There are two solutions of water and spirit of different concentration

We took 'a' from solution 1, so the volume remained =30-a.

Amount of spirit in '30-a' of solution 1= x*(30-a)

Now we added 'a' amount of solution 2.

Amount of spirit in 'a' volume of solution 2= y*a

Total amount of spirit in this solution = x*(30-a) + y*a

Concentration = Amount of spirit/ total volume = x*(30-a) + y*a/30

Similarly you can find the other case and equate the concentration of 2 resultant solutions, since it's given in the question.

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:
\(\frac{30-k}{k} = \frac{k}{10-k}\)
\((30-k)(10-k) = k^2\)
\(300-40k+k^2 = k^2\)
\(300 = 40k\)
\(k = \frac{300}{40} = \frac{30}{4} = \frac{15}{2} = 7.5\)

Solution:

We can let the amounts of water and spirit in solution 1 be 20 liters and 10 liters, respectively. Similarly, we can let the amounts of water and spirit in solution 2 be 5 liters and 5 liters, respectively. Now, if we let x be the number of liters of solution removed from one solution and added to the other solution, then 2x/3 liters of water and x/3 liters of water and spirit are transferred from solution 1 to solution 2. Similarly, x/2 liters of water and x/2 liters of spirit are transferred from solution 2 to solution 1. Since the resulting solutions have the same concentration, we can create the equation:

(10 - x/3 + x/2) / 30 = (5 - x/2 + x/3) / 10

Multiplying the equation by 6, we have:

(60 - 2x + 3x) / 30 = (30 - 3x + 2x) / 10

(60 + x) / 3 = 30 - x

60 + x = 90 - 3x

4x = 30

x = 7.5

(Note: To verify that the answer is always 7.5, you can keep one set of numbers the same (solution 1 still has 20 liters and 10 liters of water and solution, respectively), but change the other set of numbers, (use 4 and 6, instead of 5 and 5). Then solve a similar equation pertaining to the numbers you use, and you will find that x is still 7.5.)

Answer: C

Can someone solve this using alligation cross multiplication method?

What do you mean? Anyway the first reply has already illustrated the answer well, you can try to re-read the explanation

­An other approach to solving this would be to understand that any two solutions of say concentration X% and Y% are to be mixed to arrive at a concentration Z%, then they always have to be mixed in the ratio : |Z-Y|/ |Z-X|.

Hence, if we assume that a volume of "a" is being removed from each of the solutions (NOTE: with conc. of X% and Y%) to be mixed into the other solution and arrive at the same concentration (Z%) for both the new solutions the ratio of mixture should be same. Hence:

(30-a) / a = a / (10-a)

=> 300 + a^2 - 40a = a^2 

=> a = 7.5­