A two digit number ab is 60% of x. The two-digit number formed by reversing the digits of ab is 60% more than x. Find x.

Let's break this problem down:

1. We know that the two-digit number ab can be expressed as 10a + b. This is because the first digit a is in the tens place and the second digit b is in the ones place.

2. The problem states that ab is 60% of x. We can express this as an equation: 10a + b = 0.6x.

3. The two-digit number formed by reversing the digits of ab is ba, which can be expressed as 10b + a.

4. The problem states that ba is 60% more than x. We can express this as an equation: 10b + a = 1.6x.

5. Now we have a system of two equations that we can solve for x:

   10a + b = 0.6x 10b + a = 1.6x

6. To solve this system, we can multiply the first equation by 10 and the second equation by 1, then subtract the second equation from the first:

   100a + 10b = 6x 10b + a = 1.6x

   Subtracting these gives: 99a = 4.4x.

7. Solving for x gives: x = 99a / 4.4.

8. Substituting this into the first equation gives: 10a + b = 0.6 * (99a / 4.4).

9. Solving this equation for a and b gives: a = 4, b = 2.

10. Therefore, the value of x is 99 * 4 / 4.4 = 90.