Ten Most Important Box Problems in reasoning for Competitive Exams
Ten Most Important box problems in reasoning for SSC CGL, SSC CHSL , RRB NTPC and other similar competitive exams.
Problem # 1
Multiplying the numbers in first column with numbers in second column , Its product must be equal to the multiplication of numbers in third column and fourth column.
Since all the numbers in third and fourth columns are given therefore we have to start from here.
In 3rd and 4th columns
1 × 24 = 24
8 × 3 = 24
2 × 12 = 24
12 × 2 = 24
In 1st and 2nd columns
6 × 4 = 24
4 × 6 = 24
3 × 8 = 24
Similarly ? × 1 = 24
Hence ? = 24
Hence option (D) is right option
Problem # 2
Carefully study all the numbers in 1st row which are 3 ,8, 13, 18, ? , Now take the difference of two successive numbers, this is 5 in all the cases.
18 - 13 = 5
13 - 8 = 5
8 - 3 = 5
so ? - 18 = 5
⇒ ? = 5 + 18
⇒ ? = 23
Now study all the numbers in 2nd row which are 2 ,11, 20, 29, ? , Now take the difference of two successive numbers, this is 9 in all the cases.
Because 29 - 20 = 9
20 - 11 = 9
11 - 2 = 9
So ? - 29 = 9
⇒ ? = 9 + 29
⇒ ? = 38
Hence option (C) is right option
Problem # 3
If we multiply 1st column with the square root of 3rd column ,then we shall get the 2nd column.
7 × √4 = 7 × 2 = 14 (1st element in 2nd column)
4 × √9 = 4 × 3 = 12 ( 2nd element in 2nd column)
6 × √? = 24
√? = 4
√? = 4 , squaring both sides
? = 16 ( 3rd element in 2nd column)
Hence option (A) is right option
Problem # 4
Taking the sum of squares of all the elements of 1st three rows to get the element in fourth row of respective column.
4² + 2² + 3² = 16 + 4 + 9 = 29 (Last number in 1st column)
5² + 1² + 6² = 25 + 1 + 36 = 62 (Last number in 2nd column)
1² + 2² + 3² = 1 + 4 + 9 = 14 (Last number in 3rd column)
5² + 5² + 2² = 25 + 25 + 4 = 54 ,The value of question Mark (Last number in 4th column)
Hence option (A) is right option
Problem # 5
Look at the numbers in fourth row, these are large as compare to all other numbers. All the numbers in fourth row are sum of 2nd row and product of 1st row and 3rd row .
( 7 × 3 ) + ( 8 ) = 21 + 8 = 29 ( Last number in 1st column)
( 4 × 3 ) + ( 7 ) = 12 + 7 = 19 ( Last number in 2nd column)
( 5 × ? ) + ( 6 ) = 5? + 6 = 31 ( Last number in 3rd column)
⇒ 5? = 31 - 6
⇒ 5? = 25
⇒ ? = 25/5
⇒ ? = 5
Hence option (D) is right option
Problem # 6
In this box problem the difference of the numbers in 1st and 2nd columns is equal to the number in 3rd column, while the sum of the numbers in 1st and 2nd columns is equal to the number in 4th column.
5 - 2 = 3 ( Number in 3rd column of 1st row )
6 - 3 = 3 ( Number in 3rd column of 2nd row )
4 - 1 = 3 ( Number in 3rd column of 3rd row )
8 - 1 = 7 ( Number in 3rd column of 4th row )
5 + 2 = 7 ( Number in 4th column of 1st row )
6 + 3 = 9 ( Number in 4th column of 1st row )
4 + 1 = 5 ( Number in 4th column of 1st row )
8 + 1 = 9 ( Number in 4th column of 1st row ) ,
Therefore this will be the value of question mark "? "
Hence option (D) is right option
Problem # 7
In this box problem every number in 2nd column is the sum of 1st and 3rd column.
481 + 365 = 846
655 + 184 = 839
297 + 492 = 789 (The value of question mark)
Hence option (B) is right option
Problem # 8
In this box problem the difference of sum of last two columns and 1st two columns in every row is one.
( 10 + 5 ) - ( 6 + 8 ) = 15 - 14 = 1 ( 1st Row)
( 8 + 4 ) - ( 9 + 2 ) = 12 - 11 = 1 ( 2nd Row)
( 7 + 8 ) - ( 11 + 3 ) = 15 - 14 = 1 ( 3rd Row)
( ? + 2 ) - ( 5 + 8 ) = ( ? + 2 ) - ( 13) = 1 ( 4th Row)
So in order to get the difference equal to one we have to put ? = 12,
so that ( 12 + 2 ) - ( 5 + 8 ) = 14 - 13 = 1 ( 4th Row)
Hence option (B) is right option
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Problem # 9
In this box problem every number in fourth row of every column is the sum of Multiplication of all the numbers in every column and sum of all the numbers in every column.
In 1st column ( 3 × 4 × 5 ) + ( 3 + 4 + 5 ) = 60 + 12 = 72
In 2nd column ( 2 × 5 × 6 ) + ( 2 + 5 + 6 ) = 60 +13 = 73
In 3rd column ( 5 × 9 × 1 ) + ( 5 + 9 + 1 ) = 45 + 15 = 60 ( The value of question mark )
Hence option (D) is right option
Problem # 10
Multiplying the numbers in first column with numbers in second column , Its product must be equal to the multiplication of numbers in third column and fourth column.
Since all the numbers in 1st and 2nd columns are given therefore we have to start from here.
In 1st and 2nd columns
6 × 8 = 48
4 × 12 = 48
3 × 16 = 48
8 × 6 = 48
In 3rd and 4th columns
1 × 48 = 48
8 × 6 = 48
16 × 3 = 48
Similarly multiplication of 2 × ? = 48
Hence ? = 24
Hence option (D) is right option
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