Question 714794: a mixture whose weight is 100gms is composed of 25% copper, 40% tin & 35% lead. how much tin & lead must be added to obtain a mixture cotaining 10% copper,50% tin and 40% lead?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A mixture whose weight is 100gms is composed of 25% copper, 40% tin & 35% lead.
How much tin & lead must be added to obtain a mixture containing 10% copper, 50% tin and 40% lead?
:
Since there are a total of 100 gr, we can convert % to actual amt
25 gr of copper
40 gr of tin
35 gr of lead
:
let x = amt of tin and lead required to make the mixture have 10% copper
.25(100) = .10(x+100)
25 = .10x + 10
25 - 10 = .10x
15 = .1x
x = 15/.1
x = 150 total gr of tin and lead added to the 100 gr to have 10% copper.
:
How much of that 150 gr is tin, and how much is lead?
:
We know the total of the mixture is 250 gr and 25 gr is copper. therefore
225 gr is the total for tin and lead
:
Final mixture will have .50(250) = 125 gr of tin
Originally there was 40 gr of tin, therefore
125 - 40 = 85 gr of tin to be added
and
Final mixture will have .40(250) = 100 gr of lead
Originally there was 35 gr of lead, therefore
100 - 35 = 65 gr of lead to be added
:
Summing all this up. Add 85 gr of tin and 65 gr of lead
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