Question

A metallurgist has an alloy with 5

Answer 1

$\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentrated\\ amount \end{array}\\ &-----&-------&-------\\ \textit{5\% alloy}&x&0.05&0.05x\\ \textit{30\% alloy}&y&0.30&0.30y\\ -----&-----&-------&-------\\ mixture&100&0.15&(100)(0.15) \end{array}$

so hmm notice, we use the decimal format for the percent, namely 15% is just 15/100 and 5% is just 5/100 and so on

whatever the amounts of "x" and "y" are, they must add up to 100 grams
thus x + y = 100

and whatever the concentrated amounts are, they'll add up to (100)(0.15)

thus     $\bf \begin{cases} x+y=100\implies \boxed{y}=100-x\\ 0.05x+0.30y=(100)(0.15)\\ ----------\\ 0.05x+0.30\left( \boxed{100-x} \right)=(100)(0.15) \end{cases}$

solve for "x", to see how much 5% alloy will be needed

what about "y"?  well y = 100 -x