Question : In a two-digit number, the digit at unit's place is 1 less than twice the digit at the ten's place. If the digit at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is:
Option 1: $59$
Option 2: $23$
Option 3: $35$
Option 4: $47$
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Correct Answer: $47$
Solution :
Let the tens digit be $x$ so the unit digit is $(2x - 1)$.
Original number = $10x + 2x - 1= 12x - 1$.
After the interchange, new number = $10(2x - 1) + x = 20x - 10 + x = 21x –10$.
The difference between the new and the original numbers is $(21x - 10) - (12x - 1) = 9x - 9$.
According to the question,
$9x - 9 = (12x - 1) -20$
⇒ $12x - 9x = 21 - 9$
⇒ $3x = 12$
⇒ $x = 4$
Therefore, the tens digit is $4$ and the unit digit is $(2 × 4) - 1 = 7$.
Hence, the original number = $(4 × 10) + 7 = 47$.
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