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Que: A machine can be repaired for $800 and will last for one year, while the new machine would cost $3,600 ...

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Que: A machine can be repaired for $800 and will last for one year, while the new machine would cost $3,600 ...

by Max@Math Revolution » Tue Feb 16, 2021 9:11 pm

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Que: A machine can be repaired for $800 and will last for one year, while the new machine would cost $3,600 and will last for three years. The average cost per year of the new machine is what percent greater than the cost of repairing the current machine?

(A) 15%
(B) 25%
(C) 116.67%
(D) 50%
(E) 66.67%
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Re: Que: A machine can be repaired for $800 and will last for one year, while the new machine would cost $3,600 ...

by Max@Math Revolution » Thu Feb 18, 2021 9:24 pm
Solution: The cost of repairing the current machine = $800. The cost of new machine = $3,600

Since the new machine lasts for three years, the average cost per year = $\(\frac{3,600}{3}\) = $1,200.

Thus, the required percentage = \(\frac{\left(1,200-800\right)}{800}\) × 100(%) = 50%.

Therefore, D is the correct answer.

Answer D
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Re: Que: A machine can be repaired for $800 and will last for one year, while the new machine would cost $3,600 ...

by deloitte247 » Sun Feb 21, 2021 11:37 am
Total cost for repairs $$=\ 800\ dollars$$
Total cost for new purchase= $$=\ 3600\ dollars$$
Average cost of new machine
$$=\frac{\cos t\ of\ new\ machines}{\cos t\ of\ the\ years\ it\ will\ last}=\frac{3600}{3}=1200$$ $$\%\ difference=\frac{1200-800}{800}\cdot100=50\%$$
$$Answer\ is\ Option\ D$$
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