# 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

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.

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 $x$.

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/