kakz wrote:8 litres solution is removed from a 20% milk solution and 8 litres of water are added to it. The resulting solution has 16% milk in it. What was the initial quantity of the 20% milk solution?
Is is possible to do this one by the weighted averages method. problem is pretty easy but please explain using weighted averages.
Original solution = 20% milk.
Added water = 0% milk.
Mixture = 16% milk.
According to alligation:
The proportion needed of each ingredient is equal to the distance between the OTHER two percentages.
Proportion needed of original solution = |percentage in mixture - percentage in added water| = |16-0| = 16.
Proportion needed of added water = |percentage in mixture - percentage in original solution| = |16-20| = 4.
Original solution : added water = 16:4 = 4:1.
Since 8 liters of water are added, and 4:1 = 32:8, the mixture contains 32 liters of the original solution and 8 liters of added water, for a total of 32+8 = 40 liters.
Since the volume remains constant when 8 liters are removed and 8 liters are added, the original solution also was 40 liters.
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