# Mixture Problem .. Need Help

• May 28th 2012, 02:28 PM
MikeNoob
Mixture Problem .. Need Help
The Question is
Quote:

Alloy of copper and tin is 20% copper , how many pounds of copper must be added to 20 pounds of the alloy in order for resulting alloy to be 50% copper
The answer is 12 and i get 10

Here is how i constructed the table (What am i doing wrong)

 Pounds Percentage Amount of solvent Component 1 x 20%=0.2 0.2x Component 2 20-x 80%=0.8 0.8(20-x) Mixture 20 50/100 0.2x+0.8(20-x)=20.0.5

This is the equation that i get : 0.2x+0.8(20-x)=20.0.5
Could anyone please tell me what i am doing wrong ??
• May 28th 2012, 04:08 PM
emakarov
Re: Mixture Problem .. Need Help
Why do the first column and the top two elements of the second column refer to the original alloy while the bottom element of the second column refers to the modified alloy? Why do you multiply 0.2 by x? 20% = 0.2 is the ratio of copper to the total mass of the original alloy; therefore, it should be multiplied by the total mass of the original alloy. Similar point goes for 0.8(20-x). What is 0.2x+0.8(20-x)?

I am not sure if you were taught to make a table of this form. I would recommend denoting the mass of additional copper by x and writing what you can find with (or without) it: the mass of original copper, of copper in the modified alloy, total mass of the modified alloy, etc.
• May 28th 2012, 04:26 PM
skeeter
Re: Mixture Problem .. Need Help
Quote:

Alloy of copper and tin is 20% copper , how many pounds of copper must be added to 20 pounds of the alloy in order for resulting alloy to be 50% copper
let x = number of lbs of pure copper added to the 20 lbs of 20% copper alloy

x(100%) + 20(20%) = (20+x)(50%)
• May 28th 2012, 05:03 PM
MikeNoob
Re: Mixture Problem .. Need Help
I only know how to solve these problems by constructing a table.. Could anyone please tell me what the correct solution would be by using a similar table.
• May 28th 2012, 06:11 PM
Soroban
Re: Mixture Problem .. Need Help
Hello, MikeNoob!

(And you're referring to a "solvent"?)

Quote:

We have 20 pounds of alloy which is 20% copper.
How much copper must be added to get an alloy which is 50% copper?

$\displaystyle \begin{array}{c||c|c|c|} & \text{pounds} & \text{percent} & \text{copper} \\ \hline \hline \text{Start} & 20 & 0.20 & 4 \\ \hline \text{Add} & x & 1.00 & x \\ \hline \hline \text{Mixture} & x+20 & 0.50 & 0.5(x+20) \\ \hline \end{array}$

Equation: .$\displaystyle 4 + x \;=\;0.5(x+20)$
• May 28th 2012, 06:20 PM
MikeNoob
Re: Mixture Problem .. Need Help
@Soroban.. Thanks for the answer. Now i realize that I was isunderstanding the problem. Actually one component is the copper and other is the alloy which gives us a mixture. First i thought that the other component was Tin instead of the Alloy thank you for pointing that out.Also Skeeters equation helped out too..
• May 29th 2012, 02:58 AM
emakarov
Re: Mixture Problem .. Need Help
Quote:

Originally Posted by MikeNoob
Actually one component is the copper and other is the alloy which gives us a mixture. First i thought that the other component was Tin instead of the Alloy

The two components of the alloy are copper and tin.