# Odd Man Out and Series Formula: Shortcuts, Problems Tricks To Solve Easily

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**Odd Man Out and Series
Formula**

The information about

**Odd Man Out and Series Formula**is for those aspirants who are going to attend any competitive exam. In question paper, a section named as ‘Quantitative Aptitude’ consist of questions regarding reasoning, Odd Man Out and Series Formula, Distance formulas and many more. Those candidates who know the formulas, Shortcuts and problem tricks can easily solve the questions. But it is very tough for the other students to solve all the questions. Short tricks help the aspirants to complete the question paper on given time.
If you want to score good marks in exams then it is
necessary for you to cover all the important topics and learn all the series
formula. Questions asked in aptitude section are very easy so don’t take too
much time in it. Apply the correct formula, write the correct digits and don’t
get confused in similar type of questions. Some important illustrations and Odd
man out and series formula are given below.

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**Odd Man Out and Series**

__Shortcuts, Problems Tricks To Solve Easily__

To complete the question paper on a given time period,
you just need two thinks i.e. speed and accuracy. Short tricks will boost your
confidence and your speed at the time of examination. Instead of calculator try
to use your mind more, do not depend on calculator for easy calculations. Take
a look on formulas and examples that are mentioned below:

Prepare Well:

__Maths Questions and Answers__

__Series Formula__

The terms or elements follow a definite law in series
but it cannot be generalized. You should know about what is the definite
relationship between numbers which make the set of given terms in series.
Addition, subtraction, multiplication, division, transposition of terms and
series generally form such series. The different questions asked may depend
upon the following:

__Odd number/Even number/Prime numbers__

The series may consist of odd
numbers /even numbers or prime numbers except one number, which will be the odd
man out. Hence, before solving numerical on this topic must revise all basic
concepts.

__Perfect squares/Cubes__:

Squares: 9, 16, 49, 81 ….

Cubes: 27, 64, 125, 216 ….

__Multiple of numbers__:

The series contains numbers which
are multiple of different numbers.

**: 4, 8, 12, 16, 20….**

*Example*
Numbers in A.P./G.P.

Geometric progression: x, xr,
xr3, xr4

Arithmetic progression: x, x + y,
x + 2y, x + 3y are said to be in A.P.

The terms in series may be
arithmetic or geometric progression.

__Difference or sum of numbers__:

- The difference between two consecutive numbers may increase or decrease
- Cumulative series:
- In this type, the third number is the addition of previous two numbers.

**2, 4, 6, 10, 16, 26 ……**

*Example:*Take a Test Now |

__Important Illustrations__**Problem 1**

1) 13

2) 23

3) 33

4) 43

5) 53

**Solution**: Check these numbers. Have you observed any logic? Every number from the above list has 10 numbers difference from it's previous numbers. Great, we find one... But that's not the logic. So check whether the numbers are prime or not. So, here... all the given numbers are prime, except the number 33 (it is a composite number). So Option 3 is the answer.

**Problem 2**

1) 176

2) 231

3) 572

4) 473

5) 653

**Solution**: Check the above numbers carefully. You can't find the correct answer by applying the first 2 steps. So here try to find out the logic... If you observe closely, you will find out that the middle number is the sum of the first and the last numbers. Check once....

1) 176

2) 231

3) 572

4) 473

5) 653

But you are not getting middle number in 5th Option
(As 6+3 = 9, not 5). So this is the correct answer.

Read More:
How
to Prepare for Maths Exam |

**Problem 3**

1) 123

2) 246

3) 147

4) 368

5) 159

**Solution**: Here also we tried all the steps but no answer. Hey wait, have you tried the above logic?? Let's try

1) 1+3 = 4 (but the middle
number is 2)

2) 2+6 = 8 (but the middle
number is 4)

3) 1+7 = 8 (but the middle
number is 4)

4) 3+8 = 11 (but the middle
number is 6)

5) 9+1 = 10 (but the middle
number is 5)

Not
worked

Now see whether we can find anything with the newly
obtained numbers....

4, 8, 8, 11, 10.... No relation.. But you could find
out that one number except all are even numbers... So, may be this is our
answer...

Or if you want to go further, you can get another logic...
Just divide these numbers with 2.

1) 4/2 = 2

2) 8/2 = 4

3) 8/2 = 4

4) 11/2 = 5.5

5) 10/2 = 5

So, here.. Except Option 4... For remaining numbers, the middle number is
the half of the total of remaining two numbers.

Note: If you can't get answer with above step, then
you should experiment with different numbers and different mathematical
operations.

**Problem 4**

1) 29: 812

2) 37: 1332

3) 45: 1980

4) 48: 2256

5) 51: 2551

If they ask question in the above format, then it
means, there should be some relationship between the first and second number.

**Solution:**

Note: If you can't find any visible difference, then
better to find out Squares.

1) 292 = 841 (but given number
is 812)

2) 372 = 1369 (but the given
number is 1332)

3) 452 = 2025 (but the given
number is 1980)

4) 482= 2304 (but the given
number is 2256)

5) 512 = 2601 (but the given
number is 2551)

So squares are not matching with the numbers. But if
you closely observe, you can see some similarities among these numbers. So,
find out the difference, so that we could find some clue

1) 841-812 = 29 (First Number)

2) 1369-1332 = 37 (First
Number)

3) 2025-1980 = 45 (First
Number)

4) 2304-2256 = 48 (First
Number)

5) 2601-2551 = 510 (Not the
First Number)

So,
Option 5 is the odd one.

**Problem 5**

1) 126

2) 72

3) 135

4) 171

5) 162

**Solution**: At first glance we are unable to find any clues. So, check the interrelationships among these numbers (if any).

Note: People tend to mark Option 2 (72) as the answer
because it consist only two numbers where as in remaining options the numbers
are three. But please keep in mind that now a day they are not asking questions
with such a simple logic.

At a close observation, you can find that the total of
the given numbers is 9

1) 1+2+6 = 9

2) 7+2 = 9

3) 1+3+5 = 9

4) 1+7+1 = 9

5) 1+6+2 = 9

So, there might be something with this 9.

Let's subtract this 9 with the given number...

1) 126-9 = 117

2) 72-9 = 63

3) 135-9 = 126

4) 171-9 = 162

5) 162-9 = 153

You could find out some relation between option 4 and
5. But it's not enough.... So let's try division

1) 126/9 = 14

2) 72/9 = 8

3) 135/9 = 15

4) 171/9 = 19

5) 162/9 = 18

From the above answers, you can clearly say that 19
(Option 4) is a Prime Number. Where remaining numbers are not.

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Through These Important Links |

**Problem 6**

1) 48-134

2) 65-185

3) 128-374

4) 56-158

5) 81-223

Want help in Preparation? Get Tips:

__Practice Which Helps In Preparation__

**Solution**: With a close observation you can see that the second number is almost nearer to the three times of the first number.

1) 40 x 3 = 120 (given number
is 134)

2) 60 x 3 = 180 (given number
is 185)

3) 120x 3 = 360 (given number
is 374)

4) 50 x 3 = 150 (given number
is 158)

5) 80x 3 = 240 (given number is
223)

So, there should be some logic behind it

Multiply first numbers with 3

1) 48x3 = 144 (difference
between result and second number is 144-134 = 10)

2) 65x3 = 195 (195-185 = 10)

3) 128x3 = 384 (384-374 = 10)

4) 56x3 = 168 (168-158 = 10)

5) 81x3=243 (243-223= 20)

So, Option 5 is the answer.

**Problem 7**

1) 25-630

2) 36-1312

3) 16-260

4) 9-84

5) 49-2408

**Solution**:

1) 252=625 (625+5 = 630)

2) 362=1296 (1296+16 = 1312)

3) 162 = 256 (256+4 = 260)

4) 92 = 81 (81+3 = 84)

5) 492 = 2401 (2401+7=2408)

So, here the odd one is Option 2

Check
Some Other Math Preparation Links |

__Final Words__

We hope that the
above information about Shortcut tricks and Odd Man Out and Series Formula will
be helpful for you. You just need to keep practicing of it for better rank. At
last we would like to wish all the best to the students who have exams very
soon. Be confident and attempt all the questions at the time of examination.

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