A zinc-copper alloy with 3 kg of pure zinc would contain 90%

Right Answer Explanation:
x : 3
y : 100
90
10 : 90 – y
(90-y)/3 = 10/x = ………..(1)

In the second case,

x 2
y 90
84
6 84-y

Therefore (84-y)/2 = 6/x ………..(2)

Dividing (1) and (2)
=
180 – 2y = 420 – 5y
3y = 240
y = 80
x = 30/(90-80) = 30/10 = 3 kg

Quantity of zinc in x = 80% of x
= 0.8 3
= 2.4 kg
Weight and percentage of zinc is 2.4 kg or 80%

Can someone please explain this in detail? I am not sure i follow the explanation.

So we have this Zn-Cu alloy; Pure Zn and only Zn alloy.

Let the amount of Zn and Cu in the Zn-Cu alloy be x kg and y kg respectively. Thus, from the first part, we are given that :

\(\frac{x+3}{x+y+3} = 0.9\)

Thus, \(0.1*(x+3) = 0.9*y\)

or x+3 = 9y

Also, there is 2kg of 90% Zn alloy. Thus, this will give us 1.8 kg of Zn and from the second part of the question, we have :

\(\frac{x+1.8}{x+y+2} = 0.84\)

Now we need to find the percentage of Zn in the given Zn-Cu alloy. Personally, I would do this sum in GMAT by plugging in the options. You could do this by finding the value of the term \(\frac{x}{x+y}\), which is the required answer.

We need to find the above ratio. I would assume that x=2.4 kg, starting with the first answer and thus, from the first relation, we should get that this value of x, and the corresponding value of y, give the percentage, also given in the answer.

I replace the value of y, in the above ratio and I get :
\(\frac{9*x}{10*x+3}\)

I start with option A, plugin x = 2.4 , and it simplifies to 9*2.4/27 = 0.8 = 80%. Thus being Lucky I get the answer in the first shot. You can plugin all the other values given and can verify that the corresponding percentage values won't match. A 10 multiplied to "x" in the denominator makes it easier for us. If you are pluggin in, as in the above way, you wouldn't even require the second case(Only in this sum. Sometimes, you might have to verify two options giving the same answer with a second given condition).

A.

I still do not understand the question stem and the solution. can somebody help please
thanks