PMC Logical Reasoning Question – Letter And Symbol Series Questions

Solved Example – Number Series

Question 1. 20, 32, 45, 59, 74, ?
(A) 95
(B) 90
(C) 85
(D) 79

 The answer is (B).

Solution:

 The differences between  consecutive terms of  series 
12, 13, 14, 15, 16, …..
 2nd term = 1st term + 12 = 20 + 12 = 32
3rd term = 2nd term + 13 = 32 + 13 = 45
4th term = 3rd term + 14 = 45 + 14 = 59
5th term = 4th term  + 15 = 59 + 15 = 74
Thus ? = 6th term = 5th term +16 = 74 + 16 = 90




Question 2. 210, 195, 175, 150, 120, ?
(A) 75
(B) 80
(C) 85
(D) 90

The answer is (C).

Solution:

 The differences between the consecutive terms of the series are -15, -20, -25, -30, -35, …

apply the above method with subtraction



Question 3. 3, 5, 10, 12, 24, 26, ?
(A) 52
(B) 30
(C) 28
(D) 48

The answer is (A).

Solution: 

Each odd term will double the previous even term.
3th term = 2×2nd term

             =2×5=10

5th term = 2×4th term

               =2×12=24

Thus, ? = 7th term 

            = 2×6 term

            =2×26=52



Question 5. 2, 6, 10; 5, 9, 13; 10, 14, 18; ?, ?, ?.
(A) 12, 14, 16
(B) 14, 17, 20
(C) 17, 21, 25
(D) 15, 20, 25

The answer is (C).

Solution: There is  a common difference of 4 among consecutive terms

Also 

the difference between the consecutive first term of each group is 
3, 5, 7,….
Thus, ?, ?, ? = ( 10 + 7), (17 + 4), (21 + 4)

= 17, 21, 25,

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Solved Example – Wrong Number

Question 1. 1, 11, 26, 49, 80, 122, 169, 227, 290, 361
(A) 49
(B) 26
(C) 80
(D) 122

The answer is (C).

Solution: 

The differences between the consecutive terms of  series are

5, 8, 11; 5, 8, 11; ………

 2nd term = 1st term + 10 = 1 + 10 = 11
3rd term = 2nd term + 10 + 5 = 11 + 10 + 5 = 26
4th term = 3rd  term + 10 + 5 + 8 = 26 + 10 + 5 + 8 = 49
5th term = 4th term + 10 + 5 + 8 + 11 = 49 + 10 + 5 + 8 + 11 = 83
Thus, fifth term should be 83 and not 80.

Question 2. 11, 5, 20, 12, 40, 26, 74, 54
(A) 5
(B) 20
(C) 26
(D) 40

Solution: 

The differences between odd terms are 9, 18, 36, ……….

3rd term = 1st  term + 9 = 11 + 9 = 20
5th term = 3rd term + 18 = 20 + 18 = 38

7th term = 5th term + 36 = 38 + 36 = 74
Thus, 5th term should be 38 and not 40.
The answer is (D).


Question 3. 3, 6, 5, 20, 7, 42, 9, 74
(A) 6
(B) 42
(C) 74
(D) 20

The answer is (C).

Solution: 

Differences between the even terms of the series are 14, 22, 30……

4th  term = 2nd term + 14 = 6 + 14 = 20
6th  term = 4th term + 22 = 20 + 22 = 42
8th term = 6th term + 30 = 42 + 30 = 72
Thus, 8th term should be 72 and not 74.

Solved Example – Alphabet Series


Question 1.ABXW, EFTS, ?, MNLK

(A) IJOP
(B) IJPO
(C) JIOP
(D) JIPO

The answer is (B).

Solution

The first and second terms are in alphabetic sequence , while third and fourth are in reverse sequence. Therefore, missing item is IJPO.

Question 2. A Z X B V T C R ?

(A) E, O
(B) O, Q
(C) F, O
(D) P, D

The answer is (D).

Solution:

The 1st, 4th and 7th letters are in alphabetical order.Therefore, 10th letter will be after C i-e D. Also 2nd and 3rd letter are in reverse  order and so are the 5th and 6th letters.

Question 3. P3C, R5F, T8I, VI2, L, ?
(A) YI7O
(B) XI7M
(C) XI7O
(D) XI6O

The answer is (D).

Solution: The 1st letter of the terms are alternating. And the sequences are  followed by the numbers is +2, +3, +4 …The last letter of  each term are moved 3 steps forward to the last letter, so, ? = XI7O

Question 4. AK, EO, IS, ? , QA, UE
(A) LV
(B) MW
(C) NX
(D) LW
(E) MV

The answer is (B).

Solution:

so here 

From A in 1st term to E in 2nd term , count A to E then E will be on 4th number , similarly from E to I , I is also on 4th number , now count from I to next 4th number that is M so in this ? term first letter will be M

Now move to 2nd number of terms  In first term 2nd number is K now count from K to O , O is on 3 number count , now after next 3rd count their is S, next 3rd count will be W so W will be the 2nd digit of ? term.


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