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

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 .

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

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

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

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

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

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

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

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)
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Problem # 9

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

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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