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.

Volume of syrup A = 120 quarts

Volume of syrup B = 280 - 120 = 160 quarts .

Step-by-step explanation:

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

Total = 10 parts

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

Total = 8 parts .

Mixture required 280 quarts , 3 parts sugar , 7 parts water

In 280 quarts , sugar = 280 x 3  / ( 3 + 7 )

sugar = 84 parts

Let volume of syrup A be x , volume of syrup B = 280 - x

sugar A =  2 x / 10 = 0.2 x

sugar B = 3 ( 280 - x ) / 8 = .375 ( 280 - x )

Total sugar =  .2 x + .375 ( 280 - x ) = 84 .

.2 x + 105 - .375 x = 84

105 - 84 = .375 x - .2 x = .175 x

.175 x = 21

x = 21 / .175 = 120

Volume of syrup A = 120 quarts

Volume of syrup B = 280 - 120 = 160 quarts .

found it xd it ok yt

y=sec x is the correct answer

~batmans wife dun dun