The useful life of a certain piece of equipment is determine

Hi,

First post! I'm having problems with the following question.

The useful life of a certain piece of equipment is determined by the following formula: u =(8d)/h2, 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?

A 300%
B 400%
C 600%
D 700%
E 800%

I plugged the following numbers:
d=4
h=4
Formula:
u=\frac{(8d)}{h^2}

First part
u=\frac{(8*4)}{4^2}
u=\frac{32}{16}
u=\frac{2}{1}
u=2

Second part

Density doubled
d=4*2
Usage halved
h=4/2

d=8
h=2

u=\frac{(8*8)}{2^2}
u=\frac{64}{4}
u=\frac{16}{1}
u=[fraction]16[/fraction]

Third Part
So my initial value is 2 and my value after the change is 16.
16/2=8, so there is an 8 fold increase, which is 800%
This is wrong. In the spoiler is the explanation, which uses the exact same method but with other numbers and its result is different. What am I missing here?