1. ## Mixture...Copper Alloys

There were two different copper alloys of total weight 50 kg. The first contains 40% less copper than the second. Determine the percentage of copper in the first and second alloys, if it is known that there were 6kg of copper in the first alloy and 12 kg in the second.

This appears to be a system of two unknowns. I have trouble setting up the correct equation from the given information in word problems.

There are two different copper alloys. I will let x and y represent the two alloys.

Can someone set up the two system of equations for me?

2. ## Re: Mixture...Copper Alloys

I understand what you mean but if you were my student, I would mark you down for "let x and y represent the two alloys". x and y are numbers, not "alloys". What you mean is "let x and y be the weights of the two alloys in kilograms".

"There were two different copper alloys of total weight 50 kg"
So x+ y= 50.

"The first contains 40% less copper than the second. Determine the percentage of copper in the first and second alloys, if it is known that there were 6kg of copper in the first alloy and 12 kg in the second."

So if we take "p" to be the percent of copper in the first alloy, 6/x= p. Since the second had 40% more copper than the first, it has p+ .4 percent copper: 12/y= p+ .4. That gives you three equations to solve for x, y, and p

3. ## Re: Mixture...Copper Alloys

Originally Posted by HallsofIvy
I understand what you mean but if you were my student, I would mark you down for "let x and y represent the two alloys". x and y are numbers, not "alloys". What you mean is "let x and y be the weights of the two alloys in kilograms".

"There were two different copper alloys of total weight 50 kg"
So x+ y= 50.

"The first contains 40% less copper than the second. Determine the percentage of copper in the first and second alloys, if it is known that there were 6kg of copper in the first alloy and 12 kg in the second."

So if we take "p" to be the percent of copper in the first alloy, 6/x= p. Since the second had 40% more copper than the first, it has p+ .4 percent copper: 12/y= p+ .4. That gives you three equations to solve for x, y, and p

Questions:

1. Why did you divide 6 by x? Where in the problem does it say to do that?

2. Why did you divide 12 by y? Where in the problem does it say to do that?

3. I like your set up. I do not understand where 6/x and 12/y come from?

4. Is there another method we can use to find the answer?

4. ## Re: Mixture...Copper Alloys

Questions:

1. For the equation 6/x = p, can I rewrite it as 6p = x?

2. For the equation 12/y = p + 0.4, can I rewrite it as
12 (p + 0.4) = y?

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### ,there are two different copper alloys of total weight 50 kg.the 1st have 40% less copper than 2nd .what % of copper in 1st allay,if 6 kg of copper in 1st allay and 12 kg in 2nd allay solved in mixture and alligation way

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