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