Ratio problem of a beverage

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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.

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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 $\$2.75\text{ per liter}$.