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02 Mar 2021, 12:12
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There are two solutions of water and spirit of different concentrations. Solution 1 is of 30L and solution 2 is of 10L. If same amount of solution is taken from each of the solutions and added to other solutions, the concentration of both the solutions becomes same. Find the amount taken from each solutions?
A) 3.5 litres
B) 5 litres
C) 7.5 litres
D) 8 litres
E) Cannot be calculated
A) 3.5 litres
B) 5 litres
C) 7.5 litres
D) 8 litres
E) Cannot be calculated
02 Mar 2021, 12:55
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Assume concentration of sol 1 is x and of sol 2 is y.
Amount taken from each sol is a
\(\frac{x*(30-a) + y*a}{30} =\frac{ y*(10-a)+x*a}{10}\)
30x-xa +ya = 30y-3ya +3xa
30(x-y)-4a(x-y) = 0
(30-4a)(x-y) = 0
Since x-y can't be 0, 30-4a=0 or a=7.5
Amount taken from each sol is a
\(\frac{x*(30-a) + y*a}{30} =\frac{ y*(10-a)+x*a}{10}\)
30x-xa +ya = 30y-3ya +3xa
30(x-y)-4a(x-y) = 0
(30-4a)(x-y) = 0
Since x-y can't be 0, 30-4a=0 or a=7.5
ratedR123
03 Mar 2021, 08:11
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The wording of the problem is a bit off.
The following seems to reflect the intent of the problem:
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:
\(\frac{30-k}{k} = \frac{k}{10-k}\)
\((30-k)(10-k) = k^2\)
\(300-40k+k^2 = k^2\)
\(300 = 40k\)
\(k = \frac{300}{40} = \frac{30}{4} = \frac{15}{2} = 7.5\)
The following seems to reflect the intent of the problem:
Quote:
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:
\(\frac{30-k}{k} = \frac{k}{10-k}\)
\((30-k)(10-k) = k^2\)
\(300-40k+k^2 = k^2\)
\(300 = 40k\)
\(k = \frac{300}{40} = \frac{30}{4} = \frac{15}{2} = 7.5\)
