A vessel is full of a mixture of milk and water, with 9% milk. Nine li

Question

A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?

(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres

GMATinsight · Answer

Let, initial quantity of mixture = x liters
Initial quantity of milk in mixture = (9/100)*x liters

Milk in 9 litres of mixture = (9/100)*9

9 litre mixture gets replaced by pure water i.e. the net quantity of milk reduced in mixture = (9/100)*9 liters

i.e. Milk now = (9/100)*x - (9/100)*9 = (9/100)*(x-9)  = (6/100)*x (Given)

i.e. 9x - 81 = 6x
i.e. 3x = 81
i.e. x = 27 liters

Answer: Option B

Archit3110 · Answer

let total qty be x liters
so milk 9x/100 
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B

lacktutor · Answer

\frac{91}{100 } —the amount of water in 1 liter mixture.

—> ( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x

\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9

(\frac{3}{100})x = \frac{81}{100}  
—>  x = 27

Answer (B)

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HoneyLemon · Answer

--------------------------Milk--------------------------Water-----------------------------Total
Before -------------- 0.09 T (Lit)--------------------0.91T------------------------------T Lit

After----------------0.09T-9*0.09 (Lit)-------------0.91T+9---------------------------T

So 0.09T-9*0.09 = 0.06 T => T = 9* 3 = 27 Lit .

gracie · Answer

A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?

(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres

let c=vessel capacity
focus on water:
.91c-(9)(.91)+9(1)=.94c→
.03c=.81
c=27 litres
B

satya2029 · Answer

0.09*(V-9)=.0.06(V)
V=27
B:)

SiddharthR · Answer

Can this question be solved using teetor totter method ? Could someone please post the answer for that ? Instead of using milk as a quantity, I used water instead and found that the ratio of water in the (100% solution (pure water)) to the ratio of water in the (91% solution (mixture)) is 1/2 --> 1 part / 2 parts I am probably doing this all wrong. Need help

twobagels · Answer

If one thinks about it logically, removing 9 liters of solution removed 1/3 of the milk. Removing 27 liters would remove all of the milk and all the water. The container must then hold 27 liters.

adityasuresh · Answer

A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?

(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres

Replacement formula : From the pov of the element in the mixture whose concentration is being reduced
Concenentration of final = Concentration of initial *  Volume of initial/Volume of final 
 6 = 9 *  V - 9/V  
6 = 9V - 81/V 
V =27
 B

donu · Answer

Can we clarify the question to say "how much milk" does the vessel hold. Because it's confusing. Thought it was the total vessel. Thanks.

bumpbot · Answer

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A vessel is full of a mixture of milk and water, with 9% milk. Nine li

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A vessel is full of a mixture of milk and water, with 9% milk. Nine li

Prep Toolkit

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