Sum of the two digit number is less by 54 from the number. The number

1.	Sum of the two digit number is less by 54 from the number. The number obtained by interchanging the digits exceeds the given number by 27 . find the number
Answer» Answer:- LET the digit at ten's PLACE be x and digit at one's place be y. ⟶ The number = 10x + y GIVEN: Sum of the digits is 54 LESS than the number. ⟶ 10x + y - (x + y) = 54 ⟶ 10x + y - x - y = 54 ⟶ 9x = 54 ⟶ x = 54/9 ⟶ x = 6 Also given that, The number formed by reversing the digits exceeds the number by 27. ⟶ Reversed number = Original number + 27 ⟶ 10y + x = 10x + y + 27 ⟶ 10y + 6 - 10(6) - y = 27 ⟶ 9y = 27 - 6 + 60 ⟶ 9y = 81 ⟶ y = 81/9 ⟶ y = 9 ∴ The required number = 10(6) + 9 = 69.

Sum of the two digit number is less by 54 from the number. The number obtained by interchanging the digits exceeds the given number by 27 . find the number

Answer»

Answer:-

LET the digit at ten's PLACE be x and digit at one's place be y.

⟶ The number = 10x + y

GIVEN:

Sum of the digits is 54 LESS than the number.

⟶ 10x + y - (x + y) = 54

⟶ 10x + y - x - y = 54

⟶ 9x = 54

⟶ x = 54/9

⟶ x = 6

Also given that,

The number formed by reversing the digits exceeds the number by 27.

⟶ Reversed number = Original number + 27

⟶ 10y + x = 10x + y + 27

⟶ 10y + 6 - 10(6) - y = 27

⟶ 9y = 27 - 6 + 60

⟶ 9y = 81

⟶ y = 81/9

⟶ y = 9

∴ The required number = 10(6) + 9 = 69.

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