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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

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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.

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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 $x$.

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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/

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