How much of two grape drinks, one 10% grape juice and the other 40% juice, should be mixed to get 3 gallons that are 20% grape juice?

Guest Dec 21, 2018

#1**0 **

x = gallons of 10%

3-x = gallons of 40%

amount of juice in 10% = .1x

amount of juice in 40% = .4(3-x)

Final amount of juice is 3 gallons at 20% = .2 (3)

the ingredients = final

.1x + .4(3-x) = .2 (3)

-.3x + 1.2 = .6

-.3x = -.6

x = 2 2 gals of 10% 3-2 = 1 gal of 40%

ElectricPavlov Dec 21, 2018

#1**0 **

Best Answer

x = gallons of 10%

3-x = gallons of 40%

amount of juice in 10% = .1x

amount of juice in 40% = .4(3-x)

Final amount of juice is 3 gallons at 20% = .2 (3)

the ingredients = final

.1x + .4(3-x) = .2 (3)

-.3x + 1.2 = .6

-.3x = -.6

x = 2 2 gals of 10% 3-2 = 1 gal of 40%

ElectricPavlov Dec 21, 2018

#3**0 **

Yes.

Seven would not be possible when the final amount is only THREE gallons.

ElectricPavlov
Dec 21, 2018