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Ms. Wang is shopping for a new refrigerator. Brand A costs $569 and uses 635 kilowatt-hours per year. Brand B costs $647 and uses 582 kilowatt-hours per year. If electricity costs $0.18 per kilowatt-hour, how much would Ms. Wang save on electricity per year by buying Brand B?

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Answer:

Savings on electricity per year by brand B = 114.3 - 104.76 = $9.54

Explanation:

Ms. Wang is shopping for a new refrigerator. Brand A costs $569 and uses 635 kilowatt-hours per year. Brand B costs $647 and uses 582 kilowatt-hours per year.

Let's assume he buys brand B, so he already has a loss over the price of the commodity that is , loss = 647 - 569 =$78

Cost of electricity = $0.18

Now in the case of electricity,

spending on Brand A = 635
* 0.18 = $114.3

spending on Brand B = 582
* 0.18 = $104.76

Savings on electricity per year by brand B = 114.3 - 104.76 = $9.54

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