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.