This question has me completely stumped for some reason, I would appreciate a bit of help.
If you have 20 pounds of coffee for $1.80 a pound
And then add X amount of coffee for $1.44 a pound
How many pounds of coffee did you add if the final average price is $1.56 a pound.
I would really appreciate an explanation, this problem has been driving me crazy.
Set up your equation knowing $(price\space per\space pound)\times(pounds)=(total\space price)$
$$(1.56)\times(20+x)=(20\times 1.80)+(x\times 1.44)$$
\begin{array}{|c|c|c|} \hline \text{Price per pound}& \text{Pounds} & \text{Total Price}\\ \hline \text{1.80} & 20 & \text{36}\\ \hline \text{1.44} & \text{x}& 1.44\times x\\ \hline \text{1.56} & \text{20+x}& 36+(1.44\times x)\\ \hline \end{array}
Edit: To add more explanation...
You are looking to create a mixture of coffee that is \$1.56 per pound. The total value of that mixture will be $\$1.56\times(pounds\space of\space coffee)$ where 'pounds of coffee' is $20 + x$. $x$ being the amount of \$1.44 we add.
It's total value will also be the sum of the individual values of the two component coffees mixed together:$(\$1.80\times 20) + (\$1.44\times x)$ where $x$ is the amount of \$1.44 coffee we are adding.
So, setting these to values equal to each other we get $$\begin{align} (1.56)\times(20+x)&=(20\times 1.80)+(x\times 1.44)\\ 31.2+1.56(x)&=36+1.44(x)\\ .12(x)&=4.8\\ x&=40\\ \end{align}$$
Checking our work, a weighted average of $1.80\times 20+1.44\times 40 = 1.56\times60$