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

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From a cask containing y liters of milk, x liters of milk is drawn out  [#permalink]

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New post 17 Oct 2018, 00:45
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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

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Re: From a cask containing y liters of milk, x liters of milk is drawn out  [#permalink]

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New post 17 Oct 2018, 04:08
Bunuel wrote:
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


To know the fraction of milk remaining, we either need to know the relative fractions of x,y and z, or their absolute numbers.
We'll look for statements that give us this information, a Logical approach.

(1) Lacking z, we cannot say.
Insufficient.

(2) This gives us the ratios, exactly what we need!
Sufficient.

(B) is our answer.
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From a cask containing y liters of milk, x liters of milk is drawn out  [#permalink]

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New post 21 Apr 2019, 15:10
DavidTutorexamPAL wrote:
Bunuel wrote:
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


To know the fraction of milk remaining, we either need to know the relative fractions of x,y and z, or their absolute numbers.
We'll look for statements that give us this information, a Logical approach.

(1) Lacking z, we cannot say.
Insufficient.

(2) This gives us the ratios, exactly what we need!
Sufficient.

(B) is our answer.


Can you explain why the relative fractions of x, y, and z would be sufficient?
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Re: From a cask containing y liters of milk, x liters of milk is drawn out  [#permalink]

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New post 28 Apr 2019, 07:14
DavidTutorexamPAL wrote:
Bunuel wrote:
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


To know the fraction of milk remaining, we either need to know the relative fractions of x,y and z, or their absolute numbers.
We'll look for statements that give us this information, a Logical approach.

(1) Lacking z, we cannot say.
Insufficient.

(2) This gives us the ratios, exactly what we need!
Sufficient.


(B) is our answer.



Hi,

Could you please explain how the second statement is sufficient?

Regards

Ritvik
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From a cask containing y liters of milk, x liters of milk is drawn out  [#permalink]

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New post 28 Apr 2019, 07:47
menonrit wrote:

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.
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From a cask containing y liters of milk, x liters of milk is drawn out  [#permalink]

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New post 28 Apr 2019, 08:01
menonrit wrote:
DavidTutorexamPAL wrote:
Bunuel wrote:
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


To know the fraction of milk remaining, we either need to know the relative fractions of x,y and z, or their absolute numbers.
We'll look for statements that give us this information, a Logical approach.

(1) Lacking z, we cannot say.
Insufficient.

(2) This gives us the ratios, exactly what we need!
Sufficient.


(B) is our answer.



Hi,

Could you please explain how the second statement is sufficient?

Regards

Ritvik



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
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From a cask containing y liters of milk, x liters of milk is drawn out   [#permalink] 28 Apr 2019, 08:01
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