The mass of two pieces of brass is 60 kg. The first piece contains 10 kg of pure copper and the second piece contains 8 kg of pure copper. The percentage of copper in the first piece of brass if the second piece contains 15 percent more copper than the first is ____ %.

Question

The mass of two pieces of brass is 60 kg. The first piece contains 10 kg of pure copper and the second piece contains 8 kg of pure copper. The percentage of copper in the first piece of brass if the second piece contains 15 percent more copper than the first is ____ %.

Accepted Answer

Concept: The percentage of copper in the given condition would be 25%. Assume that the first and second pieces of brass have a combined mass of "x" and "y," respectively. Add the two masses together, multiply the result by 60, and use this as an equation (i) . Now, using the equation: percentage of pure copper = mass of copper total weight of brass

100,

determine the expression for the percentage of pure copper in the brass. Find the copper content of the two pieces, and then multiply the amount of copper in the first piece by the sum of the copper content of the two pieces plus 15. Find the value of "x" by substituting the value of "y" from equation I Find the value of the copper content in the first piece of brass as a result.
Assume that the first and second pieces of brass have a combined mass of "x" and "y," respectively. Next, we obtain

x + y = 60

according to the specified condition. I The amount of copper in the first component right now

\frac{mass of copper in 1 st piece}{total weight of brass} \times 100

\frac{10}{x} \times 100

Similarly, percentage of copper in 2nd piece

\frac{mass of copper in 1 st piece}{total weight of brass} \times 100

\frac{8}{y}

\times 100

The fact that the second piece has 15% more copper than the first is obvious. Therefore,

\frac{8}{y} \times 100 = 15 + \frac{10}{x} \times 100 \Rightarrow \frac{160}{y} = 3 + \frac{200}{x}

\Rightarrow \frac{160}{60 - x} = 3 + \frac{200}{x}

\Rightarrow \frac{160}{60 - x} = \frac{3 x + 200}{x}

160 x = 180 x - 3 x^{2} + 12000 - 200 x

\Rightarrow 3 x^{2} + 180 x - 12000 = 0

\Rightarrow x^{2} + 60 x - 4000 = 0

\Rightarrow x^{2} + 100 x - 40 x - 4000 = 0

\Rightarrow x (x + 100) - 40 (x + 100) = 0

\Rightarrow (x - 40) (x + 100) = 0

\Rightarrow x = 40 or x = - 100

Since weight cannot be negative, therefore x = 40.

= \frac{10}{x} \times 100 = \frac{10}{40} \times 100 = 25

%

The mass of two pieces of brass is 60 kg. The first piece contains 10 kg of pure copper and the second piece contains 8 kg of pure copper. The percentage of copper in the first piece of brass if the second piece contains 15 percent more copper than the first is ____ %.

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