Question

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?

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

Answer 1

The Correct Option is A

Let's use algebra to represent this situation and find out which statement is definitely true:
Let the two-digit number be represented as "10a + b," where "a" is the digit in the tens place, and "b" is the digit in the units place. 
According to the problem, if we halve the digit in the unit's place and double the digit in the tens place, we get: 
\((\frac{1}{2})\times\) b + 2a 
And if we interchange the digits, we get: 
10b + a The problem states that these two expressions are equal, so we can set them equal to each other: 
\((\frac{1}{2})\times\)b + 2a = 10b + a 
Now, let's solve this equation: 
First, we can subtract "a" from both sides to isolate the terms with "b": 
\((\frac{1}{2})\times\)b + 2a- a = 10b 
Simplify: 
\((\frac{1}{2})\times\)b + a = 10b 
Now, let's subtract (1/2) * b from both sides: 
a =10b- \((\frac{1}{2})\times\)b 
a =\((\frac{20}{2})\)b- \((\frac{1}{2})\times\)b 
a =\((\frac{19}{2})\)b 
Now, we can see that "a" is expressed as a fraction of "b" (19/2 times "b"). 
This means that the digit in the tens place ("a") is definitely not equal to the digit in the units place ("b"), so option (A) "Digits in the unit’s place and the ten’s place are equal" is definitely not true. Let's analyze the other options: (B) Digit in the unit’s place is twice the digit in the ten’s place: Not true, as we've just shown "a" is expressed as (19/2) times "b." (C) Sum of the digits is a two-digit number: Not necessarily true or false based on the given information. (D) Digit in the unit’s place is half of the digit in the ten’s place: Not true, as we've shown "a" is expressed as \((\frac{19}{2})\) times "b."
The correct option is (A): Digits in the unit’s place and the ten’s place are equal