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

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**Five 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.

##
**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

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**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

####
__1st Figure__

__1st Figure__

From 1st figure we can see that lower number is thrice (3 times) of sum ( addition ) of both numbers which are on upper portion of figure i. e . 18 + 9 = 27 .

Now multiply it with 3 , we get 27 ×3 =81 , which is the lower number in 1st figure.

####
__3rd Figure__

__3rd Figure__

And for third figure add both the numbers 24 and 7 and then multiply the sum with 3 like this 24 + 7 = 31 and 31 × 3 = 93

####
__2nd Figure__

Similarly ? in 2nd figure can be found by adding 23 and 8 and then Multiply it with 3 as follows __2nd Figure__

**23 +8 =31 × 3 =93 ,**Hence 93 is the required number.

#### Problem # 5

Since larger 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.**

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