SOLUTION: An alloy of silver and gold weighs 15 ounces in air and 14 ounces in
water. Assuming that the silver loses 1/10 of its weight in water and that
gold loses 1/19 of its weight, how
Question 524212: An alloy of silver and gold weighs 15 ounces in air and 14 ounces in
water. Assuming that the silver loses 1/10 of its weight in water and that
gold loses 1/19 of its weight, how many ounces of each metal are in the
alloy? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! An alloy of silver and gold weighs 15 ounces in air and 14 ounces in
water. Assuming that the silver loses 1/10 of its weight in water and that
gold loses 1/19 of its weight, how many ounces of each metal are in the
alloy?
:
Let s = amt of silver
Let g = amt of gold
:
"An alloy of silver and gold weighs 15 ounces in air"
s + g = 15
s = (15-g)
:
And 14 ounces in water.
: Assuming that the silver loses 1/10 of its weight in water and that
gold loses 1/19 of its weight,
therefore s + g = 14
Multiply by 190, to get rid of the denominators
19(9s) + 10(18g) = 190(14)
171s + 180g = 2660
:
replace s with (15-g), find g:
171(15-g) + 180g = 2660
2565 - 171g + 180g = 2660
-171g + 180g = 2660 - 2565
9g = 95
g =
g = 10.556 oz of gold
then
15 - 10.556 = 4.444 oz of silver
