Useful Life Algebraic Breakdown

[AlisonS279]

This is a question from CAT exam 2 - I would to see an algebraic breakdown rather than plugging in numbers (as the answer explanation does).

The useful life of a certain piece of equipment is determined by the following formula: u =(8d)/(h^2), where u is the useful life of the equipment, in years, d is the density of the underlying material, in g/cm3, and h is the number of hours of daily usage of the equipment. If the density of the underlying material is doubled and the daily usage of the equipment is halved, what will be the percentage increase in the useful life of the equipment?

Here is a breakdown of what I did when I did the problem:

First I found the new useful life. Let this new variable =x.
1. I doubled d since the density was doubled and got x= (16d)/(h^2).
2. Then I halved the number of hours to get x=(16d)/((h^2)/2).
3. Since this was a number divided by a fraction, I flipped the divisor and multiplied that by 16d to get x=(32d)/(h^2).

Then I subtracted the original useful life from the new useful life to get:
1. (32d)/(h^2)- (8d)/(h^2) to get (24d)/(h^2).

Then I took this number and divided it by the original useful life equation.
1. (24d)/(h^2) / (8d)/(h^2).
2. I flipped the divisor to multiply it by the numerator to get (24d)/(h^2) * (h^2)/(8d).
3. The (h^2) and the d's cancelled out leaving me with 24/8 which is 3. (So 300%).

What am I doing wrong here?

[RonPurewal]

If you cut the hours in half, you have to substitute h/2 where h used to be.

Since h is squared in the original expression, you have to square h/2, giving (h^2)/4 in that part of the expression after the time is cut in half.