Question

Two alloys a and b contain copper and zinc in the ratio 4:1 and 3:5 respectively. they are mixed in the ratio 9:4 to get a new alloy



c. what is the ratio of copper and zinc in alloy



c.

Answer 1

The ratio of copper and zinc in alloy C will be  87 : 43

Explanation

Two alloys A and B are mixed in the ratio 9 : 4

So, lets assume that the amount of A is $9x$ and the amount of B is $4x$

Now in alloy A, the ratio of copper and zinc is 4 : 1

So, the amount of copper in A $= 9x*\frac{4}{4+1}= \frac{36x}{5}$ and the amount of zinc in A $= 9x*\frac{1}{4+1}= \frac{9x}{5}$

Now in alloy B, the ratio of copper and zinc is 3 : 5

So, the amount of copper in B $= 4x*\frac{3}{3+5}= \frac{12x}{8}= \frac{3x}{2}$ and the amount of zinc in B $= 4x*\frac{5}{3+5}= \frac{20x}{8}= \frac{5x}{2}$

Thus, the total amount of copper in new alloy C  $\frac{36x}{5}+\frac{3x}{2}=\frac{72x+15x}{10}=\frac{87x}{10}$

and the total amount of zinc in new alloy C $=\frac{9x}{5}+\frac{5x}{2}= \frac{18x+25x}{10}=\frac{43x}{10}$

So, the ratio of copper and zinc in alloy C will be.....

$\frac{87x}{10}:\frac{43x}{10}= 87x:43x= 87:43$