Question

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

Answer 1

Answer:

The correct answer is E.

Step-by-step explanation:

First I look for the relationships between the coefficients. The first term corresponds to the base value.

$\frac{8*d}{h^{2}}  = \frac{8}{1} * \frac{d}{h^{2} }$

The second term corresponds to the value obtained when the density of the underlying material is doubled and the daily usage of the equipment is halved.

$\frac{2*8*d}{(0,5*h)^{2}}  = \frac{16*d}{0,25 * h^{2}}  = \frac{16}{0,25} * \frac{d}{h^{2} }$

With the I get the ratio of coefficients of $\frac{d}{h^{2}}$ :

$\frac{8}{1}$ = 8

$\frac{16}{0,25}$ = 64

Now I calculate  the percentage increase in the useful life of the equipment as:

% = $\frac{64}{8}$ * %100 = %800