The wording of the problem is a bit off.
The following seems to reflect the intent of the problem:
Quote:
Tank 1 contains 30 liters of a solution that is x% alcohol. Tank 2 contains 10 liters of a solution that is y% alcohol, where x≠y. How many liters must be exchanged between the two tanks so that the resulting concentration of alcohol in each tank is the same?
A) 3.5 litres
B) 5 litres
C) 7.5 litres
D) 8 litres
E) Cannot be calculated
Tank 1 contains 30 liters of x% solution and 0 liters of y% solution.
Tank 2 contains 0 liters of x% solution and 10 liters of y% solution.
For the exchange of volume to yield the same concentration of alcohol in the two tanks, each tank must end with the same ratio of x% solution to y% solution.
We can PLUG IN THE ANSWERS, which represent the volume that must be exchanged between the two tanks.
B: 5 liters
When 5 liters are exchanged between the two tanks, we get:
Tank 1 --> \(\frac{new-x}{new-y} = \frac{30-5}{0+5} = \frac{25}{5} = \frac{5}{1}\)
Tank 2 --> \(\frac{new-x}{new-y} = \frac{0+5}{10-5} = \frac{5}{5} = \frac{1}{1}\)
In this case, the resulting ratio in Tank 1 is GREATER than that in Tank 2.
Eliminate B.
D: 8 liters
When 8 liters are exchanged between the two tanks, we get:
Tank 1 --> \(\frac{new-x}{new-y}= \frac{30-8}{0+8} = \frac{22}{8} = \frac{11}{4}\)
Tank 2 --> \(\frac{new-x}{new-y} = \frac{0+8}{10-8} = \frac{8}{2} = \frac{4}{1}\)
In this case, the resulting ratio in Tank 1 is LESS than that in Tank 2.
Eliminate D.
Since B yields a ratio for Tank 1 that is too GREAT, while D yields a ratio for Tank 1 that is too SMALL, the correct answer must be BETWEEN B AND D.
C: 7.5 liters
When 7.5 liters are exchanged between the two tanks, we get:
Tank 1 --> \(\frac{new-x}{new-y} = \frac{30-7.5}{0+7.5} = \frac{22.5}{7.5} = \frac{45}{15} = \frac{3}{1}\)
Tank 2 --> \(\frac{new-x}{new-y} = \frac{0+7.5}{10-7.5} = \frac{7.5}{2.5} = \frac{75}{25} = \frac{3}{1}\)
Success!
The ratio in each tank is the same.
Algebra:
When k liters are exchanged between the two tanks, we get:
Tank 1 --> \(\frac{new-x}{new-y} = \frac{30-k}{k}\)
Tank 2 --> \(\frac{new-x}{new-y} = \frac{k}{10-k}\)
Since the two ratios must be equal, we get:
\(\frac{30-k}{k} = \frac{k}{10-k}\)
\((30-k)(10-k) = k^2\)
\(300-40k+k^2 = k^2\)
\(300 = 40k\)
\(k = \frac{300}{40} = \frac{30}{4} = \frac{15}{2} = 7.5\)
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