# 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. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. –  Gerry Myerson Jun 15 '12 at 1:53

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.

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Thanks that did the trick. I was way off. –  Rajeshwar Jun 15 '12 at 1:57
No problem. If you need any clarification, let me know. –  Joe Jun 15 '12 at 1:57

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?

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Hint $\rm\ 20\,(7.8 = 0.75 a + 0.8 b)\:\Rightarrow\: 156 = 15\,(a\!+\!b)\!+\!b = 150 + b\:\Rightarrow\: b = 6\:\Rightarrow\: a=4.$

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Nice one-liner, yet again. Never cease to amaze me with your succinctness Bill. –  Joe Jun 15 '12 at 4:44

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 $B = 6, A = 4$.

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A great short "To the point" answer ! –  Rajeshwar Jun 15 '12 at 1:57
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.