Question

The sum of the digits of a two-digit number is 8.The number obtained by interchanging its digits is 18 more than the original number. Find the original number.

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Answer 1

Let the no. be 10x+ y
x+ y= 8 ......... (i)
10y+ x = 10x+ y + 18
9x- 9y= - 18
x- y= -2.........(ii)
i + ii
2x= 6
x= 3
y= 5
No= 35

Answer 2

Solution -

Let the digit in ones place be x

So, the digit in tens place be 8 - x

∴ Original no. = 10(8 - x) + 1(x)

= 80 - 10x + x

= 80 - 9x

∴ New no. = 10(x) + 1(8 - x)

= 10x + 8 - x

= 9x + 8   [ ∵ By reversing the digits ]

According to Question,

New no. - Original no. = 18

⇒ (9x + 8) - (80 - 9x) = 18

⇒ 9x + 8 - 80 + 9x = 18

⇒ 18x - 72 = 18

⇒ 18x = 18 + 72   [∵ By transposition]

⇒ 18x = 90

⇒ x = 90/18

⇒ x = 5

Required numbers -

Original no. = 80 - 9x = 80 - 9(5)

= 80 - 45 = 35

New no. = 53 [∵ By reversing the digits]

∴ The required number is either 35 or 53

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