Therefore, the pattern is: a a b c a a b c a a b c a a b c a a b c
• Thus, we will get with the help of option(d) – abccbb.

Type - 4

Miscellaneous Questions   

In the exam one can also ask questions to make a meaningful word, or by arranging letters in ascending or descending order, or by getting the same gap in between two letters as in the alphabet series.
This can be clearly explained by the following examples:

Example 1. If the letters of the word “ BEAUTIFUL” are arranged in Ascending/increasing order. How many letters are there whose position will not change?
(a) 4
(b) 3
(c) 1
(d) 2
Ans. (c)
Explanation.

In ascending order:

As we can see only letter “U” is on the same position after rearrangement. Therefore, answer is option – (c).

Example 2. How many pairs are there in the word “BEAUTIFUL”, which have the same number of letters between them as in the alphabet series?
(a) 1
(b) 4
(c) 3
(d) none
Ans. (b)
Explanation. 

Therefore, ans is option - (b)

Example 3. With the help of letters “ E , S, R , O” each letter used
Ones, how many meaningful words can be formed?
(a) 1
(b) 2
(c) 3
(d) none
Ans. (a)
Explanation. With the help of these letters only one word is formed: ROSE.
Hence, the answer is option – (a)

Note: 

These types of questions are just to check your General intelligence.

Example 4. With the help of the letters “A, D, E, R” each letter used
Ones, how many meaningful words can be formed?
(a) 2
(b) 3
(c) more than 3
(d) none
Ans. (c)
Explanation. With the help of these letters words formed are: READ, DEAR, DARE, RADE
Hence, the answer is option- (c).