Question

question 10 of 14 step 1 of 1 a metallurgist has one alloy containing 30% aluminum and another containing 64% aluminum. how many pounds of each alloy must he use to make 52 pounds of a third alloy containing 42% aluminum? (round to two decimal places if necessary.) pounds of the alloy containing 30% aluminum pounds of the alloy containing 64% aluminum answer how to enter your answer (opens in new window) 5 points keyboard shortcuts keypad 00:33:22

Explanation:

Step1: Let the amount of 30% - aluminum alloy be $x$ pounds and the amount of 64% - aluminum alloy be $y$ pounds.

We know two equations based on the total - weight and the amount of aluminum:

1. Total - weight equation: $x + y=52$ (since the total weight of the new alloy is 52 pounds), so $x = 52 - y$.
2. Aluminum - content equation: $0.3x+0.64y = 0.42\times52$.

Step2: Substitute $x = 52 - y$ into the aluminum - content equation.

$0.3(52 - y)+0.64y=0.42\times52$.
Expand the left - hand side: $15.6-0.3y + 0.64y=21.84$.
Combine like terms: $15.6 + 0.34y=21.84$.
Subtract 15.6 from both sides: $0.34y=21.84 - 15.6$.
$0.34y = 6.24$.

Step3: Solve for $y$.

$y=\frac{6.24}{0.34}\approx18.35$ pounds.

Step4: Solve for $x$.

Since $x = 52 - y$, then $x = 52-18.35 = 33.65$ pounds.

Answer:

Pounds of the alloy containing 30% aluminum: 33.65
Pounds of the alloy containing 64% aluminum: 18.35