Ten Latest tricky and logical Questions in box and circle reasoning for SSC CGL And SSC CHSL Exams

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Ten Most Important tricky and logical Questions of Reasoning of box and circles problems with solutions are discussed in this post. . These types of problems are very helpful for cracking competitive exams like ssc cgl, ssc chsl and various Bank exams and many other similar exams .

 Problem #1

Questions in box and circle reasoning

This circle is divided into eight parts and each part is composed of three numbers . In each numbers are in outer part of the circle and one number is in the inner part of the circle. If we multiply both the numbers which are in the outer part of the circle and then add one to it then this resultant number will in the inner part of the opposite circle.

( 2 × 2 ) + 1 = 4 + 1 = 5 (In the inner part of opposite sector ).
 ( 9 × 2 ) + 1 = 18 + 1 = 19 (In the inner part of opposite sector ).
( 4 × 3 ) + 1 = 12 + 1 = 13 (In the inner part of opposite sector ).
( 5 × 2 ) + 1 = 10 + 1 = 11 (In the inner part of opposite sector ).
( 6 × 1 ) + 1 = 6 + 1 = 7  (In the inner part of opposite sector ).
( 3 × 7 ) + 1 = 21 + 1 = 22 (In the inner part of opposite sector ).
( 4 × 2 ) + 1 = 8 + 1 = 9 (In the inner part of opposite sector ).
( 3 × ? ) + 1 =  3? + 1 = 25
After solving  
⇒3? =25 - 1
⇒ ? = 24 ÷3
⇒ ? = 8 (In the inner part of opposite sector )
Therefore Correct option is (3) 

Problem # 2

Questions in box and circle reasoning
This circle is divided into four parts and each part is composed of three numbers . In each part two numbers are in outer part of the circle and one number is in the inner part of the circle. 

1st Method 

If we multiply both the numbers which are in the outer part of the circle and then add two to it then this resultant number will in the inner part of the opposite circle.
(8 × 6 ) + 2 = 48 + 2 = 50 (Number in the inner part of the circle in 4th quadrant )
(5 × 7 ) + 2 = 35 + 2 = 37 (Number in the inner part of the circle in 3rd quadrant )
(8 × 10 ) + 2 = 80 + 2 = 82 (Number in the inner part of the circle in 2nd quadrant )
(9 × 11 ) + 2 = 99 + 2 = 101 = ? (Number in the inner part of the circle in 1st quadrant ). 
Therefore Correct option is (4) 

2nd Method 

(8² + 6² ) ÷ 2 = ( 64  + 36 ) ÷ 2 = 100 ÷ 2 = 50 (Number in the inner part of the circle in 4th quadrant )
(5² + 7² ) ÷ 2 = ( 25  + 49 ) ÷ 2 = 74 ÷ 2 = 37 (Number in the inner part of the circle in 3rd quadrant )
(8² + 10² ) ÷ 2 = ( 64  + 100 ) ÷ 2 = 164 ÷ 2 = 82 (Number in the inner part of the circle in 2nd quadrant )
(9² + 11² ) ÷ 2 = ( 81  + 121 ) ÷ 2 = 202 ÷ 2 = 101 = ? (Number in the inner part of the circle in 1st quadrant )

Therefore Correct option is (4) 

Problem # 3

Questions in box and circle reasoning

This circle is also divided into eight parts and  there is also one number written between  any  two numbers towards outer side of these two numbers .  If we  add these  two simultaneous  numbers then reverse the order of the result so obtained then this number will be equal to the number in the small circle opposite to both these numbers .
5  + 17 =  22 ↔️  22 ( Reversing the order of digits)
17 + 12 = 29 ↔️  92 ( Reversing the order of digits)
12 + 31 = 43 ↔️  34 ( Reversing the order of digits)
31 + 10 = 41 ↔️  14 ( Reversing the order of digits)
10 + 11 = 21 ↔️  12 ( Reversing the order of digits)
11+ 6 = 17 ↔️  71  ( Reversing the order of digits)
6  + 8  = 14 ↔️  41 ( Reversing the order of digits) 
8  + 5 = 13 ↔️  31 = ? ( Reversing the order of digits)
Hence the value of  "? " will be 31 
Therefore Correct option is (C) 

Problem # 4


Questions in box and circle reasoning

Taking the difference of two consecutive terms in clockwise direction and analyse the difference , this difference always increases with the increment of 4.

10 -  4 = 6
20 - 10 = 10  with an increment of 4 as compare to last difference which is 6
34 - 20 = 14  with an increment of 4 as compare to last difference which is 10
52 - 34 = 18  with an increment of 4 as compare to last difference which is 14
? - 52 = 22  must be an increment of 4 as compare to last difference which is 18
? = 22 + 52
? = 74
100 - ? = 26  must be an increment of 4 as compare to last difference  which is 22
? = 100 - 26
? = 74 
130 - 100 = 30  with an increment of 4 as compare to last difference  which is 30
130 - 100 = 30  with an increment of 4 as compare to last difference  which is 30
Therefore Correct option is (2) 

Problem # 5


Questions in box and circle reasoning
This circle is divided into four parts and each part is composed of three numbers . In each part two numbers are in outer part of the circle and one number is in the inner part of the circle. If we  check the greater of the numbers which are in the outer part of the circle and  then this resultant number will in the inner part of the opposite circle.

Max (17,14) = 17  (Number in the inner part of the circle in 4th quadrant )
Max (18,13) = 18  (Number in the inner part of the circle in 3rd quadrant )
Max (5,3) = 5  (Number in the inner part of the circle in 2nd quadrant )
Max (4,8) = 8 = ? (Number in the inner part of the circle in 1st quadrant )

Therefore Correct option is (2) 

Problem # 6


Questions in box and circle reasoning

Here 1st circle consist of four parts and if we divide sum of all the digits in its outer part with number of parts then result will be the middle number.
( 6 + 2 + 9 + 7  )/4 =  24/4 = 6 ( Middle number)
2nd circle consist of five parts and if we divide sum of all the digits in its outer part with number of parts then result will be the middle number. 
( 7 + 2 + 5 + 8 + 3 )/5 = 25/5 = 5 ( Middle number)
3rd circle consist of six parts and if we divide sum of all the digits in its outer part with number of parts then result will be  the middle number. 
(  8 + 6 + 7 + 5 + 7 + 9 )/6 = 42/6 = 7 ( Middle number)
Hence option (3) is correct answer

Problem # 7

Questions in box and circle reasoning

In all these triangles the difference of the product of  all the numbers which are in outside of triangle and sum of these numbers is equal to the numbers in the middle of each triangles.

In 1st triangle

(5  × 3 × 2 ) - ( 5 + 3 + 2 ) = 30 - 10 = 20 ( Number in centre of the 1st triangle)

In 2nd triangle

(8  × 5 × 5 ) - ( 8 + 5 + 1 ) = 40 - 14 = 26 ( Number in centre of the 2nd triangle)

In 3rd triangle

(9  × 2 × 3 ) - ( 9 + 2 + 3 ) = 54 - 14 = 40 ( Number in centre of the 3rd triangle)

Therefore Correct option is (4) 

Problem # 8

Questions in box and circle reasoning
This figure consist of four squares around  one big square. Look carefully in this big square every number in it is perfect cube. And if we multiply the cube roots of  two adjacent numbers then we shall have the number attached to big square between these two numbers whose cube root had been multiplied.
Cube root of 512  ×  cube root of 216  = 8  × 6  =  48 ( The number at bottom line in the box)
Cube root of 216  ×  cube root of 343 = 6 × 8 = 42 ( The number at rightmost box )
Cube root of 343  ×  cube root of  64 ( Choose this number because its cube root is 4 ) = 7 × 4  = 28 ( The number at uppermost box )
Cube root of 512  ×  cube root of  64 ( Choose this number because its cube root is 4 ) = 8 × 4  = 32 ( The number at leftmost box )
Therefore Correct option is (B) 
Also read these posts on Reasoning

Problem # 9

Questions in box and circle reasoning

Taking the difference of two consecutive terms in clockwise direction and analyse the difference , this difference always increases with the increment of 5.
30 - 15 = 15 
50 - 30 = 20 with an increment of 5 as compare to last difference which is 15
75 - 50 = 25 with an increment of 5 as compare to last difference which is 20
?  -  75 =  30 with an increment of 5 as compare to last difference which is 25
140 -  ? = 35 with an increment of 5 as compare to last difference which is 30
180 - 140 = 40.  with an increment of 5 as compare to last difference which is 35
Because in every two numbers taken in clockwise order from least number , the difference between two numbers   will be  an increment of 5 . So if we put 105 instead of question mark then we shall have complete pattern discussed above.  
Therefore Correct option is (1)  

Problem # 10

 Questions in box and circle reasoning
This circle is divided into four parts and each part is composed of three numbers . In each part two numbers are in outer part of the circle and one number is in the inner part of the circle. If we multiply both the numbers which are in the outer part of the circle and then divide the resultant so obtained with 4  then this resultant number will in the inner part of the opposite circle.
(7 × 4 ) ÷ 4 = 28 ÷ 4 = 7 (Number in the inner part of the circle in 4th quadrant )
(5 × 8 ) ÷ 4 = 40 ÷ 4 = 10  (Number in the inner part of the circle in 3rd quadrant )
(4 × 20 ) ÷ 4 = 80 ÷ 4 = 20 (Number in the inner part of the circle in 2nd quadrant )
(6 × 8 ) ÷ 4 = 48 ÷ 4 = 12  = ? (Number in the inner part of the circle in 1st quadrant )

Ten Most Important tricky and logical Reasoning of box and circles problems with solutions were discussed in this post. . These types of problems are very helpful for cracking competitive exams like ssc cgl, ssc chsl and various Bank exams and many other similar exams . Feel free to share your opinion regarding this post in comment box.

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Reasoning questions with answers for competitive exams, Reasoning questions with answers for bank exams

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Reasoning questions with answers for competitive exams,  Reasoning questions with answers for bank exams. These questions of reasoning in latest reasoning questions with answers are very very important for upcoming competitive exams  like Bank PO , SSC CGL etc . So let us start solving and understanding these Maths logical reasoning questions with answers.


Reasoning questions with answers for competitive exams and bank exams


Problem # 1


Reasoning questions with answers for bank exams
 
Every number in any upper row is the average of two numbers in the lower row.

Last row /Bottom Row

(9 + 7) ÷ 2  = 16 ÷2 = 8 ( Number in 2nd last row ), 
(7 + 5) ÷ 2  = 12 ÷ 2 =  6 ( Number in 2nd last row ), 
(5 + 3) ÷ 2  = 8 ÷ 2 = 4 ( Number in 2nd last row ), 

2nd last row

(8 + 6) ÷ 2  = 14 ÷ 2 =  7 ( Number in 3rd last row ), 
(6 + 4) ÷ 2  = 10 ÷ 2 = 5 ( Number in 3rd last row ), 

3rd last row

(7 + 5) ÷ 2  = 12 ÷ 2 =  6 ( Number in Top most row ), And this will be the value of question mark ?
Hence option (1)6 is correct option

Problem # 2



Reasoning questions with answers for bank exams

This problem of reasoning have three figures, And every figure consist of five numbers .Out of five numbers one is at centre and four numbers are around the central number.
In every figure central number is the sum of the cube of remaining four numbers.

 33  +   23  +   43  +   63   = 27 + 8  + 64 + 216 =  315 (In 1st figure)
 113  +   73  +  83  +    63  = 1331 + 343  + 512 +216 =  2402 (In 2nd figure)
 13  +  53  +   ?3  +    43   = 1 + 125  +  ?3 + 64 =  1190  (In 3rd figure)
⇒   ?3  = 1190 - 64 - 125 - 1 
 ⇒   ?3  = 1190 - 190
⇒   ?3  = 1000
⇒   ? = 10
Hence option (3)10 is correct option


Problem # 3


Reasoning questions with answers for bank exams

Take the cube of product of 1st two numbers of every row to get the 3rd number in the box.
 (2 × 3)3  =   63  = 216 (last number in 1st row)
 (3 × 4)3   =   123  = 1728 (last number in 2nd row)
 (4 × 5)3  =   203  = 8000 (last number in 3rd row)
  (5 × 6)3  =   303  = 27000 (last number in 4th row)
or 
Cube of ( 2 × 3 )  =  cube of 6 = 216 (last number in 1st row) 
Cube of ( 3 × 4) = cube of 12 = 1728 (last number in 2nd row) 
Cube of ( 4 × 5)  =  cube of 20 = 8000 (last number in 3rd row) 
Cube of ( 5 × 6)  =  cube of 30 = 27000  (last number in 4th row) 
Option (B)27000 is correct option

Problem # 4


Reasoning questions with answers for bank exams

Take the difference of every number with its preceding number
4 - 2  = 2 ( 2nd number - 1st number ) , 
9 - 4 = 5,  ( 3rd number - 2nd number ) , 
20 - 9 = 11 , ( 4th number - 3rd number ) , 
40 -20 = 20, ( 5th number - 4th number ) , 
? - 40 ,   ( 6th number - 5th number ) 
Now this new series of difference is here
2 , 5 , 11 , 20 , ? - 40 
Again take the difference of every number with its preceding number
5 - 2 = 3 , 
11 - 5 = 6 ,
 20 - 11 = 9 
( ? - 40 )  - 20 = 12 (If the pattern follows)
Now study the difference , this difference is multiple of 3, Hence proceeding in the same manner the next difference will be 12
So  ( ? -  40 )  - 20 = 12 
? = 12 + 20 + 40
?  = 72
Option (2)72 is correct option

Problem # 5


Reasoning questions with answers for bank exams

This problem of reasoning  have also three figures, And every figure consist of four numbers .Out of four numbers one is at centre and remaining three numbers are around the central number.
In every figure central number is the sum of the number in base line and product of other two numbers . This implies 1st we have to multiply the two numbers which are in the top row then we have to add this sum to the number in bottom line to get the central number in each figure.
(5 × 2) + 10 = 10 + 10 = 20 (1st Figure)
(6 × 3) + 6 = 18 + 6 = 24  ( 2nd Figure)
(7 × 5) + 6 = 35 + 6 = 41 ( 3rd Figure)
Option (1)24 is correct option

Also read this article

Problem # 6


Reasoning questions with answers for bank exams
This circle has been divided into four parts and each parts consist of three numbers , Two numbers are written in the outer part of the quadrant and one is at the inner part of the quadrant.
(12 - 8 )³  - 1 =  4³  - 1 =  64 - 1 = 63
 (4 - 3 )³  - 1  =  1³  - 1 =  1 - 1 =  0
 (9 - 6 )³  - 1 =  3³  - 1 =  27 - 1 =  26
 (12 - 10 )3  - 1 =  23  - 1 =  8 - 1 =  7
                                or
Cube(12 - 8 ) -  1 = 64 - 1 =  63
Cube (4 - 3 ) -  1 = 1 - 1 =  0
Cube( 9 - 6 ) - 1 = 27 - 1 = 26
Cube(12 - 10 ) - 1 = 8 - 1 = 7
Option (1)7 is correct option


Problem # 7


Reasoning questions with answers for bank exams

This series is the difference of cube  and square of  the number starting from 2.
  23  -   22   = 8 - 4 =  4   ( 1st number in the series) 
 33  -   32   = 27 - 9 =  18   ( 2nd number in the series) 
  43  -   42   = 64 - 16 =  48   ( 3rd number in the series) 
 53  -   52   = 125 - 25 =  100   ( 4th number in the series) 
  63  -   62   = 216 - 36 =  180   ( 5th number in the series) 
  73  -   72   = 343 - 49 =  294   ( 6th number in the series) 
 83  -   82   = 512 - 64 =  448   ( 7th number in the series) 
or
Cube of 2 -  square of 2 = 8 - 4 = 4
Cube of 3 -  square of 3 = 27 - 9 = 18
Cube of 4 - square of 4 = 64 - 16 = 48
Cube of 5 -  square of 5 = 125 - 25 = 100
Cube of 6-  square of 6 = 216 - 36 = 180
Cube of 7 -  square of 7 = 343 - 49 = 294
Option (3)48 is correct option

Problem # 8


Reasoning questions with answers for bank exams

Take the sum of cube roots of two adjoining numbers from the big box in the centre to get the number in outer part of small box .
  (343)⅓ +   (27)⅓  = 7 + 3 = 10 ( The number in the outer small box)
 (27)⅓+  (125)⅓  = 3 + 5 = 8 ( The number in the outer small box)
 (125)⅓ +  (?)⅓  = 5  + (?)⅓  = 7 ( The number in the outer small box)
⇒   (?)⅓  = 7 - 5 
⇒   (?)⅓  = 2  , cubing both  sides
⇒  ?  = 8
 ( ?)⅓ + (343)⅓  =  (?)⅓  + 7 = 9 ( The number in the outer small box)
⇒  (?)⅓  = 9 - 7 
⇒ (?)⅓  = 2  , cubing both  sides
⇒  ?  = 8
Hence we get 2 in both the cases 
Hence Option (C)8 is correct option

Problem # 9


Reasoning questions with answers for bank exams 
This problem of reasoning also have three figures, And every figure consist of five numbers .Out of five numbers one is at centre and remaining four numbers are around the central number.

In this problem central number is 2 less than the sum of  all the number which are around the middle number.

1st figure

(0 + 6 + 4 + 2) - 2 = 12 - 2 = 10 (middle number in first figure) 

2nd figure

( 6 + 2 + 10 + 8) - 2 = 26 - 2 = 24 (middle number in 2nd figure) 

3rd figure

(4 + 14 + 12 + 10 ) - 2 = 40  - 2 = 38 (middle number in 3rd figure) 
Option (C)38 is correct option

Problem # 10


Reasoning questions with answers for bank exams

This circle has been divided into four parts and each parts consist of three numbers , Two numbers are written in the outer part of the quadrant and one is at the inner part of the quadrant.
Difference of digits of numbers in outer circle of any Quadrant - (difference of digits of numbers in inner circle of any Quadrant ) = one in every case
(12 - 8 ) - (6 - 3)  = 4 - 3 =1
(4  - 3 ) - (1 - 1)  = 1 - 0 =1
(9 - 6 ) - (6 - 4)  = 3 - 2 = 1
Similarly we have to proceed like this and choose the correct option whose difference of digits is 1. Look at the option all other option except 2nd option will not satisfies the condition required.
Hence    (12 - 10 ) - (4 - 3)  = 2 - 1 = 1
Option (2)43 is correct

In this post I discussed Ten Tricky logical reasoning questions+answers , Latest reasoning questions with answers  . Comment your valuable suggestion for further improvement.














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Ten Tricky logical reasoning questions+answers , Latest reasoning questions with answers

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Ten latest and Tricky logical reasoning questions+answers are discussed in this post . These questions of reasoning in latest reasoning questions with answers are very very important for upcomig competitive exams  like Bank PO , SSC CGL etc . So let us start solving and understanding these Maths logical reasoning questions with answers.

Problem #1

Ten Tricky logical reasoning questions+answers

Magnitude of {(7 + 5) (7 - 5 ) - cube of 5 } = 125 - 12 *2 = 125 - 24 = 101
Magnitude of {(8 + 3) ( 8 - 3 ) - cube of 3 } = 11 * 5 - 27 = 55 - 27 = 28
Magnitude of {(6 + 2) ( 6 - 2 ) - cube of 2  } = 8 * 4  - 8 = 32 - 8 = 24
Magnitude of {(5 + 4) ( 5  - 4 )  - cube of 4 } =  64 -  9*1 =  64 - 9  = 55
 Note :- Magnitude of  difference of two numbers means , we have to consider the difference of two numbers irrespective of their positions, And the magnitude of difference of two numbers will always be a positive value .The magnitude of  3 - 2 and 2 - 3 will be same and will be equal to 1 ( not -1)  and  magnitude of 7 - 9  and 9 - 7 will be same and equal to  2 ( not -2 ), Similarly the magnitude of 11 - 2 and 2 - 11 will same and equal to 9 (  not -9 ).

Alternate Method 

Magnitude  { { 72  -   52  } -  (  53 ) } =  (  53 ) -  { 72  -   52  } = 125 - ( 49 - 25) =125 - 24 = 101
Magnitude  { { 82  -   32  } -  (  33 ) } = { 82  -   32  }  -  (  33 )  = ( 64 - 9 ) - 27 = 55 - 27 =  24
Magnitude  { { 62  -   22  } -  (  23 ) } = {62  -   22  }  -  (  23 )  =  ( 36  - 4) - 8 = 32 - 8 = 24
Magnitude  { { 52  -   42  } -  (  43 ) } =  (  43 ) -  { 52  -   42  } = 64 - ( 25 - 16 ) = 64  - 9 = 55

Option (2) is correct

Problem #2

Ten Tricky logical reasoning questions+answers

Difference of both the digits of numbers in outer circle of any Quadrant - (difference of digits of numbers in inner circle of any Quadrant ) = two  in every case
(55 - 52 ) - (9 - 8)  = 3 - 1 = 2
(29  - 21 ) - (9 - 3)  = 8 - 6 = 2
(18 - 12) - (8 - 4)  = 6 - 4 = 2
(7 - 5) - (4 - 4)  = 2 - 0 =2
Option (4) is correct

Problem #3

Ten Tricky logical reasoning questions+answers

1st quadrant 
4 ×   (7 - 3)2   = 4 ×  42  =  4  × 16 = 64
2nd quadrant 
4 ×  (5 - 5)2  = 4 × 02  =  4 × 0  = 0
3rd quadrant 
4 ×  (11 - 8)2  = 4 ×  32  =  4 × 9 = 36
4th quadrant 
4 ×   (8 - 2)2  = 4 ×  62  =  4×  36 = 144
Option  (3) is correct option

Problem #4


Ten Tricky logical reasoning questions+answers
In this figure the sum of the digits of the numbers obtained from the multiplication of both the numbers which are outer part of that particular quadrant.
8 × 2 = 16 = 1 + 6 = 7 digit in middle left box
6 × 5 = 30 = 3 + 0 = 3 digit in middle left box
6 × 9 = 54 = 5 + 4  = 9 digit in middle right box
4 × 1 = 04 = 0  + 4 = 4 digit in middle right box
Option  (2) is correct option

Problem #5

Ten Tricky logical reasoning questions+answers
 (11 - 7 )3  =  43  = 64  (1st number in 2nd row ) 
  (14 - 11 )3 =  33 = 27  ( 2nd number in 2nd row ) 
  (64 - ?)2  =   43  = 8  ( It will be 3rd number in 2nd row ) 
So ? = 66
Alternate Method 

Add the cube root of respective number  in 2nd row to the number in 1st row to get number in 3rd row. 
7 +  64⅓ = 7 + 4 = 11 ( 1st number in 3rd row) 
11 +  27⅓ = 11 + 3 = 14 ( 2nd number in 3rd row) 
64 +  8⅓ = 64 + 2 = 66 ( 3rd number in 3rd row) 
Option  (B) is correct option

Problem #6

Ten Tricky logical reasoning questions+answers
 Pick all prime number up to 11
1st prime number = 2
   Multiply it with next number
2 × 3 = 6 ( 2nd  number in the series) 
2nd prime number = 3
   Multiply it with next number
3 × 4 = 12 ( 3rd  number in the series) 
3rd prime number = 5
   Multiply it with next number
5 × 6 = 30 ( 4th  number in the series) 
4th prime number = 7
   Multiply it with next number
7 × 8 = 56 ( 5th  number in the series) 
5th prime number = 11
   Multiply it with next number
11 × 12 = 132 ( 6th   number in the series) 
Option  (D) is correct option

Problem #7

Ten Tricky logical reasoning questions+answers

Add the twice of the number in 1st column to the half of the number in 2nd column to get the 3rd number in every row. 

( 2 × 6 ) + ( 8 ÷ 2 ) = 12 + 4 = 16 ( 1st number in 3rd column) 
( 2 × 1 ) + ( 6 ÷ 2 ) = 2 + 3 = 5 ( 2nd number in 3rd column) 

( 2 × 3 ) + ( 10 ÷ 2 ) = 6 + 5 = 11 ( 3rd number in 3rd column) 
Option  (A) is correct option
Also read these posts on Reasoning
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Problem #8

Ten Tricky logical reasoning questions+answers
Divide with 10 the Multiplication/Product of  all the numbers which are in the outer part of triangles to get the number in the centre of the triangle. And this method will be applicable to all the three  triangles.
1st triangle 
(5 × 6 × 4 ) ÷ 10 = 120 ÷10 =12
2nd triangle
(6 × 7 × 5 ) ÷ 10 = 210 ÷10 =21
3rd triangle
(4 × 8 × 10 ) ÷ 10 = 320 ÷10 =32
Option  (3) is correct option

Problem #9


Ten Tricky logical reasoning questions+answers
Here every figure consist of three numbers in 1st line and one number in 2nd line. And 2nd number written in every figure is the (HCF) Highest common Factor of all the three numbers in the 1st line in each figure. 
H C F ( Highest common Factor ) of 9 ,15 , 18 = 3 (1st figure )
HCF ( Highest common Factor ) of 16 ,28 , 32 = 4 ( 2nd figure )
HCF ( Highest common Factor ) of 24 ,36 , 48 = 12 (3rd figure )
Therefore option (4) is correct option .

Problem #10

Ten Tricky logical reasoning questions+answers
Take reverse of sum of both the numbers in every quadrant to get the number attached to middle of this figure. 

9 + 8  = 17 ↔️71 ( The number in the middle of  4th Quadrant)
8 + 3  = 11 ↔️11 (  The number in the middle of  1st Quadrant)
5 + 7  = 12 ↔️21 ( The number in the middle of   2nd Quadrant)
7 + 7  = 14 ↔️41 ( The number in the middle of   3rd Quadrant)
Therefore option (2) is correct option .

In this post I discussed Ten Tricky logical reasoning questions+answers , Latest reasoning questions with answers  . Comment your valuable suggestion for further improvement.

Also Read these posts on Reasoning













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