Most Important box Problems of Reasoning for different competitive Exams
Reasoning of missing number in box problems will be discussed with the help of most important examples. Some of these examples are of 3 × 3 order and other are of different orders .
Most Important Box Problems of Reasoning for different competitive Exams
Problem # 1
This reasoning problem consists of three figures and every figure have five numbers associated to it . Four numbers are on the corner of each box and one number is in the middle of every box. Look at last figure , it have ?(question mark) in its centre . So the solution of this problem is to find the value of question mark using other four numbers associated to it .
But the main aim is how to utilised these two numbers to get the value of question mark?.
We have to find or search the formula for these four numbers in each figure to utilised them in any possible way to get number in each box .
Formula :-
Left most number in 1st row of any box + { Product of all other numbers in same box } = Middle number in that box.
1st Box
3 + (4 × 5 × 3) = 3 + 60 = 63
2nd Box
6 + (7 × 3 × 5) = 6 + 105 = 111
3rd Box
2 + (6 × 5 × 4) = 2 + 120 = 122
Therefore option (3) 122 is correct.
Problem # 2
This reasoning problem consists of two figures and every figure have five numbers associated to it . Four numbers are on the corner of each box and one number is in the middle of both the box. Look at 2nd figure , it have ? in its centre . So the solution of this problem is to find the value of question mark using other four numbers associated to it .
We have to find the formula for these four numbers in both figures to utilised them in any possible way to get number in each box .
Formula
Product of both the numbers in 1st row - Product of both the numbers in 3rd row = Number in central row
1st Box
(11 × 12) - ( 6 × 9) = 132 - 54 = 78.
2nd Box
(14 × 10) - ( 7 × 8) = 140 - 56 = 84
Therefore option (1) 84 is correct.
Problem # 3
This box problem consist of three rows and three columns . And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 3rd row of 3rd column.
Formula:-
Square of Number in 2nd row ÷ Number in 1st row = Number in 3rd row
8² ÷ 4 = 64 ÷ 4 = 16
6² ÷ 3 = 36 ÷ 3 = 12
10² ÷ 2 = 100 ÷ 2 = 50
Therefore option (2) 50 is correct.
Problem # 4
_11zon.jpg "Reasoning for competitive Exams")
This box problem consists of six rows and two columns. And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 4th row of 2nd column.
Formula:-
The sum of both the numbers in any row = Number in 1st column of lower row.
6 + 9 = 15 (The number in 1st column of 2nd row)
15 + 7 = 22 (The number in 1st column of 3rd row)
22 + 5 = 27 (The number in 1st column of 4th row)
27 + ? = 36 ⇒ ? = 36 - 27⇒ = 9 (The number in 1st column of 5th row)
36 + 3 = 39 (The number in 1st column of 6th row)
Therefore option (1) 9 is correct.
Problem # 5
_11zon.jpg "Reasoning for competitive Exams")
This box problem also consist of six rows and two columns . And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 2nd row of 2nd column.
1st Column
Formula :-
Sum of two numbers in consecutive rows of same column = Number in Succeeding row of same column taken in order from top to bottom.
3 + 6 = 9 (Sum of the numbers in 1st and 2nd rows is equal to number in 3rd row of 1st column).
6 + 9 = 15 (Sum of the numbers in 2nd and 3rd rows is equal to number in 4th row of 1st column).
9 + 15 = 24 (Sum of the numbers in 3rd and 4th rows is equal to number in 5th row of 1st column).
15 + 24 = 39 (Sum of the numbers in 4th and 5th rows is equal to number in 6th row of 1st column).
2nd Column
Formula:-
Sum of two numbers in consecutive rows of same column = Number in Succeeding row of same column taken in reverse order from bottom to top.
4 + 7 = 11 (Sum of the numbers in 6th and 5th rows is equal to number in 4th row of 2nd column).
7 + 11 = 18 (Sum of the numbers in 5th and 4th rows is equal to number in 3rd row of 2nd column).
11 + 18 = 29 (Sum of the numbers in 4th and 3rd rows is equal to number in 2nd row of 2nd column).This is also the value of question mark.
18 + 29 = 47 (Sum of the numbers in 3rd and 2nd rows is equal to number in 1st row of 2nd column).
Therefore option (1) 29 is correct.
Problem # 6
This box problem consist of three rows and three columns . And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 2nd row of 3rd column.
To find the value of question mark. we shall divide this box into two parts vertically then we can have the formula for these numbers written in this box . Because after careful observation we can see that the product of both the numbers in left half in any particular row is equal to sum of both the numbers in right half in that particular row.
This reasoning problem can be solved by two different methods.
1st Method
Column wise
(1 × 2 ) + 1 = 2 + 1 = 3 (1st Column)
(7 × 14 ) + 7 = 98 + 7 = 105 (2nd Column)
(9 × ? ) + 9 = 117 (3rd Column)
(9 × ? ) = 117 - 9
(9 × ? ) = 108
? = 108/9
? = 12
2nd Method
(2 + 1 ) × 1 = 3
(14 + 1 ) × 7 = 15 × 7 = 105
(? + 1 ) × 9 = 117
(? + 1 ) = 117/9
(? + 1 ) = 13
? = 13 - 1
? = 12
Therefore option (2) 12 is correct.
Problem # 7
This box problem consist of three rows and four columns .And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 3rd row of 3rd column.
This reasoning problem can be solved by two different methods.
1st Method
Formula:-
{R4 ÷ R3} × R2 = R1
(8 /4) × 3 =2 × 3 = 6
(27 /3) × 2 = 9 × 2 = 18
(9/?) × 5 = 15
(9/?) = 15/5 = 3
(?/9) = 1/3
? = 9/3
? = 3
2nd Method
Multiplication of 1st and 3rd columns = Multiplication of 2nd and 4th columns.
Formula:-
{R1 × R3} = {R2 × R4}
6 × 4 = 8 × 3
18 × 3 = 2 × 27
15 × ? = 5 × 9
? = 45/15
? = 3
Therefore option (4) 3 is correct.
Problem # 8
This reasoning problem consists of three figures and every figure have five numbers associated to it . Four numbers are on the outer side of each box and one number is in the middle of both the box. Look at 3rd figure , it have ? in its centre . So the solution of this problem is to find the value of question mark using other four numbers associated to it.
We have to find the formula for these four numbers in 1st two figures to utilised them in any possible way to get number in each box .
1st method
Formula :-
Product of all the numbers in outer parts ÷ 10 = Middle number
(5 × 3 × 4 × 2)/10 = 120/10 = 12 (1st Box)
(5 × 6 × 2 × 3)/10 = 180/10 = 18 (2nd Box)
(5 × 2 × 2 × 9)/10 = 180/10 = 18 (3rd Box)
2nd Method
Since 5 and 2 are common in all the three figures, so ignore these two numbers and check the multiplication of other two numbers to get middle number in all the three figures.
3 × 4 = 12(1st Box)
6 × 3 = 18(1st Box)
9 × 2 = 18(3rd Box)
Therefore option (3) 18 is correct.
Problem # 9
1st Method
Formula :-
Any Term = { (Previous Term) × 2 } + n, where n is natural number less than 6
1st Term = (4 × 2) + 1 = 8 + 1 = 9
2nd Term = (9 × 2) + 2 = 18 + 2 = 20
3rd Term = (20 × 2) + 3 = 40 + 3 = 43
4th Term = (43 × 2) + 4 = 86 + 4 = 90
5th Term = (90 × 2) + 5 = 180 + 5 = 185(The value of question mark).
2nd Method
Now taking difference of two consecutive terms
Again taking difference of two consecutive terms. From last row we are seeing that differences are double of previous difference , Hence next difference must be 48.
Therefore
x - 47 = 48
x = 48 + 47
x = 95
Hence differences in 2nd row would be 5 , 11 , 23 , 47, 95
And from 1st row
? - 90 = 95
? = 95 + 90
? = 185
Therefore option (3) 185 is correct.
Problem # 10
Since question mark in this figure is in the outer circle of this figure. To find the value of question mark (?). We can combine two numbers present in the inner circle with the help of any mathematical operation/s to find the value of number in the outer circle .
Starting from 3 and moving clockwise , taking two numbers at a time in inner circle will be equal to number in outer circle.
Formula :-
{2 × 1st Number} + {2nd Number} = Number in outer circle
(2 × 3) + 4 = 6 + 4 = 10
(2 × 4) + 6 = 8 + 6 = 14
(2 × 6) + 8 = 12 + 8 = 20
(2 × 8) + 3 = 16 + 3 = 19 = ?(The value of question mark)
Therefore option (1) 19 is correct.
Also Reads these Articles
Reasoning for competive exams
Box and circle reasoning
Reasoning for bank exams
Ten Tricky logical reasoning
Missing number series questions
Box and circle reasoning
Reasoning for bank exams
Ten Tricky logical reasoning
Missing number series questions
Reasoning questions with answers
Circle Reasoning
Box Problems
15 Questions Circle Problems
SSC CGL Reasoning
Circle Reasoning
Box Problems
15 Questions Circle Problems
SSC CGL Reasoning
Reasoning of missing number in box problems with solutions 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. please feel free to comment your opinions regarding this post.
