To solve this problem, let's assume the two-digit number as "ab", where a represents the tens digit and b represents the units digit.

According to the given conditions:

1. "The square of a two-digit number is divided by half the number. After 36 is added to the quotient, this sum is then divided by 2." This can be represented as: $\frac{ab^2}{\frac{ab}{2}} + 36 = \frac{\frac{ab^2}{\frac{ab}{2}} + 36}{2}$

2. "The digits of the resulting number are the same as those in the original number, but they are in reverse order." This implies that the resulting number is "ba".

3. "The ten's place of the original number is equal to twice the difference between its digits." This can be represented as: a = 2(b - a)

Now, let's solve the equation:

$\frac{ab^2}{\frac{ab}{2}} + 36 = \frac{\frac{ab^2}{\frac{ab}{2}} + 36}{2}$

Simplifying the equation, we get:

$ab^2 + 72ab = ab^2 + 36$

Cancelling out the common terms, we get:

$72ab = 36$

Dividing both sides by 36, we get:

$2ab = 1$

Since a and b are digits, the only possible value for ab is 46.

Now, let's check if this value satisfies the third condition:

a = 2(b - a) 4 = 2(6 - 4) 4 = 2(2) 4 = 4

Since the value of a satisfies the third condition, the number is 46.

Therefore, the correct answer is option B) 46.

• Thanks
• Wrong explanation