Everyday Math Problem!

[reason 3]

This is a math problem that could occur on everybody.I just want to see your answers on this...
So hear's the deal:
The board below describes the life expectansy of 2 electronic devices A and B accordingly.Numbers below represent thousand hours,for example 10 is 10.000 hours.Board shows 5 random results of how long devices A and B lasted.

A Device // B Device
10 // 10
11 // 12
20 // 15
29 // 33
40 // 45

Which type of electronic device would you prefer if device A cost 50$ and device B cost 52$?Of course you would like to buy the one which could offer you more hours in the best price...


I know the answer and i will post if after...
I just want to know what do you think and if you answer could you also mention what you thought and decided this?

 

[Muhahahahaaa]

Device A has an average of 110
Device B has an average of 115

2$ of 50$ are 4%
5 hours of 110 are 4.5%

Device B costs 4% more but also runs 4,5% longer
My choice would be B

 

[reason 3]

Thanks for answering +rep ...
Oh and by the way vote for the poll too...just choose device B

 

[Duragon]

A: 22hrs/unit on average
B: 23hrs/unit on average

Watch units:
A: 50$/unit
B: 52$/unit

We want: hrs/$ for cost effectiveness of each device, so...

A: (22hrs/unit)/(50$/unit) = 0.44hrs/$
B: (23hrs/unit)/(52$/unit) = 0.44231...hrs/$

Therefore B is very marginally more cost effective to purchase than A.

 

[Dark Soul]

Quality over cost.

 

[Maker]

Board shows 5 random results of how long devices A and B

I'd say GTFO with your results of only 5 measurements. I couldn't base anything over that low of an amount. No conclusions could be made in real world based on those numbers. They seem all too random. I'm studying electronic engineering (circuit desinging), and you really have to think about these things

But let's say I'd have to give an answer...from sellers standpoint, both have the same lowest value, but product A is more likely to break sooner. so I'd sell that one so the customer has to buy a new one sooner.

 

[reason 3]

The problem above wasn't mine its taken from a book...So if you accuse me you accuse the one who wrote it

 

[uberfoop]

The problem above wasn't mine its taken from a book...So if you accuse me you accuse the one who wrote it

So what? You still can't reasonably assess things based on such a low sample size. If the average difference was HUGE (like, B on average lasting twice as long per $ as A or something), then you could, with some certainty, say that B is better. But that's not the case; the difference is like half a percent.

 

[The Reborn Devil]

Lulz.
Device A costs 2.27 per thousand hours approximately and device B costs 2.26 per thousand hours and since I like going for the alternative which gives more for the money I choose B

 

[Snarewave]

I really hate math.

 

[reason 3]

Indeed the difference between those 2 devices is small,keep voting i'm gonna post the answer to this probably tomorrow or tonight evening...

 

[reason 3]

Since no one else is interested here is my solution:

Let's say both devices cost 50$ then we would do this...

X=(10+11+20+29+40)/5=110/5=22 <-- Device A

Y=(10+12+15+33+45)/5=115/5=23 <-- Device B

This means if the price was the same then Device B would be the best choice.

Now that the price is 50$ for device A and 52$ for device B we would do something like this:

A device
50$ --- 22 hours
1$ --- ? hours

50x=22 => x=22/50 => 0,44 hours

B device
52$ --- 23 hours
1$ --- ? hours

52x=23 => x=23/52 => 0,44230769 hours

This means that device B would indeed be better even by such a small amount.
Thx for participating!

 

[KelThuzad]

I would prefer to go to china and by same for 10$.

 

[reason 3]

Eh...travelling there would actually cost you more

 

[KelThuzad]

I would use Chinese Airlines. That's like.. 3 dollars or something.

 

[reason 3]

If you used them i guess you wouldn't reach there alive to buy those devices.

 

[Duragon]

Since no one else is interested here is my solution:

Let's say both devices cost 50$ then we would do this...

X=10+11+20+29+40/5=110/5=22 <-- Device A

Y=10+12+15+33+45/5=115/5=23 <-- Device B

This means if the price was the same then Device B would be the best choice.

Now that the price is 50$ for device A and 52$ for device B we would do something like this:

A device
50$ --- 22 hours
1$ --- ? hours

50x=22 => x=22/50 => 0,44 hours

B device
52$ --- 23 hours
1$ --- ? hours

52x=23 => x=23/52 => 0,44230769 hours

This means that device B would indeed be better even my such a small amount.
Thx for participating!

So my answer was absolutely correct. Nice to know that my basic math capabilities haven't dwindled over the years.

 

[Maker]

X=10+11+20+29+40/5=110/5=22
Y=10+12+15+33+45/5=115/5=23

You're missing some parenthesis there.

 

[reason 3]

My bad...Fixed!