AKHIL KUMAR TRIPATHI wrote:A container contained 80kg of milk.From this container 8 kg of milk was taken out and replaced by water.This process was further repeated two times.How much milk is now contained by the container?
This problem is too complex for the GMAT.
That being said, here's a solution:
Each time 8 kg are removed from the 80kg container, the volume of every ingredient in the container -- including the volume of milk -- decreases by 8/80 = 1/10.
If an amount
x DECREASES by fraction
a/b exactly
n times, we can use the following formula:
Final amount = x * (1 - a/b)^n.
In the problem at hand:
x = 80 kg of milk
a/b = 1/10
Since the volume decreases by 1/10 exactly 3 times, n = 3.
Plugging these values into the formula, we get:
F = 80 * (1 - 1/10)³ = 80 * (9/10)³ = (80 * 729)/1000 = 58320/1000 = 58.32 kg.
Note the following:
If an amount
x INCREASES by fraction
a/b exactly
n times, the formula is as follows:
Final amount = x * (1 + a/b)^n.Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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