From a cask containing y liters of milk, x liters of milk is drawn out

Hi,

Could you please explain how the second statement is sufficient?

Regards

Ritvik

Hey menonrit,

First, apologies for missing your first query. PM me if you want to be sure I see it...

A more technical way to think of it is like this (we call this a Precise approach):
at the start: y liters of milk out of y liters
remove x: y - x liters of milk out of y - x liters
add z: y - x liters of milk out of y - x + z liters.
so far the fraction is (y - x)/(y - x + z)
Do this again and you will have (y - 2x)/(y - 2x + 2z)
So far we haven't used the statements at all...

Via (2), substitute x = 0.2y and z = 0.1y and you have your answer.

Maybe now it is easier for you to understand the Logical solution suggested above:
Since all we need is a fraction, then all we need is the ratio between the 'final volume of milk' to the 'final volume of milk + water'. If we know the ratio between x to y to z, we know the ratio between the 'original volume of milk' to the 'final volume of milk' (because this is directly determined by the ratio of x to y) and the ratio between the 'original volume of milk' to the 'final volume of milk + water' (which is directly determined by the ratio of y to x and z). In other words, we know the ratio between the numerator and denominator of the fraction so can calculate it.

Hi menonrit and jachavez06.

I am not an expert but let me try to clear your doubts.
Let's look at st2 closer. So we are given only ratios. We can approach st2 by two methods: either algebraically or numerically. I would say we try both and you guys check whichever method is either for you.
1.Algebra. We know that we had originally y liters. Ok, now we are told that 20% of y or 0.2y is drawn out, while 10% of y or 0.1y is added. So, we have y-0.2y+0.1y=0.9y. This process is repeated one more time. 0.9y-0.18y+0.09y=0.81y. So we have 0.62y of milk remaining after both actions.
2. Plug-in. Let's pick some easy numbers to work with. Say, y = 10 liters. Consequently, 2 liters drawn out and 1 liter of water is poured. At the end of this process we have 9liters of mixture. When the process is repeated once again, we get 9liters-1.8liters+0.9water=8.1 liters of mixture. We are left with 10-2-1.8=6.2 out of 10 liters of milk. Now, lets try another number, say 5. 5-1+0.5=4.5 left. Then, 4.5-0.9+.045=4.05 left. So 5-1-0.9=3.1/5 or 62%. Same answer. So B

From a cask containing y liters of milk, x liters of milk is drawn out and z liters of water are then added to the cask. This process is repeated one more time. What is the fraction of milk finally present in the mixture in the cask?

(1) x = 20, y = 100
(2) x and z form 20% and 10% of y, respectively

Formula for replacement :
F/IF/I = (1 - B/AB/A)^N
F = Final concentration of the element in the mixture that is being reduced
I = Initial concentration of the element in the mixture that is being reduced
B = Amount of mixture that is being replaced
N = No of iterations

(F1) X = 20 Y = 100
Stem : X liters of milk is drawn out and is replaced by z liters of water. Since the amount drawn out and amount replaced are different we don't know the final volume of the solution
\(F/I \)= (1 -B/A)^N
F/100 = (1 - \(20/100 - X + Z\))^2
2 unknowns Not sufficient

(F2) x and z form 20% and 10% of y, respectively
Since percentage is given we can assume volume to be 100. the question asks for fraction of milk finally present not the exact value
\(F/100\) = 1 - \(20/100 - 20 + 10\)^2
1 unknown sufficient

B

Hi Menonrit,

When the process is repeated again, at that time, the content of the solution contains both the milk and water, and not just milk, as it says "x liters of content is drawn". Hence, I have doubt if we can use this equation (y - 2x)/(y - 2x + 2z), as it would eliminate only the content of the milk and not the water.