SOLUTION: i am not getting mixture problems so can someone help me. the question is The radiator of an automobile already contains 12 quarts of a 10% solution. How much alchohal must be adde

Question 25563: i am not getting mixture problems so can someone help me. the question is The radiator of an automobile already contains 12 quarts of a 10% solution. How much alchohal must be added to make a mixture of 20% alcohal. How much alcohol must be added to make a mixture containing 25% alcohal. I would appreciate it so much if someone could help
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
(a)---------------------------------------------------------
This is how I see it.
I'm starting out with 12 quarts and adding alcohol to it, so I'm going
to end up with a little over 12 quarts of mixture with a little more than twice as much
alcohol in it than I started.
I'll call the amount of alcohol I need to add x.
So, I'm going to end up with (12 + x) quarts
How much alcohol did I have in the radiator to start with?
I had 10% of 12 quarts, which is 1.2 quarts
Now I can write an equation in words
(alcohol started out with) + (alcohol to be added)
_________________________________________________ = 20%
(12 quarts of mixture) + (alcohol to be added)
putting in values:
(1.2 + x)/(12 + x) = .20
multiply both sides by (12 + x) and expand
1.2 + x = 2.4 + .2x
subtract .2x from both sides
1.2 + .8x = 2.4
subtract 1.2 from both sides
.8x = 1.2
divide both sides by .8
x = 1.5
so, the answer is - add 1.5 quarts to make a 20% mixture
check the answer:
(1.2 + 1.5)/(12 + 1.5) =20%
2.7 / 13.5 = .20
27/135 = .20
this is true- answer checks
(b)--------------------------------------------------------------
How much alcohol needs to be added to make a 25% mixture?
If I'm starting with 12 quarts of 10% mixture, then the equation is:
(1.2 + x) / (12 + x) = 25%
(c)---------------------------------------------------------------
But if I'm starting out with the 20% mixture I just made, I need to ask
new questions
Whats's the total amount of mixture I'm starting with?
It's (12 + 1.5)
What's the total amount of mixture I'm going to end up with?
It's (12 + 1.5 + x) where x is what I have to add to get a 25% mixture
How much alcohol am I starting out with?
It's (1.2 + 1.5)
How much will I end up with?
It's (1.2 + 1.5 + x)
So, what's my equation?
(1.2 + 1.5 + x) / (12 + 1.5 + x) = .25
(2.7 + x) / (13.5 + x) = .25
multiply both sides by (13.5 + x) and expand
2.7 + x = 3.375 + .25x
subtract .25x from both sides
2.7 + .75x = 3.375
subtract 2.7 from both sides
.75x = .675
divide both sides by .75
x = .9
So, if I start with 13.5 quarts of a 20% mixture, I'll have to add .9 quarts
to make it a 25% mixture.
check the answer:
(2.7 + .9)/(13.5 + .9) = .25
3.6 / 14.4 = .25
This is true, so answer checks
The (c) approach is probably what's required, because it takes a little more thought. Hope all this helps.
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