# Thread: Mixture Problem .. Need Help

1. ## Mixture Problem .. Need Help

The Question is
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 ??

2. ## 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.

3. ## Re: Mixture Problem .. Need Help

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%)

4. ## 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.

5. ## Re: Mixture Problem .. Need Help

Hello, MikeNoob!

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

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

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

Equation: . $4 + x \;=\;0.5(x+20)$

6. ## 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..

7. ## Re: Mixture Problem .. Need Help

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.