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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%

User Madeleine
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2 Answers

1 vote

Answer: E. 800%

Explanation:

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.

Assuming d = 1 and h = 1, then

u = (8 × 1)/1^2 = 8

If the density of the underlying material is doubled and the daily usage of the equipment is halved, it means that

d = 2 and h = 1/2 = 0.5, therefore,

u = (8 × 2)/0.5^2 = 16/0.25 = 64

64/8 = 8

The percentage increase in the useful life of the equipment is

8 × 100 = 800%

User Charles Gagnon
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4.6k points
2 votes

Answer:

E. 800%

Explanation:

Since,

u = 8d/h² __________ eqn (1)

Now, density (d) is doubled and usage (h) is halved.

Hence the new life (u'), becomes:

u' = 8(2d)/(0.5h)²

u' = 8(8d/h²)

using eqn (1), we get:

u' = 8u

In percentage,

u' = 800% of u

In words, the percentage increase in useful life of the equipment is 800%.

User Austin Fatheree
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4.9k points