Question

A 15 ounce alloy containing 30% gold was mixed with an alloy containing 5% gold to get an alloy containing 20% gold. How many ounces of the 5% alloy were used?

Answer 1

Answer:

4.5 % alloy

Step-by-step explanation:

Let x = amt of 30% alloy

The resulting total is to be 25 oz, therefore:

(25-x) = amt of 5% alloy:

A typical mixture equation:

.30x + .05(25-x) = .20(25)

.30x + 1.25 - .05x = 5 .30x - .05x = 5 - 1.25

.25x = 3.75

x = 3.75%2F.25

x = 15 oz of 30% alloy required

then

25-15 = 10 oz of 5% alloy:

Check solution

.30(15) + .05(10) = .20(25)

4.5 + .5 = 5