The CWC meeting of NUGDS AP circle was held at Sattupalli town in Khammam Division from 24.11.12 to 25.11.12 under the presidentship of Sri M.Murali. Sri G.Sankargoud Asst Secretary General FNPO, Sri G.Nageswararao President NUPE Gr-c AP circle, Sri V.Sivaji C/s NUPE Gr-c, and Sri M.Tirumalarao ACS NUPE Gr-c attended as guests. Sri G.Satyanarayanareddy explined the problems facing by the GDS officials and the CWC unanimously resolved to bring preasure on Federation and All india union to give a indefinate strike call on GDS demands only. Sattupalli GDS branch conference held on 25.11.12 in a grandmanner.
National Association of Postal Employees Group-C, Andrapradesh Circle, Vijayawada.
Sunday, 25 November 2012
DPC for promotion to LSG APMs (Genl. & Accounts line)
It is learnt that various DPCs for promotion to the posts of LSG APMs both General Line and Accounts line are in final stages at Circle Office, Hyderabad and results will likely be announced in the next week before 30.11.2012. The approximate no of Vacancies for which DPCs are being held are as follows.
A.P.M (General line) --- 75 to 80
A.PM (Accounts line) -- 35 to 40Wednesday, 21 November 2012
MTS/POSTMAN/GDS TO PA/PA RECRUITMENT MATERIAL
This material prepared and compiled by
Akula. Praveen Kumar, SPM, Papannapet
Sub Office-502 303, Medak Division,
AndhraPradesh,9849636361, 8019549939
Disclaimer:- All Material/Questions/Information provided in this post are Compiled by A. Praveen Kumar for in good faith of Departmental Employees. The types of questions, number of questions and standard of questions may be vary in actual examination. This is my predictions only. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of the content
AVERAGE
To Download Full Chapter Click below link
Definition
Central Tendency of data is called an Average. Thus an average is a single value that indicates a group of values. Average can be calculated with the help of mean, median and mode. The most commonly used average is the Arithmetic Mean (AM). While some other average geometric mean and harmonic mean are also quite popular.
Arithmetic Mean is calculated by dividing the sum of all the number/quantities by the number of numbers/quantities. The mean of n numbers
x1, x2, x3, … , xn, is denoted by X and Calculated as
X =
= or
= ∑xi / n
where ∑ (sigma) denotes the sum of the term of type I from 1 to n.
Average is a very simple but effective way of representing an entire group by a single value.
STIPES
(Standard Types In Problem Execution And Solving)
STIPE 01: To find the average when the number of quantities and their sum is given. We have the following formula, Average = Sum of the quantities / Number of quantities.
STIPE 02: To find the sum, when the number of quantities and their average is given. We have the following, Sum of quantities = Average × Number of quantities.
STIPE 03: To find the number of quantities, when the sum of quantities and average are given. We have the following formula, Number of quantities = Sum of quantities / Average.
AVERAGE SHORTCUT METHODS
1. Average = [Total of observations / No. of observations]
2. (i) When a person joins a group in case of increasing average Age
weight of new comer = [ (Previous Age + No. of persons) * Increase in Average ](ii) In case of decreasing Average, Age (or) weight of new comer = [ (Previous Age - No. of persons) * Decrease in Average ] 3. When a person leaves a group and another person joins the group in the place of person left, then (i) In case of increasing average, Age (or) weight of new comer = [ (Age of person left + No. of persons) * Increase in Average ] (ii) In case of decreasing Average, Age (or) weight of new comer = [ (Age of person left - No. of persons) * Decrease in Average ] 4.When a person leaves the group but nobody joins this group, then (i) In the case of increasing Average, Age (or) weight of man left = [ (Previous Age - No. of present persons) * Increase in Average ] (ii) In case of decreasing Average, Age (or) weight of new comer = [ (Previous Age + No. of present persons) * Decrease in Average ]5. If a person travels a distance at a speed of x Km/hr returns to the original place of y Km/hr then average speed is [ 2.x.y / ( x + y ) ] 6.If half of the journey is travelled at speed of x km/hr and the next half at a speed of x km/hr. Then average speed during the whole journey is [ 2.x.y / (x + y) ] 7. If a person travels 3 equal distances at a speed of x Km/hr, y Km/hr,z km/hr. Then average speed during whole journey is [ 3.x.y / (x.y +y.x +z.x) ] EXCERCISE
1. Find the average of all prime numbers between 30 and 50.
Sol. There are five prime numbers between 30 and 50.
They are 31, 37, 41, 43 and 47. Required average = [(31 + 37 + 41 + 43 + 47) / 5] = 199/5 = 39.8.
2. Find the average of first 40 natural numbers.
Sol. Sum of first n natural numbers = 40 * 41 / 2 = 820.
Required average = 820 / 40 = 20.5
3. Find the average of first 20 multiples of 7.
Sol. Required average = 7(1+2+3 ... + 20) / 20 = [7 * 20 * 21 / 20 * 2] = [147 / 2] = 73.5.
4. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53kg is replaced by a new man. find the weight of the new man
Sol. Total weight increased = (1.8 * 10)kg = 18kg.
Weight of the new man = (53 + 18)kg = 71kg.
5. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. find his average after 17th inning.
Sol. Let the average after 17th inning = x.
Then, average after 16th inning = (x - 3) 16(x - 3) + 87 = 17x or x = (87 - 48) = 39.
Sol: Therefore age of the mother
= ( 12 × 7 – 7 × 6) = 42 years
7. The age of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is
Sol: Therefore excluded number
= (27 × 5) - ( 25 × 4) = 135 – 100 = 35.
8. Three years ago, the average age of A and B was 18 years. With C joining them, the average age becomes 22 years. How old is C now?
Sol: Present age of (A + B) = (18 × 2 + 3 × 2) years = 42 years.
Present age of (A + B + C) = (22 × 3) years = 66 years. Therefore C’s age = (66 – 42) years = 24 years.
9. The average age of 36 students in a group is 14 years. When teacher’s age is included to it, the average increases by one. What is the teacher’s age in years?
SoL : Age of the teacher = ( 37 × 15 – 36 × 14 ) years = 51 years
10. The average runs of a cricket player of 10 innings was 32. How many runs must he makes in his next innings so as to increase his average of runs by 4?
Sol: Average after 11 innings = 36.
Therefore required number of runs = (36 × 11) – (32 × 10) = 396 - 320 = 76.
11. The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is
Sol: Let the total number of workers be x. Then,
8000x = (12000 × 7) + 6000 ( x – 7) ‹=› 2000x = 42000 ‹=› x = 21.
Sol: Average
= 76 + 65 + 82 + 67 + 85 / 5) = (375 / 5) = 75.
13. In the first 10 over’s of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 0vers to reach the target of 282 runs?
Sol: Required run rate = 282 – (3.2 × 10 / 40)
= 240 / 40 = 6.25.
14. The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is
Sol: Weight of the teacher
= (35.4 × 25 – 35 × 24)kg = 45 kg.
15. After replacing an old member by a new member, it was found that the average age of five numbers of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member?
Sol: Age decrease = (5 × 3) years = 15 years.
So, the required difference = 15 years
16. The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
Sol: Total weight increased
= (8 × 2.5) kg = 20 kg. Weight of new person = (65 + 20) kg = 85 kg.
Sol: Runs scored in the first 10 overs = 10×3.2=32
Total runs = 282, remaining runs to be scored = 282 - 32 = 250 remaining over’s = 40, Run rate needed = 250 40 =6.25
Sol: Let the sale in the sixth month = x
Then 6435+6927+6855+7230+6562+x 6 =6500 => 6435 + 6927 + 6855+ 7230 + 6562 + x = 6 \times 6500 = 39000 => 34009 + x = 39000, x = 39000 - 34009 = 4991
Sol: Average of 20 numbers = 0
=>Sum of 20 numbers 20 =0 => Sum of 20 numbers = 0 Hence at the most, there can be 19 positive numbers. (Such that if the sum of these 19 positive numbers is x, 20th number will be -x)
20. The average age of a class is 15.8 years. The average age of the boys in the class is 16.4 yrs while that of the girls is 15.4 years. What is the ratio of boys to girls in the class?
21. If a, b, c, d, e are five consecutive odd numbers, their average Is:
22. The average of first five multiples of 3 is
Since the month begins with a Sunday, to there will be five Sundays in the month.
Weight of new person = (65 + 20) kg = 85 kg.
26. The average (arithmetic mean) of 3 numbers is 60. If two of the numbers are 50 and 60, what is the third number?
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Sunday, 18 November 2012
MTS / POSTMAN / GDS TO PA / POSTAL ASSISTANT MATERIAL
This material prepared and compiled by Akula. Praveen Kumar, SPM, Papannapet Sub Office-502 303,MedakDivision,AndhraPradesh,9849636361, 8019549939
Disclaimer:- All Material/Questions/Information provided in this post are Compiled by A. Praveen Kumar for in good faith of Departmental Employees. The types of questions, number of questions and standard of questions may be vary in actual examination. This is my predictions only. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of the content
3. SERIES COMPLETION
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 series. These terms follow a certain pattern throughout the series.The candidate is required to study the given series,identify the pattern followed in the series and either complete the given series with the most suitable alternative or find the wrong term in the series.
Number Series
Prime Number Series:
Example 1. 4, 9, 25, 49, 121, 169,…
(a) 324 (b) 289 (c) 225 (d) 196
Solution. (b) The given series is a consecutive square of prime number series. The next prime number is 289.
Example 2. 5, 7, 13, 23, …
(a) 25 (b) 27 (c) 29 (d) 41
Solution. (d) The difference between prime numbers is increasing. 7 is next prime to 5; 13 is second to next prime to 7; 23 is third to next to 13. Hence, next should be fourth to next prime to 23. Hence, required number is 41.
Multiplication Series:
Example 3. 4, 8, 16, 32, 64… 256
(a) 96 (b) 98 (c) 86 (d) 106
Solution. (a) The numbers are multiplied by 2 to get the next number.
64 × 2 = 128
Example 4. 5, 20, 80, 320, … 1280
(a) 5120 (b) 5220 (c) 4860 (d) 3642
Solution. (a) The numbers are multiplied by 4 to get the next number.
1280 × 4 = 5120
Difference Series:
Example 5. 3,6,9,12,15,…. 21
(a) 16 (b) 17 (c) 20 (d) 18
Solution. (d) The difference between the numbers is 3.
15 + 3 = 18
Example 6. 55, 50, 45, 40,….30
(a) 33 (b) 34 (c) 35 d) 36
Solution. (c) The difference between the numbers is -5.
40 – 5 = 35
Division Series:
Example 7. 5040, 720, 120, 24, ….2,1
(a) 8 (b) 7 (c) 6 (d) 5
Solution. (c)
Example 8. 16, 24, 36,… 81
(a) 52 (b) 54 (c) 56 (d) 58
Solution. (b) Previous number × = Next number
n2 Series
Example 9. 4, 16, 36, 64, …. 144
(a) 112 (b) 78 (c) 100 (d) 81
Solution. (c) The series is square of consecutive even numbers. 22, 42,62, 82
Next number is 102 = 100
Example 10. 1, 4, 9, 16, 25, 36, 49, … 81
(a) 100 (b) 121 (c) 64 (d) 144
Solution. (c) The series is 12, 22, 32, 42, 52,62, 72,….
The next number is 82 = 64
(n2 + 1) Series
Example 11. 17, 26, 37, 50, 65,….101
(a) 82 (b) 75 (c) 78 (d) 90
Solution. (a) The series is 42 + 1, 52 +1, 62 + 1, 72 + 1, 82 + 1.
The next number is 92 + 1 = 82
Example 12. 101, 401, 901, 1601, 2501, …. 4901
(a) 2201 (b) 3301 (c) 4401 (d) 3601
Solution. (d) The series is 102 + 1, 202 +1, 302 + 1, 402 + 1, 502 + 1, etc.
The next number is 602 + 1 = 3601
(n2 -1) Series
Example 13. 3, 8, 15, 24,…48
(a) 32 (b) 33 (c) 34 (d) 35
Solution. (d) The series is 22 – 1, 32 –1, 42 – 1,52 – 1. etc.
The next number is 62 – 1 =35
Example 14. 99, 80, 63,….35
(a) 48 (b) 84 (c) 46 (d) 64
Solution. (a) The series is 102 -1, 92 -1, 82 -1, etc.
The next number is 72 – 1 = 48
(n2 + n) Series
Example 15. 2, 6, 12, 20, 30,…. 56
(a) 32 (b) 34 (c) 42 (d) 24
Solution. (c) The series is 12 + 1, 22 + 2, 32 + 3, 42 + 4, 52 + 5, etc.
The next number is 62 + 6 = 42
Example 16. 110, 132, 156, 182,….
(a) 212 (b) 201 (c) 211 (d) 210
Solution. (d) The series is 102 + 10, 112 + 11, 122 + 12, etc.
The next number is 142 + 14 = 210
(n2 – n) Series
Example 17. 0, 2, 6, 12, 20,….42
(a) 25 (b) 30 (c) 32 (d) 40
Solution. (b) The series is 12 – 1 = 0, 22 – 2 = 2, 32 – 3 = 6, etc.
The next number is 62 – 6 = 30
Example 18. 90, 380, 870, 1560,…..
(a) 2405 (b) 2450 (a) 2400 (d) 2455
Solution. (b) The series is 102 – 10, 202 – 20, 302 – 30, etc.
The next number is 502 – 50 = 2450
n3 Series
Example 19. 1, 8, 27, 64,…. 216
(a) 125 (b) 512 (c) 215 (d) 122
Solution. (a) The series is 13, 23, 33 , 43, etc.
The next number is 53 = 125
Example 20. 1000, 8000, 27000, 64000,….
(a) 21600 (b) 125000 (c) 152000 (d) 261000
Solution. (b) The series is 103 , 203, 303, 403, etc.
The next number is 503 = 125000
(n3 + 1) Series
Example 21. 2, 9, 28, 65,…217
(a) 123 (b) 124 (c) 125 (d) 126
Solution. (d) The series is 13 +1, 23 + 1, 33 + 1, etc.
The next number is 53 + 1 = 126
Example 22. 1001, 8001, 27001, 64001, 125001,….
(a) 261001 (b) 216001 (c) 200116 (d) 210016
Solution. (b) The series is 103 + 1, 203 + 1, 303 + 1, etc.
The next number is 603 + 1 = 216001
(n3 -1) Series
Example 23. 0, 7, 26, 63, 124,…
(a) 251 (b) 125 (c) 215 (d) 512
Solution. (c) The series is 13 – 1, 23 – 1, 33 – 1, etc.
The next number is 63 – 1 = 215
Example 24. 999, 7999, 26999, 63999,….
(a) 199924 (b) 124999 (c) 129994 (d) 999124
Solution. (b) The series is 103 – 1, 203 – 1, 303 – 1, etc.
The next number is 503 – 1 = 124999
(n3 + n) Series
Example 25. 2, 10, 30, 68,….222
(a) 130 (b) 120 (c) 110 d) 100
Solution. (a) The series is 13 + 1, 23 + 2, 33 + 3, etc.
The next number is 53 + 5 = 130
Example 26. 1010, 8020, 27030, 64040,….
(a) 125500 (b) 125050 (c) 100255 (d) 120055
Solution. (b) The series is 103 + 10 = 1010, 203 + 20 = 8020, etc.
The next number is 503 + 50 = 125050
(n3 – n) Series
Example 27. 0, 6, 24, 60,…. 210
(a) 012 (b) 210 (c) 201 (d) 120
Solution. (d) The series is 13 – 1 = 0, 23 – 2 = 6, 33 – 3 = 24, etc.
The next number is 53 – 5 = 120
Example 28. 990, 7980, 26970, 63960,….
(a) 124500 (b) 124005 (c) 120045 (d) 124950
Solution. (d) The series is 103 – 10, 203 – 20, 303 – 30 etc.
The next number is 503 – 50 = 124950
Case 1 : Completing the given series by finding the missing terms.
Directions : Find the missing terms in each of the following series.
Ex. 1. 1,6,15,?,45,66,91
(a)25 (b) 26 (c) 27 (d) 28
Soln. Clearly, the given sequence follows the pattern :
+5,+9,+13,+17,+21,+25,……
Thus, 1+5=6,6+9=15,…..
So, missing terms=15+13=28.
Hence, the answer is (d).
Ex. 2. 2,5,9,19,37,?
(a) 73 (b) 75 (c) 76 (d) 78
Soln. Clearly, we have : 2x2+1=5,5x2-1=9,9x2+1=19,19x2-
1=37,.
So, missing terms=37x2+1=75.
Hence, the answer is (b).
Ex. 3. 4,8,28,80,244,?
(a) 278 (b) 428 (c) 628 (d) 728
Soln. The terms of the given series are : 31+1,32-2,33+1,34-
1,35+1
So,missing terms=36-1=729-1=728
Hence, the answer is (d).
Ex. 4. 10000,11000,9900,10890,9801,?.
(a) 10241 (b) 10423 (c) 10781 (d) 10929
Soln. Clearly,alternately we add and subtract 10% of a term to
obtain the next term of the series .
Thus, 10000+(10% of 10000)=11000;
11000-(10% of 11000)=9900
9900+(10% of 9900)=10890;
10890-(10% of 10890)=9801;
So,missing term=9801+(10% of 9801)=9801+980=10781.
Hence, the answer is (c).
Ex.5. 0,6,24,60,120,210,?
(a) 240 (b) 290 (c) 336 (d) 504
Soln. Clearly,the given series is : 13-1,23-2,33-3,43-4,53-5,63-6
So, Missing term =73-7=343-7=336
Hence, the answer is (c).
Ex.6. 1,4,27,16,?,36,343
(a) 25 (b) 87 (c) 120 (d) 125
Soln. Clearly,the given series consists of cubes of odd number
and square of even number. i.e. 13,22,33,42,…..
So, missing term=53=125
Hence, the answer is (d).
Ex.7. 4,6,12,14,28,30,?
(a) 32 (b) 60 (c) 62 (d) 64
Soln. The given sequence is a combination of two series :
I. 4,12,28,? And II. 6,14,30,…..
Now, the pattern followed in each of the two above series is : +8, +16, +32,…….
So, missing no =(28+32)=60.
Hence ,the answer is (b).
Ex.8. 1,3,3,6,7,9,?,12,21
(a) 10 (b) 11 (c) 12 (d) 13
Soln . Clearly, the given sequence is a combination of two series :
I.1,3,7,?,21 and II. 3,6,9,12
The pattern followed in I is +2,+4,……. And the pattern in
II is +3,.
So, missing no is =7+6=13.
Hence , the answer is (d).
Ex.9. which fraction comes next in the sequence
1/2,3/4,5/8,7/16,?
(a) 9/32 (b)10/17 (c) 11/34 (d) 12/35
Soln. Clearly, the numerators of the fractions in the given
sequence form the series 1,3,5,7 in which each
term is obtained by adding 2 to the previous terms.
The denominators of the fractions form the series
2,4,8,16 i.e. 21,22,23,24
So,the numerator of the next fraction will be (7+2) i.e. 9
and the denominator will be 25
Thus, the next term is 9/32.
Hence , the answer is (a).
CASE II: Finding the Wrong term in the Given Series.
Ex1.Find the wrong number in the series :
7, 28, 63, 124, 215, 342, 511
(a) 7 (b)28 (c)124 (d)215 (e)342
Sol. Clearly, the correct sequence is :
23-1,33-1,43-1,53-1,63-1,73-1,83-1
So, 28 is wrong and should be replaced by 33-1.
Hence, the answer is (b).
Ex2. Find the wrong number in the series :
3, 8, 15, 24, 34, 48, 63
(a) 15 (b)24 (c)34 (d)48 (e)63
Sol. The difference between consecutive terms of the given
series 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).
Ex3. Identity the wrong number in the series :
69, 55, 26, 13, 5
(a)5 (b)13 (c)26 (d)55
Sol. Clearly in the given series, each term is one more than the
product of the digits of the preceding term. Thus,
(6x9)+1=55, (5x5)+1=26, (2x6)+1=13.
So, 5 is wrong and must be replaced by (1x3)+1 i.e.4.
Hence, the answer is (a).
EXERCISE -1
Directions : In each of the following questions, a number series is given with one term missing.Chosse the correct alternative that will continue the same pattern and replace the questions mark in the given series.
1. 1,9,25,49,?,121.
(a) 64 (b) 81 (c) 91 (d) 100
Ans B, Series consists of square of ODD numbers.
--> 1^2, 3^2, 5^2, 7^2, 9^2, 11^2
--> 1 , 9 , 25 , 49 , 81 , 121
1, 9, 25, 49, 81, 121
2. 4,7,12,19,28,?.
(a) 30 (b) 36 (c) 39 (d) 49
Ans : C, The sequence in n^2 + 3
1+3, 4+3, 9+3, 16+3, 25+3, Hence 36+3 = 39
3. 11,13,17,19,23,25,?
(a) 26 (b) 27 (c) 29 (d) 37
Ans: C , The pattern is + 2, + 4, + 2, + 4, .....
So, missing term = 25 + 4 = 29.
4. 6,12,21,?,48.
(a) 33 (b) 38 (c) 40 (d) 45
Ans A, Pattern is +6, +9, +12, +15...
6 +6=12, 12+9=21
Missing term : 21 +12=33
5. 2,5,9,?,20,27
(a) 14 (b) 16 (c) 18 (d) 24
Ans. A, it is increased in the order of no.+3,no.+4,no.+5,no.+6.
i.e=> 2+3=5 , 5+4=9 , 9+5=14 , 14+6=20 , 20+7=27.
6. 6,11,21,36,56,?
(a) 42 (b) 51 (c) 81 (d) 91
Ans: C, The pattern is + 5, + 10, + 15, + 20,....
So, missing term = 56 + 25 = 81.
7. 10,18,28,40,54,70,?
(a) 85 (b) 86 (c) 87 (d) 88
Ans: D, the pattern is + 8, + 10, + 12, + 14, .....
So, missing term = 70 + 18 = 88.
8. 120,99,80,63,48,?
(a) 35 (b) 38 (c) 39 (d) 40
Ans: A, The pattern is - 21, - 19, - 17, - 15,.....
So, missing term = 48 - 13 = 35.
9. 22,24,28,?,52,84.
(a) 36 (b) 38 (c) 42 (d) 46
Ans: A
The difference of each successive term is 4,8,16,32
Hence 28+8 = 36 or 52-16= 36
10. 4832,5840,6848,?
(a) 7815 (b) 7846 (c) 7856 (d) 7887
Ans:C
The pattern is + 1008. So, missing term - 6848 + 1008 = 7856
11. 10,100,200,310,?
(a) 400 (b) 410 (c) 420 (d) 430
Ans: D
This pattern is +90, +100, +110, +120
Hence 310+120 =430
12. 0,2,8,14,?,34
(a) 20 (b) 23 (c) 24 (d) 25
Ans : C, 0+2=2, 2+(2+4)=8, 8+(2+4)=14, 14+(2+4+4)=24, 24+(2+4+4)=34
13. 28,33,31,36,?,39.
(a) 32 (b) 34 (c) 38 (d) 40
Ans:B. , Series like +5,-2,+5,-2.......
28 +5=33, 33 -2=31, 31 +5=36, 36 -2=34
14. 125,80,45,20,?.
(a) 5 (b) 8 (c) 10 (d) 12
Ans: A, Series decreased by 5 *9,5 *7,5 *5,5 *3(i.e, 45,35,25,15)
125-45 =80, 80 -35 =45, 45 -25 =20, 20 -15 =5
Final sequence = 125,80,45,20,5
15. 1,5,13,25,41,?
(a) 51 (b) 57 (c) 61 (d) 63
Ans: C 1+4 = 5, 5+8 = 13, 13+12 = 25, 25+16 = 41,41+20 = 61
nth term is, 2n(n-1) + 1
16. 2,15,41,80,?
(a) 111 (b) 120 (c) 121 (d) 132
Ans: D, The pattern is like = + 13, + 26, + 39,...
Missing number = 80 + 52 = 132.
17. 6,17,39,72,?
(a) 83 (b) 94 (c) 116 (d) 127
Ans: C., 17-6=11, 39-17=22, 72-39=33, 116-72=44.
they are increased by 11.
18. 325,259,204,160,127,105,?.
(a) 94 (b) 96 (c) 98 (d) 100
Ans : A
19. 1,4,10,22,46,?
(a) 64 (b) 86 (c) 94 (d) 122
Ans: C , 4*2+2=10, 10*2+2=22, 22*2+2=46, 46*2+2=94, 94*2+2=190, 190*2+2=382
20. 0.5,0.55,0.65,0.8,?
(a) 0.9 (b) 0.82 (c) 1 (d) 0.95
Ans: C, The pattern is + 0.05, 0.10, + 0.15,....
Missing number = 0.8 + 0.20 = 1.
21. 5,6,9,15,?,40
(a) 21 (b) 25 (c) 27 (d) 33
Ans: B
5 6 9 15 25 40, 1 3 6 10 15, 2 3 4 5
22. 2,3,5,7,11,?,17
(a) 12 (b) 13 (c) 14 (d) 15
Ans: B.,It is series contains consecutive prime numbers.
Final = 2,3,5,7,11,13,17
23. 4,9,25,?,121,169,289,361
(a) 49 (b) 64 (c) 81 (d) 87
Ans: A, because each consecutive prime number is squared.
24. 1,9,25,49,81,?
(a) 100 (b) 112 (c) 121 (d) 144
Ans: C.
the series is +8,+16,+32,+40.....
therefore 81+40=121
25. 1,1,4,8,9,27,16,?
(a) 32 (b) 64 (c) 81 (d) 256
Ans: B.
The sequence is (1^2)(1^3)(2^2)(2^3)(3^2)(3^3)(4^2)(4^3)
26. 4,12,36,108,?
(a) 144 (b) 216 (c) 30 (d) 324
Ans: D The Pattern is
4X3=12, 12x3=36, 36x3=108, 108x3= 324
27. 1,1,2,6,24,?,720.
(a) 100 (b) 104 (c) 108 (d) 120
Ans:D 20/6=120, 120/5=24, 24/4=6
6/3=2, 2/2=1, 1/1=1
28. 240,?,120,40,10,20
(a) 180 (b) 240 (c) 420 (d) 480
Ans: B., From the back, series is like
2 *5 = 10, 10 *4 = 40, 40 *3 = 120, 120*2 = 240, 240*1 = 240
So, missing term is 240
29. 4,6,9,13.1/2,?
(a) 17.1/2 (b) 19 (c) 20.1/4 (d) 22.3/4
30. 5760,960,?,48,16,8.
(a) 120 (b) 160 (c) 192 (d) 240
Ans:C., The pattern is 6, 5, 4, 3, 2.
So, missing term = 960 - 5 = 192.
31. 1,2,6,7,21,22,66,67,?
(a) 70 (b) 134 (c) 201 (d) 301
Ans: C, The pattern is + 1, x 3, + 1, x 3, + 1, x 3, + 1,.....
So, missing term = 67 x 3 = 201.
32. 48,24,96,48,192,?
(a) 76 (b) 90 (c) 96 (d) 98
Ans: C The Pattern is
48/2= 24, 96/2=48, 192/2=96
33. 1,2,3,6,9,18,?,54.
(a) 18 (b) 27 (c) 36 (d) 81
Ans: B The pattern is x 2, x 3/2, x 2, x 3/2, x 2,.....
So, missing term = 18 x 3/2 = 27.
34. 165,195,255,285,345,?.
(a) 375 (b) 390 (c) 420 (d) 435
Ans: D , 15*11=165, 15*13=195, 15*17=255, 15*19=285, 15*23=345, 15*29=435
35. 9,27,31,155,161,1127,?
(a) 316 (b) 1135 (c) 1288 (d) 2254
Ans: B , (9*3)27..(27+4)31..(31*5)155..(155+6)161..(161*7)1127..(1127+8 )1135
36. 2,3,3,5,10,13,?,43,172,177
(a) 23 (b) 38 (c) 39 (d) 40
Ans: C, 43*4= 172,
so series becomes
2, 3, 3, 5, 10, 13, 39, 43,172,177
37. 3,15,?,63,99,143
(a) 27 (b) 35 (c) 45 (d) 56
Ans: B
The difference between , 3,15 - 12
15,x - Unknown 1
x,63 - - Unknown 2
63,99 - 36
99,143 - 44
143,195 - 52
The difference between them increases by 8
So the *unknown difference* are 20,28.
15 + 20 = 35 (or) 63 - 28 = 35
38. 7,26,63,124,215,342,?.
(a) 391 (b) 421 (c) 481 (d) 511
Ans: D
The sequence is:
U[n] = (n + 1)³ - 1 = n³ + 3n² + 3n = n (n² + 3n + 3)
39. 3,7,15,?,63,127.
(a) 30 (b) 31 (c) 47 (d) 52
Ans: B
3x2=6+1=7, 7x2=14+1=15, 15x2=30+1=31, 631x2=62+1=63 so on
40. 4,10,?,82,244,730
(a) 24 (b) 28 (c) 77 (d) 218
Ans: B, the difference is tripled, each term.
41. 6,13,25,51,101,?.
(a) 201 (b) 202 (c) 203 (d) 205
42. 8,28,116,584,?.
(a) 1752 (b) 3502 (c) 3504 (d) 3508
Ans: D 8*3+4=28, 28 *4 +4 =116, 116 * 5 +4 =584, 584*6+4=3508
43. 6,13,28,59,?.
(a) 111 (b) 113 (c) 114 (d) 122
Ans: D., The pattern is x 2 + 1, x 2 + 2, x 2 + 3,.....
So, missing term = 59 x 2 + 4 = 122.
44. 3,7,23,95,?
(a) 62 (b) 128 (c) 479 (d) 575
Ans: C 3 * 2 +1 = 7
7 * 3 +2 =23
23 *4 +3 =95
95 *5+4 = 479
see the number you multiply it goes up by one, as does the number you add
45. 2,3,8,27,112,?
(a) 226 (b) 339 (c) 452 (d) 565
Ans is D
ie is 2*1+1=3, 3*2+2=8, 8*3+3=27, 27*4+4=112, 112*5+5=565
46. 1,5,14,30,55,91,?.
(a) 130 (b) 140 (c) 150 (d) 160
Ans: B, Series is increasing +2^2, +3^2, +4^2, +5^2, +6^2, +7^2,
1 +(2^2) = 5, 5 +(3^2) = 14, 14+(4^2) = 30, 30+(5^2) = 55, 55+(6^2) = 91
91+(7^2) = 140
47. 198,194,185,169,?.
(a) 92 (b) 112 (c) 136 (d) 144
Ans: D
48. 2,2,5,13,28,?.
(a) 49 (b) 50 (c) 51 (d) 52
Ans: D, There difference b/w series is 0,3,8,15,24,...(2-2,5-2,..)
ie,
2*0=0, 3*1=3, 4*2=8, 5*3=15 so then
6*4=24.
so 28+24=52 is the answer
49. 2,7,27,107,427,?
(a) 1262 (b) 1707 (c) 4027 (d) 4207
Ans: B
1707------ 2,7,27,107,427,1707------difference between them is 5,20,80,320----and differrence btwn dis is all multiplication to 4 each ... So the result comes nxt is = 1707
50. 24,60,120,210,?
(a) 300 (b) 336 (c) 420 (d) 525
Ans B , The pattern is+ 36, + 60, + 90, ....
i.e., + [ 6 * (6 + 0) ], + [ 6 * (6 + 4) ], + [ 6 * (6 + 9) ],.....
.'. Missng number = 210 + [ 6 * (6 + 15)] = 210 + 126 = 336.
EXERCISE-2
Directions : In the following questions, one term in the number series is wrong. Find out the wrong term.
1. Find out the incorrect number 121, 143, 165, 186, 209
A. 143 B. 165 C. 186 D. 209
Answer: C
Each term of the series is increased by 22 to obtain the next term.
So, 186 is wrong and must be replaced by (165 +22) i.e. 187.
2. Find out the incorrect number 1, 2, 4, 8, 16, 32, 64, 96
A. 4 B. 32 C. 64 D. 96
Answer: D
Each term of the series is obtained by multiplying the preceding term by 2.
So, 96 is wrong and must be replaced by (64 * 2) i.e. 128.
3. Find out the incorrect number 13, 21, 32, 47, 63, 83
A. 13 B. 21 C. 32 D. 47
Answer: D
The sequence is + 5, + 8, + 11,...
.'. 47 is wrong and must be replaced by (32 + 14) i.e. 46.
4. Find out the incorrect number 380, 188, 92, 48, 20, 8, 2
A. 188 B. 92 C. 48 D. 20
Answer: C
Each term in the series is four more than two times the next term.
So, 48 is wrong and must be replaced by (20 * 2 + 4) i.e. 44.
5. Find out the incorrect number 1, 3, 7, 15, 27, 63, 127
A. 7 B. 15 C. 27 D. 63
Ans: C, Go on multiplying the number by 2 and adding 1 to it to get the next number.
So, 27 is wrong.
6. Find out the incorrect number 5, 10, 17, 24, 37
A. 10 B. 17 C. 24 D. 37
Answer: C , The sequence is + 5, + 7,.....
So, 24 is wrong and should be replaced by (17 + 9) i.e. 26.
7. Find out the incorrect number 1, 3, 10, 21, 64, 129, 256, 778
A. 10 B. 21 C. 129 D. 256
Answer: D, The sequence is * 2 + 1, * 3 + 1, * 2 + 1, * 3 + 1,....
So, 256 is wrong and must be replaced by (129 * 2 + 1) i.e. 259.
8. Find out the incorrect number 15, 16, 22, 29, 45, 70
A. 16 B. 22 C. 45 D. 70
Answer: B , The pattern is + 1, + 4, + 9, + 16, + 25,.... i.e. + 12, + 22, + 32, + 42, + 52,...
So, 22 is wrong and must be replaced by (16 + 4) i.e. 20.
9. Find out the incorrect number 6, 14, 30, 64, 126
A. 6 B. 14 C. 64 D. 126
Answer: C , Each term is multiplied by 2 and then increased by 2 to obtain the next term. So, 64 is wrong and must be replaced by (30 * 2 + 2) i.e. 62.
10. Find out the incorrect number 10, 26, 74, 218, 654, 1946, 5834
A. 26 B. 74 C. 218 D. 654
Answer: D , Each term is 4 less than thrice the preceding number.
So, 654 is wrong and must be replaced by (218 * 3 - 4) = 650.
11. Find out the incorrect number 3, 7, 15, 39, 63, 127, 255, 511
A. 15 B. 39 C. 63 D. 127
Answer: B, Each number in the series is multiplied by 2 and the result increased by 1 to obtain the next number.
So, 39 is wrong and should be replaced bu (15 * 2 + 1) i.e. 31.
12. Find out the incorrect number 445, 221, 109, 46, 25, 11, 4
A. 25 B. 46 C. 109 D. 221
Answer: B, 3 is subtracted from each number and the result is divided by 2 to obtain the next number of the series.
So , 46 is wrong and must be replaced by 109 - 3 / 2 i.e. 53.
13. Find out the incorrect number 1236, 2346, 3456, 4566, 5686
A. 1236 B. 3456 C. 4566 D. 5686
Answer: D ,The first digits of the numbers from the series 1, 2, 3, 4, 5;
the second digits from the series 2, 3, 4, 5,6;
the third digits from the series 3, 4, 5, 6;
while the last digit in each of the numbers is 6.
So, 5686 is wrong and must be replaced by 5676.
14. Find out the incorrect number 5, 10, 40, 80, 320, 550, 2560
A. 80 B. 320 C. 550 D. 2560
Answer: C , The sequence is * 2, * 4, * 2, * 4,....
So, 550 is wrong and must be replaced by (320 * 2) i.e. 640.
15. Find out the incorrect number 3, 2, 8, 9, 13, 22, 18, 32, 23, 42
A. 8 B. 9 C. 13 D. 22
Answer: B , The given sequence is a combination of two series :
I. 3, 8, 13, 18, 23 and
II 2, 9, 22, 32,42
The pattern in I is + 5 and the pattern in II is + 10.
So, in II, 9 is wrong and must be replaced by (2 + 10) i.e. 12
16. Find out the incorrect number 8, 27, 125, 343, 1331
A. 8 B. 343 C. 1331 D. None of these
Answer: D The numbers are cubes of prime numbers i.e. 23, 33, 53, 73, 113.
Clearly, none is wrong.
17. Find out the incorrect number 8, 27, 125, 343, 1331
A. 8 B. 343 C. 1331 D. None of these
Answer: D ,The numbers are cubes of prime numbers i.e. 23, 33, 53, 73, 113.
Clearly, none is wrong.
18. Find out the incorrect number 10, 14, 28, 32, 64, 68, 132
A. 28 B. 32 C. 64 D. 132
Answer: D , Alternately, the numbers are increased by four and doubled to get the next number>Thus, 10 + 4 = 14; 14 * 2 = 28; 28 + 4 = 32; 32 * 2 =64 and so on.
So, 132 is wrong and must be replaced by (68 * 2) i.e. 136.
19. Find out the incorrect number 1, 5, 5, 9, 7, 11, 11, 15, 12, 17
A. 11 B. 12 C. 17 D. 15
Answer: B , The given sequence is a combination of two series :
I. 1, 5, 7, 11, 12 and
II. 5, 9, 11, 15, 17
The pattern in both I and II is + 4, + 2, + 4, + 2.
So, 12 is wrong and must be replaced by (11 + 2) i.e. 13.
20. Find out the incorrect number 11, 2, 21, 3, 32, 4, 41, 5, 51, 6
A. 21 B. 11 C. 32 D. 51
Answer: C , The given sequence is a combination of two series :
I. 11, 21, 32, 41, 51 and
II. 2, 3, 4, 5, 6.
Clearly, the pattern in I is + 10.
So, 32 is wrong and should be replaced by (21 + 10) i.e. 31
21. Find out the incorrect number 11, 5, 20, 12, 40, 26, 74, 54
A. 5 B. 20 C. 40 D. 26
Answer: C , The given sequence is a combination of two series :
I. 11, 20, 40, 74 and
II. 5, 12, 26, 54, The pattern in I becomes + 9, + 18, + 36, ... if 40 is replaced by 38.
So, 40 is wrong.
22. Find out the incorrect number 56, 72, 90, 110, 132, 150
A. 72 B. 90 C. 110 D. 150
Answer: D , The numbers are 7 * 8, 8 * 9, 9 * 10, 10 *11, 11 * 12, 12 * 13.
So, 150 is wrong and must be replaced by (12 * 13) i.e. 156.
23. Find out the incorrect number 8, 13, 21, 32, 47, 63, 83
A. 13 B. 32 C. 47 D. 63
Answer: C , The sequence is + 5, + 8, + 11,....
So, 47 is wrong and must be replaced by (32 + 14) i.e. 46.
24. Find out the incorrect number 89, 78, 86, 80, 85, 82, 83
A. 83 B. 82 C. 86 D. 78
Answer: C , The sequence is - 11, + 9, - 7, + 5, - 3, + 1.
So, 86 is wrong and should be replaced by (78 + 9) i.e. 87.
25. Find out the incorrect number 25, 36, 49, 81, 121, 169, 225
A. 36 B. 49 C. 169 D. 225
Answer: A , The correct sequence is 52, 72, 92, 112, 132, 152.
So, 36 is wrong.
26. Find out the incorrect number 2, 5, 10, 17, 26, 37, 50, 64
A. 17 B. 26 C. 37 D. 64
Answer: D, The numbers are 12 + 1, 22 + 1, 32 + 1 and so on.
So, 64 is wrong. The correct term is (82 + 1) i.e. 65.
27. Find out the incorrect number 1, 5, 9, 16, 25, 37, 49
A. 9 B. 15 C. 25 D. 37
Answer: B The given sequence is a combination of two series :
I. 1, 9, 25, 49 and
II. 5, 15, 37, The pattern in I is + 8, + 16, + 24.
The sequence in II is 22 + 1, 42 + 1, 62 + 1.
So, 16 is wrong and must be replaced by (42 + 1) i.e. 17.
28. Find out the incorrect number 2, 5, 10, 50, 500, 5000
A. 5 B. 10 C. 50 D. 5000
Answer: D , Each term of the series is the product of the preceding two terms.
So, 5000 is wrong and must br replaced by (50 * 500) i.e 25000.
29. Find out the incorrect number 46080, 3840, 384, 48, 24, 2, 1
A. 384 B. 48 C. 24 D. 2
Answer: C , The terms are successfully divided by 12, 10, 8, 6,....
So, 24 is wrong and must be replaced by (48 / 6) i.e. 8.
30. Find out the incorrect number 105, 85, 60, 30, 0, -45, -90
A. 105 B. 60 C. 0 D. -45
Answer: C , The sequence is - 20, - 25, - 30,....
So, 0 is wrong and must be replaced by (30 - 35) i.e. -5
31. Find out the incorrect number 325, 259, 202, 160, 127, 105, 94
A. 94 B. 127 C. 202 D. 259
Answer: C , The sequence is - 66, - 55, - 44, - 33, - 22, -11.
So, 202 is wrong.
The correct term is (259 - 55) i.e 204.
32. Find out the incorrect number 125, 126, 124, 127, 123, 129
A. 126 B. 124 C. 123 D. 129
Answer: D , The sequence is + 1, - 2, + 3, - 4, + 5.
So, 129 is wrong and and must be replaced by (123 + 5) i.e. 128.
33. Find out the incorrect number 3, 4, 10, 32, 136, 685, 4116
A. 10 B. 32 C. 685 D. 4116
Answer: B ,The sequence is as follows :
2nd term = (1st term + 1) * 1
3rd term = (2nd term + 1) * 2
4th term = (3rd term + 1) * 3 and so on.
So, 32 is wrong and must be replaced by (10 + 1) * 3 i.e. 33.
34. Find out the incorrect number 3, 10, 27, 4, 16, 64, 5, 25, 125
A. 3 B. 4 C. 10 D. 27
Answer: C, The correct sequence is 3, 32, 33, 4, 42, 43, 5, 52, 53.
So ,10 is wrong and should be replaced by 32 i.e. 9.
35. Find out the incorrect number 5, 27, 61, 122, 213, 340, 509
A. 27 B. 61 C. 122 D. 509
Answer: A ,The correct sequence is 23 - 3, 33 - 3, 43 - 3, 53 - 3, 63 - 3, 73 - 3, 83 - 3.
So, 27 is wrong and should be replaced by 33 - 3 i.e. 24.
ALPHABET SERIES
In this type of questions, a series of single , pairs or group of letters of combination of letters and numeral is given .The terms of the series form a certain pattern as regards the position of the letter in the English alphabet. The candidate is required to decipher this pattern and accordingly find the missing term or the wrong term in the given series.
ILLUSTRATIVE EXAMPLES.
Ex. 1. Find the next two terms in the series : A,C,F,J,?,?,
(a) L,P (b) M,O (c) O,U (d) R,V
Sol. Clearly,the 1st,2nd,3rd,…..letters of the series are respectively moved 2,3,4 steps forward to obtain the successive terms of the series.
Thus, the 5th term in the series must be a letter which is 5 steps ahead of J i.e. O, while the 6th term must be a letter six steps a head of O i.e. U.
Thus,we have the following pattern:
+2 +3 +4 +5 +6
A---Ã C---Ã F----Ã J----Ã O----Ã U
So, the missing term are O and U.
Hence ,the answer is (c).
Ex.2. Which term comes next I the sequence :AC,FH,KM,PR,?
(a) UW (b) VW (c) UX (d) TV
Soln : Clearly, the first and the 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. 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).
Ex.3. Find the next term in the series : BMO,EOQ,HQS,?
(a) KSU(b) LMN(c) SOV(d) SOW
Soln : Clearly,we observe the following pattern :
+3 +3 +3
The 1st letter follows the pattern +3 i.e B----Ã E-----Ã H----Ã K
+2 +2 +2
The 2nd letter follows the pattern +2 i.e M----Ã O-----Ã Q----Ã S
+2 +2 +2
The 3rd letter follows the pattern +2i.e O----Ã Q-----Ã S--Ã U
Thus,the missing term is KSU.
Hence,the answer is (a).
Ex.4. Which term comes next in the series : YEB,WFD,UHG,SKI,?
(a) QOL (b) QGL (c) TOL (d) QNL
Sol. Clearly,we observe the following pattern :
-2 -2 -2 -2
1st letter : Y----Ã W-----Ã U----Ã S--Ã Q
+1 +2 +3 +4
2nd letter: E---Ã F----Ã H-----Ã K----Ã O
+2 +3 +2 +3
3rd letter : B----Ã D-----Ã G----Ã I---Ã L
Thus the missing term is QOL.
Hence, the answer is (a).
Ex.5. Which term will replace the ? in the series:
ABD,DGK,HMS,MTB,SBL,?
(a) ZKU (b) ZKW (c) ZAB (d) XKW
Soln. : Clearly,we observe the following pattern :
+3 +4 +5 +6 +7
1st letter : A----Ã D-----Ã H----Ã M---Ã S---Ã Z
+5 +6 +7 +8 +9
2nd letter: B----Ã G-----Ã M----Ã T-----Ã B----Ã K
+7 +8 +9 +10 +11
3rd letter : D----Ã K-----Ã S----Ã B-----Ã L------Ã W
Thus,the missing term is ZKW.
Hence, the answer is (b).
EXERCISE-3
Directions: In each of the following questions ,various terms of an alphabet series are given with one or more terms missing as shown by (?). Choose the missing terms out of the given alternatives.
1. U, O, I, ?, A
A. E B. C C. S D. G
Answer: A, The series consists of vowles A, E, I, O, U written in a reverse order.
2. Y, W, U, S, Q, ?, ?
A. N, J B. M, L C. J, R D. L, M E. O, M
Answer: E ,
The series consits of alternate letters in reverse order.
3. A, B, D, G, ?
A. M B. L C. K D. H
Answer: C
The first, second, third,.... letters of the series are respectively moved one, two, three,... steps forward to obtain the successive terms.
4. Z, U, Q, ?, L
A. I B. K C. M D. N
Answer: D
The first, second, third,... letters of the series are respectively moved one, two, three,... steps forward to obtain the successive terms.
5. A, C, F, H, ?, M
A. L B. K C. J D. I
Answer: B .
The letters are alternately moved two and three steps forward to obtain the successive terms.
6. A, Z, X, B, V, T, C, R, ?, ?
A. P, D B. E, O C. Q, E D. O, Q E. Q, O
Answer: A
The first, fourth and seventh letters are in alphabetical order.So, tenth letter would be the letter after C i.e. D.Also, the second and third letters are alternate and in reverse order and so are the fifth and sixth letters and the eighth and ninth letters.
7. R, M, ?, F, D, ?
A. C, B B. J, H C. B, H D. H, C E. I, C
Answer: E
Letter are in reverse order in which from the last 0, 1, 2, 3 and 4 letters are missing between two consecutive letters.
8. R, M, ?, F, D, ?
A. C, B B. J, H C. B, H D. H, C E. I, C
Answer: E
Letter are in reverse order in which from the last 0, 1, 2, 3 and 4 letters are missing between two consecutive letters.
9. Z, L, X, J, V, H, T, F, ?, ?
A. R, D B. R, E C. S, E D. Q, D
Answer: A .
They given sequence consists of two series -- Z, X, V, T, ? and L, J, H, F, ?, both consisting of alternate letters in a reverse order.
10. Z, S, W, O, T, K, Q, G, ?
A. N, C B. N, D C. O, C D. O, D
Answer: A
The given sequence consists of two series :
I. Z, W, T, V, Q, ? in which each letter is moved three steps backward to obtain the next term.
II. S, O, K, G in which each letter is moved four steps backward to obtain the next term.
11. W, V, T, S, Q, P, N, M, ?, ?
A. I, J B. J, I C. J, K D. K, J
Answer: D
The letters are alternately moved one and two steps backward to obtain the successive terms.
12. Z, Y, X, U, T, S, P, O, N, K, ?, ?
A. H, G B. H, I C. I, H D. J, I
Answer: D
The given series consists of three consecutive letters from the end, then two letters skipped, then again three consecutive letters from the end so on.
13. b, e, d, f, ?, h, j, ?, l
A. i, m B. m, i C. i, n D. j, m
Answer: A
The series may be divided into groups as shown :
b e d / f i h / j m l
In each group, first letter is moved two steps forward to obtain the third letter
while the third letter is moved one step forward to obtain the second letter.
14. AZ, BY, CX, ?
A. EF B. GH C. IJ D. DE E. DW
Answer: E .
The first letter of each term is moved one steps forward and the second letter is moved one steps backward to obtain the corresponding letters of the next term.
15. AZ, CX, FU, ?
A. IR B. IV C. JQ D. KP
Answer: C
The first letter of the first, second, third, .... terms are respectively moved two, three, four, ... steps forward to obtain the first letter of the successive term.The second letter of the first, second, third.... terms are respectively moved two, three, four, ... steps ackward to obtain the second letter of the successive terms.
16. AZ, GT, MN, ?, YB
A. KF B. RX C. SH D. TS
Answer: C
The first letter of each term is moved six steps forward while the second letter is moved six steps backward to obtain the corresponding letter of the next term.
17. BF, CH, ?, HO, LT
A. DN B. EL C. EK D. EM E. FJ
Answer: C
The first letter of the first, second, third, .... terms are respectively moved one, two, three,... steps forward while the second letter are respectively moved two, three, four,... steps forward to obtain the corresponding letters of the successive terms.
18. CE, GI, KM, OQ, ?
A. TW B. TV C. SU D. RT E. UW
Answer: C
The letters of each term are alternate and also the last letter of each term and the first letter of the next term are alternate.
19. BD, GI, LN, QS, ?
A. TV B. UW C. WX D. WY E. VX
Answer: E
Each term of the series consists of two alternate letters and there is a gap of two letters between the last letter of each term and the first letter of the next term.
20. AD, EH, IL, ?, QT
A. LM B. MN C. MP D. OM
Answer: C
The first and second letter of each term are moved four steps forward to obtain the corresponding letters of the next term.
21. JE, LH, OL, SQ, ?
A. WV B. WX C. VW D. VX E. XW
Answer: E
The first letter of the first, second, third,... terms are respectively moved two, three, four,... steps forwardwhile the second letters of three terms are respectively moved three, four, five,... steps forward to obtain the corresponding letters of the successive terms.
22. DF, GJ, KM, NQ, RT, ?
A. UW B. YZ C. XZ D. UX E. YA
Answer: D
There is a gap of one letter between both the letters of first term, a gap of two letters between both the letters of second term and again a gap of one and two letters between the letters of third and fourth terms respectively.Beside, the last letter of each term and the first letter of next term are in alphabetical order.
23. cx, fu, ir, ?, ol, ri
A. lo B. mn C. no D. op E. or
Answer: A
The first letter of each term is moved three steps forward andthe second letter is moved three steps backward to obtain the corresponding letters of the next term.
24. OTE, PUF, QVG, RWH, ?
A. SYJ B. TCI C. SXJ D. SXI E. TYJ
Answer: D
The first letters of the terms are in alphabetical order,and so are the second and third letters.
25. eac, gce, ieg, ?
A. jhi B. jgi C. kgi D. khi E. kij
Answer: C
The first letters of the terms are alternate and so are the second and third letters.
26. ejo, tyd, ins, xch, ?
A. nrw B. mrw C. msx D. nsx E. nsw
Answer: B
There is a gap of four letters between the first and second, the second and third letters of each term,and also between the last letter of a term and the first letter of the next term.
27. CAT, FDW, IGZ, ?
A. KJA B. KTC C. LHD D. LJC
Answer: D
All the letters of each term are moved three steps forward to obtain the corresponding letters of the next term.
28. BEH, KNQ, TWZ, ?
A. IJL B. CFI C. BDF D. ADG
Answer: B
All the letters of each term are moved nine steps forward to obtain the corresponding letters of the next term.
29. deb, ijg, nol, ?, xyv
A. rsp B. stp C. rsq D. stq E. sto
Answer: D
All the letters of each term are moved five steps forward to obtain the corresponding letters of the next term.
30. ?, siy, oeu, kaq, qwm, cri
A. wnc B. wnb C. vnc D. vmc E. wmc
Answer: E
The letters in each term are moved four steps backward to obtain the corresponding letters of the next term.
31. QPO, SRQ, UTS, WVU, ?
A. XVZ B. ZYA C. YXW D. VWX E. AZY
Answer: C
Each term in the series consists of three consecutive letters in reverse order.
The first letter of each term and the last letter of the next term are the same.
32. ?, ayw, gec, mki, sqo
A. zxw B. bzw C. usq D. may E. xyv
Answer: C
Each term in the series consists of alternate letters in reverse order.The first letter of each term and the last letter of the next term arealso alternate.
33. dfe, jih, mln, ?, vut
A. oqp B. psr C. prq D. rsp E. oqr
Answer: C
There is a gap of three letters between the first letter of each term and the last letter of the next term.
34. DEF, HIJ, MNO, ?
A. STU B. RST C. RTV D. SRQ E. TUV
Answer: A .
The letters in each term are consecutive. There is a gap of one letter between the last letter of the first term and the first letter of the second term and a gap of two letters between the last letter of the second term and the first letter of third term. So, there would be a gap of three letters between the last letter of the third term and the first letter of the fourth term
35. FLP, INS, LPV, ?
A. ORY B. UXZ C. VXY D. SVW
Answer: A
The first and third letters of each term are moved three steps forward and the second letters is moved two steps forward to obtain the corresponding letters of the next term.
36. shg, rif, qje, pkd, ?
A. ole B. olc C. nmc D. nlb
Answer: B
The first and third letters of each term are moved one step backward andthe second letter is moved one step forward to obtain the corresponding letters of the next term.
37. LXF, MTJ, NPN, OLR, ?
A. PHV B. PIU C. PJW D. PKX E. PRV
Answer: A
The first letter of each term is moved one step forward, the second letter is moved four steps backward andthe third letter is moved four steps forward to obtain the corresponding letters of the next term.
38. MHZ, NIW, OKT, PNQ, ?
A. RRN B. QRN C. QRM D. QQN
Answer: B
The first letters of the terms are consecutive letters. The third letter of eac h term is moved three steps backward to obtain the third letter of the successive term. The middle letters of the first, second, third and fourth terms are moved one, two, three and four steps forward respectively to obtain the middle letter of the successive terms.
39. AYD, BVF, DRH, ?, KGL
A. FMI B. GMJ C. HLK D. GLJ
Answer: D
The first letters of the first, second, third and fourth are moved one, two, three and four steps forward respectively to obtain the first letter of the successive terms.The second letters of the first, second, third and fourth terms are moved three, four, five and six steps backward respectively to obtain the second letters of the successive terms.The last letters of the terms are alternate
40. AB, BA, ABC, CBA, ABCD, ?
A. ACBD B. BACD C. CABD D. DBAC E. DCBA
Answer: E
The first group of letters is reversed to obtain the second group. The second group is reversed and the next consecutive letter is added to it to obtain the subsequent group.
CONTINUOUS PATTERN SERIES
This type of questions consists of a series of small letters which follow a certain pattern. However, some letters are missing from the series . These missing letters are then given in a proper sequence as one of the alternatives. The candidate is requires to choose this alternative as answer.
Example : aab_aaa_bba_
(a) Baa (b) abb (c) bab (d) aab (e) bbb
Solution : we Proceed step by step as shown below :
1. The first blank space should be filled in by ‘b’ so that we have two a’s followed by two b’s .
2. The second blank space should be filled in either by ‘a’ so that we have four a’s followed by two b’s , or ‘b’ so that we have three a’s followed by three ‘b’.
3. The last space must be filled in by ‘a’.
4. Thus, we have two possible answer: ‘baa’ and ‘bba’. But, only ‘baa’ appear in the alternative. So, the answer is (a).
5. In case, we had both the possible answer in the alternatives, we would have chosen the one that forms a more prominent pattern, which is aabb/aaabbb/aa.Thus,our answer, would have been ‘bba.’
CORRESPONDENCE SERIES
This type of series consists of three sequence with three different elements(usually capital letters, digits and small latters).on th basis of the similarity in positions in the three sequence, a capital letter is found to correspond with a unique digit and a unique small letter, whenever it occurs. The candidate is requires to trace out this correspondence and accordingly choose the elements to be filled in at the desired places.
Consider the following example:
Ex. In the following series, choose the alternative which contains the numeral to be filled in the marked places,in the correct order :
B_ _ D _ _ C A B D A C B
- - 4 1 3 2 - - - ? ? ? ?
a _ a _ b c _ c _ _ _ _ _
(a) 1,2,3,4 (b) 2,3,1,4 (c) 1,2,4,3 (d) 2,1,4,3
Soln. Clearly, in the second series,1 occures at the same positions as D occurs in the first
Series.So, 1 corresponds to D.Thus, the first questions marks below D is to be replaced by 1. Now, in the third series, c at the 8th place corresponds to A in the first series, while c at the 6th place corresponds to 2 in the second series . so, 2 corresponds to A. Thus, the second questions marks below A is to be replaced by 2. In the third series, a at the first place corresponds to B in the first series and a at the third place corresponds to 4 in the Second series. so, 4 corresponds to B.Thus, the questions mark below B is to be replaced by 4.Now, only 3 remains. So, 3 correspond to C. Thus, the questions mark below C is to be replaced by 3.Thus, DACB corresponds to 1,2,3,4.
Hence, the answer is (a).
EXERCISE-4
Directions: In each of the following letter series, some of the letters are missing which are given in that order as one of the alternatives below it. Choose the correct alternative
1. _ _ aba _ _ ba _ ab
A. abbba B. abbab C. baabb D. bbaba
Answer: B , The series is ab / ab / ab / ab / ab / ab.
Thus, the pattern ab is repeated.
2. ab _ _ _ b_ bbaa _
A. abaab B. abbab C. baaab D. babba
Answer: C , The series is abb / aab / abb / aab.
Thus, the pattern abb, aab is repeated.
3. _ baa _ aab _ a _ a
A. aabb B. aaba C. abab D. baab
Answer: C , The series is aba / aba / aba / aba.
Thus, the pattern aba is repeated.
4. _ _ babbba _ a _ _
A. ababb B. baaab C. bbaba D. babbb
Answer: D ,The series is bababb / bababb.
Thus, the pattern bababb is repeated.
5. aa _ ab _ _ aaa _ a
A. aaab B. aabb C. abab D. baaa
Answer: A , The series is aaaaba / aaaaba.
Thus, the pattern aaaaba is repeated.
6. a _ bbc _ aab _ cca _ bbcc
A. bacb B. acba C. abba D. caba
Answer: B , The series is aabbcc / aabbcc / aabbcc.
Thus, the pattern aabbcc is repeated.
7. ab _ aa _ bbb _ aaa _ bbba
A. abba B. baab C. aaab D. abab
Answer: B , The series is abb / aaabbb / aaaabbbb / a.
Thus, the letters are repeated twice, then thrice, then four times and so on.
8. bc _ b _ c_ b _ ccb
A. cbcb B. bbcb C. cbbc D. bcbc
Answer: A , The series is bccb / bccb / bccb.
Thus, the pattern bccb is repeated.
9. abb _ baa _ a _ bab _ aba
A. abba B. abab C. ccac D. aabb
Answer: A , The series is abba / baab / abba / baab / a.
Thus, the pattern abba, baab is repeated.
10. abca _ bcaab _ ca _ bbc _ a
A. ccaa B. bbaa C. abac D. abba
Answer: C , The series is abc / aabc / aabbc / aabbcc / a.
11. _ bbca _ bcca _ ac _ a _ cb
A. abcba B. acbab C. bacab D. bcaab
Answer: B ,The series is abbc / ac / bcca / bc / caab / cb.
12. _ bcc _ ac _ aabb _ ab _ cc
A. aabca B. abaca C. bacab D. bcaca
Answer: C , The series is bbccaa / ccaabb / aabbcc.
The letter pairs move in a cycle order.
13. a _ bccb _ ca _ cca _ baab _ c
A. ababc B. abcaa C. accab D. bacaa
Answer: A ,The series is aabcc / bbcaa / ccabb / aabcc.
The letters move in a cyclic order and in each group, the first and third letters occur twice.
14. ab _ aa _ caab _ c _ abb _ c
A. bbcaa B. bcbca C. cabac D. cbbac
Answer: D ,The series is abc / aabc / aabbc / aabbcc.
First all the letters occur once, then a occurs twice, then both a and b occur twice and finally all the three letters appear twice.
15. c _ baa _ aca _ cacab _ acac _ bca
A. acbaa B. bbcaa C. bccab D. cbaac
Answer: A , The series is cab / aa / cacab / cacab / aa / cacab / ca.
Thus, the pattern cacab, cacab, aa is repeated.
16. _ aba _ cabc _ dcba _ bab _ a
A. abdca B. bcadc C. abcdd D. cbdaa
Answer: A .The series is aababcabcddcbacbabaa.
Thus, the letters equidistant from the beginning and the end of series are the same.
17. a _ cdaab _ cc _ daa _ bbb _ ccddd
A. bdbda B. bddca C. dbbca D. bbdac
Answer: D , The series is abca / aabbccdd / aaabbbcccddd.
Thus, each letter of first sequence is repeated two times in the second sequence and three times in the third sequence.
18. a _ abbb _ ccccd _ ddccc _ bb _ ba
A. abcda B. abdbc C. abdcb D. abcad
Answer: C The series is aaa / bbbb / cccc / dddd / cccc / bbbb / a.
19. _ bcdbc _ dcabd _ bcdbc _ dc _ bd
A. aaaaa B. ccccc C. bbbbb D. ddddd
Answer: A , The series is abcd / bcad / cabd / abcd / bcad / cabd.
Thus, the pattern abcd / bcad / cabd is repeated twice.
20. adb _ ac _ da _ cddcb _ dbc _ cbda
A. bccba B. cbbaa C. ccbba D. bbcad
Answer: B , The series is adbc acbd abcd dcba dbca cbda.
Thus, the letters equidistant from the beginning and the end of series are the same.
21. c _ bbb _ _ abbbb _ abbb _
A. aabcb B. abccb C. abacb D. bacbb
Answer: B ,The series is cabbbb / cabbbb / cabbbb.
Thus, the pattern cabbbb is repeated.
22. b _ abbc _ bbca _ bcabb _ ab
A. acaa B. acba C. cabc D. cacc
Answer: C , The series is bcab / bcab / bcab / bcab / bcab.
Thus, the pattern bcab is repeated.
23. ac _ cab _ baca _ aba _ acac
A. aacb B. acbc C. babb D. bcbb
Answer: A , The series is acac / abab / acac / abab / acac.
Thus, the pattern acac, abab is repeated.
24. _ acca _ ccca _ acccc _ aaa
A. acca B. caaa C. ccaa D. caac
Answer: B , The series is ca / ccaa / cccaaa / ccccaaaa.
25._ bc _ _ bb _ aabc
A. acac B. babc C. abab D. aacc
Answer: A ,The series is abc / cab / bca / abc.
26. aa _ aaa _ aaaa _ aaaa _ b
A. baaa B. bbaa C. bbbb D. bbba
Answer: D , The series is aab / aaab / aaaab / aaaaab.
Thus, the number of a`s is increasing by one in the successive sequence.
27. aba _ baca _ ba _ bacaabac _ aca
A. cacb B. ccab C. cabc D. abcc
Answer: A , The series is abac / baca / abac / baca / abac / baca.
Thus, the pattern abac, baca is repeated.
28. ab _ bc _ c _ ba _ c
A. baac B. aabb C. caab D. aaab
Answer: C ,The series is abc / bca / cab / abc.
Thus, the letters are written in a cyclic order.
29. a _ ca _ bc _ bcc _ bca
A. bbaa B. bbab C. aabb D. baba
Answer: A , The series is abcab / bcabc / cabca.
30. ab _ bcbca _ _ c _ bab
A. acbc B. baaa C. abcc D. ccaa
Answer: D ,The series is abcbc / bcaca / cabab.
Thus, the series consists of three sequences.
The first sequence begins with a, the second with b and the third with c.
Each sequence consists of a letter followed by other two letters repeated twice.
31. a _ cacbc _ baca _ _ b
A. baba B. babc C. abac D. cacb
Answer: B , The series is abcac / bcaba / cabcb.
Thus, the series consits of three sequence.
The first three letters of each sequence are in a cyclic order and the last two letters of each sequence are the same as the first and third letters of the sequence.
32. _ aaba _ bba _ bba _ abaa _ b
A. aabab B. ababa C. baaba D. bbaba
Answer: A , The series is aaab / aabb / abbb / aaab / aabb.
33. ab _ bbc _ c _ ab _ ab _ b
A. ccaac B. cbabc C. cacac D. bccab
Answer: C , The series is abc / b / bca / c / cab / a / abc / b.
34. _ bca _ cca _ ca _ b _ c
A. aaaaa B. bbbab C. aabaa D. bbabb
Answer: B , The series is bbca / bcca / bcaa / bcaa / bbc.
35. b _ ac _ cc _ cb _ ab _ ac
A. cbaba B. bbaac C. abbbc D. aabba
Answer: D ,The series is baac / accb / cbba / baac.
36. c _ ac _ aa _ aa _ bc _ bcc
A. cabba B. ccbbb C. bbbbb D. cbacb
Answer: B ,The series is ccacc / aabaa / bbcbb / cc.
37. abc _ d _ bc _d _ b _ cda
A. bacde B. cdabe C. dacab D. decdb
Answer: C , The series is abcdd / abccd / abbcd / a.
38. ba _ b _ aab _ a _ b
A. abaa B. abba C. baab D. babb
Answer: B , The series is baab / baab / baab.
Thus, the pattern baab is repeated.
39. gfe _ ig _ eii _ fei _ gf _ ii
A. eifgi B. figie C. ifgie D. ifige
Answer: C , The series is gfeii / gfeii / gfeii / gfeii.
Thus, the pattern gfeii is repeated.
40. mnonopqopqrs _ _ _ _ _
A. mnopq B. oqrst C. pqrst D. qrstu
Answer: C , The series is mno / nopq / opqrs / pqrst.
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