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In a two-digit number, digit at the ten’s place is twice the digit at unit’s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.

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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 placeDigit in tens placeNumber
Original Numberx2x(2x × 10) + x
= 20x + x
= 21x
2xx(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.

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