Question

Brass is an alloy containing copper and zinc, bronze is an alloy containing 80 percent copper, 4 of zinc, and 16 of tin. A fused mass of brass and bronze is found to contain 74 percent of copper, 16 of zinc, and 10 of tin: Find the ratio of copper to zinc in the composition of brass.

Answer 1

Hint: We will first assume that the proportion of copper and zinc in brass is x: y. And, we will also assume that the amount of brass and bronze in the fused mass is m and n. So, we can say that percentage of tin in the fused mass = $m\times 0+n\times 16=\left( m+n \right)\times 10$
And, the percentage of copper in fused mass = $m\times x+n\times 80=\left( m+n \right)\times 74$.
Similarly, the percentage of zinc in the fused mass = $m\times y+n\times 4=\left( m+n \right)\times 16$.
We will solve all the equations and then get the value of x and y.

Complete step-by-step solution:
We will first assume that the ratio of copper to zinc in brass is x:y.
 And, let us assume that the amount of brass in the fused mass as ‘m’ and the amount of bronze in the fused mass as ‘n’.
So, we can say that the total fused mass is m + n.
Now, we know from the question that bronze contains 16 percent of tin and brass does not contain tin. And, the fused mass which is containing both brass and bronze contains 10 percent tin.
So, we can say that amount of tin in fused mass = $\left( m+n \right)\times 10=m\times 0+n\times 16$
$\Rightarrow 10m+10n=16n$
$\Rightarrow 10m=6n$
$\therefore \dfrac{m}{n}=\dfrac{3}{5}............(1)$
And, we also know that bronze contains 80 percent of copper and we have also assumed that the amount of copper in brass is x and the fused mass which is containing both brass and bronze contains 74 percent of copper.
So, we can say that amount of bronze infused mass = $\left( m+n \right)\times 74=m\times x+n\times 80$
$\Rightarrow 74m+74n=m\times x+n\times 80$
$\Rightarrow \left( 74-x \right)\times m=6n$
$\therefore \dfrac{m}{n}=\dfrac{6}{74-x}..............(2)$

Similarly, we also know that bronze contains 4 percent of zinc and we have also assumed that the amount of zinc in brass is y and the fused mass which is containing both brass and bronze contains 16 percent of zinc.
So, we can say that amount of zinc in fused mass = $\left( m+n \right)\times 16=m\times y+n\times 4$
$\Rightarrow \left( y-16 \right)m=12n$
$\Rightarrow \dfrac{m}{n}=\dfrac{12}{y-16}...............(3)$
Now, after equating equation (1) and (2), we will get:
$\Rightarrow \dfrac{3}{5}=\dfrac{6}{74-x}$
$\Rightarrow 3\left( 74-x \right)=30$
$\Rightarrow 74-x=10$
$\therefore x=64$
Now, after equating equation (1) and (3), we will get:
$\Rightarrow \dfrac{3}{5}=\dfrac{12}{y-16}...............(3)$
$\Rightarrow 3\left( y-16 \right)=12\times 5$
$\Rightarrow y-16=20$
$\therefore y=36$
Hence, ratio of x and y, that is x : y = 64 : 36 = 16 : 9
This is our required solution.

Note: Students are required to first read the question and understand and then convert the statement into mathematical form by assuming the variable as the value of things which is required to be found. Students must not interchange the percentages in a hurry to solve the problem. Also, students have to find the ratio of copper to zinc and not vice versa.