How many oz. of a metal containing 40% gold must be combined with 6 oz. of a metal containing 30% gold to form an alloy containing 36% gold?

cubanwongux

cubanwongux

Answered question

2022-09-11

How many oz. of a metal containing 40% gold must be combined with 6 oz. of a metal containing 30% gold to form an alloy containing 36% gold?

Answer & Explanation

alinearjb

alinearjb

Beginner2022-09-12Added 10 answers

Let x oz metal contains 40 % gold.
Then metal contains 30% of gold = 6 oz
Also alloy contains 36% of gold = ( x + 6 )
Grold in 6 oz of metal = 30 %  of  6 = 30 100 × 6  oz = 180 100  oz = 36 ( x + 6 ) 100  oz
Therefore,
40 x 100 + 180 100 = 36 ( x + 6 ) 100 40 x + 180 100 = 36 ( x + 6 ) 100 40 x + 180 = 36 x + 216 40 x 36 x = 216 180 4 x = 36 x = 36 4 = 9

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