Tricky Question from GRE using Ratios 
Of two kinds of alloy, silver and copper are contained in the ratio of $5:1$ and the other in $7:2$. What weights of the two alloys should be melted and mixed together so as to makeup a $5$ lb mass with $80\%$ silver.

I am stuck with the $5$ lb mass with $80\%$ silver as to what I means here
 A: Let $a_1$ be the amount of alloy 1 used and $a_2$ be the amount of allow two used.
Clearly $a_1 + a_2 = 5$. This is one equation.
The first alloy is $\frac{5}{6}$ silver and the other is $\frac{7}{9}$ silver. We want the final allow to be $80\%$ silver. 
This gives us a second equation (the weight of silver from $a_1$ combined with the weight of silver from $a_2$ is the weight of silver from the resulting mixture:
$\frac{5}{6}a_1 + \frac{7}{9}a_2 = \frac{8}{10}(5)$.
Then you can solve the system by substitution or addition method. 
Note that the question is a bit poorly worded because the ratio is meant to be a weight/weight ratio, but I guess we can assume that.
A: Call the two weights $x$ and $y$.  Then
\begin{align}
x+y & = 5, \tag{total weight} \\[8pt]
\frac 5 6 x + \frac 7 9 y & = \frac 4 5 (x+y). \tag{silver weight}
\end{align}
The common denominator in the second equation is $90$.  Multiplying both sides of that equation by $90$ yields
$$
75 x + 70 y = 72 (x+y).
$$
Then
$$
3x - 2 y = 0.
$$
The first equation tells you that you can then substitute $5-x$ for $y$ in the second equation.  Then solve that for $x$.
A: Lb is a unit of mass so it will not effect our answer.
Look, 
in the final mixture the ratio of silver to copper = $80\%$ $: 20\%$ $= 4:1$
in mixture one the ratio of silver to copper $= 5:1$
in 2nd one of silver to copper $=7:2 $
we can see here that ($5+7):(1+2) = 12:3 = 4:1$
thus both mixture should be mixed in the ratio of $(5+1):(7+2) = 6:9 = 2:3$ 
thus weights of alloy will be $2$ lb and $3$ lb respectively. 
for more such problem : https://www.handakafunda.com/ratio-and-proportion-concepts-properties-and-cat-questions/
