Thursday, 16 April 2020

Ten most important problems of Reasoning for competitive exam Part1

Ten most important problems of Reasoning for competitive exam which includes 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 Reasoning for competitive exams


Problem # 1


6  :  29   ::   24  :  ?

(a)109  (b)129  (c)119   (d)99


Solution 

6 is related to 29 in the same way 24 will be related to ? , It means we have to apply same mathematical operations to 24 to get ?.
So if we multiply 6 with 5  and then subtract 1 from result obtained in previous step like this ( 6 × 5 )  - 1 which will be equal to 29.
Same operation we have to apply  to 24.
( 24 × 5 ) - 1 = 120 - 1 =  119

So Correct  option is ( c ) 119

  Problem # 2


5  :  100   ::   7  :  ?

(a) 135   (b) 91   (c) 196  (d) 49


Solution

Because in 1st case if we take Square of  5 then multiply it with 4 we shall have 100.
5² × 4 = 25 × 4 = 100
Same procedure will be applied in 3rd number by taking square of 7 then multiply it with 4
7² × 4 = 49 × 4 = 196

Correct  option is  ( c  ) 196

Problem # 3

6  :  18   ::   4  :  ?

(a)4     (b)6     (c)8    (d)10


Solution 

Divide  6² by 2
6² ÷ 2 = 36 ÷ 2 = 18
Similarly divide 4² with 2
i.e.  4² ÷ 2 = 16 ÷2 = 8
Divide the square of 1st number by 2 to get 2nd number. Similarly take square of third number and then divide it by 2 to to get the number equal to? Since square of 4 is 16 then divided by 2 to get it.
6² ÷ 2 = 36 ÷ 2 = 18

4² ÷ 2 = 16 ÷ 2 = 8 

Correct  option is (  c ) 8


Problem # 4

18  :  30   ::   36  :  ?

(a)64   (b)  62   (c)54    (d) 66


Solution  

 (18 × 2 ) - 6  = 36 - 6 = 30
 (36 × 2 ) - 6  = 72 - 6 = 66
Multiply 1st number (18) with 2 and then subtract 6 from it
18 × 2 =  36 - 6 = 30
Similarly Multiply 3rd number ( 36 ) with 2 and then subtract 6 from it ,

36 × 2 = 72 - 6 = 66

correct  option is ( d) 66


Problem # 5

12  :  20   ::   30  :  ?

(a)48  (b)42    (c)15   (d)35


Solution 
Method 1  

Split 12  =  3 × 4 
Split  20  = 4 × 5, 
Split  30 =  5 × 6,  
Here study these factors (3 , 4 ) , (4 ,5 ) ,(5 ,6 ) so next pair will be  ( 6 , 7 ) , It means ? Will be replaced by the number 6 × 7 = 42

Method 2 


12 will be written as square of 3 plus 3 , 20 will be written as square of 4 plus 4 ,30 will be written as square of 5 plus 5, so next number will be written as square of 6 plus 6 which is equal to 42 , Therefore   required option will be 42
                         Or
Make continuous factors 3 and 4  , 4 and 5 ,5 and 6 and 6 and 7 of 12, 20, 30 and 42 respectively 
12 = 3 × 4  , 
20 = 4 × 5  ,
30 = 5 × 6 so 
42 = 6 × 7

Add same number to its square to get next number 
Or multiply next number to the number
 9  = 3² + 3   ,
 20 = 4² +4 ,
 30 = 5² + 5
42 = 6² + 6
Correct  option is ( b ) 42

 Problem # 6


12  :  54   ::   15  :  ?

(a)64   (b)69   (c)56  (d)67


Solution

{(1st number ) × 5} - 6 = 2nd number
(12 × 5) - 6 = 54
Multiply 3rd number with 5 then subtract 6 from it
{(3rd number ) × 5} - 6 = 4th  number
(15 × 5) - 6 = 69
Multiply first number ( 12 ) with 5 then subtract 6 from it to get 2nd number ( 54 ) ,   Similarly   Multiply third  number ( 15 ) with 5 then subtract 6 from it.
( 1 2 × 5  ) - 6 = 60 - 6 = 54
( 1 5 × 5  ) - 6 = 75 - 6 = 69
Correct  option is ( b) 69

Problem # 7

6  :  5   ::   8  :  ?

(a) 6      (b)10        (c) 2     (d)4 


Solution
Add 4 to 1st number and divide it with 2 to get 2nd number( 6 + 4 )/2 = 5
Similarly in 2nd case Add 4 to 3rd number then divide the resultant with 2 to get 4th number
( 8 + 4 )/2 = 6
Correct  option is ( a ) 6


Problem # 8

29  :  319  ::    23 :  ?


    (a)115  (b)252 (c)151  (d)46

Solution

Add both the digits of first number  i.e. 2 and 9 then multiply it with first number to get second number. similarly in the third number add both the digits I.e  2 and 3  and multiply it with 3rd number to get fourth number.
29 × ( 2 + 9 ) = 29 × 11 = 319
23 × ( 2 + 3 ) = 23 ×  5  = 115

So 253 is the right option (a) 115

Problem # 9

6  :  64   ::   11  :  ?


(a) 127      (b) 124     (c) 144    (d) 169


Solution

Add 2 to 1st given number and take its Square to get 2nd number. Similarly add 2 to 3rd number and take it square to get fourth number.
(6 + 2 )² = 8² = 64
(11 + 2)² = 13² = 169

Correct  option is ( d ) 169

Problem # 10

36  :  50   ::   64  :  ?

(a)70   (b) 82     (c)78      (d) 72


Solution

Take Square Root of 1st number and add 1 to it then add 1 to  its Square to get 2nd number.
Similarly take square root of 64 add 1 to its square  root and take it square and add 1  to get 4th number.
36 = 6² --->  (6 + 1)² + 1 = 50 ,

 64 = 8² ------> ( 8 + 1)² + 1 = 82
√36 = 6 add 1 to it = 7 , square this number = 49 + 1 = 50
√64 = 8 add 1 to it = 9 , square this number = 81 + 1 = 82

So  Correct option is ( b ) 82



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