## Ten most important problems of missing numbers in Reasoning Analogy part 2

Series of missing numbers and missing terms and their solutions in reasoning analogy . These questions are very very important for upcoming competitive exams like SSC CGL ,SSC CHSL and RRB NTPC Etc.

Ten most important problems of missing numbers in Reasoning Analogy

This series consists of two alternate series 1st series consist of number 1,3, 5 and 7 and second series consist of the series 2 ,5 ,8,?

So in 1st series there is difference of 2, so next number in the series must be 2 greater than 7 that is 9 .

In the 2nd series we observed a difference of 3 show the next number must be 11 that is 3 greater than 8 the last term therefore the last two terms of the series 11 and 9.

#### The correct option is (1) 11,9

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__Problem # 2__

__Problem # 2__

This series is also the combination of two series one with the number 2, 48 ,16 ,32 and second series consist of these numbers 6, 9 13 ,18 , ? . So ? Will be filled out using 2nd series not from 1st series.

Observing 2nd series carefully ,In second series numbers are increased by 3 ,4 and 5 ,so next Increment should be of 6. This implies next number should be 24 with an increment of 6.

#### The correct option is (1)24

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__Problem # 3__

__Problem # 3__

#### Observe carefully all the numbers in the series are prime number , so next prime number after 37 will be 41 .This implies that ? will be replaced by 41.

#### The correct option is (3)41

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__Problem # 4__

__Problem # 4__

#### All the numbers in this series are consisting of three digits numbers and Middle digit is the sum of first and last digit in all the numbers except one that is 342 in this case 3 + 2 is equal to 5 not equal to 4 .So correct option is option is (c) ,this is the wrong number in the series .

#### The correct option is (3)342

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__Problem # 5__

__Problem # 5__

#### Every number in this series is the sum of its two preceeding number plus 3.

14 = ( 7 + 4 ) + 3

24 = ( 14 + 7 ) + 3

41 = ( 24 +14 ) + 3

? = ( 41 + 24 ) + 3 = 65+3 = 68

#### The correct option is (3)68

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__Problem # 6__

__Problem # 6__

#### This series consists of triplet ,that is every third number is product its two preceding numbers

#### As third number 10 is the product of 2 and 5, sixth number 18 is the product of fourth (3) and fifth (6 ) numbers .

So ninth number of the series must be the product of seventh and eighth numbers. 4 × 7 = 28 instead of 30

#### The correct option is (3) 30

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__Problem # 7__

__Problem # 7__

Here second number is the sum of the digits of first number the forth number is the sum of digits of third number similarly every even positioned number is the sum of its preceding number's digits question mark will be replaced by the sum of 5 + 3 that is 8

#### The correct option is (1)8

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__Problem # 8__

__Problem # 8__

Writing 3 = 2² - 1

#### Writing 8 = 3² - 1

#### Writing 15 = 4² - 1

#### Writing 24 = 5² - 1

Show next number should be 6² - 1 (6 square minus 1 ) that is 35

#### Writing 6² - 1 = 35

#### The correct option is (2)35

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__Problem # 9__

__Problem # 9__

Every number in this series is the double of its preceding number .So double of 1 is 2 , double of 2 is 4 ,double of 4 should be 8 ( instead of 7 ) , double of 8 is 16 and double of 16 should be 32 . So 7 is the wrong number in this series.

#### The correct option is (1)7

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__Problem # 10__

__Problem # 10__

#### In the series every number is the the cube of its position plus 1 .

#### Since

#### 2 = 1³ + 1

#### 9 = 2³ + 1

#### 28 = 3³ + 1

#### 65 = 4³ + 1

#### 126 = 5³ + 1

#### In the same way next number of the series must be 6³ + 1 = 216 + 1 = 217

#### The correct option is (1) 217

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