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?
+36916 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%
+36916 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%
+36916 Yes.
Seven would not be possible when the final amount is only THREE gallons.
