3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jun 15 '12 at 1:53
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks that did the trick. I was way off. $\endgroup$ – Rajeshwar Jun 15 '12 at 1:57
  • $\begingroup$ No problem. If you need any clarification, let me know. $\endgroup$ – Joe Jun 15 '12 at 1:57
3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Nice one-liner, yet again. Never cease to amaze me with your succinctness Bill. $\endgroup$ – Joe Jun 15 '12 at 4:44
1
$\begingroup$

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 $B = 6, A = 4$.

$\endgroup$
  • $\begingroup$ A great short "To the point" answer ! $\endgroup$ – Rajeshwar Jun 15 '12 at 1:57
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Nice clear solution, with the intuition in control all the way. $\endgroup$ – André Nicolas Jun 15 '12 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.