John has two different sugar syrups. Syrup A is 2 parts of sugar and 8 parts of water, and Syrup B is 3 parts of sugar and 5 parts of water. How much of each should john mix together to get 280 quarts of syrup that is 3 parts sugar and 7 parts water.

John mixes 120 quarts of syrup A and 160 quarts of syrup of B together to get 280 quarts of syrup that is 3 parts sugar and 7 parts water.

Given that,

John has two different sugar syrups.

Syrup A is 2 parts of sugar and 8 parts of water, and Syrup B is 3 parts of sugar and 5 parts of water.

We have to determine,

How much of each should john mix together to get 280 quarts of syrup that is 3 parts sugar and 7 parts water.

According to the question,

Syrup A: 2 parts sugar, 8 parts of water.

Total number of parts = 10 parts

Syrup B : 3 parts sugar, 5 parts of water.

The total number of parts = 8 parts.

The mixture required 280 quarts, 3 parts sugar, 7 parts water,

In 280 quarts, parts of sugar are,

[tex]= \dfrac{280 \times 3 }{ ( 3 + 7 )}\\\\= \dfrac{280 \times 3}{10}\\\\= 84[/tex]

Let the volume of syrup A be x,

And the volume of syrup B be 280 - x

[tex]Sugar\ A = \dfrac{2x}{10} \\\\Sugar \ A = 0.2 x[/tex]

[tex]Sugar \ B = \dfrac{3 ( 280 - x ) }{8 }\\\\Sugar \ B= .375 ( 280 - x )[/tex]

The total amount of sugar is,

[tex]0.2 x + 0.375 ( 280 - x ) = 84 \\\\0.2 x + 105 -0 .375 x = 84\\\\105 - 84 = 0.375 x - 0.2 x = 0.175 x\\\\ 0.175 x = 21\\\\x = \dfrac{21}{0.175}\\\\x = 120[/tex]

Therefore,

The volume of syrup A = 120 quarts

Volume of syrup B = 280 - 120 = 160 quarts

Hence, John mixes 120 quarts of syrup A and 160 quarts of syrup of B together to get 280 quarts of syrup that is 3 parts sugar and 7 parts water.

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https://brainly.com/question/11921451