Mixture...4

[nycmathguy]

You have 6 liters of water that have 20 percent strawberry juice. How many liters of a 80 percent strawberry juice should be added to the mixture to make 75 percent strawberry juice?

Let s = liters of strawberry juice that must be added to the mixture.

6(0.20) + s(0.80) = 0.75(s + 6)

You say?

 

[MathLover1]

Le x be number of liters of water that contain 20% strawberry juice.

Let y represent number of liters of water that contain 80% strawberry juice.

x + y = 6............eq.1

0.20x + 0.80y = 0.75 * 6............eq.2
________________________________

Solve by substitution

x = 6 - y

0.20 × (6 - y) + 0.80y = 4.5

0.20 × 6 - 0.20 × y + 0.80y = 4.5

1.2 + 0.80y - 0.20y = 4.5

1.2 + 0.60y = 4.5

1.2 - 1.2 + 0.60y = 4.5 - 1.2

0.60y = 3.3

0.60y / 0.60 = 3.3 / 0.60

y = 5.5

then

x = 6 - 5.5 = 0.5

Therefore, if you want your juice to contain 75% strawberry juice, do the following:

Mix 5.5 liters of water that has 80% strawberry juice with 0.5 liter of water that has 20% strawberry juice.

 

[nycmathguy]

Le x be number of liters of water that contain 20% strawberry juice.

Let y represent number of liters of water that contain 80% strawberry juice.

x + y = 6............eq.1

0.20x + 0.80y = 0.75 * 6............eq.2
________________________________

Solve by substitution

x = 6 - y

0.20 × (6 - y) + 0.80y = 4.5

0.20 × 6 - 0.20 × y + 0.80y = 4.5

1.2 + 0.80y - 0.20y = 4.5

1.2 + 0.60y = 4.5

1.2 - 1.2 + 0.60y = 4.5 - 1.2

0.60y = 3.3

0.60y / 0.60 = 3.3 / 0.60

y = 5.5

then

x = 6 - 5.5 = 0.5

Therefore, if you want your juice to contain 75% strawberry juice, do the following:

Mix 5.5 liters of water that has 80% strawberry juice with 0.5 liter of water that has 20% strawberry juice.

Did I get every mixture word problem set up wrong? Wow! Sad. Should I really be studying precalculus? I don't know what to say.