If a jeweller wishes to form 12 ounces oif 75%pure gold from substances that arev 60% AND 80% pure gold .How much of each substances must be mixed together to produce this?
STILL not following directions, I see.
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To determine the amounts of 60% and 80% pure gold needed, we can set up a system of equations based on the given information.
Let's assume that the jeweler needs x ounces of the 60% pure gold (Substance A) and y ounces of the 80% pure gold (Substance B) to form the desired mixture.
We know that the total amount of gold in the mixture is 12 ounces. Therefore, our first equation is:
x + y = 12
Next, we know that the resulting mixture should be 75% pure gold. This means that 75% of the total weight of the mixture should be pure gold. So, the second equation is:
(0.60x + 0.80y) / (x + y) = 0.75
To solve the system of equations, we can use substitution or elimination. Let's use substitution:
From the first equation, we can express x as 12 - y, and substitute it into the second equation:
(0.60(12 - y) + 0.80y) / (12 - y + y) = 0.75
(7.2 - 0.60y + 0.80y) / 12 = 0.75
(7.2 + 0.20y) / 12 = 0.75
Now, we can simplify the equation:
7.2 + 0.20y = 0.75 * 12
7.2 + 0.20y = 9
0.20y = 9 - 7.2
0.20y = 1.8
y = 1.8 / 0.20
y = 9
Therefore, the jeweler needs 9 ounces of the 80% pure gold (Substance B) and 12 - 9 = 3 ounces of the 60% pure gold (Substance A) to form the desired mixture of 12 ounces containing 75% pure gold.