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Hi,
Let's tackle this problem using a strategic, non-equation approach that aligns well with GMAT problem-solving techniques. :)
We'll use a back-solving method, starting with option C and eliminating choices until we find the correct answer.
Start by analyzing Choice C (q = 5 liters):
Our target is to have 3.2 liters of Chemical X in the final 8-liter solution, which is 40% of the total volume. If we consider q=5 liters, this implies we're using 5 liters of the 50% solution, providing us with 2.5 liters of Chemical X. Adding to this, we already have 2 liters of a 20% solution, contributing 0.4 liters of Chemical X. This leaves us with 1 liter for the 10% solution (as 8−(5+2)=1 liter), equating to 0.1 liter of Chemical X. Summing these up we now have 0.4+2.5+0.1=3 liters of Chemical X, which falls short of the required 3.2 liters. Eliminate C

Eliminate Options A and B:
Logically, options A and B offer even less volume of the 50% solution than option C. Therefore, they will result in a total Chemical X content less than 3 liters, which is insufficient. Thus, we can safely eliminate options A and B.

Evaluate Choice D (q = 5.5 liters):
For option D, with q = 5.5 liters of the 50% solution, we get 2.75 liters of Chemical X. This leaves 8−(2+5.5)=0.5 liters for the 10% solution. The 0.5 liters of the 10% solution contribute 0.05 liters of Chemical X. Adding these values: 0.4+2.75+0.05=3.2 liters, perfectly matching our requirement of 3.2 liters of Chemical X in the final mixture.

Option D is our answer.
Hope you are clear ! On the GMAT, it's not just what you find as an answer, but how you think ! :upsidedown

Devmitra Sen
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Did anyone try to do this using the teeter-totter method? Is it possible with those 3 parts?
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Charise made 8 liters of a 40% solution of Chemical X. If she began with 2 liters of a 20% solution of Chemical X, and then added p liters of a 10% solution of Chemical X and q liters of a 50% solution of Chemical X, what is the value of q ?

A. 4
B. 4.5
B. 5
D. 5.5
E. 6

We can PLUG IN THE ANSWERS, which represent the amount of 50% solution.
When the correct answer is plugged in, the average percentage for all 8 liters witll be 40.

B --> 4.5 liters of 50% solution, implying 1.5 liters of 10% solution, along with the original 2 liters of 20% solution
Average percentage for all 8 liters \(= \frac{(4.5 * 50) + (1.5 * 10) + (2 * 20)}{8} = \frac{(225+15+40)}{8}= \frac{280}{8} = 35\)
Since the average percentage is TOO SMALL, more of the 50% solution is needed.

D --> 5.5 liters of 50% solution, implying 0.5 liters of 10% solution, along with the original 2 liters of 20% solution
Average percentage for all 8 liters \(= \frac{(5.5 * 50) + (0.5 * 10) + (2 * 20)}{8} = \frac{(275+5+40)}{8} = \frac{320}{8} = 40\)
Success!

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fmendes
Did anyone try to do this using the teeter-totter method? Is it possible with those 3 parts?
­@fmendes it's possible, but could take a bit longer as one will first need to need to treat it as two solutions being mixed at a time. Here it is using the similar alligation method:

We are mixing 2l of a 20% solution with an unknown solution (let the percentage = a) in the ratio of 2:6 (simplified to 1:3) and get a solution of 40%.

20          a
       40
x            3x


From this we know that \(40-20 = 20 = 3x\) and thus \(x =  \frac{20}{3}\).

As \(a - 40 = x = \frac{20}{3}\), this means that \(a = 40 + \frac{20}{3} = \frac{140}{3}\)

We now know that the 10% solution and 50% solution mix to create a \(\frac{140}{3}\)% solution. To make things easier, multiply through by 3 to get rid of the fraction: (as one is multiplying through, the ratio will remain the same).

The two solutions are are now 30, 150 and the resulting solution is 140.

\(140 - 30 = 110\) 
\(150 - 140 = 10\)

30          150
      140
10         110


The ratio of p:q simplifies to 1:11, and as we know that \(p+q = 6\) we know then that every unit of the ratio is \(0.5\) litres. Therefore, \(q = 11*0.5 = 5.5\) litres


ANSWER D
 ­
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0.4(2)+0.1(p)+0.5(q) = 0.4(8) => p + 5q = 28 ---(1)

Given, 2+p+q = 8 => p+q = 6 ---(2)

Solve (1) & (2) ->> q = 11/2 = 5.5 i.e. Option D
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