Question

Randall wants to mix 50 lb of nuts worth $2 per lb with some nuts worth $7 per lb to make a mixture worth $6 per lb. How many pounds of $7 nuts must he use ?

Answer 1

A = nuts of $2/lb

B = nuts of $7/lb

let's say we'll mix "x" lbs of B to get a mixture with A of "y" lbs, therefore

$\bf \begin{array}{lcccl} &\stackrel{lbs}{quantity}&\stackrel{\textit{cost}}{per~lb}&\stackrel{\textit{cost}}{total}\\ \cline{2-4}&\\ A~ nuts&50&2&100\\ B~nuts&x&7&7x\\ \cline{2-4}&\\ mixture&y&6&6y \end{array} \\\\\\ \begin{cases} 50+x=\boxed{y}\\ 100+7x=6y \end{cases}\qquad \qquad \implies 100+7x=6\left( \boxed{50+x} \right) \\\\\\ 100+7x=300+6x\implies \blacktriangleright x=200 \blacktriangleleft$