Two parts of first ratio and one part of second ratio

Question

I was preparing for Quantitative aptitude exams and I came across this question of ratios

An alloy contains copper and zinc in ratio 5:2 and another alloy contains zinc and tin in the ratio 3:2. If 2 parts of 1st alloy and one part of second alloy are melted together to form a new alloy of copper, zinc and tin, the ration of the metals will be?

What I've understood is that 2 parts of 1st ratio(lets call it A) means 2*A=10:4 and 1 part of the second ratio(lets call it B) means 1*B=3:2. Now when these parts are melted together the ratios will be copper:zince:tin=10:4+3:2 Did I get this right or not?

Answer 1

So $A_1$ has the following composition:

$$\frac{5}{7} \text{ Cu, } \frac{2}{7} \text{ Zn } $$

$A_2$ has the following composition:

$$\frac{3}{5} \text{ Zn, } \frac{2}{5} \text{ Pb } $$

Now when they are mixed in the ratio $2:1$, the overall mixture will have Cu:Zn:Pb in the ratio :

$$ \frac{2}{3}\times\frac{5}{7}:\frac{2}{3}\times\frac{2}{7} + \frac{1}{3}\times\frac{3}{5}:\frac{1}{3}\times\frac{2}{5}$$

Answer 2

In the first alloy, copper : zinc = 5:2 in the 2nd , zinc : tin = 3:2 Let the 70x of first alloy is mixed with 35x of 2nd alloy ( 35 being LCM of (5+2 = 7 and 3+2 =5)) so amount of zinc and copper in the mixture , 70x = 50x -copper , 20x -zinc 35 x of 2nd will contain 21x - zinc and 14x tin so required ratio of copper : zinc : tin = 50x :(20x+21x):14x = 50:41 : 14

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