Reasoning of missing number in box problems will be discussed with the help of 10 most important examples. Some of these examples are of 3 × 3 order and other are of 3 × 4 orders .
Reasoning of missing number in box problems
PROBLEM # 1
Step 1.
Subtract the number in first column from the number in third column of every row. i. e. C3 - C1 = S
Step 2 .
Multiply the the above difference with 3 to get the numbers in second column of every row i. e. C2 = 3S.
1st row 3 (48 - 28 ) = 3 × 20 = 60
2nd row 3 (7 - 5) = 2 × 3 = 6
3rd row 3(27 - 14) = 3 × 13 = 39
4th row 3(16 - 7 ) = 3 × 9 = 27
Hence required number is 27
Option (A) 27 is correct option
PROBLEM # 2
Every number in fourth row has been written as the square of all the numbers in 1st three rows in any particular column.
Take the square of addition of the elements of 1st three rows of 1st column like this
(1 + 4 + 2 )² = 7² = 49
Take the Take the square of addition of the elements of 1st three rows of 1st column like this
(4 + 2 + 2 )² = 8² = 64
Take the square of addition of the elements of 1st three rows of 1st column like this
(? + 5 + 3 )² = (? + 8 )² = 169
(? + 8 )² = (13)² = 169 , Taking Square root
? + 8 = 13
? = 13 - 5
? = 5
So required number will be 5
Option (C) 5 is correct optionAlso Read Latest Post of Reasoning Questions
PROBLEM # 3
Multiply all the numbers in the first three rows of every column and then divide the product by 2 to get the number in last row like this
(5×2×8)÷2 = 80÷2 = 40 ( Last number in 1st column )
(5×4×3)÷2 = 60÷2 = 30 ( Last number in 2nd column )
(2×1×10)÷2 = 20÷2 = 10 ( Last number in 3rd column )
So required number will be 10
Option (D)10 is correct option.
PROBLEM # 4
Adding both the numbers in second and third rows then multiply it with the number in First row to get the number in last row ,the process will be repeated for all the three columns
( 6 + 7 ) × 5 = 65
( 3 + 2 ) × 4 = 20
( ? + 4 ) × 9 = 45
This implies ( ? + 4 ) = 45/9 = 5
? = 5 - 4 = 1
Hence required number is 1
Option (A) 1 is correct option.
PROBLEM # 5
Decrease the numbers in first columns of every row by 1 then multiply it to get the number in third column of every row.
1st Row ( 8 - 1) × 3 = 7 × 3 = 21
2nd Row ( 6 - 1) × 5 = 5 × 5 = 25
3rd Row ( 12 - 1) × 2 = 11 × 2 = 22
Hence required answer is 22
Option (B) 22 is correct option.PROBLEM # 6
Add 1st two rows of each column to multiply with the number in 3rd column to get fourth number in every row.
(5 + 4 ) × 2 = 9 × 2 = 18 (1st column 3rd row )
(6 + 3 ) × 3 = 9 × 3 = 27 ( 2nd column 3rd row )
(12 + 4 ) × ? = 16 × ? = 96 (3rd column 3rd row )
16 × ? = 96
? = 6
Hence required number is 6
Option (D) 6 is correct option.
PROBLEM # 7
First of all multiply 1st three numbers in every Column then add the sum of 1st three numbers in every row to get the numbers in 4th row. i. e. formula for this problem is [ if a, b ,c and d are numbers in any column then d = abc + (a+b+c) ]
[ ( 3 + 4 + 5) + (3 × 4 × 5 ) ] = 12 + 60 = 72
[ ( 2 + 5 + 6) + (2 × 5 × 6 ) ] = 13 + 60 = 73
[ ( 5 + 9 + 1) + (5 × 9 × 1 ) ] = 15 + 45 = 60
Hence required number is 60
Option (D)60 is correct option.
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PROBLEM # 8
Decrease both the numbers in first and second columns of every row then multiply the reduced numbers to get the numbers in third column of every row
{ 8 - 1 }×{ 6 - 1 } = 7 × 5 = 35
{ 6 - 1 }×{ 4 - 1 } = 5 × 3 = 15
{ 7 - 1 }×{ 5 - 1 } = 6 × 4 = 24
Hence required number is 24.
Option (A)24 is correct option.
PROBLEM # 9
Decrease the numbers in first column of every row and increase the numbers in second column of every row when multiplying both the reduced number of first and second column to get the numbers in third column of every row
{ 6 - 1 } × { 8 + 1 } = 5 × 9 = 45
{ 4 - 1 } × { 6 + 1 } = 3 × 7 = 21
{ 7 - 1 } × { 5 + 1 } = 6 × 6 = 36
Hence required number is 36
Option (C)36 is correct option.PROBLEM # 10
Total of 1st row is 3 + 6 + 8 = 17
Total of 2nd row is 5 + 6 + 8 = 17
Total of 3rd row is 4 + 7 + ? = 17
Therefore replacing ? with 6 to get required answer.
Option (D)6 is correct option
So these were the most important box problems for many competitive exams
So these were the most important box problems for many competitive exams
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