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The problem has been reversed on me and I can't figure it out?
There are 4 problems like this one. Can someone explain the steps to reach the conclusion.
An alloy weighing 30 lbs. is 11% tin. The alloy was made by mixing a 15% tin alloy and a 9% tin alloy. How many pounds of each alloy were used to make the 11% alloy?
? lbs. of the 15% alloy and ? lbs. of the 9% alloy.
Hi Duke,
Thank you for your response.
I'm feeling really stupid now. I worked the problem as laid out but must not be doing something correctly because I can't seem to come up with the answers you provided. Could you please extend your explanation so I might see what I am doing wrong.
Please someone help me understand how to work this problem.
Alright will some one please let me know if I'm being blind and not seeing the answer in front of me. I honestly am not getting how to do this. I don't want to just copy down the answers I get here I truly would like to understand what it is I'M DOING.
Maybe I'm just not doing these mixture problems correctly. This is what I have done.
An alloy of copper is 10% copper and weighs 25 pounds. A second alloy is 18% copper. How much (to the nearest tenth lb.) of the second alloy must be added to the first alloy to get a 13% mixture.
=15 pounds
(.10)(25)+(.18x)=(.13)(25+x)
(2.5)+(.18x)=(3.25)+(.13x)
3.25 -2.5=.18x -.13x
.75=.05x
.75/.05=x
15=x
2 Answers
- David in KenaiLv 61 decade agoFavorite Answer
Wow, someone who's actually trying to learn instread of just get answers to a problem set. Good on you.
Duke got the right answer but used two equations to do it. Your (Michelle)'s approach in your second problem is correct. Here's the same approach for your first problem:
x= pounds of 9% alloy
pounds of tin = 30(.11)=.09(x)+.15(30-x) Because the amount of 15% alloy must be 30 (total weight) - x (weight of 9%)=(30-x)
3.3 = 4.5 - .15x + .09x
-1.2 = -.06x
20 = x = pounds of 9% alloy
So 30-20=10 = pounds of 15% alloy.
Again, your approach works. Duke's approach works.
Your's is easier for this problem.
But you should get familiar with his approach of substituting an equation for one variable into a second equation. Not for stuff like this but for more complicated problems.
Source(s): Algebra taken long ago. Material Science at Berkeley too, I suppose. But it could have been a fudge recipe. You'd use the same algebra to get the answer. - duke_of_urlsLv 71 decade ago
There are 2 equations here. One is a tin-only equation and one is a total weight equation.
The amount of tin in the final alloy is (30)(.11)
the amount of tin in the first alloy is F(.15)
the amount of tin in the second alloy is S(.09)
So, the first equation is:
(30)(.11)=F(.15)+S(.09)
The second equation is the total weight of metal which is:
30=F+S where F is the weight of the first alloy and S is the weight of the second alloy.
Rearranging the second equation gives an equation for F:
F=30-S
Substituting the right side of this equation for F in the first equation gives:
(30)(.11)=(.15)(30)-(.15)S + (.09)S
Solving for S gives S=20
The weight of the first alloy is 30-20=10
