Question

The sum of the digits in a two-digit number is 7. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.

Let the units place digit in original number bey and tens place be x.

So, Original number = `square`

Now, from the first condition, x + y = `square`  ...(i)

From the second condition, 10y + x = `square` + `square`

10y + x – `square` = `square`

`square` – `square` = `square`

9(y – `square`) = `square`

y – `square = square/9 = square`  ...(ii)

Adding equation (i) and (ii), we get

2y = `square`

⇒ y = `square`

Putting y = `square` equation (i), we get

x + `square` = 7

⇒ x = `square`

So, Original number = 10x + y = 10 × `square` + `square`

= `square`

Answer 1

Let the units place digit in original number be y and tens place be x.

So, Original number = 10x + y

Now, from the first condition, x + y = 7  ...(i)

From the second condition, 10y + x = 10x + y + 27

10y + x – 10x + y = 27

9y – 9x = 27

9(y – x) = 27

y – x = `bb27/9` = 3  ...(ii)

Adding equation (i) and (ii), we get

2y = 10

⇒ y = 5

Putting y = 5 equation (i), we get

x + 5 = 7

⇒ x = 2

So, Original number = 10x + y = 10 × 2 + 5

= 25