A chemist combined 'a' milliliters of a solution that contained 20 percent substance S, by volume, with 'b' milliliters of a solution that contained 8 percent substance S, by volume, to product 'c' milliliters of a solution that was 10 percent substance S, by volume. What is the value of a?
(1) b = 20
(2) c = 24
I'm happy to help with this.
Before we look at the statements, let's process what's in the prompt.
First of all, the total amount of liquid "a" mL of one plus "b" mL of the other make "c" mL of the result. This must mean:
(A) a + b = c
That's one equation, the volume equation
, that relates the three variables. Now, consider the concentration of the substance S.
The first solution was "a" mL, and it was 20% S, so that contained 0.20*a mL of S
The second solution was "b" mL, and it was 8% S, so that contained 0.08*b mL of S.
Those two were combined.
The result solution was "c" mL, and it was 10% S, so that contained 0.10*c of S
The concentration equation
(B) 0.20*a + 0.08*b = 0.10*c
That's two unknowns and three equations. We currently couldn't solve for values, but if we got down to two unknowns with two equations, then we could solve for everything. Whenever (# of variables) is less then or equal to (# of equations), as long as the equations are truly different (independent), then you can solve for values.
Now, look at the DS statements. Each one gives a value for one of the three variables. If any one of the variables gets a value, then we are down to two unknowns with two equations, and we can solve for everything. Each DS statement individually allows me to solve for everything and answer the prompt question definitively. Therefore, both statements are individually sufficient, and answer = D.
Notice how little actually algebra-math I had to do once my equations were set up. This is the point of DS question --- you often don't have to do a lot of calculations, and in fact, it's often a mistake to do so. In fact, if you already knew about the two equations --- the volume equation and the concentration equation --- then you could have jumped directly to the deductions without having done any algebra-math at all. That would be a very elegant way to dispatch this problem!
Does all this make sense? Please let me know if you have any further questions.