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# help

+1
223
3

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?

Dec 21, 2018

#1
+23557
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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%

Dec 21, 2018

#1
+23557
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
#2
0

I got 7. did I do something wrong?

Guest Dec 21, 2018
#3
+23557
0

Yes.

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

ElectricPavlov  Dec 21, 2018