If the digit in the unit's place of a two-digit number is halved and the digit in the 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?

### Concept & Logic Manipulating place values algebraically. We must carefully construct the expression for the "new" number based on the fractional and multiplicative changes to its individual digits. ### Step-by-Step Solution Let the ten's digit be $t$ and the unit's digit be $u$. The original number is $10t + u$. Condition 1: Unit's digit is halved ($u/2$) and ten's digit is doubled ($2t$). The new number formed is: $10(2t) + (u/2) = 20t + 0.5u$. Condition 2: This new number is equal to the number obtained by interchanging the original digits ($10u + t$). $20t + 0.5u = 10u + t$ Rearrange to find the relationship between $t$ and $u$: $20t - t = 10u - 0.5u$ $19t = 9.5u$ Divide both sides by 9.5: $2t = u$ This equation directly translates to: The unit's digit ($u$) is twice the ten's digit ($t$). ### Exam Strategy & Shortcut Plug in options to create a valid number. For (d), if unit is twice ten's, let's pick $t=2, u=4$. Original number = 24. Halve unit (2), double ten's (4). New digits: tens=4, units=2. New number = 42. Interchanged original number (24) is also 42. $42 = 42$. The condition holds perfectly. ### Common Pitfall Constructing the new number incorrectly. A common mistake is writing the new number as $2t + u/2$ instead of properly multiplying the new ten's digit ($2t$) by its place value of 10. ### Final Answer Therefore, the correct answer is **Digit in the unit's place is twice the digit in the ten's place.**.