MATH SOLVE

4 months ago

Q:
# Mike has coffee worth $4 per pound that he wishes to mix with 20 pounds of coffee worth $7 per pound to get a mixture that can be sold for $5 per pound. How many pounds of the cheaper coffee should he use?

Accepted Solution

A:

[tex]\bf \begin{array}{lccclll}
&\stackrel{lbs}{amount}&\stackrel{per~lb}{price}&\stackrel{amount}{price}\\
&------&------&------\\
\textit{\$4/lb coffee}&x&4&4x\\
\textit{\$7/lb coffee}&20&7&140\\
------&------&------&------\\
mixture&y&5&5y
\end{array}[/tex]

so, we know the mixture is the sum of both types of coffee, thus x + 20 = y, and 4x + 140 = 5y.

[tex]\bf \begin{cases} x+20=\boxed{y}\\ 4x+140=5y\\ ----------\\ 4x+140=5\left( \boxed{x+20} \right) \end{cases} \\\\\\ 4x+140=5x+100\implies 140-100=5x-4x\implies 40=x[/tex]

so, we know the mixture is the sum of both types of coffee, thus x + 20 = y, and 4x + 140 = 5y.

[tex]\bf \begin{cases} x+20=\boxed{y}\\ 4x+140=5y\\ ----------\\ 4x+140=5\left( \boxed{x+20} \right) \end{cases} \\\\\\ 4x+140=5x+100\implies 140-100=5x-4x\implies 40=x[/tex]