SOLUTION: A jeweler plans to combine two silver alloys to make 50 grams of alloy that is 75% silver and contains 37.5 grams of pure silver (75% of 50 grams). If some silver alloy is 80% silv

Question 213861This question is from textbook Algebra 2
: A jeweler plans to combine two silver alloys to make 50 grams of alloy that is 75% silver and contains 37.5 grams of pure silver (75% of 50 grams). If some silver alloy is 80% silver and the other is 60% silver, how many grams of each are needed? This question is from textbook Algebra 2

Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = grams of 80% alloy needed
Let = grams of 60% alloy needed
given:
(1) grams
= grams of 80% silver in 75% alloy
= grams of 60% silver in 75% alloy
----------------------------------
In words:
(grams of silver from a & b) / (total grams of 75% alloy) = 75%

(2)
Multiply both sides of (1) by and subtract from (2)
(2)
(1)

grams
and

grams
37.5 grams of 80% alloy and
12.5 grams of 60% alloy are needed
check answer:
I should end up with 37.5 grams of silver in the 75% alloy

OK
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A jeweler plans to combine two silver alloys to make 50 grams of alloy that is
75% silver and contains 37.5 grams of pure silver (75% of 50 grams).
If some silver alloy is 80% silver and the other is 60% silver, how many grams of each are needed?
:
Let x = amt of 80% silver required
Resulting amt is to be 50, therefore:
(50-x) = amt of 60% silver
:
.8x + .6(50-x) = .75(50)
:
.8x + 30 - .6x = 37.5
:
.8x - .6x = 37.5 - 30
:
.2x = 7.5
:
x =
x = 37.5 gr of 80% silver
and
50 - 37.5 = 12.5 gr of 60% silver
:
Check solutions in the amt of silver equation
.8(37.5) + .6(12.5) = .75(50)
30 + 7.5 = 37.5

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