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Example A jeweler is using an alloy, a mixture of metals. He has 10 pounds of the alloy. 2% is the rare metal robium. He has another source which is 5% robium. He wishes to combine these mixtures to form a mixture that is 3% robium. Solution: 1. Identify the variables. (0.02)(10) The quantity of robium in material 1 (5%)(x) The quantity of robium in material 2 (0.03)(x+10) The quantity of the desired mixture 2. Write an equation. 5% x + (0.02)(10) = (0.03)(x+10) The equation required. 3. Perform the Solution. 0.05x + 0.2 = .03x + 0.3 = 0.02x = 0.1 4. Write the answer. x = 5 The desired amount.

Mathematics
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And i need help with these problem. An alloy of tin is 15% tin and weighs 20 pounds. A second alloy is 10% tin. How much (to the nearest tenth lb.) of the second alloy must be added to the first alloy to get a 12% mixture. ______ pounds
(0.15)(20) (10%)(x) (0.12)(x+20)
so far right???

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Other answers:

yess :)
so next 10%x(0.15)(20)=(0.12)(x+20) right
yess. seems like you have it figured out.
im confuse on last part
What do you need now Claire?
perform the solution
hrmmm i got .7 ?
you mean solve the following: \[.1x(.15)(.20)=(.12)(x+20)?\]
.10x+3=.12x+2.4=
.10=.1 its 10, you don't need the 0, but it could be easier. to solve. \[.1x+3=.12x+2.4\]\[.1x+.6=.12x\]\[.6=.11x\]\[x=1.833\infty(3)\]
so how did you get .6
.06* my bad 1 sec. \[.1x+.06=.12x\][\.06=.11x\]\[1.833\infty(3)=x\] oddly,its the same answer.
1.8 pounds???
you could always use the percent bar method. that's easiest.
so im confused how many pounds.
A jeweler is using an alloy, a mixture of metals. He has 10 pounds of the alloy. 2% is the rare metal robium. He has another source which is 5% robium. He wishes to combine these mixtures to form a mixture that is 3% robium. |dw:1349373397808:dw|
And i need help with these problem. An alloy of tin is 15% tin and weighs 20 pounds. A second alloy is 10% tin. How much (to the nearest tenth lb.) of the second alloy must be added to the first alloy to get a 12% mixture. ______ pounds
i really hate percents just so you know. lol but ill help. just post ina new question, and don't forget to givethe best response medal.
i will give bunch of them if you help me lol

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