# Thread: Basic KWH question from exam prep course seems to be broken or I'm a total moron.....

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Decided to rewrite my post. Wayne
Last edited by wwhitney; 08-06-17 at 04:44 PM.

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So, the question is underspecified, since the number of operating hours per year isn't given. However, among the possible answers presented, the only reasonable possibility is (B) \$275, as follows:

The power savings while operating is 1.1 kW, and with a cost of \$0.10/kWh, that makes a savings of \$0.11/hr. We can reasonably assume that whatever number of operating hours the test writer had in mind, it was not an integer multiple of 1/11. That is, it may be an integer, or possibly half an integer, or a third of an integer, but not 1/11 of an integer. [More precisely, it will be rational, and the denominator in reduced form won't be divisible by 11.]

That means when we multiply 11 cents by the number of operating hours, we are going to get a multiple of 11. Dollars versus cents is immaterial, as 100 is not a multiple of 11. Of the possible dollar values given, only \$275 is divisible by 11. So the answer is (B).

BTW, that means the test writer had in mind 2500 operating hours/year, e.g. 10 hours/day * 5 days/week * 50 weeks/year.

Cheers, Wayne

3. Originally Posted by wwhitney
So, the question is underspecified, since the number of operating hours per year isn't given. However, among the possible answers presented, the only reasonable possibility is (B) \$275, as follows:

The power savings while operating is 1.1 kW, and with a cost of \$0.10/kWh, that makes a savings of \$0.11/hr. We can reasonably assume that whatever number of operating hours the test writer had in mind, it was not an integer multiple of 1/11. That is, it may be an integer, or possibly half an integer, or a third of an integer, but not 1/11 of an integer. [More precisely, it will be rational, and the denominator in reduced form won't be divisible by 11.]

That means when we multiply 11 cents by the number of operating hours, we are going to get a multiple of 11. Dollars versus cents is immaterial, as 100 is not a multiple of 11. Of the possible dollar values given, only \$275 is divisible by 11. So the answer is (B).

BTW, that means the test writer had in mind 2500 operating hours/year, e.g. 10 hours/day * 5 days/week * 50 weeks/year.

Cheers, Wayne
Yes, but it shouldn't be up to the student to guess what the question setter had in mind.

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Originally Posted by Besoeker
Yes, but it shouldn't be up to the student to guess what the question setter had in mind.
Sure, I agree, the question should specify 2500 operating hours/year.

Cheers, Wayne

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If they pay \$0.10/KWH, how much will they save in a year?

The answer is obviously purple, because bacon spoils on Tuesday.

Less obviously, I would assume given the total lack of context and sufficient detail that a 75W incandescent bulb would be an exterior floodlight that runs from dusk to dawn. I'd figure 12 hrs a day 365, giving 1/2 of 8760 or 4380 hours a year. 20 bulbs x 75W each is 1.5kw/hr. 4380 x 1.5 x .1 = \$657 a year. New bulbs are 400W/hr. 4380 x .4 x .1 = \$175.2, difference of ~\$480. I'd go with D.

That said, Wayne's answer seems more correct just on the math that B is the only answer evenly divisible by \$0.11, and his reverse engineered math makes sense. Without multiple choice, nobody would get the correct answer

Still, such a s***** question I would hand write in "purple, because bacon spoils on Tuesday."

6. My guess is the training material gives a baseline hours/year for a school
my guess is 2500
which is \$275

https://www.energizect.com/sites/def...#_Toc145406301

7. Given the choices, it can not be C or D since thy exceed the theoretical maximum when you're working within the constraints provided.

I would call A or B correct. It doesn't say where the lamps are changed. No schools I am aware of leave the entire premise pitch dark at any given moment of the year. There are lights that remain on 24/7 or at least dusk to dawn.

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Originally Posted by wwhitney
So, the question is underspecified, since the number of operating hours per year isn't given. However, among the possible answers presented, the only reasonable possibility is (B) \$275, as follows:

The power savings while operating is 1.1 kW, and with a cost of \$0.10/kWh, that makes a savings of \$0.11/hr. We can reasonably assume that whatever number of operating hours the test writer had in mind, it was not an integer multiple of 1/11. That is, it may be an integer, or possibly half an integer, or a third of an integer, but not 1/11 of an integer. [More precisely, it will be rational, and the denominator in reduced form won't be divisible by 11.]

That means when we multiply 11 cents by the number of operating hours, we are going to get a multiple of 11. Dollars versus cents is immaterial, as 100 is not a multiple of 11. Of the possible dollar values given, only \$275 is divisible by 11. So the answer is (B).

BTW, that means the test writer had in mind 2500 operating hours/year, e.g. 10 hours/day * 5 days/week * 50 weeks/year.

Cheers, Wayne
If that was the mindset of the author of the question, my opinion is that he (or she) should look for another job. The purpose of the prep test is to prepare the student for the real world exam; if what you describe is the "correct" method of determining the answer to the question, the author has missed the point by a light year or two.

9. Originally Posted by Ingenieur
My guess is the training material gives a baseline hours/year for a school
my guess is 2500
which is \$275

https://www.energizect.com/sites/def...#_Toc145406301
2500 is not even in the above document.

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Simplifying, the cost saving per hour =0.11. If total usage hours in a year is X, then total saving is 0.11*X. Then answer is 275, provided X can assume only integral values:1,2,3,4,5,.........

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