SOLUTION: Please help me to find the solution and answer in linear equation. How much tin and how much lead must be added to 700 lbs of an alloy containing 50 percent tin and 25 percent lead

Question 491696: Please help me to find the solution and answer in linear equation. How much tin and how much lead must be added to 700 lbs of an alloy containing 50 percent tin and 25 percent lead to make an alloy which is 60 percent tin and 20 percent lead.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
You have to approach these problems in terms of how much "pure" stuff you have on hand and how much you will need to solve the problem.
.
You have 700 lb of an alloy that contains:
50% tin = 350 lb of pure tin
25% lead = 175 lb of pure lead
25% other stuff = 175 lb of other stuff
.
You need to add enough pure lead and enough pure tin to make an alloy that is:
60% tin
20% lead
20% other stuff
.
Assuming you add only pure lead or tin, then the amount of "other stuff" will not change.
That means, your final amount of alloy will contain 175 lb of other stuff.
20% of new alloy = 175 lb
new alloy = 175/.2 = 875
Which means the final amount of total alloy will weigh 875 lb.
.
.6*875 = total amount of tin = 525 lb
You know you already have 350 lb of tin in the mix, so you need to add 175 lb of pure tin.
.
.2*875 = total amount of lead = 175
Well, you already have 175 lb of lead in the mix, so you don't need to add any lead.
.
So the proposed answer is to add 175 lb of pure tin to the 700 lb of alloy to produce the needed alloy.
.
Check your work:
700 lb + 175 lb = 875 lb total
525 lb of tin = 60%
175 lb of lead = 20%
175 lb of other stuff = 20%
.
Done.
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