LOGICAL REASONING - LETTERS AND SYMBOL SERIES

Question Bank 20 Pages

Aptitude

Logical Reasoning

Contributed by

Isha Sama

Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

1
LETTERS AND SYMBOL SERIES
INTRODUCTION
This chapter consists of questions in which series of numbers or alphabetical letters or
combinations of both are given, which are generally called the terms of the series. These
terms follow a certain pattern throughout the series. Now we have to study the given
series, identify the pattern followed in the series and either find the missing term or
wrong term in the series.
TYPES
Following are the types of series completion:
Type - 1: Number series 8, 16, 32, 64, 128, ?
Type - 2: Alphabetical series B, D, F, H, J, L, ? BZ, EW, HT, KQ, ?
Type - 3: Alpha-Numeric series 6, H, 9, K, 11, M, 14, P, 16, R, 19, U, 21, ?
TYPE - 1: NUMBER SERIES
CASE I: COMPLETING THE GIVEN SERIES BY FINDING THE MISSING TERM
ILLUSTRATIVE EXAMPLES (ADDITION)
Directions: Find the missing term in each of the following sequence:
Example-1. 3, 6, 9, 12, 15, ?
(a) 16 (b) 17 (c) 18 (d) 19
Solution:
Clearly, the given sequence follows the pattern:
3, 3, 3, 3,...++++

Thus,
3 3 6, 6 3 9,...+ = + =

So, missing term
15 3 18.= + =

Hence, the answer is (c).
Example-2. 2,5,10,17, 26, ?
(a) 29 (b) 31 (c) 35 (d) 37
Solution:
Clearly, the given sequence follows the pattern:
3, 5, 7, 9, 11,...+ + + + +

Thus,
2 3 5, 5 5 10,...+ = + =

So, missing term
26 11 37.= + =

Hence, the answer is (d).
Example-3.
1,3,7,?,21,31,43

(a) 11 (b) 13 (c) 15 (d) 17
Solution:
Clearly, the given sequence follows the pattern:
2, 4, 6, 8, 10, 12,...+ + + + + +

Thus,
1 2 3, 3 4 7,...+ = + =

So, missing term
7 6 13.=+=

Hence, the answer is (b).
Example
Example
Example
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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

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Example-4.
12,14,34,36,56,58,78,?

(a) 98 (b) 80 (c) 90 (d) 82
Solution:
Clearly, the given sequence follows the pattern:
2, 20, 2, 20,...+ + + +

Thus,
12 2 14, 14 20 34,...+ = + =

So, missing term
78 2 80.= + =

Hence, the answer is (b).
Example-5.
6,10,18,34,66,?

(a) 98 (b) 126 (c) 130 (d) 132
Solution:
Clearly, the given sequence follows the pattern:
4, 8, 16, 32,...+ + + +

Thus,
6 4 10, 10 8 18,...+ = + =

So, missing term
66 64 130.= + =

Hence, the answer is (c).
Example-6.
1,3,3,6,7,9,?,12,21

(a) 15 (b) 11 (c) 12 (d) 13
Solution:
Clearly, the given sequence is a combination of two sequence:
I.
1,3,7,?,21
and II.
3,6,9,12

The pattern followed in I is
2, 4,...++
and the pattern followed in II is + 3.
So, missing number
7 6 13.=+=

Hence, the answer is (d)
PRACTICE EXERCISE
NUMBER SERIES (ADDITION)
Directions: In each of the following questions, a number sequence is given with one
term missing. Choose the correct term that will continue the same pattern and
replace the question mark in the given sequence.
(1)
7,12,19,28,?

(a) 39 (b) 38 (c) 40 (d) 41
(2)
0,7,15,24,34,?

(a) 55 (b) 45 (c) 46 (d) 50
(3)
87,102,119,138,159,?

(a) 184 (b) 180 (c) 182 (d) 185
(4)
2,14,32,56,86,?

(a) 120 (b) 116 (c) 128 (d) 122
(5)
2,20,42,68,98,?

(a) 132 (b) 130 (c) 134 (d) 136
(6)
1,2,4,7,?,16

(a) 8 (b) 11 (c) 9 (d) 13
(7)
45,54,47,?,49,56,51,57,53

(a) 48 (b) 50 (c) 55 (d) 60
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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

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(8)
9,11,17,35,89,?

(a) 143 (b) 197 (c) 250 (d) 251
(9)
6,8,10,14,18,26,34,50,66,?

(a) 98 (b) 82 (c) 90 (d) 100
(10)
1 3 6 10 15
, , , , ,?
3 7 13 21 31

(a)
21
43
(b)
43
21
(c)
17
35
(d)
20
41

(11)
71,25,83,37,95,49,?

(a) 61 (b) 107 (c) 59 (d) 105
(12)
2,5,9,?,20,27

(a) 14 (b) 16 (c) 18 (d) 24
(13)
4,6,12,14,28,30,?

(a) 32 (b) 60 (c) 62 (d)64
(14)
11,13,17,19,23,25,?

(a) 26 (b) 27 (c) 29 (d) 37
(15)
6,11,21,36,56,?

(a) 42 (b) 51 (c) 81 (d) 91
NUMBER SERIES (SUBTRACTION)
(16)
––5,5,3, 1, 7,?

(a) –15 (b) 15 (c) 0 (d) –13
(17)
17,20,19,22,21,24,23,?

(a) 22 (b) 24 (c) 26 (d) 2
(18)
2,7,5,5,8,3,11,?

(a) 1 (b) 0 (c) 2 (d) 3
(19)
−−2, 1,2, 1,2,?

(a) 2 (b) –1 (c) 5 (d) 4
(20)
180,140,179,138,177,134,?

(a) 120 (b) 174 (c) 128 (d) 175

1
a
6
b
11
b
16
a
2
b
7
c
12
a
17
c
3
c
8
d
13
b
18
a
4
d
9
a
14
c
19
b
5
a
10
a
15
c
20
b

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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

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ILLUSTRATIVE EXAMPLES (MULTIPLICATION)
Example-1.
1,5,25,125,?

(a) 625 (b) 650 (c) 600 (d) 725
Solution:
The terms are obtained by multiplying 5 with previous term.
So, missing term
125 5 625=  =

PRACTICE EXERCISE
NUMBER SERIES (MULTIPLICATION)
Directions: In each of the following questions, a number sequence is given with one
term missing. Choose the correct term that will continue the same pattern and
replace the question mark in the given sequence.
(1)
8,16,32,64,128,?

(a) 256 (b) 384 (c) 250 (d) 192
(2)
,
1 1 1 1
, , , ,2,4 ?
32 16 4 2

(a) 8 (b) 16 (c) 12 (d) 20
(3)
2,6,18,54,162,?

(a) 486 (b) 324 (c) 243 (d) 450
(4)
4,7,15,29,59,117,?

(a) 230 (b) 233 (c) 234 (d) 235
(5)
2,6,15,34,73,?

(a) 152 (b) 150 (c) 151 (d) 153
(6)
7,13,24,45,86,?

(a) 167 (b) 169 (c) 170 (d) 172
(7)
2,4,8,32,?

(a) 256 (b) 512 (c) 1024 (d) 128
(8)
3,10,33,104,?

(a) 310 (b) 312 (c) 319 (d) 315
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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

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(9)
4,12,36,108,?

(a) 144 (b) 216 (c) 304 (d) 324
(10)
1,1,2,6,24,?,720

(a) 100 (b) 104 (c) 108 (d) 120
(11)
1000,500,250,?

(a) 50 (b) 125 (c) 100 (d) 150
(12)
240,?,120,40,10,2

(a) 120 (b) 180 (c) 200 (d) 240
(13)
156,76,36,16,?

(a) 6 (b) 8 (c) 10 (d) 12
(14)
48,24,96,48,192,?

(a) 76 (b) 90 (c) 96 (d) 98
(15)
6,13,28,59,?

(a) 111 (b) 113 (c) 114 (d) 122

1
a
6
a
11
b
2
b
7
a
12
d
3
a
8
c
13
a
4
d
9
d
14
c
5
a
10
d
15
d

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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

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ILLUSTRATIVE EXAMPLES (SQUARES/CUBES)
Example-1.
1,4,9,16,25,36,?

(a) 64 (b) 49 (c) 81 (d) 100
Solution:
Clearly, the given sequence consists of squares of numbers 1, 2, 3,… i.e.,
2 2 2
1 ,2 ,3 ,...

So, missing term
2
7 49.==

Hence, the answer is (b)
Example-2.
1,4,27,16,?,36,343

(a) 25 (b) 87 (c) 120 (d) 125
Solution:
Clearly, the given sequence consists of cubes of odd numbers and squares of even
numbers, i.e.,
3 2 3 2
1 ,2 ,3 ,4 ,...

So, missing term
3
5 125.==

Hence, the answer is (d)
PRACTICE EXERCISE
NUMBER SERIES (SQUARES/CUBES)
Directions: In each of the following questions, a number sequence is given with one
term missing. Choose the correct term that will continue the same pattern and
replace the question mark in the given sequence.
(1)
63,37,15,5,?

(a) –1 (b) 0 (c) 1 (d) 2
(2)
0,3,8,15,?

(a) 25 (b) 24 (c) 35 (d) 36
(3)
0,6,24,60,120,210,?

(a) 336 (b) 343 (c) 340 (d) 315
(4)
1,4,27,16,?,36,343

(a) 25 (b) 87 (c) 120 (d) 125
(5)
7,26,63,124,?

(a) 216 (b) 215 (c) 248 (d) 210
(6)
5,10,17,26,37,50,?

(a) 64 (b) 65 (c) 63 (d) 66
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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

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(7)
3,9,15,?

(a) 32 (b) 33 (c) 31 (d) 30
(8)
1,5,14,30,55,91,?

(a) 130 (b) 140 (c) 150 (d) 160
(9)
198,194,185,169,?

(a) 92 (b) 112 (c) 136 (d) 144
(10)
2,2,5,13,28,?

(a) 49 (b) 50 (c) 51 (d) 52

1
a
6
b
2
b
7
b
3
a
8
b
4
d
9
d
5
b
10
d

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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

8
ILLUSTRATIVE EXAMPLES (WRONG TERM)
Case II: FINDING THE WRONG TERM IN THE GIVEN SERIES
Example - 1. Find the wrong number in the sequence:
7,26,63,126,215,342,511

(a) 7 (b) 26 (c) 126 (d) 215
Solution:
Clearly, the correct sequence is:
3 3 3 3 3 3 3
2 1,3 1,4 1,5 1,6 1,7 1,8 1.− − − − − − −

So, 126 is wrong and should be replaced by
( )
3
51−
i.e. 124.
Hence, the answer is (c)
Example - 2. Find the wrong number in the sequence:
3,8,15,24,34,48,63

(a) 15 (b) 24 (c) 34 (d) 48
Solution:
The difference between consecutive terms of the given sequence are respectively
5,7,9,11.
13 and 15.
Clearly, 34 is a wrong number and must be replaced by
( )
24 11+
i.e. 35.
Hence, the answer is (c)
PRACTICE EXERCISE
NUMBER SERIES (WRONG TERM)

Directions: In each of the following questions, one term in the number sequence is
wrong. Find out the wrong term.
(1) 196, 169, 144, 121, 101
(a) 101 (b) 121 (c) 169 (d) 196
(2) 2, 5, 10, 17, 26, 37, 50, 64
(a) 17 (b) 26 (c) 37 (d) 64
(3) 5, 27, 61, 122, 213, 340, 509
(a) 27 (b) 61 (c) 122 (d) 509
(4) 125, 126, 124, 127, 123, 129
(a) 126 (b) 124 (c) 123 (d) 129
(5) 105, 85, 60 30, 0, – 45, – 90
(a) 105 (b) 60 (c) 0 (d) – 45
(6) 2, 5, 10, 50, 500, 5000
(a) 5 (b) 10 (c) 50 (d) 5000
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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

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(7) 3, 7, 15, 39, 63, 127, 255, 511
(a) 15 (b) 39 (c) 63 (d) 127
(8) 3, 2, 8, 9, 13, 22, 18, 32, 23, 42
(a) 8 (b) 9 (c) 13 (d) 22
(9) 1, 2, 5, 10, 17, 28
(a) 30 (b) 28 (c) 27 (d) 17
(10) 2, 3, 5, 8, 13, 34
(a) 21 (b) 25 (c) 29 (d) 34

1
a
6
d
2
d
7
b
3
a
8
b
4
d
9
b
5
c
10
a

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Unit 1
Letters and Symbol Series
LOGICAL REASONING (PMC)

10
TYPE - 2: ALPHABETICAL SERIES
INTRODUCTION
In this type of questions, a sequence of single, pairs or groups of letters is given. The
terms of the sequence form a certain pattern according to the position of the letters in the
English alphabet. So, in this type we have to identify the pattern of alphabetical letters
and find the missing term in the given sequence.
ILLUSTRATIVE EXAMPLES
Example - 1. Find the next two terms in the sequence: A, C, F, J, ?, ?
(a) L, P (b) M, O (c) O, U (d) R, V
Solution:
Clearly, the first, second, third,… letters of the sequence are respectively moved two,
three, four,… steps forward to obtain the successive terms of the sequence.
Thus, the fifth term in the sequence must be a letter which is five steps ahead of J i.e. O,
while the sixth term must be a letter six steps ahead of O i.e. U.
Thus, we have the following pattern:
+2 +3 +4 +5 +6
A C F J O U⎯⎯→ ⎯⎯→ ⎯⎯→ ⎯⎯→ ⎯⎯→

So, the missing terms are O and U. Hence, the answer is (c)
Example - 2. Which term comes next in the sequence: AC, FH, KM, PR,?
(a) UW (b) VW (c) UX (d) TV
Solution:
Clearly, the first and second letters of each term are moved five steps forward to obtain
the corresponding letters of the next term.
Thus, the first letter of the missing term must be five steps ahead of P i.e. U, while the
second letter must be five steps ahead of R i.e. W.
So, the missing term is UW. Hence, the answer is (a)
Example - 3. Which term comes next in the sequence: YEB, WFD, UHG, SKI, ?
(a) QOL (b) QGL (c) TOL (d) QNL
Solution:
Clearly, we observe the following pattern in the first, second and third letters of the given
sequence:
1
st
letter:
2 2 2 2
Y W U S Q
− − − −
⎯⎯→ ⎯⎯→ ⎯⎯→ ⎯⎯→

2
nd
letter:
1 2 3 4
E F H K O
+ + + +
⎯⎯→ ⎯⎯→ ⎯⎯→ ⎯⎯→

3
rd
letter:
2 3 2 3
B D G I L
+ + + +
⎯⎯→ ⎯⎯→ ⎯⎯→ ⎯⎯→

Thus, the missing term is QOL. Hence, the answer is (a)
MDCAT SOCIETY FOR FUTURE DOCTORS
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LOGICAL REASONING - LETTERS AND SYMBOL SERIES

LOGICAL REASONING - LETTERS AND SYMBOL SERIES

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