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I'm given the following data on a problem:

A company wants to produce a beverage AZ which is a mixture of beverage A and Z.

Cost/liter of A = 3.00$

Cost/liter of Z = 2.00$

I'm asked to determine the quatities of A and Z required to make 600 liters of AZ at 2.75$ per liter.

How would I go about finding the exact ratio for the ammount required?

I've tried going with a 50% to 50% ratio but it gets me to 2.5$/liter. I can go trial by error on ratios until I get it right, but something tells me that there is a way to get the exact solution required. If someone could help me get the logic going that would be nice. Thanks in advance.

user149031
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  • Let A be the number of litres of A and Z the number of litres of Z. So, A + Z = 600 and 3A + 2Z = 600(2.75). So, solve the equations to find A and Z. – user137481 May 09 '14 at 20:44
  • I can't understand very well your notation. 3A won't get me 3 times the number of A liters? How would I go about solving for A or Z when they are both "related"? – user149031 May 09 '14 at 20:48
  • First let X be the number of litres you use of A and Y the number of litres you use of Z. Do you agree that we must have X + Y = 600? – user137481 May 09 '14 at 20:52
  • Yes. So far I'm understanding. I think the price/liter is what is confusing me. – user149031 May 09 '14 at 20:54
  • Next, if you use X litres of A and Y litres of Z, then the total cost must be 3X + 2Y right? – user137481 May 09 '14 at 20:55
  • Yes. if I want 1 liter of A and Z that would be equal to 5 (from 3+2), but still it would give me 2 liters of AZ in the end. – user149031 May 09 '14 at 20:59
  • Now, you agree that X + Y = 600 and 3X + 2Y = 600(2.75), right? – user137481 May 09 '14 at 21:05
  • Yes, the left part I agree, what I don't understand now is the right part where you put 600 x 2.75. Are you using the same relation as you did for 3X for example but instead of "X" you put 600 and 2.75 respectivly? – user149031 May 09 '14 at 21:07
  • The total cost to make 600 litres at 2.75 per litre is 600 x 2.75. You understand that, right? – user137481 May 09 '14 at 21:18
  • Yep, I do. that gives me a value of 1650$ for the 600 right? – user149031 May 09 '14 at 21:20
  • Right. Now if you use X litres of A and Y litres of Z, then the total cost is 3X + 2Y. And you want 3X + 2Y = 1650 – user137481 May 09 '14 at 21:21
  • Yep, but if I want to solve for X that would give me I think: $ X=\frac{1650-2Y}3$ no? And solving for Y would give me $Y=\frac{1650-3X}2$ no? how can I be able to solve one without the other? I can only think of system in that case but that would complicate things a bit or is there a simpler way? – user149031 May 09 '14 at 21:26
  • You forgot that X + Y = 600. So, Y = 600 - X. So, 3X + 2Y = 3X + 2(600-X) = 1650. So, 3X + 1200 - 2X = 1650. So, X = 450 and Y = 150 – user137481 May 09 '14 at 21:29
  • You lost me on 3X + 2Y = 3X + 2(600-X) = 1650. Did you replace Y with 600-X? – user149031 May 09 '14 at 21:38
  • yes, i did. Ok, now? – user137481 May 09 '14 at 21:39

1 Answers1

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What percent of $A$ and what percent of $Z$ do you need to create something that costs $\$2.75\text{ per liter}$?

Let $x$ be the percentage of $A$ you use. The cost for $1$ liter would be:

$$x\times\$3.00+(1-x)\times\$2.00$$

Solve for $x$ that makes the cost $\$2.75$.

$$3x+2(1-x)=2.75\\3x+2-2x=2.75\\x=.75$$

$75\%$ of $A$ and $25\%$ of $Z$ gets you a drink that costs $\$2.75\text{ per liter}$.

user137794
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