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Alligation?

by madhan_dc » Wed Feb 23, 2011 3:00 pm
A container contains 40 litres of milk. From this container 4 litres of milk was taken away and replaced by water.
This process was repeated further 2 times. How much milk is contained by the container.

This is what i thought:
Milk water
begin 40 0
1st 36 4
2nd 32 8
3rd 28 12

so the answer is 28.

But the answer is 29.16 and i cant seem to understand how. It would be great if someone can shed some light.

thanks.
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by Night reader » Wed Feb 23, 2011 4:25 pm
10% decrease in milky concentration three times makes --> 1*0.9*0.9*0.9
40 lt milk *1*0.9*0.9*0.9 = 29.16 :)
simple as my five fingers' count/says process...
madhan_dc wrote:A container contains 40 litres of milk. From this container 4 litres of milk was taken away and replaced by water.
This process was repeated further 2 times. How much milk is contained by the container.

This is what i thought:
Milk water
begin 40 0
1st 36 4
2nd 32 8
3rd 28 12

so the answer is 28.

But the answer is 29.16 and i cant seem to understand how. It would be great if someone can shed some light.

thanks.
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by stormier » Wed Feb 23, 2011 4:31 pm
madhan_dc wrote:A container contains 40 litres of milk. From this container 4 litres of milk was taken away and replaced by water.
This process was repeated further 2 times. How much milk is contained by the container.

This is what i thought:
Milk water
begin 40 0
1st 36 4
2nd 32 8
3rd 28 12

so the answer is 28.

But the answer is 29.16 and i cant seem to understand how. It would be great if someone can shed some light.

thanks.
When you remove 4 liter the second time - you will remove some of the milk as well - and that's the mistake in your calculation

Let's look at just the water in the container - remember that the total volume remains 40.

Water after Step 1 = 4 _______________________ (36M,4W)
Water after Step 2 = 4 - 0.1*4 + 4 = 7.6__________ (32.4M,7.6W)
Water after Step 3 = 7.6 - 0.1*7.6 + 4 = 10.84 ___ (29.16M,10.84W)

When you remove 4 liter (or 10 % of 40 liter) of a 36:4 mixture, you remove 10% of 4 liter water.

Thus milk = 40 - 10.84 = 29.16
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by Rahul@gurome » Fri Feb 25, 2011 9:44 pm
Solution:
The container has 40 liters of milk.
From this, 4 liters of milk was taken away and replaced with water.
So, now we have 36 liters of milk and 4 liters of water.
Hence the new mixture has milk and water in the ratio 36:4 or 9:1.
So, now if we withdraw 4 liters, the remaining 36 liters has 36*9/10 liters milk and 36*1/10 liters water.
Again, when we add 4 liters of water, amount of milk is 36*9/10 and the amount of water is 36*1/10 + 4 or 76/10 liters.
Hence, now the mixture has milk and water in the ratio 36*9 : 76 or 81 : 19.
Again, 4 liters is being withdrawn.
Or 36 liters of mixture is left.
So now amount of milk is 36*81/100 = 29.16 liters.
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by gary391 » Sun Mar 13, 2011 8:00 am
We also have a formula for solving these type of problems:

Formula: If a Jar contains 'a' litres of liquid 'A', and if 'b' litres is withdrawn and replaced by liquid 'B',... Now again a if 'b' litres of mixture is withdrawn and replaced by liquid B, and the operation is repeated 'n' number of times in all, then:

LIQUID A LEFT IN JAR AFTER Nth OPERATION/INITIAL QUANTITY OF LIQUID A IN JAR= (a-b/a)^n

Gaurav
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by Liarish » Fri Sep 07, 2012 2:09 am
You are right Gary. We could also simply use compound interest formula. without going through so many steps.
A=P(1+r/100)^t

A=40(1-10/100)^3 We are subtracting here instead of adding as we are taking out a fixed quantity every month. So the final Amount will actually be less than 40. When solved, A= 29.16.

Oops, just realized my post came about a year and a half after the previous post. Never mind, hope someone will find it useful
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