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by vinay1983 » Wed Aug 21, 2013 3:46 am
6 litres from a 30 litre mixture of milk and water, which are in the ratio 3:2, is replaced with milk. The operation is repeated once. What will be the ration of milk and water in the resultant mixture?

The OA is a weird ratio [spoiler]22.32/7.68[/spoiler]
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by ganeshrkamath » Wed Aug 21, 2013 7:07 am
vinay1983 wrote:6 litres from a 30 litre mixture of milk and water, which are in the ratio 3:2, is replaced with milk. The operation is repeated once. What will be the ration of milk and water in the resultant mixture?

The OA is a weird ratio [spoiler]22.32/7.68[/spoiler]
30 liters of mixture has milk and water in the ratio 3:2
So amount of milk = 3/5*30 = 18 liters
Amount of water = 2/5*30 = 12 liters

6 liters of the mixture is removed => 3/5*6 liters of milk and 2/5*6 liters of water is removed
3.6 liters of milk and 2.4 liters of water is replaced with 6 liters of milk

Now the mixture contains 18+2.4 = 20.4 liters of milk
and 12-2.4 = 9.6 liters of water

Again, 6 liters of the mixture is removed => 9.6/30 * 6 = 1.92 liters of water is replaced by milk

So the mixture now contains (20.4 + 1.92) liters of milk and (9.6 - 1.92) liters of water

Required ratio = [spoiler]22.32/7.68[/spoiler]

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by GMATGuruNY » Wed Aug 21, 2013 8:57 am
vinay1983 wrote:6 litres from a 30 litre mixture of milk and water, which are in the ratio 3:2, is replaced with milk. The operation is repeated once. What will be the ration of milk and water in the resultant mixture?

The OA is a weird ratio [spoiler]22.32/7.68[/spoiler]
The operation consists of removing 6 liters from the container.
Each time this operation is performed, the fraction REMAINING in the container = 24/30 = 4/5.
Implication:
Of the total amount of water in the container PRIOR to each operation, 4/5 will remain in the container AFTER each operation.

Since the question stem ask for a RATIO, we can plug in our own value for the total volume, as long we satisfy the constraint that M:W = 3:2.
Since the TWO operations require that we multiply the original amount of water by 4/5 TWO TIMES, the following values will make the math easy:
Let M:W = (3*5*5)/(2*5*5).
Total volume = M+W = (3*5*5) + (2*5*5) = 125 liters.

After the two operations, the amount of water remaining = (4/5)(4/5)(2*5*5) = 32 liters.
Thus, the amount of milk in the container after the two operations = 125-32 = 93 liters.
Resulting ratio:
M:W = 93:32.
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