Easy Reasoning methods for figure problems of S S C Exams

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 

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 .

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


 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

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 
     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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