Question

Dr. Bennington took an alloy containing 10% silver and mixed it with an alloy containing 40% silver to get a 60-ounce alloy containing 30% silver. How many ounces of the 10% alloy did he use?

Answer 1

He used 20 ounces of the 10% alloy.

Explanation:

Let x represent the number of ounces of alloy containing 10%

silver that was used.

Let y represent the number of ounces of alloy containing 40%

silver that was used.

The weight of the alloy made is 60 ounces. It means that

x + y = 60

The amount of silver contained in the first alloy is 10/100 × x = 0.1x

The amount of silver contained in the second alloy is 40/100 × x = 0.4x

The amount of silver contained in the mixture is 30/100 × 60 = 18

Therefore,

0.1x + 0.4y = 18- - - - - - - - - - - - -1

Substituting x = 60 - y into equation 1, it becomes

0.1(60 - y) + 0.4y = 18

6 - 0.1y + 0.4y = 18

- 0.1y + 0.4y = 18 - 6

0.3y = 12

y = 12/0.3

y = 40

x = 60 - 40

x = 20