SOLUTION: Bubba has 48 ml of a solution that is 25% acid. How much water should he add to obtain a solution is only 15% acid?

Question 1100474: Bubba has 48 ml of a solution that is 25% acid. How much water should he add to obtain a solution is only 15% acid?
Found 4 solutions by Alan3354, ikleyn, greenestamps, josgarithmetic:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Bubba has 48 ml of a solution that is 25% acid. How much water should he add to obtain a solution is only 15% acid?
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Add W ml of water.
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48*25 + W*0 = 48*(25+W)

Answer by ikleyn(52884)   (Show Source): You can put this solution on YOUR website!
.

Let V be the volume of water to add, in milliliters. Then in the new mixture you have 0.25*48 mL of the pure acid, same amount of the pure acid as in the original 25% solution. You add the volume V of water, and then your new mixture has the volume (48+V). The concentration of acid in the new solution is the ratio of the pure acid in the new solution to its volume. This ratio must be 15%, as the condition requires. It gives you the "concentration equation" = 0.15. <<<---=== It is your basic equation, and it expresses the fact that the concentration of the new solution is 15%. It is your basic equation, and as soon you got it, the setup step is completed. To solve the equation, multiply its both sides by (48+V). You will get 0.25*48 = 0.15*(48+V). Simplify it further and solve for V: 12 = 7.2 + 0.15V ====> 0.15V = 12-7.2 = 4.8 ====> V = = 32. Answer. 32 mL of water must be added. Check. = 0.15. ! Correct: the concentration condition is SATISFIED !

Solved.

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There is entire bunch of introductory lessons covering various types of mixture problems
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
in this site.

Read them and become an expert in solution mixture word problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13209)   (Show Source): You can put this solution on YOUR website!

There are many ways to solve this kind of mixture problem. One tutor has shown you the setup of the problem using the traditional algebraic approach; another has shown you a detailed solution which I believe is how it might be taught in a science class.

Let me show you another method that I think is, if you understand it, much easier and faster than either of those methods. Then I will show you still another method that can make this problem easy because of the fact that you are adding only water to the original 25% acid solution.

I will not go into a detailed description of how this general method works. I will instead just show you the calculations and tell you how they are used to solve the problem.

Start with a "tic-tac-toe" board, with the acid percentages of the two ingredients (25 and 0) in the top and bottom of the first column, and the acid percentage of the mixture in the center square:

Working diagonally, write the positive difference between the top left square and the middle square in the bottom right square, and the positive difference between the bottom left square and the middle square in the top right square:

The numbers in the right column (15 and 10) tell you the ratio in which the two ingredients must be mixed. You have 48ml of the 25% acid solution; so if x is the number of ml of water you need to add, then

You need to add 32ml of water to get a 15% acid solution.

And since you are adding water (0% acid), you can also solve this particular problem as follows:

You have 48ml of 25% acid, so you have 12ml of actual acid.
That 12ml of acid won't change when you add the water; so that 12ml now needs to be 15% of the total volume. So the total volume x can be found by solving this equation:

The total volume is 80ml, and you started with 48ml; the amount you added was 80-48=32ml.

Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
Just addition of water to dilute.

x, how much water to add

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