Solution mixture word problem

Question

If drink A contains 15% water and drink B contains 81% water, how much of each solution should be mixed in order to create 60 milliliters of a solution that contains 37% water?

Answer 1

40 ml of drink A and 20 ml of drink B should be mixed together.

(60-x)= amount of drink A . x= amount of drink B

60 ml (0.37 H2O) = 22.2 H2O by composition

22.2 = (.15)(60-x) + (.81)(x)

22.2 = 9 - 0.15x + .81x

22.2 =9 + .66x

13.2 = 0.66x

x=20 = drink B

60-x = 40 = drink A

Answer 2

Hey Sunkissed!

These problems can be quite tricky for us chemistry and biology students, because we're so used to working with moles, molarity, and grams that percentages can throw us for a loop. The logic behind this problem is very similar to that of standard molarity dilution problems though, so let's take a look.

First, let's establish how many milliliters of water are in 60mL of each solution (A, B and the Final solution, which I will call solution F for clarity). This step is unnecessary, but I find it helps me work through the logic of the rest of the problem.

Solution A is 15% water, so 60mL of solution A would have: 0.15 • 60mL = 9mL of H₂O

Solution B is 81% water, so 60mL of Solution B would have: 0.81 • 60mL = 48.6mL of H₂O

The desired percent H₂O for solution F is 37%. If we multiply 0.37 • 60mL we get 22.2mL of H₂O

From the above calculations, we can see that what we want is a solution with a total volume of 60mL, where 22.2mL of that volume is water. We can also see that 60mL of solution A gives us only 9mL of water (too little) and 60mL of solution B gives us 48.6mL of water (too much). From this point, we need to set up a mathematical statement that will tell us in what proportions solution A and solution B should be mixed to give us a 60mL solution with 22.2mL of water in it.

The way I think about setting up these mathematical statements is like this:

We want a certain (unknown) volume of solution A. Since the volume is unknown, let's call it "x".

We also want a certain unknown volume of solution B. We can call that volume "y" for now.

We want a mixture of Solution A and Solution B to combine in a way that gives us the final solution, so we can say:

0.15x + 0.81y = 0.37 • 60mL

If we can solve this equation, it will give us the answers we seek. The only problem is that there are two unknown variables, which makes the problem impossible to solve. We need to get down to one unknown variable. We can do this by realizing that whatever volume "x" and volume "y" are, they must add up to 60mL. If they did not, it would violate the parameters of the problem, which states that our final solution has to be 60mL. This allows us to say that:

Volume soln. A + Volume Soln. B = 60mL

Now we can just replace those long names of the solutions with our variables from earlier:

x + y = 60mL

Rearranging the equation to solve for "y" gives:

y = 60mL - x

Since "y" is our volume of solution B, we can substitute it out in our final equation, like this:

0.15x + 0.81(60ml - x) = 0.37 • 60mL

The rest is algebra. We distribute 0.81 inside the parentheses and multiply 0.37 • 60mL to get:

0.15x + 48.6mL - 0.81x = 22.2mL

then we subtract 48.6mL from both sides:

0.15x - 0.81x = -26.4mL

We subtract 0.81x from 0.15x:

-0.66x = -26.4mL

then we divide both sides by -0.66 to get:

x = 40mL which is our volume for solution A.

we said earlier that our volume for solution B was just 60 - x, so now we plug in x and get:

60mL - 40mL = 20mL and that's our volume for solution B. we have solved the problem, but to do a quick double check, we can multiply our volume values with our percent values for solutions A and B, to see if they match what we wanted.

Recall our final solution is supposed to be 60mL in volume, with 22.2mL of that volume comprised of H₂O

0.15 • 40mL = 6mL of H₂O

0.81 • 20mL = 16.2mL of H₂O

16.2mL + 6mL = 22.2mL. This is the answer we wanted, so our math checks out.

I'm sorry this answer is so long! I tend to error on the side of too much information rather than too little, and I try to explain all of my math because I know when a see a bunch of equations my eyes just glaze over. I hope it helps, let me know if anything needs clarified more!

-Bryce