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 .

third option: 6x+8

step-by-step explanation:

we can see that the given question is a sum of series and the values of n vary from 0 to 3

the first term will be:

when n=0, [(0)*x+2] => (0+2) => 2

when n=1, [(1)*x+2] => (x+2) => x+2

when n=2, [(2)*x+2] => 2x+2

when n=4, [(3)*x+2] => 3x+2

so adding up all the terms will give us:

=2+x+2+2x+2+3x+2

=x+2x+3x+2+2+2+2

=6x+8

so,

third option is the correct answer ..

goog may be able to you find that answer