Can you please help me

Question

A biologist has a 25% acetic solution and a 40% acetic solution. How many cubic centimeters of the 40% solution should be mixed to obtain 240 cc of a 30% solution?

Answer 1

Alright I can understand your confusion since the wording is a bit tricky. But here's how you can think of the problem. You have a container of 25% solution and another container of 40% solution that will both be used to fill up a 240 cubic centimeter container. This final mixed solution will be 30% acetic. When I say (some number)% solution I'm referring to the fraction of the substance that is acetic...So that's the background of the problem. How you go about solving it is setting up what is called a system of equations. I will first show you the equations and then explain.

 

Eq. 1: x+y=240

Eq. 2: .25x+0.4y=(0.3)(240)

 

so first lets go over the variables. I'll let x be the volume of the 25% acetic solution and y be the volume of the 40% acetic solution. when looking at any problem it's important to understand what you don't know, and in this case it's the volumes of the two original solutions...Now here's the reason why I used two equations: For a certain number of unknown variables you need the same number of independent equations to figure out the values of those variables. (Say that previous sentence in your head until it makes sense. Also, when I say independant equations, I mean that both sides of one equation can't be multipliced by a constant to produce any of the other equations. As you can see, that is not the case in this problem and the two equations are certainly independent). So, you might write out equation 2 after analyzing the problem for a long time. This equation basically says that the fraction of one solution that is acedic plus the fraction of the other solution that is acedic equals the fraction of the mixed substance (this is given in the problem) that is acedic. This is a valid equation. (Remember that percentages can be modeled with fractions. Another subtle point is that x and y is the amount (which is measured as a volume) of each solution that we use in the mixed solution. These volumes could be different then the original volumes of the two containers of solutions, but this distinction is irrevelant in solving the problem because of the reason I will soon state). If equation 2 was our only equation used to solve for x and y we would end up with a bunch of possible values for these two variables. So we need another equation, and that's where equation 1 comes in, and this equation is just stating the obvious, which is that the volume used of one solution plus the volume of the second solution equals the volume of the mixed solution. Now we have two equations and two variables. (But, these actually aren't variables after all. Another way to think of what I said above is that when you have the same number of variables as independant equations there will only be one possible value that each of these 'variables' can have. So a more appropriate name for x and y would be unknown values). Now what you have to do, is algebraicly solve for y (the volume of the 40% solution) using the two equations. Try that yourself. If you get overwhelmed with the algebra, try using one of the equations to make x as a function of y (put the x to one side of the eqn and then the rest of the stuff to the other side). And then substitute in this equation for the x of the other equation. 

 

Hope all of this helps and feel free to follow up this response with any additional questions you may have,

 

Steven B.

Answer 2

Algebraic solution:  Two equations with 2 unknowns

 

x + y = 240cc   or   x = 240-y

 

.25x + .40y = .30(240)

 

.25(240-y)   +  .40y  =  72

 

60 - .25y    +  .40y   =  72

 

.15y   =  12

 

y=  80cc

 

 

OR the general ratio method

 

Since the final solution (30%) is one third of the way from the lower to the higher (5 of the 15% difference),  then you need two thirds of the lower (25%) and one third of the more concentrated (40%), so 1/3 of 240cc is 80cc.