## Easy Reasoning analogy methods for problems of S S C Exams

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**Reasoning problems with their Solutions**

#### Problem # 1

Starting from 5 and moving clockwise we have these numbers 5, 10,?,50,122.

so 5 can be written as 2^2+1 starting from 1st prime number i. e. Square of 1st prime number and plus 1

10 can be written as 3^2+1 , Square of 2nd prime number and plus 1.

?

50 can be written as 7^2 +1,Square of 4th prime number and plus 1.

Similarly 122 can be written as 11^2 +1, Square of 5th prime number and plus 1.

Hence ? can be replaced as 5^2 +1, Square of 3rd prime number ( 3 ) and plus 1.

So

Therefore Required number is 26.

#### Problem # 2

In this figure largest numbers are appearing in last row. so we should search the relation column wise .

In 1st column we have to search the relationship between 9 and 6 to give 45. so in 2nd column we have to find relationship between 12 and 7 to give 95. And same logic we shall apply in 3rd columns.

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**1st Column **

In 1st columns if we add and subtract both numbers then multiply it then we shall have 45 as follows

we have two numbers in 1st column 9 and 6 .

(9+6)×(9-6) = (15 ) × ( 3 ) = 45

##
**2nd Column**

And again in 2nd columns if we add and subtract both numbers of 2nd column and then multiply these with each others then we shall have 45 like this

we have two numbers in 1st column 9 and 6 .

(12+7 ) × ( 12 - 7) = (19 ) × ( 5 ) = 95

##
**3rd Column**

Same logic can be applied for 3rd column ,we add and subtract both numbers of 3rd column and then multiply these with each others

( 8 + 3 ) × ( 8 -3 ) = (11 × 5 ) = 55

**Hence 55 shall replace "?" in the given figure**

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**Test your Reasoning ability**

#### Problem # 3

Since largest numbers in all the three rows lie in 1st column of the given figure.

It means we have to search a relation between 20 and 7 to give 28 in 1st row. and similar relation must be between 35 and 12 to give 84 .

##
**1st Row**

So in 1st row if we multiply both numbers with each others and divide the result with 5 we shall get 28 like this 20 × 7 =140

Now divide 140 with 5 such that 140/5 = 28

##
**2nd Row**

So in 2nd row if we have to multiply both numbers with each others and divide the result with 5 we shall get 28 like this 35 × 12 = 420 .

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**3rd Row**

Now divide 420 with 5 such that 420/5 = 84

Same logic we have to apply in 3rd row to get answer 45 . So If we multiply 9 with 45 we shall get 275 then we have to divide 405 with 5 to give , It will replace ? question mark as follows

9 × 25 = 225 / 5 = 45

**So 45 shall be the right number to replace question mark .**

#### Problem # 4

Since largest numbers appears on 3rd columns therefore the solution must be row wise, if we treat a and b as 1st and 2nd numbers then 3rd number which is our desired numbers must be equal to (a-1)×(b) i . e. product/ multiplication of (a-1) and b. Hence 3rd element of 1st row must be

(8-1)×(3) = 7×3 = 21

3rd element of 2nd row must be

(6-1)×(5) = 5×5 = 25

So 3td element of 3rd row must be

(12-1)×(2) = 11×2= 22

**So 22 will replace "?" Question mark.**

#### Problem # 5

In 1st column 5 × ( 6 +7 ) = 5 × 13 = 65

In 2nd column 4 × ( 3 +2 ) = 4 × 5 = 20

So same formula will be used in 3rd column

In 3rd column. 9 × ( ? +4 ) = 45

( ? +4 ) = 45/9 =5

? = 5 - 4 = 1

So required answer is " 1 ".

#### Problem # 6

Multiplying 1st three elements of all the columns to get 4th elements

1 × 8 × 9 = 72

3 × 6 × 5 = 90

2 × 7 × ? = 56

This implies ? = 4

So ? will be replaced by 4

#### Problem # 7

In 1st column square of three numbers i.e.

( 1 + 4 + 2 )^2 = 7^2 = 49

In 2nd column square of three numbers i.e. ( 4 + 2 +2 ) ^2 = ( 8)^2 = 64

In 3rd column the square of ( ? + 5 + 3 ) ^2 must be 169

this implies ( ? + 8 )^2 = 169

( ? + 8 ) = 13

? = 5

Required answer is 5

#### Problem # 8

If we calculate the sum of 1st column ,2nd column , 2nd row and 3rd row ,then it is found 25 in all the cases .so total of all rows and all the columns must be 25 . It can be seen that if we put "11" in place of question mark then total of all the rows and columns is 25

#### Problem # 9

Since biggest numbers are in the fourth columns of every row. So if we mutiply 1st three numbers and then add 4th number to it ,we shall have 5th number in every row .

( 4 × 3 × 2 ) + 8 = 24 + 8 = 32

( 5 × 3 × 1 ) + 9 = 15 + 9 = 24

( 7 × 3 × 3 ) + 7 = 63 + 7 = 70

Similarly when we multiply first three numbers and then adding force number to it in the last row we shall have fourth number in the last row like this

( 2 × 9 × 4 ) + 12 = 72 + 12 = 84

So 84 will replace question mark " ?"

**So required answer is 84**

#### Problem # 10

**1st Row**

multiply 3 and 2 then add one less than the 2nd number to it

( 3 × 2 ) + ( 2 - 1 ) = 6 + 1 = 1

**2nd Row**

multiply 5 and 4 then add one less than the 2nd number to it

**( 5 × 4 ) + ( 4 - 1 ) = 20 + 3 = 23**

**3rd Row**

multiply 7 and 6 then add one less than the 2nd number to it

**( 7 × 6 ) + ( 6 - 1 ) = 42 + 5 = 47**

**4th Row**

( 9 × 8 ) + ( 8 - 1) = 72 + 7 = 79

**5th Row**

( 10 × 9 ) + ( 9 - 1 ) = 90 + 8 = 98

**So 98 will replace. " ? "**

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