Allegations

[coolhabhi]

A can contains a mixture of Milk & Water in the ratio 7:5. When 9 litres of mixture are drawn off & the can is filled with the same quantity of water, the ratio of Milk & Water becomes 7:9. How many litres of Milk were contained in the can initially?
A) 10
B) 20
C) 21
D) 28

[GMATGuruNY]

A can contains a mixture of Milk & Water in the ratio 7:5. When 9 litres of mixture are drawn off & the can is filled with the same quantity of water, the ratio of Milk & Water becomes 7:9. How many litres of Milk were contained in the can initially?
A) 10
B) 20
C) 21
D) 28

Let W = the pure water, S = the original solution, and M = the mixture.

The following approach is called ALLIGATION -- a very good way to handle MIXTURE PROBLEMS.
Alligation can be performed only with fractions or percentages.

Step 1: Convert any ratios to FRACTIONS.
W:
Here, water/total = 1/1.
S:
Since milk:water = 7:5, and 7+5=12, water/total = 5/12.
M:
Since milk:water = 7:9, and 7+9=16, water/total = 9/16.

Step 2: Put the fractions over a COMMON DENOMINATOR.
W = 1/1 = 48/48.
S = 5/12 = 20/48.
M = 9/16 = 27/48,

Step 3: Plot the 3 numerators on a number line, with the numerators for W and S on the ends and the numerator for M in the middle.
W 48-----------------27------------------20 S

Step 4: Calculate the distances between the numerators.
W 48-------21--------27---------7--------20 S

Step 5: Determine the ratio in the mixture.
The ratio of W to S in the mixture is equal to the RECIPROCAL of the distances in red.
W:S = 7:21 = 1:3.

Since W:S = 1:3 = 9:27, the resulting mixture is composed of 9 liters of replacement water and 27 liters of original solution, for a total volume of 36 liters.

Since a portion of the original solution was drawn off and REPLACED with pure water, the resulting volume is the same as the original volume.
Thus, the volume of the original solution is also 36 liters.
In the original solution, water/total = 5/12, implying that milk/total = 7/12.
Thus:
Milk in the original solution = (7/12)(36) = 21 liters.

The correct answer is C.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583

[GMATGuruNY]

A can contains a mixture of Milk & Water in the ratio 7:5. When 9 litres of mixture are drawn off & the can is filled with the same quantity of water, the ratio of Milk & Water becomes 7:9. How many litres of Milk were contained in the can initially?
A) 10
B) 20
C) 21
D) 28

An alternate approach is to PLUG IN THE ANSWERS, which represent the amount of milk in the original solution.
Given that M:W = 7:5 in the original solution, the amount of milk in the original solution is probably a multiple of 7.

Answer choice C: 21 liters
Since M:W = 7:5 = 21:15, the original solution is composed of 21 liters of milk and 15 liters of water, implying that the volume of the can = 21+15 = 36 liters.
After 9 liters are drained off, the remaining volume in the can = 36-9 = 27 liters.
Since M:W = 7:5, and 7+5 = 12, the remaining amount of milk must constitute 7/12 of the remaining volume:
(7/12)(27) = 63/4.
In the resulting solution, M:W = 7:9, implying that the remaining amount of milk must constitute 7/16 of the volume of the can.
Check whether the remaining 63/4 liters of milk is equal to 7/16 of the 36-liter can:
(63/4)/36 = 7/16.
Success!

The correct answer is C.