Question

If the digit in the unit's place of a two-digit number is halved and the digit in ten's place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true?

• A. Digits in the units place and the tens place are equal.
• B. Sum of digits is a two-digit number
• C. Digit is the units place is half of the digit in the tens place
• D. Digit in the unit's place is twice the digit in the tens place

Answer 1

The Correct Option is D

To solve this problem, let's denote the two-digit number as \(10x + y\), where \(x\) is the digit in the ten's place, and \(y\) is the digit in the unit's place.

1. According to the problem, when the digit in the unit's place is halved and the digit in the ten's place is doubled, the number becomes \(10(2x) + \frac{y}{2}\). This simplifies to \(20x + \frac{y}{2}\). 
2. The problem also states that this number is equal to the number obtained by interchanging the digits. When digits are interchanged, the number becomes \(10y + x\).
3. We need to equate these expressions as follows:

\(20x + \frac{y}{2} = 10y + x\)

1. Multiply the entire equation by 2 to eliminate the fraction:

\(40x + y = 20y + 2x\)

1. Rearrange terms to solve for the relationship between \(x\) and \(y\):

\(40x - 2x = 20y - y\)\(38x = 19y\)\(2x = y\)

1. This shows that the digit in the unit's place \(y\) is twice the digit in the ten's place \(x\).

Thus, option Digit in the unit's place is twice the digit in the ten's place is definitely true.