The sum of the digits of a two digit number is 5. If the digits are reversed, the number is reduced by 27. Find the number.

Let digit at unit's place be x and ten's place be y. Number = 10 × y + x = 10y + x Given, Sum of the digits of a two digit number is 5. ∴ x + y = 5 ⇒ x = 5 - y ........(1) If the digits are reversed, then number = 10x + y. Given, If the digits are reversed, the number is reduced by 27. ∴ 10x + y = 10y + x - 27 ⇒ 10x - x = 10y - y - 27 ⇒ 9x = 9y - 27 ⇒ 9x = 9(y - 3) ⇒ x = y - 3 ........(2) From (1) and (2), we get : ⇒ y - 3 = 5 - y ⇒ y + y = 5 + 3 ⇒ 2y = 8 ⇒ y = = 4. Substituting value of y in equation (2), we get : ⇒ x = 4 - 3 = 1. Number = 10y + x = 10(4) + 1 = 40 + 1 = 41. Hence, the number = 41.

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The sum of the digits of a two digit number is 7. If the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.

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The ten's digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is 32. Find the number.

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Mathematics

The sum of the digits of a two digit number is 5. If the digits are reversed, the number is reduced by 27. Find the number.

Linear Equations

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The sum of the digits of a two digit number is 7. If the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.

The ten's digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is 32. Find the number.