SOLUTION: Assume that a jeweler needs 45 grams of a alloy that is 70% gold. He has avaiable one alloy that is 35% gold, and another that is 80% gold; he will mix some of each.
If x = numbe
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: Assume that a jeweler needs 45 grams of a alloy that is 70% gold. He has avaiable one alloy that is 35% gold, and another that is 80% gold; he will mix some of each.
If x = numbe
Log On
Question 664988: Assume that a jeweler needs 45 grams of a alloy that is 70% gold. He has avaiable one alloy that is 35% gold, and another that is 80% gold; he will mix some of each.
If x = number of grams of the first alloy ( so the number of grams of the second is 45-x). Find x.
Im having some trouble in how I should proceed this type of math problem and how to find the correct answer,. Thanks for the help to anyone that can answer. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Assume that a jeweler needs 45 grams of a alloy that is 70% gold.
He has available one alloy that is 35% gold, and another that is 80% gold;
he will mix some of each.
:
A typical mixture equation
.35x + .80(45-x) = .70(45)
:
.35x + 36 - .80x = 31.5
:
.35x - .80x = 31.5 - 36
:
-.45x = -4.5
x =
x = +10 grams of 35% gold
:
You can find the amt of 80% gold
