beater wrote:There are two bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver, and the other has 3 parts of gold to 7 parts of silver. If both bars are melted into a 8-kg bar with the final ratio of 5:11 (gold to silver), what was the weight of the first bar?
a. 1 kg
b. 3 kg
c. 5 kg
d. 6 kg
e. 7 kg
There are a few ways to do this problem. It is a weighted average problem, so you could use a weighted average approach here. We have:
first bar: 2/5 gold
second bar: 3/10 gold
mixture: 5/16 gold
We should get a common denominator:
first: 32/80 gold
second: 24/80 gold
mix: 25/80 gold
It's now clear we need a lot more of the second bar than the first to get a mix with such a small amount of gold; indeed they should be mixed in a 7 to 1 ratio (to see this quickly, I'm using weighted average principles: the ratio of the distances to the combined average is the reverse of the ratio of the amounts of the two components of the mixture; that is, the ratio of the two parts is (32 -25) to (25 - 24), or 7 to 1). So the answer should be A: we want 1 kg of the first bar, 7 kg of the second.
That explanation may not make much sense if you haven't worked much with weighted averages. You could also do this algebraically:
say we have x kg of bar 1. We then have 8-x kg of bar two. The first bar is 2/5 gold, the second 3/10 gold, so when we mix them, we will have this much gold in the resulting mixture:
(2/5) * x + (3/10)*(8 - x) = (x/10) + 2.4
We know that the resulting bar weighs 8 kg and the ratio of gold to the total weight is 5/16:
[(x/10) + 2.4]/8 = 5/16
x/5 + 4.8 = 5
x = 1
Or you could likely backsolve this one, though I haven't tried it to see how fast it is.
