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Algebra / Systems of two linear equations in two variables Difficulty: Hard

A sample of a certain alloy has a total mass of $50.0$ grams and is $50.0 percent sign$ silicon by mass. The sample was created by combining two pieces of different alloys. The first piece was $30.0 percent sign$ silicon by mass and the second piece was $80.0 percent sign$ silicon by mass. What was the mass, in grams, of the silicon in the second piece?

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Explanation

Choice B is correct. Let $x$ represent the total mass, in grams, of the first piece, and let $y$ represent the total mass, in grams, of the second piece. It's given that the sample has a total mass of $50.0$ grams. Therefore, the equation $x + y = 50.0$ represents this situation. It's also given that the sample is $50.0 percent sign$ silicon by mass. Therefore, the total mass of the silicon in the sample is $0.500 left parenthesis 50.0 right parenthesis$ , or $25.0$ , grams. It's also given that the first piece was $30.0 percent sign$ silicon by mass and the second piece was $80.0 percent sign$ silicon by mass. Therefore, the masses, in grams, of the silicon in the first and second pieces can be represented by the expressions $0.300 x$ and $0.800 y$ , respectively. Since the sample was created by combining the first and second pieces, and the total mass of the silicon in the sample is $25.0$ grams, the equation $0.300 x + 0.800 y = 25.0$ represents this situation. Subtracting $y$ from both sides of the equation $x + y = 50.0$ yields $x = 50.0 - y$ . Substituting $50.0 minus y$ for $x$ in the equation $0.300 x + 0.800 y = 25.0$ yields $0.300 (50.0 - y) + 0.800 y = 25.0$ . Distributing $0.300$ on the left-hand side of this equation yields $15.0 - 0.300 y + 0.800 y = 25.0$ . Combining like terms on the left-hand side of this equation yields $15.0 + 0.500 y = 25.0$ . Subtracting $15.0$ from both sides of this equation yields $0.500 y equals 10.0$ . Dividing both sides of this equation by $0.500$ yields $y equals 20.0$ . Substituting $20.0$ for $y$ in the expression representing the mass, in grams, of the silicon in the second piece, $0.800 y$ , yields $0.800 left parenthesis 20.0 right parenthesis$ , or $16.0$ . Therefore, the mass, in grams, of the silicon in the second piece is $16.0$ .

Choice A is incorrect. This is the mass, in grams, of the silicon in the first piece, not the second piece.

Choice C is incorrect. This is the total mass, in grams, of the second piece, not the mass, in grams, of the silicon in the second piece.

Choice D is incorrect. This is the total mass, in grams, of the first piece, not the mass, in grams, of the silicon in the second piece.