## Missing number in box, Reasoning problem

## Reasoning of missing number in box problems

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**PROBLEM # 1**

Step 1. Subtract the number in first column from the number in third column of every row. i. e. C3 - C1 = Step 2 . Multiply the the above difference with 3 to get the numbers in second column of every row i. e. C2 = 3S1st row 48 - 28 = 20 × 3 = 602nd row 7 - 5 = 2 × 3 = 63rd row 27 - 14 = 13 × 3 = 394th row 16 - 7 = 9 × 3 = 27 Hence required number is 27

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**PROBLEM # 2**

Take the square of addition of the elements of 1st three rows of 1st column like this

Square(1 + 4 + 2 ) = sqr (7) = 49

Take the Take the square of addition of the elements of 1st three rows of 1st column like this

Square(4 + 2 + 2 ) = sqr (8) = 64

Take the square of addition of the elements of 1st three rows of 1st column like this

Square(? + 5 + 3 ) = sqr (? + 8 ) = 169

sqr (? + 8 ) = sqr (13)

? + 8 = 13

? = 5

So required number will be 5

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**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

(5×4×3)÷2 = 60÷2 = 30

(2×1×10)÷2 = 20÷2 = 10

So required number will be 10

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**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 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

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**PROBLEM # 5**

Decrease the numbers in first two 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

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**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 = 18

(6 + 3 ) × 3 = 27

(12 + 4 ) × ? = 96

16 × ? = 96

? = 6

Hence required number is 6

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**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 [ 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

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**Also study these missing number series questions**

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**PROBLEM # 8**

Decrease both the numbers in first and second columns of every row then multiply the reduced number to get the numbers in third column of every row

{ 8 - 1 }×{ 6 - 1 } = 35

{ 6 - 1 }×{ 4 - 1 } = 15

{ 7 - 1 }×{ 5 - 1 } = 24

Hence required number is 24.

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**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 } = 45

{ 4 - 1 }×{ 6 + 1 } = 21

{ 7 - 1 }×{ 5 + 1 } = 36

Hence required number is 36

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**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.

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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