Question

what is the sum of the original number and the new number if the digits 1 and 9 are interchanged in the number 1009?
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Answer 1

Let The Unit Digit Be Y And Tens Digit Be X

• Number formed = 10x + y

• Reverse number = 10y + x

Given On Equation 1 :-

⇒x + y = 9

Given On Equation 2 :-

⇒10y + x = 10x + y + 27

⇒9y - 9x = 27

Given On Equation 3 :-

⇒y - x = 3

Solving Equation 1,2 And 3:-

• We Get :-

⇒x = 3 and y = 6

• Original Number = 36 Reversed Number = 63

Answer 2

Step-by-step explanation:

Given that the sum of the original number and the new number if the digits 1 and 9 are interchanged in the number 1009.

Let's say the tens digit number be a and ones digit number be b.

So, original number is 10a + b While interchanged number is 10b + a.

As per given statement in the question,

→ a + b = 9

→ a = 9 - b --------- (1)

→ 10b + a = 10a + b = 27

→ 10b - b + a - 10a = 27

→ 9b - 9a = 27

→ b - a = 3

Substitute value of a in above equation,

→ b - (9 - b) = 3

→ b - 9 + b = 3

→ 2b = 12

→ b = 6

Substitute value of b in (1)

→ a = 9 - 6

→ a = 3

So, original number = 10a + b = 10(3) + 6 = 36

Interchanged number = 10b + a = 10(6) + 3 = 63