Ten most important problems of Reasoning for competitive exam Part1



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 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 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  the right option is (b) 82

 Correct option is ( b ) 82






Do you like Missing number in box, Reasoning problem

To buy cell phones and books for competitive Exams click here



Share:

No comments:

Post a comment

Recent Posts