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Bunuel
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. What part of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A. 1/5

B. 2/7

B. 7/11

C. 6/7

E. 7/8


\(\frac{Final}{Initial} = (1 - \frac{b}{a})^n\)


Final = The Final Quantity of that component who's concentration is being reduced.

Initial = The Initial Quantity of that component who's concentration is being reduced.

b = Amount of liquid replaced

a = Final Volume in the container after the replacement of quantity b.

n = number of times the operation is done


here the concentration of syrup is being reduced.

Initial quantity = 5/8 Final quantity = 1/2 (Since the final ratio is 1 : 1)

Therefore \(\frac{\frac{1}{2}}{\frac{5}{8}} = 1 - \frac{b}{a}\)

\(\frac{4}{5} = 1 - \frac{b}{a}\)

\(\frac{b}{a} = 1 - \frac{4}{5} = \frac{1}{5}\)


Option A

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Let the solution be 8 litres.
So, 5 litres is syrup, 3 litre is water.
For the solution to have equal quantity of syrup and water, it needs to have 4 litre syrup and 4 litre water.
4 litre syrup means (8/5)*4 litre solution.
Which is 6.4litres of solution.
So the remaining 1.6 litres has to be removed and same quantity of water has to be added.
So the ratio/part of solution to be removed is 1.6/8 = 0.2 = 1/5

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Bunuel
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. What part of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A. 1/5

B. 2/7

B. 7/11

C. 6/7

E. 7/8


Let the total quantity be \(16\), let \(x \) be withdrawn , now we have left with \(16-x\)

So syrup before adding water \( =\frac{5}{8}* (16-x)\)

Syrup after adding water\(= \frac{1}{2} *16 =8\)

After we add water back, again we have total quantity \(16\) , but the VOLUME or QUANTITY of syrup remains unchanged.

NOTE : we are adding ONLY water, hence QUANTITY of syrup in the soln. before and after water addition remains unaltered.

so syrup before water addition = syrup after water addition

\(\frac{5}{8}* (16-x) = 8\)

\(x= \frac{16}{5}\) , this is \(\frac{1}{5} \) of the total initial soln.

\(\frac{16}{5} * \frac{1}{16}= \frac{1}{5} \)

Ans-A

Hope it's clear.
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Bunuel / CrackverbalGMAT - just a quick question on the formal used here
a = Final Volume in the container after the replacement of quantity b.
FQ/IQ = [1-x/c], isn't the c here not the total capacity?
this is how i approached but didn't get the answer,

since half and half the final quantity should be 4 and 4
acc to formula,
4/5 = [1-x/8]
4/5 = 8-x/8
32 = 40-5x
5x = 8
x = 8/5 is what i got
a little help please - the mistake is taking c as 8 i think but i wanna know the gap in my logic
thanks in advance,
swetha

CrackverbalGMAT



\(\frac{Final}{Initial} = (1 - \frac{b}{a})^n\)


Final = The Final Quantity of that component who's concentration is being reduced.

Initial = The Initial Quantity of that component who's concentration is being reduced.

b = Amount of liquid replaced

a = Final Volume in the container after the replacement of quantity b.

n = number of times the operation is done


here the concentration of syrup is being reduced.

Initial quantity = 5/8 Final quantity = 1/2 (Since the final ratio is 1 : 1)

Therefore \(\frac{\frac{1}{2}}{\frac{5}{8}} = 1 - \frac{b}{a}\)

\(\frac{4}{5} = 1 - \frac{b}{a}\)

\(\frac{b}{a} = 1 - \frac{4}{5} = \frac{1}{5}\)


Option A

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A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. What part of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A. 1/5
B. 2/7
B. 7/11
C. 6/7
E. 7/8

Let the total mixture be 1.

Initial water = 3/8
Initial syrup = 5/8

Suppose fraction x of the mixture is drawn off and replaced with water.

After drawing off x, the syrup left is:

5/8 * (1 - x)

For the final mixture to be half syrup:

5/8 * (1 - x) = 1/2

5(1 - x) = 4

1 - x = 4/5

x = 1/5

Answer: A.


SwethaReddyL
Bunuel / CrackverbalGMAT - just a quick question on the formal used here
a = Final Volume in the container after the replacement of quantity b.
FQ/IQ = [1-x/c], isn't the c here not the total capacity?
this is how i approached but didn't get the answer,

since half and half the final quantity should be 4 and 4
acc to formula,
4/5 = [1-x/8]
4/5 = 8-x/8
32 = 40-5x
5x = 8
x = 8/5 is what i got
a little help please - the mistake is taking c as 8 i think but i wanna know the gap in my logic
thanks in advance,
swetha



In the formula,

\(\frac{FQ}{IQ}=1-\frac{b}{a}\)

\(b\) is the amount of mixture drawn off and replaced.

\(a\) is the total quantity of the mixture in the vessel after replacement. Since the vessel remains full, this is also the total capacity of the vessel.

So using 8 as the total quantity is fine.

Your equation gives:

\(\frac{4}{5}=1-\frac{x}{8}\)

so \(x=\frac{8}{5}\). But this \(x\) is the amount drawn off, not the fraction of the whole mixture drawn off.

Since the whole mixture is 8 parts, the fraction drawn off is:

\(\frac{8/5}{8}=\frac{1}{5}\)
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A shortcut is to realize that when you draw off and replace, the total volume doesn’t change
Suppose you start with 8 equal parts of liquid , 5 of them syrup. You want to end up with 4 parts , so get rid of 1 of the 5 syrup parts

1/5 is the answer
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