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?
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?
Updated On: Oct 11, 2024
- Digits in the unit’s place and the ten’s place are equal
- Digit in the unit’s place is twice the digit in the ten’s place
- Sum of the digits is a two-digit number
- Digit in the unit’s place is half of the digit in the ten’s place
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The Correct Option is B
Solution and Explanation
Let's assume the two-digit number be \(10a + b\),
Here, \(a\) = the digit in the tens place
\(b\) is the digit in the units place.
From Condition 1:
b is halved = \(\frac{b}{2}\)
tens place a is doubled = \(2a\)
Number Formed = \(10 × 2a + \frac{b}{2} = 20a + \frac{b}{2}\)
From Condition 2 number changed
New number = \(10b + a\)
From the question, the number obtained by halving and doubling the digits is equal to the number obtained by interchanging the digits
= \(20a + \frac{b}{2} = 10b + a\)
= \(40a + b = 20b + 2a\)
After simplifying we get,
\(40a - a = 20b - b\)
\(38a = 19b\)
\(2a = b\)
The correct option is (B): Digit in the unit’s place is twice the digit in the ten’s place
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