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.
 A: 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.
A: Instead of trying to solve these in an ad hoc way, try to look at these problems as systems of equations.  For example, let $a$ be the pounds of coffee A in the mix and $b$ be the pounds of coffee B.  Then, we can write two equations that describe the problem: 
$$.75a +.8b = .78*10=7.8$$
$$a + b=10$$
Can you take the problem forward from there?
A: Hint $\rm\ 20\,(7.8 = 0.75 a + 0.8 b)\:\Rightarrow\: 156 = 15\,(a\!+\!b)\!+\!b = 150 + b\:\Rightarrow\: b = 6\:\Rightarrow\: a=4.$
A: Since there are ten lbs of coffee, $A + B = 10$.  Notice that both sides are unitted in pounds.
Next
$$.75A + .80 B = 7.80.$$
This is unitted in dollars.  Now solve these two equations.
We get $B = 6, A = 4$.  
A: With these problems, there is a shortcut we can use that solves it faster.
Notice that $78$ is $3/5$ the way from $75$ (Coffee A) to $80$ (Coffee B).  Therefore, the ratio of Coffee B to total coffee will be $3/5$.
So, since there are $10$ pounds of coffee total, there are $6$ pounds of Coffee B and $4$ pounds of Coffee A.
A: Coffee A normally costs 100 taka per pound. It is mixed with coffee B, which normally costs 70 taka per pound, to form a mixture which costs 88 taka per pound. If there are 10 pounds of the mix, how many pounds of coffee A are used in the mix? ###################
ANSWER::
Let, the amount of coffee A = X pound
and, the amount of coffee B = 10-X pound
So, the cost of coffee A = 100X
Cost of coffee B =(10-X)70
Total cost: 100x + 70 (10-x) = 88 *10 ⇒ 100x +700 - 70x = 880
⇒ 30x =880-700
⇒ 30x = 180
∴ X = 180/30 =6
Answer: 6
