Let the digit at unit’s place be x.
The digit at the ten’s place is twice the digit at unit’s place.
∴ The digit at ten’s place = 2x
|
Digit in units place |
Digit in tens place |
Number |
Original Number |
x |
2x |
(2x × 10) + x
= 20x + x
= 21x |
New
Number |
2x |
x |
(x × 10) + 2x
= 10x + 2x
= 12x |
Since, the sum of the original number and the new number is 66.
∴ 21x + 12x = 66
∴ 33x = 66
∴ x = \(\cfrac{66}{33}\)
∴ x = 2
∴ Original number = 21x = 21 × 2 = 42
∴ the original number is 42.