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Math Question... (pg. 3)
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| ChemEnhanced |
| quote: | Originally posted by DigiNut
That really depends whether or not Billy is running an industrial assembly line. I mean, it's pretty unlikely that he has the equipment sitting around in his basement to melt and fuse copper with perfect efficiency and without introducing any impurities. So I'd have to assume he's working at a plant somewhere.
Assuming that, we still need to know if it's a clean room, like a fab, or a regular factory where the copper slabs have no doubt picked up all sorts of dirt, and the melting process probably leaves a copper residue behind.
Now, since the question explicitly states "the resulting piece", it also clearly suggests a freezing process. Does it go into a freezer, sealed, or is it just left sitting around until it solidifies? If memory serves, copper doesn't sublimate, but the impurities might at the right atmospheric temperature, which means that the resulting solid might actually be slightly more pure than the original liquid.
Finally, how is the determination of purity being made? It's easy enough to weigh the thing, but in order to determine the purity you usually have to melt it down or distill it, which seems to be a pointless exercise here after making the new alloy. And if Billy had to melt down the original two samples just to find out the purity, how can we be sure that he didn't lose some copper in that process, before he even started combining them? Are we assuming a specific loss rate here or are these ideal, perfect (i.e. imaginary) conditions?
More to the point, if Billy already has a scale, which he would have needed to weigh the final sample, why didn't he just weigh the original two samples in the first place instead of going through all this ridiculous hassle? |
I could just imagine some kid at school writing that down on a test because he didn't know the answer. |
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| VDub |
| quote: | Originally posted by DigiNut
That really depends whether or not Billy is running an industrial assembly line. I mean, it's pretty unlikely that he has the equipment sitting around in his basement to melt and fuse copper with perfect efficiency and without introducing any impurities. So I'd have to assume he's working at a plant somewhere.
Assuming that, we still need to know if it's a clean room, like a fab, or a regular factory where the copper slabs have no doubt picked up all sorts of dirt, and the melting process probably leaves a copper residue behind.
Now, since the question explicitly states "the resulting piece", it also clearly suggests a freezing process. Does it go into a freezer, sealed, or is it just left sitting around until it solidifies? If memory serves, copper doesn't sublimate, but the impurities might at the right atmospheric temperature, which means that the resulting solid might actually be slightly more pure than the original liquid.
Finally, how is the determination of purity being made? It's easy enough to weigh the thing, but in order to determine the purity you usually have to melt it down or distill it, which seems to be a pointless exercise here after making the new alloy. And if Billy had to melt down the original two samples just to find out the purity, how can we be sure that he didn't lose some copper in that process, before he even started combining them? Are we assuming a specific loss rate here or are these ideal, perfect (i.e. imaginary) conditions?
More to the point, if Billy already has a scale, which he would have needed to weigh the final sample, why didn't he just weigh the original two samples in the first place instead of going through all this ridiculous hassle? |
LOL... |
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| VDub |
My math...
I worked the percentages...
Copper...
60% of 51% is 30.6% of 400g is 122.4g...
40% of 51% is 20.4% of 400g is 81.6g...
And the alloy...
40% of 49% is 19.6% of 400g is 78.4g...
60% of 49% is 29.4% of 400g is 117.6g...
Add the totals...
122.4 + 78.4 = 200.8
81.6 + 117.6 = 199.2
200.8g + 199.2g = 400g |
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| Search&Rescue |
x = mass of alloy w/ 60% copper
y = mass of allow w/ 40% copper
Total mass of copper = 51% of 400g = 204g
Mass of copper in x = 60% of x = 0.60x
Mass of copper in y = 40% of y = 0.40y
Assuming, no constituent of the alloy is lost,
x + y = 400g
and
0.60x + 0.40y = 204
Solving for x and y gives:
x = 220g
y = 180g |
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| VDub |
| quote: | Originally posted by Search&Rescue
x = mass of alloy w/ 60% copper
y = mass of allow w/ 40% copper
Total mass of copper = 51% of 400g = 204g
Mass of copper in x = 60% of x = 0.60x
Mass of copper in y = 40% of y = 0.40y
Assuming, no constituent of the alloy is lost,
x + y = 400g
and
0.60x + 0.40y = 204
Solving for x and y gives:
x = 220g
y = 180g |
Yes that answer does seem a lot easier to swallow...
But where my math is faulted is what I want to know.... |
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| DigiNut |
| quote: | Originally posted by ChemEnhanced
I could just imagine some kid at school writing that down on a test because he didn't know the answer. |
In all seriousness, it's a better answer than the obvious one. I'd give bonus marks for it. In the real world, you don't work under ideal conditions; more importantly, I wish more people could look at a complicated, long-winded process and say "Hey, there's a WAY simpler solution here."
This is why I always hated school. Contrived problems, oversimplified solutions. (And I was an engineer) |
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| Search&Rescue |
| quote: | Originally posted by ChemEnhanced
now look in the back of the math book and get the answer
Copper = 204 Alloy = 196
60%copper (122.4) + 40%alloy (78.4) = 200.8
40%copper (81.6) + 60%alloy(117.6) = 199.2 |
in every number you have in parentheses, you have assumed the original mass of the alloy to be either 100% copper (when you're calculating the mass of the copper in each) or 0% copper (when you're calculating the mass of contituents other than copper in each) |
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| ChemEnhanced |
| quote: | Originally posted by Search&Rescue
x = mass of alloy w/ 60% copper
y = mass of allow w/ 40% copper
Total mass of copper = 51% of 400g = 204g
Mass of copper in x = 60% of x = 0.60x
Mass of copper in y = 40% of y = 0.40y
Assuming, no constituent of the alloy is lost,
x + y = 400g
and
0.60x + 0.40y = 204
Solving for x and y gives:
x = 220g
y = 180g |
your formula looks good to me....can you break down the calculations to get x and y.....its been way too long since I've had to do this type of math....and I actually use to be good at it :D |
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| Search&Rescue |
| quote: | Originally posted by ChemEnhanced
can you break down the calculations to get x and y..... |
are you being serious? Because I can't tell |
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| ChemEnhanced |
| quote: | Originally posted by Search&Rescue
are you being serious? Because I can't tell |
I am being serious....I haven't done problem solving like this probably 15 years
I am trying to work through it...but keep getting stuck. |
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| Search&Rescue |
| quote: | Originally posted by ChemEnhanced
I am being serious....I haven't done problem solving like this probably 15 years
I am trying to work through it...but keep getting stuck. |
well, in that case:
you re-arrange any one unknown and express the other in terms of this
e.g. x + y = 400
so x = 400 - y
then you got your second equation in terms of x and y
substitute x = 400 - y into your second equation, re-arrange the terms and solve for y. Then solve for x which is just = 400 - y |
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| Search&Rescue |
| ..and that's just one way of solving linear equations |
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