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

[docj67]

First, let me say that the exam prep school i'm working with is not good and I have found another.
While Ive gotten some value from their materials, the material is rife with typos, formatting errors and is based on 2005/2008 NEC. Boo. I also have extreme strengths in some areas and weakness' in others due to the unorthodox method I entered and worked in the field. Note that the CSLB has accepted my application irregardless. I'm weak in safety regs, big jobsite protocols and employment labor laws, etc.

However, I'm very strong technically, so I'm mystifed by this question:

	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-2][SIZE=-1]Question 10[/SIZE][/SIZE][/FONT][FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-1][/SIZE][/FONT][FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-1]:[/SIZE][/FONT]​ [FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-1][/SIZE][/FONT][FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-2][SIZE=-1][/SIZE][/SIZE][/FONT]	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-2][SIZE=-1]In a school, twenty 75-watt incandescent lights were replaced with twenty 20-watt fluorescent bulbs. If they pay $0.10/KWH, how much will they save in a year? [/SIZE][SIZE=-1][/SIZE][SIZE=-1][/SIZE][SIZE=-1][/SIZE][/SIZE][/FONT]
	​	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-2] [/SIZE][/FONT]​	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-1]$100[/SIZE][/FONT]​
	​	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-2] [/SIZE][/FONT]​	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-1]$275[/SIZE][/FONT]​
	​	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-2] [/SIZE][/FONT]​	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-1]$375[/SIZE][/FONT]​
	​	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-2] [/SIZE][/FONT]​	[FONT=Verdana, Arial, Helvetica, sans-serif][SIZE=-1]$475 [/SIZE][/FONT]​

By my calculations, the existing light array is 1.5kwh so it costed $.15 per hr so yearly costs were $.15 x24x365 assuming lights on 24x7. = $1314 per year

The New array is .4kw so costs .4x $.10 per hr or $.04 so yearly costs are $.04 x24x365=$350 a year

~ Approx 1/4 power, 1/4 the cost, so the savings should be greater then any of those values unless I'm making a mistake. Any help would be appreciated. Thx, Humble Hank.

 

[GoldDigger]

You are assuming that classroom and other lights are on 24 hours per day.
See what happens when you revise that assumption.

 

[docj67]

I ran it for 12 hrs a day

 

[docj67]

thx

thx

Oh you are in Placerville! Cool! I ve been spending lots of time up that way-ish

I'll re run it for 8 hrs a day

Yah it's really hard to figure out what the baseline contexts are for some of these questions

thx

 

[docj67]

@!

@!

I ran it for 8 and 12 hrs and none of the answers appear as a choice. They were close tho, close enough to have me feel sane again:roll:

 

[GoldDigger]

If eight hours by seven days does not work, try five days.

Sent from my XT1585 using Tapatalk

 

[charlie b]

The person who wrote the test question failed in his or her responsibility to provide sufficient information to allow the student to obtain an answer. I would not concern myself over finding how they came up with one of the four possible answers. You can't get it right because the question's author got it wrong. You should find another school. Oh wait, you already did that. Good decision.

 

[K8MHZ]

You are assuming that classroom and other lights are on 24 hours per day.
See what happens when you revise that assumption.

And for 12 months a year.

 

[ggunn]

First, let me say that the exam prep school i'm working with is not good and I have found another.
While Ive gotten some value from their materials, the material is rife with typos, formatting errors and is based on 2005/2008 NEC. Boo. I also have extreme strengths in some areas and weakness' in others due to the unorthodox method I entered and worked in the field. Note that the CSLB has accepted my application irregardless. I'm weak in safety regs, big jobsite protocols and employment labor laws, etc.

However, I'm very strong technically, so I'm mystifed by this question:

	[SIZE=-2][SIZE=-1]Question 10[/SIZE][/SIZE][SIZE=-1]:[/SIZE]​	[SIZE=-2][SIZE=-1]In a school, twenty 75-watt incandescent lights were replaced with twenty 20-watt fluorescent bulbs. If they pay $0.10/KWH, how much will they save in a year? [/SIZE][/SIZE]
		[SIZE=-2] [/SIZE]​	[SIZE=-1]$100[/SIZE]​
		[SIZE=-2] [/SIZE]​	[SIZE=-1]$275[/SIZE]​
		[SIZE=-2] [/SIZE]​	[SIZE=-1]$375[/SIZE]​
		[SIZE=-2] [/SIZE]​	[SIZE=-1]$475 [/SIZE]​

By my calculations, the existing light array is 1.5kwh so it costed $.15 per hr so yearly costs were $.15 x24x365 assuming lights on 24x7. = $1314 per year

The New array is .4kw so costs .4x $.10 per hr or $.04 so yearly costs are $.04 x24x365=$350 a year

~ Approx 1/4 power, 1/4 the cost, so the savings should be greater then any of those values unless I'm making a mistake. Any help would be appreciated. Thx, Humble Hank.

Without usage data this question is nonsensical. Since it's a school it might be 8 hours/day, 5 days/week, 9 months/year, but who knows? We know the differential in steady state load is 55W/light = (55W)(20) = 1.1kW = 1.1kWh per hour, but for how many hours per year?

 

[Besoeker]

The person who wrote the test question failed in his or her responsibility to provide sufficient information to allow the student to obtain an answer. I would not concern myself over finding how they came up with one of the four possible answers. You can't get it right because the question's author got it wrong. You should find another school. Oh wait, you already did that. Good decision.

Totally agree.

 

[wwhitney]

Decided to rewrite my post. Wayne

 

[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

 

[Besoeker]

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.

 

[wwhitney]

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

 

[JFletcher]

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."

 

[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/de...ing Baseline FINAL 09-13-06.doc#_Toc145406301

 

[Electric-Light]

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.

 

[ggunn]

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.

 

[K8MHZ]

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/de...ing Baseline FINAL 09-13-06.doc#_Toc145406301

2500 is not even in the above document.

 

[Sahib]

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,.........