Solve.
At least how many grams of copper must be alloyed with 387 g of pure silver to produce an alloy that is no more than 90% pure silver?
2007-05-13 03:28:29 · 5 answers · asked by linter 1 in Science & Mathematics ➔ Mathematics
43
2007-05-13 03:37:51 · answer #1 · answered by suuppp123 2 · 0⤊ 0⤋
The minimum copper content needed is 10% of the alloy and let us say x grams of copper is needed. So, x + 387 is such that 387 is 90% of that quantity y
0.9 y = 387 or y = 387 / 0.9 = 430 grams
So, the minimum amount of copper needed is 430 - 387 = 43 grams and this is assuming no losses in alloying.
2007-05-13 10:35:09 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
Let g = grams of copper.
387/(387 + g) = .90
387 = .90(387 +g)
387 = 348.3 + .9g
38.7 = .9g
g = 38.7/.9 = 43
At least 43 grams of copper must be alloyed with 387 g of silver to produce an alloy of no more than 90% silver.
2007-05-13 10:33:36 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Let x be the number of grams of copper.
Then 387 / (x + 387) <= 90 / 100
Multiply by x + 387:
90(x + 387) >= 387 * 100
Divide by 90:
x + 387 >= 38700 / 90
Subtract 387:
x >= 38700 / 90 - 387
x >= 43
At least 43gm of copper is required.
2007-05-13 10:38:44 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Im assuming this is a straight math question (no chemistry needed):
387g = .9*x (where x is the total mass of alloy)
x - y = 387g (where y is the mass of copper)
solve x.........x = 430g
solve y.........y = 43g
43g of copper
2007-05-13 10:37:32 · answer #5 · answered by thomas7399 2 · 0⤊ 0⤋