Analyzing a mixture issue. I am having a problem with this question:


Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?


According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$
so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound.
So for ten pounds it would be $\frac{780}{155}$ which is still not the answer.

I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$
$$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$ 75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4 $$
If you need to me to add some details on how I set up the original two equations, let me know and I will gladly add in some details.
