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. What is the percentage of copper in the first piece of brass if the second piece contains 15 percent more copper than the first?

Answer 1

Hint: Assume that the total mass of first and second pieces of brass are ‘x’ and ‘y’ respectively. Add these two masses and equate this with 60 and consider this as equation (i). Now, find the expression for the percentage of pure copper in the brass, using the formula: percentage of pure copper = $ \dfrac{\text{mass of copper}}{\text{total weight of brass}}\times 100 $ . Find the percentage of copper in the two pieces and then equate the percentage of copper in the first piece with the sum of percentage of copper in the second piece and 15. Substitute the value of ‘y’ from equation (i) and find the value of ‘x’. Hence, find the value of the percentage of copper in the first piece of brass.

Complete step-by-step answer:
Let us assume that the total mass of first and second pieces of brass are ‘x’ and ‘y’ respectively. Then according to the given condition, we have,
x + y = 60………………..(i)
Now, percentage of copper in first piece
 $ \begin{align}
  & =\dfrac{\text{mass of copper in 1st piece}}{\text{total weight of brass}}\times 100 \\
 & =\dfrac{10}{x}\times 100 \\
\end{align} $
Similarly, percentage of copper in 2nd piece
 $ \begin{align}
  & =\dfrac{\text{mass of copper in 2nd piece}}{\text{total weight of brass}}\times 100 \\
 & =\dfrac{8}{y}\times 100 \\
\end{align} $
It is given that the second piece contains 15 percent more copper than the first. Therefore,
 $ \begin{align}
  & \dfrac{8}{y}\times 100=15+\dfrac{10}{x}\times 100 \\
 & \Rightarrow \dfrac{160}{y}=3+\dfrac{200}{x} \\
\end{align} $
Substituting the value of ‘x’ from equation (i), we get,
 $ \begin{align}
  & \dfrac{160}{60-x}=3+\dfrac{200}{x} \\
 & \Rightarrow \dfrac{160}{60-x}=\dfrac{3x+200}{x} \\
\end{align} $
By cross-multiplication, we get,
 $ \begin{align}
  & 160x=180x-3{{x}^{2}}+12000-200x \\
 & \Rightarrow 3{{x}^{2}}+180x-12000=0 \\
 & \Rightarrow {{x}^{2}}+60x-4000=0 \\
\end{align} $
Splitting the middle term, we get,
 $ \begin{align}
  & {{x}^{2}}+100x-40x-4000=0 \\
 & \Rightarrow x\left( x+100 \right)-40\left( x+100 \right)=0 \\
 & \Rightarrow \left( x-40 \right)\left( x+100 \right)=0 \\
 & \Rightarrow x=40\text{ or }x=-100 \\
\end{align} $
Since weight cannot be negative, therefore x = 40.
Now, percentage of copper in the first piece of brass
 $ \begin{align}
  & =\dfrac{10}{x}\times 100 \\
 & =\dfrac{10}{40}\times 100 \\
 & =25 \\
\end{align} $

Note: One may note that we can also solve this question by another approach. We can assume the percentage of pure copper in the first piece of brass as ‘x’ and that in the second piece as ’15 + x’. We have to form two equations according to the given conditions and then by using the substitution method we have to solve the obtained quadratic equation for the value of ‘x’, which will be our answer.