Types of Questions in Letter Series Notes | EduRev

Logical Reasoning (LR) and Data Interpretation (DI)

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CAT : Types of Questions in Letter Series Notes | EduRev

The document Types of Questions in Letter Series Notes | EduRev is a part of the CAT Course Logical Reasoning (LR) and Data Interpretation (DI).
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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.
Q.(1-4). If the first half and second half of the alphabet series are Arranged in reverse order, then find out the answers of given questions:

(1) Which letter is 5th to the right of 12th letter from let hand  side on the series ?
Ans. 5th to the right of 12th letter from left means
12+5 = 17th letter of the series, because series is in reverse order for that the trick is
(i) to add first and last position of the group in which this position fall.
(ii) deduct the given position from the sum we will get position of same letter in the real series. Hence,
Letter 17th fall in 2nd half which starts from position 14th  and ends on 26th .
Therefore,
14 + 26 = 40.
40 - 17 = 23,
which means letter which is on 17th position in the given series is 23rd in the real series. Hence, letter is “W”.

(2) Which letter is 5th to the left of 12th letter from the left hand side on the series?
Ans. 5th to the left of 12th letter from left means,
12 – 5 = 7
Which falls in first half of the series which starts from position 1st and ends on 13th. Therefore,
1 + 13 = 14, 14 – 7 = 7th
Which is letter “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 Changed even after letter get reversed)

(3) Which letter is 5th to left of 12th letter from the right hand side on the series?
Ans. 5th to the left of 12th letter from the right.
In this case first of all we will convert 12th letter from right to its position from left. (27 – 12 = 15)
Means, we have to find 5th letter to left of 15th letter from left. Hence,  15 – 5 = 10,
Which falls in first half of the series.
Therefore,
1 + 10 = 14,   14 – 10 = 4th
Which is letter “d”.

(4) Which letter is 5th to the right of 12th letter from right hand side of the series?
Ans. 5th to the right of 12th letter from right end.
12th from right end means 27-12 = 15th from left.
Means,  15 + 5 = 20,
Which falls in second half of the series. Therefore,
14 + 26 = 40,  40 – 20 = 20th
Which is letter “t”.
(because position 20 is in middle of 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.1B, F, ? , N, R
(a) G
(b) K
(c) J
(d) L
(e) None of The Above
Solution :-
Option C

Types of Questions in Letter Series Notes | EduRev

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

Types of Questions in Letter Series Notes | EduRev


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

Types of Questions in Letter Series Notes | EduRev

 

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

Types of Questions in Letter Series Notes | EduRev

 

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

Types of Questions in Letter Series Notes | EduRev


TYPE-3 Fill in the blank series:
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 following examples:

Example 1:    a _ c a a _ b c c _ a a b bb _ c c
(a) bbca
(b) abac
(c) bbac
(d) bacb
We can solve questions of these types by following steps:
1. Check out total number of positions in the statement which is 18 here.
2. Now,18 can be divided by 2, 3, 6, 9 if the pattern here is of groups of equal size.
3. We can’t divide them in group of 2 or 3 letters here because of repeated letters in the last of series.
4. If we divide them in group of 6 or 9 letters each then also we are not getting a particular pattern here.
5. Hence, groups here are in of different size.
6. Considering this part _ a a b bb _ c c, we can see that letter “b” comes thrice here, which means 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 .
7. That means , in the previous groups letters are repeated twice- a a _ b c c. if we fill it with “b”, we will get that.
8. So, groups here are formed by all the three letters first one time then twice and at last thrice.
9. By looking at the options we will get option (c) – bbac is our ans.

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
We can solve this by following steps:
1. If we look at the total no. of positions that are 19
2. Because, 19 is a prime number that means we can’t divide this, so we can use either 18 or 16 letters to divide in groups
3. If we are taking 18 letters , we can divide it in groups of 3, 6, and 9 letters from which we will get optimum numbers and size of group if we divide by 6. But , it will also not lead to particular pattern.
4. If we are taking 16, we can divide it in groups of 2, 4, and 8 letters of which we will get optimum number and size of group if we divide it by 4. therefore,-a _ b b a a _ b a a _ b a a b _ a a b,
5. 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
Which 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
We can solve this by following steps:
1. Total no. of postions here are 20 which can be divided by 2, 4, 5 and 10
2. To make optimum group of optimum size either we can divide it by 4 or by 5.
3. 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.
4. Therefore, the pattern is –
a a b c a a b c a a b c a a b c a a b c ,
which we will get with the help of option(d) – abccbb.

TYPE-4  Miscellaneous Questions   
In the exam one can also ask questions to make 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.
Let us discussed one example of each type.

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
Solutions: 
2     5     1       21       20     9    6     21      12
B     E     A       U         T       I     F      U        L
In ascending order :
1     2      5     6     9    12    20    21    21
A     B      E    F     I     L      T     U    U
As we can see only letter “U” is on same position after rearrangement. Therefore, answer is option – (c).

Example 2: How many pairs are there in the word “BEAUTIFUL”
Which have same number of letters between them as
In the alphabet series?
(a) 1
(b) 4
(c) 3
(d) none
Solutions:
Types of Questions in Letter Series Notes | EduRev
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
Solutions: With the help of these letters only one word is formed-
ROSE. Hence, 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
Solutions: with the help of these letters words formed are :
1-READ,  2-DEAR, 3-DARE, 4-RADE.
Hence, answer is option- (c).

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