Question 58274: How many ounces of pure gold must be added to five ounces of something that is 1% gold in order that the mixture is 6% gold?
Found 3 solutions by ankor@dixie-net.com, stanbon, ptaylor:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! How many ounces of pure gold must be added to five ounces of something that is 1% gold in order that the mixture is 6% gold?
:
Let x = amt of pure gold required
:
Write a per cent gold equation. (100% gold is 1.0(x))
The final amt will be be (x+5)
:
.01(5) + 1.0(x) = .06(x+5)
.05 + 1x = .06x + .3
1x - .06x = .3 - .05
.94x = .25
x = .266 oz of pure gold required
:
Check:
.05 + .266 = .06(5.266)
.05 + .266 = .316, checks out
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! How many ounces of pure gold must be added to five ounces of something that is 1% gold in order that the mixture is 6% gold?
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Let # of ounces of pure gold be "x".
1% gold DATA:
Amount = 5 ounces ; amount of gold is 0.01x ounces
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100% gold DATA:
Amount = x ounces ; amount of gold is x ounces
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Mixture DATA:
Amount = x+5 ounces; amount of gold is 0.06(x+5)=0.06x+0.3 ounces
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EQUATION:
gold + gold = gold
0.01x + x = 0.06x+0.3
1.01x-0.06x = 0.3
0.95x=0.3
x=0.3158 ounces of 100% gold must be added
Cheers,
Stan H.
Answer by ptaylor(2198)  (Show Source):
You can put this solution on YOUR website! Mixture problems generally have several features in common, for example:
(1) They usually involve two or more substances
(2) They usually have an initial condition that, after some manipulation, results in a final condition.
(3) There usually exists more than one approach in solving the problems and
(4) Frequently, one of the substances does not change throughout the process and this is the first thing I look for!
This problem consists of a mixture of gold and, for lack of a better word, "otherstuff". We'll solve this problem using two approaches. NOTE: THE OTHERSTUFF DOES NOT CHANGE so lets look at the otherstuff first:
Let x=amount of pure gold added to the original 5 oz.
initial amt of otherstuff (5(.99))+amt of otherstuff added (x(0))=amt of otherstuff in final mixture ((x+5)(.94))
Thus, our equation is:
5(.99)=(x+5)(.94), or
.94x=.25; x=.2659oz
Now lets look at the gold:
Amt of gold in the original 5 oz (5(.01))+amt of pure gold added (x(1))=amt of gold in final mixture ((x+5)(.06))
Thus, our equation is:
5(.01)+x=(x+5)(.06),or
.94x=.25; x=.2659oz
Hope this helps----ptaylor
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