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The boys mixed 10 gallons of the 20% pure lemon juice mix and 10 gallons of the 70% pure lemon juice mix, because they wanted 20 gallons of lemonade at a 40% lemon juice mixture. They thought that because 40% was almost halfway between 20% and 70%, they should just mix equal parts of both, but the lemonade turned out too tart. How much of each should they have used to get a final mixture of 20 gallons at 40% lemon juice? Write your answer in complete sentences. Show all work.

if you can help please show the work and have at least 1-2 paragraphs.

Respuesta :

Answer:

The correct quantities are 8 gallons of 70% and 12 gallons of 20%.

Step-by-step explanation:

Boys have to prepare 20 gallons of 40% mixture using 20% and 70% mixtures.

Therefore, lemon in final mixture = 40% of 20 gallons = 8 gallons.

Let number of gallons taken from 20% mixture be x

So, Lemon contribution by this mixture = 20% of x gallons.

                                                             [tex]=\frac{x}{5}\thinspace gallons.[/tex]

Now, since the mixture has to be 20 gallons, we have to take (20-x) gallons of 70% mixture.

Lemon contribution by this mixture =70% of (20-x)

                                          [tex]=14-(\frac{7\cdot x}{10})\thinspace gallons[/tex]

But, these both mixture should be added to get the final mixture :

[tex]\implies\frac{x}{5} +14-(\frac{7\cdot x}{10}) =8\\\\\implies (2-7)\cdot \frac{x}{10} = -6\\\\\implies x = 6\times \frac{10}{5}\\\\\implies x = 12\thinspace gallons.[/tex]

Thus, 20% mixture = 12 gallons and 70% mixture = 8 gallons

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