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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

Hmmm. I see I was mistaken.

You are correct. My equations add up the wrong stuff. I see I should have checked my answer.