Mixture problem?

Question

I have no clue how to set up this problem to even start on it:

How much rum (40% alcohol) must be added to 2 Liters of punch to get a 15% alcohol happy punch mixture?

Answer 1

Okay, letting a be number of liters of run, then
(.4a)/(2+a) = .15

.3+.15a = .4a

.3=.25a

a=1.2

So, you need 1.2 liters of rum.

Answer 2

For any and all mixture problems:

V1 x C1 = V2 x C2

V=volume C=concentration

so we have:

V1 (.40) = (V1+2liters)(.15)

solve for V1 and thats the amount of rum

Answer 3

Let the volume of rum = x liter
Strength of rum = 40%=0.4
the volume of punch = 2 lit re
Strength of punch = 0%=0.0
the volume of mixture = (2 +x ) lit re
Strength of punch = 15%=0.15
When rum and punch are added we get mixture
Rum + Punch = Mixture
Volume of rum x strength of rum + volume of punch x strength of punch = volume of mixture x strength of mixture
Therefore
(x)(0.4)+2(0.0) = (2+x)(0.15)
0.4 x + 0 = 0.3 + 0.15 x
(0.4 - 0.15)x =0.3
0.25 x = 0.3
x = 0.3/(0.25)=1.2 liters

Answer 4

Let x equal the amount of rum.

0.4x+2liters = 0.15(2+x)

solve for x.

Understand that this assumes concentration is by volume percent.

Answer 5

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