Please show me how to solve this problem. Step by step please.?

Question

Your aunt calls to ask for your help in making a decision about buying a new refrigerator.
She says that she found two that seem to fit her needs, and both are supposed
to last at least 14 years, according to Consumer Reports. The initial cost for
one refrigerator is $712, but it only uses 88 kilowatt-hours (kWh) per month. The
other refrigerator costs $519 and uses an estimated 100 kWh/per month. You do
not know the price of electricity per kilowatt-hour where your aunt lives, so you
will have to decide what in cents per kilowatt-hour will make the first refrigerator
cheaper to run during its 14 years of expected usefulness. Write your aunt a letter
explaining what you did to calculate this cost, and tell her to make her decision
based on how the kilowatt-hour rate she has to pay in her area compares with your
estimation.

This is not one of my assigned questions, I just would like to know how to do it if needed.

Answer 1

We will do some algebra first. We won't send the algebra to your aunt. We will use the algebra to draw conclusions, and send only the conclusions.

Let the cost of a kw-hr be k. The number of months in 14 years is 12 * 14.

The total cost of refrig. no. 1 is 712 + 12 * 14 * 88k
The total cost of refrig. no. 2 is 519 + 12 * 14 * 100k

Now, the statement of the problem asks for the first to be cheaper than the second, so

712 + 12 * 14 * 88k < 519 + 12 * 14 * 100k
193 < 12 * 14 * 12k
0.095734127 < k

Dear Auntie,

If one kilowatt-hour costs more than $0.095734127, the first refrig. is cheaper over the 14 years.

Your loving niece,
Rene

We might take a moment to look at Egan's analysis and see why it is wrong. If the cost of electricity were very, very low, almost zero, the more expensive refrigerator would cost more over the 14 years. If the cost of electricity were very, very high, thousands of dollars per month, the refrigerator that uses less electricity would be the better buy. The solution to this problem most definitely depends on the cost of electricity.

Answer 2

The first thing you do translate the power usage of each fridge fromkwh/month to kwh/year. By multiplying the number of months you expect the fridge to work in its lifetime.

14 years= 128 months

Fridge #1: 88kwh/ mo. * 128 mo./14 yr.=11264 kwh in lifetime
Fridge #2: 100 kwh/mo. * 128 mo./14 yr = 12800 kwh in lifetime

Then write equations for both fridges so that you may determine the price at which both fridges will cost the same amount of money in 14 years. The will be linear equations that follow y=ax+b where the constant b is the initial price of the fridge, b is the amount of kilowatt hours used in the fridge's lifetime and x is the cost of the kwhs which you will solve for.

Your equations are 712+11264x and 519+12800x respectively.

Solve for x:
712+11264x = 519+12800x
-11264x -11264x
------------------------------------
712 = 519 + 636x
-519 -519
------------------------------------
193 = 636x
-------------------------------------
193/636 = 636x/636
.303 = x

WHEW! So the fridges cost the same with 14 years of use when the cost of the fridge is about .303 dollars or 30 cents!

Now the final test is to test which fridge (#1 or 2) will cost MORE when the price of a kwh is less than 30 cents. Lets say 25 cents for example, or .25 dollars.

Fridge #1: 712+11264(.25)= 3528
Fridge #2: 519+12800(.25)= 3719

Fridge #2 costs more money if both fridges live up to their 14 years! Go for Fridge #1.

Answer 3

First refrigerator is $712, use 88 kWh/month
Second refrigerator is $519, use 100 kWh/month
For first refrigerator to be cheaper for 14 years,
assume x = dollar per kWh
712 + 88x(12 month)(14 years) < 519 +100x(12 month)(14 years)
712 + 14784x < 519 + 16800x
2016x > 193
x > $0.0957 / kWh

Answer 4

the first step is to subtract the cost of the cheap fridge from the exp one. (712-519=193). In order to make it cheaper, it will have to make up this difference.

Now the exp one uses 12 kwh/month less than the cheaper one. They should both last 14 year, and there are 12 months a year.

So over their whole life expectancy, there will be 168 (12*14) months. Therefore, over the 14 years, the exp fridge will use 2016 (168 * 12) KWH less than the cheaper fridge.

Now divide the cost difference of the 2 fridges ($193) by this amount, we get just under 0.0096. So if KWH is 9.6 cents or above, the more exp fridge is cheaper over 14 years.

Answer 5

the energyconsumption of refrigerator 1 in 14 years
=88*12*14
the enerdyconsumption of refrigerator 2 in 14 years
=100*12*14
the differencein energy consumption
=12*14(100-88)
=12*14*12 kwh
=2016 units
the difference in price=$712-$519
$193
no of units of energy=2016 units
distributed over $193=9.57 cents
if the cost of electricity where youraunt lives is more than
say 9.57 cents she might go in for the refrigerator 1

Answer 6

Since I don't know your cost per kwh, let us assume $0.50 per kwh. It doesn't matter how big or small your per kwh cost if you use it in both cases you'll arrive at the same conclusion.

First refrigerator:
Given; 14 years of life, $712 initial cost, and 88 kwh/mo.

sol'n. {($0.50/kwh)(88 kwh/mo)(12mo/yr)(14yrs)} + $712
answer = $8,104 (cost in 14 yrs.)

Second refrigerator:
Given; 14 years of life, $519 initial cost, and 100 kwh/mo.

Sol'n. {($0.50/kwh)(100kwh/mo)(12mo/yr)(14yrs)} + $519
Answer = $8,919 (cost in 14 rys)

Conclusion; Basing from this calculations it is cheaper to buy the first refrigerator no matter what price your kwh is.

Dear Auntie,

I hope this solutions can help you decide.

Answer 7

first, artwork on the left edge. we are able to easily multiply fractions right now around the final and backside 1x1=a million, and 3x4=12, so a million/3 x a million/4 = a million/12. only image it pizza chop up into 4 aspects, then splitting those products into 3 greater each and each. the final edge. this is a million/6 divided through 8/3, genuine? first re-write it as i only did. then, invert the backside and multiply it. so it somewhat is now a million/6 x 3/8. only like dividing something through two, and multiplying through a million/2 are a similar element, dividing through 8/3 is comparable to multiplying 3/8. so multiply right now during back so we get 3/40 8, which reduces to a million/sixteen. then upload a million/12 + a million/sixteen through finding a common denominator, which as a result's 40 8. 4/40 8 + 3/40 8 = 7/40 8

Answer 8

i don't know if this is what you wanted. but if you set them equal to each other
712+88x=519+100x
solve
x=173/20852=.0083

the first refrigerator is only cheaper when the cents per kilowatt per hour is less than .0083 cents
whenever it is more, the second refrigerator is more profitable

Answer 9

In going over the answers so far, I agree with Raj's calculation. I could not do it simpler, and this puts it in a context that the 'aunt ' can understand. It is concise and easily understood. Way to go Raj! Amazing isn't it, so many different interpretations to one problem. Sorry I couldn't solve it for you, I was beaten to it. All I can do is add my agreement to what I think is the best answer.