Asked by Anne B.
The amounts by weight of gold, silver and lead in three alloys of these metals are in the ratio 1:5:3 in the first alloy, 2:3:4 in the second and 5:2:2 in the third. How many kg of the first alloy must be used to obtain 10kg of an alloy containing equal amounts by weight of gold, silver and lead?
Answers
Answered by
Steve
If there are x,y,z kg of the alloys, then we want
x+5y+3z = 10
2x+3y+4z = 10
5x+2y+2z = 10
Just solve those to find x,y,z. Since the x value is from the 1st alloy, in the ratios 1:5:3, that is how many kg of that alloy we need.
x+5y+3z = 10
2x+3y+4z = 10
5x+2y+2z = 10
Just solve those to find x,y,z. Since the x value is from the 1st alloy, in the ratios 1:5:3, that is how many kg of that alloy we need.
Answered by
Anonymous
I don't think they should be equal to 10.
kg of 1st alloy + kg of 2nd alloy + kg of 3rd alloy = 10 kg
I got an answer, my answer is 2kg.
kg of 1st alloy + kg of 2nd alloy + kg of 3rd alloy = 10 kg
I got an answer, my answer is 2kg.
Answered by
Steve
Hmmm. If 2 kg of alloy 1 are used, then that means there are
2 kg of gold
10kg of silver
6 kg of lead
You are already over 10 kg of the mixture. Add in to that any amount of the other alloys and things are even worse.
2 kg of gold
10kg of silver
6 kg of lead
You are already over 10 kg of the mixture. Add in to that any amount of the other alloys and things are even worse.
Answered by
Anonymous
The question is how many kg of the first alloy must be used...
So my answer is 2kg of first alloy, 4kg of 2nd alloy and 4kg of 3rd alloy resulting to 10kg.
So my answer is 2kg of first alloy, 4kg of 2nd alloy and 4kg of 3rd alloy resulting to 10kg.
Answered by
Steve
Hmmm. I see I was mistaken.
You are correct. My equations add up the wrong stuff. I see I should have checked my answer.
You are correct. My equations add up the wrong stuff. I see I should have checked my answer.
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