Need Help Word Problems

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I need some help setting up these word problems.

10. a druggist has prure alcohol and a 20% solution. How many gallons of each shoud he use in forming 4 gallons of a 30%

11. A radiator with a capacity of 16 quarts is filled with 20% antifreeze. How much of the solution should be drained and replace with pure antifreeze to produce a 50% solution

12. how many gallons of water mucst be evaporated from 40 gallons of a 3% salt solution to ro produce a 5% solution

If anyone could set up equations to solve any one of these three problems that would be great. You don't have to solve it just set it up plz.
 
Anarkt Tearial said:
I need some help setting up these word problems.

10. a druggist has prure alcohol and a 20% solution. How many gallons of each shoud he use in forming 4 gallons of a 30%

11. A radiator with a capacity of 16 quarts is filled with 20% antifreeze. How much of the solution should be drained and replace with pure antifreeze to produce a 50% solution
Let x=the number of qts of mixture to be drained.

Original 20% mixture minus amount drained: \(\displaystyle .20(16-x)\)

Pure antifreeze:\(\displaystyle 1.00(x)=x\)qts

New mixture: \(\displaystyle .50(16)=8\) qts.

We have the equation: \(\displaystyle .20(16-x)+x=8\)

12. how many gallons of water mucst be evaporated from 40 gallons of a 3% salt solution to ro produce a 5% solution

If anyone could set up equations to solve any one of these three problems that would be great. You don't have to solve it just set it up plz.
Click to expand...
 
10) Use the standard mixture set-up:

. . .gallons solution:
. . . . .pure: x
. . . . .20%: 4 - x
. . . . .mix: 4

. . .percent alcohol:
. . . . .pure: 1
. . . . .20%: 0.2
. . . . .mix: ???

. . .gallons alcohol:
. . . . .pure: 1(x)
. . . . .20%: 0.2(???)
. . . . .mix: ???

Complete the table. Add the input amounts; set equal to the output amount. Solve for the value of the variable.

The other two exercises work similarly.

Eliz.
 
Thanks very much, I didn't think the solutions would be that easy.

But I got something like this for the table

Gal Sol
P x
20% 4-x
Mix 4

% alc
P 1
20% .2
Mix .3

Gal Alc
P x
20% .2(4-x)
Mix x+.2(4-x)

Is this correct? I;m a lilttle confused about where I go from here.
 
Sum what you're putting in to the mixture, and set equal to what you're getting out. (In other words, look very closely at your last row.)

Don't feel bad: these are only "real easy" after you've done a few! :wink:

Eliz.
 
So, would this be the correct final equation?

x+.2(4-x)=x+.2(4-x)+4
 
i really don't understand how to set up the final equation. Can you do it and show how to step by step.
 
Think of 4 gallons of solution. Let x=pure alcohol and y=the 20% solution.

Both must equal 4 gallons. See?. x+y=4. This is the total amount of solution.

Now, think of the alcohol. How much in each solution?. The pure will just be x;

The 20% solution will be .20y.

The total amount of alcohol is .30(4)=1.2 gallons of alcohol.

Try two equations to see what is going on, one for the total solution and one for the amount of alcohol:

x+y=4
x+.20y=1.2

x=4-y, so you have:

(4-y)+.20y=1.2, Solving for y, we get y=3.5 gallon of 20% solution. Well, it's easy to see you only need .5 gallon of pure stuff to make 4 gallon.

Now, after you get your answers, check and see if they make sense.

.20(3.5)=.7 gallon of alcohol. Adding .5 gallon of pure alcohol gives 1.2 gallon of alcohol. 1.2/4=.3. Check. 1.2 gallon of alcohol is 30% of 4 gallon, so we have our solution mixed correctly.

These three problems are, pretty much, mirrors of one another. Do as Stapel suggested and treat them all about the same way and you'll be fine. Does this help?. You can get #3, can't you?.
 
Be careful with terminology.

x + y = 4
x + 0.20y = 1.2

In these equations, x and y are volumes.

The first equation represents the sum of the two volumes being 4 gallons.

The second equation represents amount being conserved.

concentration = amount / volume --> amount = concentration * volume
 
In case this helps you, this is the way I do these:
Code: 
  a  : x
  b  : y
========
a+b  : z

And general equation is: z = (ax + by) / (a + b)
So this one:
"10. a druggist has prure alcohol and a 20% solution.
How many gallons of each shoud he use in forming 4 gallons of a 30%?"
would be:
x : 20
4-x : 100
======
4 : 30

30 = (20x + 100(4-x)) / 4
120 = (20x + 400 - 100x)
120 = 400 - 80x
80x = 280
x = 280/80 = 3 1/2 : so 3 1/2 gallons of 20% solution; hope that helps...
 
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