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A sample of x litres from a container having a 60 litre mixture of mil

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A sample of x litres from a container having a 60 litre mixture of mil  [#permalink]

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New post 29 Jan 2019, 07:52
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A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

A. 6 litres

B. 10 litres

C. 15 litres

D. 20 litres

E. 30 litres

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A sample of x litres from a container having a 60 litre mixture of mil  [#permalink]

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New post 29 Jan 2019, 08:03
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Bunuel wrote:
A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

A. 6 litres

B. 10 litres

C. 15 litres

D. 20 litres

E. 30 litres


Conc of milk in mixture (C1) = 2/5
Conc of milk in pure milk (C2) = 1
Conc of milk in new mix (Cavg) = 1/2

w1/w2 = (1 - 1/2) / (1/2 - 2/5) = (1/2) / (1/10) = 5/1

So original mixture was 5 parts and pure milk was 1 part to get equal proportions of milk and water. Total 6 parts is actually 60 litres which means the 5 parts are 50 litres and 1 part is 10 litres.
Hence 10 litres of original mix was replaced with pure milk.

Answer (B)

Check this post: https://www.veritasprep.com/blog/2012/0 ... -mixtures/
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Re: A sample of x litres from a container having a 60 litre mixture of mil  [#permalink]

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New post 29 Jan 2019, 09:44
Bunuel wrote:
A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

A. 6 litres

B. 10 litres

C. 15 litres

D. 20 litres

E. 30 litres


use weighted avg formula
w1/w2 = ( a2-aavg)/ ( aavg-a1)
here a2= 1
aavg= 1/2
and a1= 2/5

w1/w2= ( 1-1/2) / ( 1/2-2/5)
w1/w2 = 5/1
total 60 liters
5/6 * 60 = 50 ltrs w1 and w2 = 10 liters
so x= 10
IMO B
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Re: A sample of x litres from a container having a 60 litre mixture of mil  [#permalink]

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New post 29 Jan 2019, 10:02
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Bunuel wrote:
A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

A. 6 litres

B. 10 litres

C. 15 litres

D. 20 litres

E. 30 litres


A great weighted average method above by VeritasKarishma..
But let me give you another method..
M:W=2:3 means M = \(\frac{2}{5}\) and W=\(\frac{3}{5}\)
Total 60 liters means M = \(\frac{2}{5}*60=24\) and W=\(\frac{3}{5}*60=36\)
To be equal both should be 30 each, so a decrease of 6 liters of water from initial 36 liters...

Now if reduction of 1 liter would mean reduction of \(\frac{3}{5}\) of water, reduction of x liter would mean \(\frac{3}{5}*x\), but this should be equal to 6..
thus \(\frac{3}{5}*x=6.....x=\frac{5}{3}*6=10\)

B
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Re: A sample of x litres from a container having a 60 litre mixture of mil  [#permalink]

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New post 29 Jan 2019, 11:39
Bunuel wrote:
A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

A. 6 litres

B. 10 litres

C. 15 litres

D. 20 litres

E. 30 litres


2/5*(60-x)+x=1/2*60
x=10 litres
B
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Re: A sample of x litres from a container having a 60 litre mixture of mil  [#permalink]

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New post 31 Jan 2019, 17:37
Bunuel wrote:
A sample of x litres from a container having a 60 litre mixture of milk and water containing milk and water in the ratio of 2 : 3 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

A. 6 litres

B. 10 litres

C. 15 litres

D. 20 litres

E. 30 litres


Since milk and water is in a ratio of 2 to 3, we can create the following equation:

2n + 3n = 60

5n = 60

n = 12

So originally, the container has 24 litres of milk and 36 liters of water. We are told that x liters of the mixture are removed. Note that these x liters that are removed will consist of 3x/5 liters of water and 2x/5 liters of milk. . The x liters of liquid that were removed are replaced by x liters of pure milk. Therefore, we can create the equation:

(24 - 2x/5 + x)/(36 - 3x/5) = 1

24 + 3x/5 = 36 - 3x/5

120 + 3x = 180 - 3x

6x = 60

x = 10

Answer: B
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Re: A sample of x litres from a container having a 60 litre mixture of mil   [#permalink] 31 Jan 2019, 17:37
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