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Eight litres are drawn off from a vessel full of water and s
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12 Jun 2010, 14:17
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47
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E
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95% (hard)
Question Stats:
48% (03:00) correct 52% (02:50) wrong based on 683 sessions
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Eight litres are drawn off from a vessel full of water and substituted by pure milk. Again eight litres of the mixture are drawn off and substituted by pure milk. If the vessel now contains water and milk in the ratio 9:40, find the capacity of the vessel.
A. 21 liters B. 22 liters C. 20 liters D. 14 liters E. 28 liters
Eight litres are drawn off from a vessel full of water and s
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13 Jun 2010, 05:49
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virupaksh2010 wrote:
Eight litres are drawn off from a vessel full of water and substituted by pure milk. Again eight litres of the mixture are drawn off and substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity of the vessel.
A. 21 liters B. 22 liters C. 20 liters D. 14 liters E. 28 liters
Let the capacity of the vessel be \(x\).
After the first removal there would be \(x-8\) liters of water left in the vessel. Note that the share of the water would be \(\frac{x-8}{x}\);
After the second removal, the removed mixture of 8 liters will contain \(8*\frac{x-8}{x}\) liters of water, so there will be \((x-8)-8*\frac{x-8}{x}=\frac{(x-8)^2}{x}\) liters of water left.
As the ratio of water to milk after that is \(\frac{9}{40}\), then the ratio of water to the capacity of the vessel would be \(\frac{9}{40+9}=\frac{9}{49}\).
So \(\frac{\frac{(x-8)^2}{x}}{x}=\frac{9}{49}\) --> \(\frac{(x-8)^2}{x^2}=\frac{9}{49}\) --> \(\frac{x-8}{x}=\frac{3}{7}\) --> \(x=14\).
Re: Eight litres are drawn off from a vessel full of water and s
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12 Jun 2010, 19:42
the equation you'd get once 8 lts are drawn the second time is: (w-16)/16 = 9/40 where w is the capacity of the vessel. Solving for w, w = 19.6. So, its C
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Re: Eight litres are drawn off from a vessel full of water and s
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12 Jun 2010, 22:37
iambroke wrote:
the equation you'd get once 8 lts are drawn the second time is: (w-16)/16 = 9/40 where w is the capacity of the vessel. Solving for w, w = 19.6. So, its C
Ya but second time 8 liters are removed from the mixture of milk and water not only water.
Re: Eight litres are drawn off from a vessel full of water and s
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13 Jun 2010, 22:39
After the second removal, the removed mixture of 8 liters will contain 8 * (x-8)/x liters of water, so there will be x-8-8 * (x-8)/8 = (x-8)^2/x liters of water left.
Re: Eight litres are drawn off from a vessel full of water and s
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14 Jun 2010, 06:49
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bibha wrote:
After the second removal, the removed mixture of 8 liters will contain 8 * (x-8)/x liters of water, so there will be x-8-8 * (x-8)/8 = (x-8)^2/x liters of water left.
Please explain this part
Step1 :-
Water in vessel = x
8 liters is removed and replaced with milk
so now,
Water = x-8 Milk = 8
Step 2 :-
Now again 8 liters of mixture is removes , so water in that mixture is w = (x-8/x) * 8
So Total water removed = x-8-w = x-8-(x-8/x)*8 = (x^2-8x-8x-64)/x = (x^2-16x-64)/x
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Re: Eight litres are drawn off from a vessel full of water and s
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05 Sep 2010, 03:00
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virupaksh2010 wrote:
Eight litres are drawn off from a vessel full of water and substituted by pure milk. Again eight litres of the mixture are drawn off and substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity of the vessel.
Possible AnswersSelected Possible Answer A. 21 litrers
B. 22 litres
C. 20 litres
D. 14 litres
E. 28 litres
Using Wine formula:
{ (P-8) / P }^2 = 9/49
P solves to 14 - Ans - D
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Re: Eight litres are drawn off from a vessel full of water and s
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13 Feb 2012, 14:12
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nonameee wrote:
Bunuel, could you please pin point to a mistake in my reasoning?
Let T = total capacity of vessel
1.) the concentration of milk after the first round is 8/T 2.) After the second round the concentration of milk is \(40/49\)
\(8/T\) * (T-8) + 8 = \(40/49\) T
Now, I can't solve this equation. And even substituting answer choices won't help.
Where's the mistake?
Thank you very much.
Equation is correct.
\(\frac{8}{t}(t-8)+8=\frac{40}{49}t\) --> \(\frac{t-8}{t}+1=\frac{5t}{49}\). Now, if you substitute t=14 you'll see that it'll hold true.
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Re: Eight litres are drawn off from a vessel full of water and s
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13 Feb 2012, 14:19
Bunuel, thanks a lot. I must have made some miscalculation.
Conceptually the question is interesting. However the numbers are quite "difficult" to work with. Do you think that a real GMAT question would have better numbers?
Re: Eight litres are drawn off from a vessel full of water and s
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13 Feb 2012, 14:25
nonameee wrote:
Bunuel, thanks a lot. I must have made some miscalculation.
Conceptually the question is interesting. However the numbers are quite "difficult" to work with. Do you think that a real GMAT question would have better numbers?
If you and up with an equation shown in my post above then it won't be that difficult to solve it, for your equation it's better to plug numbers of course. So I think that the question, though tough, is still a GMAT type.
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15 Feb 2012, 01:58
10
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nonameee wrote:
Bunuel, thanks a lot. I must have made some miscalculation.
Conceptually the question is interesting. However the numbers are quite "difficult" to work with. Do you think that a real GMAT question would have better numbers?
Actually, the numbers are quite suitable for a very efficient, quick and oral solution. This is what I thought of when I came up with the answer in 20 secs. Mind you, you need to go through the link provided below to understand this theory. Else the 20 sec solution will probably not make sense to you.
We are substituting milk so we should work with water.
Final concentration of water = 9/49 There were two iterations. So, 9/49 = (100%)*(Vi/Vf)^2 Vi/Vf = 3/7 Since we are putting 8 liters of water but difference between Vi and Vf is 4, final volume (which is also equal to volume of the vessel) must be twice too i.e. 7*2 = 14 liters.
Re: Eight litres are drawn off from a vessel full of water and s
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31 Dec 2012, 12:01
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nonameee wrote:
Bunuel, thanks a lot. I must have made some miscalculation.
Conceptually the question is interesting. However the numbers are quite "difficult" to work with. Do you think that a real GMAT question would have better numbers?
This is not a GMAT like question. Answers in gmat are usually sorted. as for as calculation is concerned it is easy. See below
Following formulae works in such scenario. More about its derivation later: If out of x, y is taken repeatedly after nth trial amount left is x(1-y/x)^n
Here asuming x ltr for water water left after two attempts is : x(1-8/x)^2
=> (x(1-8/x)^2)/x = 9/(9+40) => 1-8/x = 3/7 => 8/x = 4/7 or x = (7/4)*8 = 14ltr
Re: Eight litres are drawn off from a vessel full of water and s
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25 May 2013, 11:28
Bunuel wrote:
virupaksh2010 wrote:
Eight litres are drawn off from a vessel full of water and substituted by pure milk. Again eight litres of the mixture are drawn off and substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity of the vessel.
Possible AnswersSelected Possible Answer A. 21 litrers
B. 22 litres
C. 20 litres
D. 14 litres
E. 28 litres
Let the capacity of the vessel be \(x\).
After the first removal there would be \(x-8\) liters of water left in the vessel. Note that the share of the water would be \(\frac{x-8}{x}\);
After the second removal, the removed mixture of 8 liters will contain \(8*\frac{x-8}{x}\) liters of water, so there will be \(x-8-8*\frac{x-8}{x}=\frac{(x-8)^2}{x}\) liters of water left.
As the ratio of water to milk after that is \(\frac{9}{40}\), then the ratio of water to the capacity of the vessel would be \(\frac{9}{40+9}=\frac{9}{49}\).
So \(\frac{\frac{(x-8)^2}{x}}{x}=\frac{9}{49}\) --> \(\frac{(x-8)^2}{x^2}=\frac{9}{49}\) --> \(\frac{x-8}{x}=\frac{3}{7}\) --> \(x=14\).
Answer: D.
if we are removing 8 litres of mixture,does that mean we are removing water share ?? because in mixture,how can we remove water now,as it contains milk too now. why do we use 8*water share??Pls clarify..
Re: Eight litres are drawn off from a vessel full of water and s
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26 May 2013, 04:59
up4gmat wrote:
Bunuel wrote:
virupaksh2010 wrote:
Eight litres are drawn off from a vessel full of water and substituted by pure milk. Again eight litres of the mixture are drawn off and substituted by pure milk.If the vessel now contains water and milk in the ratio 9:40, find the capacity of the vessel.
Possible AnswersSelected Possible Answer A. 21 litrers
B. 22 litres
C. 20 litres
D. 14 litres
E. 28 litres
Let the capacity of the vessel be \(x\).
After the first removal there would be \(x-8\) liters of water left in the vessel. Note that the share of the water would be \(\frac{x-8}{x}\);
After the second removal, the removed mixture of 8 liters will contain \(8*\frac{x-8}{x}\) liters of water, so there will be \(x-8-8*\frac{x-8}{x}=\frac{(x-8)^2}{x}\) liters of water left.
As the ratio of water to milk after that is \(\frac{9}{40}\), then the ratio of water to the capacity of the vessel would be \(\frac{9}{40+9}=\frac{9}{49}\).
So \(\frac{\frac{(x-8)^2}{x}}{x}=\frac{9}{49}\) --> \(\frac{(x-8)^2}{x^2}=\frac{9}{49}\) --> \(\frac{x-8}{x}=\frac{3}{7}\) --> \(x=14\).
Answer: D.
if we are removing 8 litres of mixture,does that mean we are removing water share ?? because in mixture,how can we remove water now,as it contains milk too now. why do we use 8*water share??Pls clarify..
We are calculating how much water is left in the vessel after two removals.
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Re: Eight litres are drawn off from a vessel full of water and s
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03 Jun 2013, 13:24
VeritasPrepKarishma wrote:
nonameee wrote:
Bunuel, thanks a lot. I must have made some miscalculation.
Conceptually the question is interesting. However the numbers are quite "difficult" to work with. Do you think that a real GMAT question would have better numbers?
Actually, the numbers are quite suitable for a very efficient, quick and oral solution. This is what I thought of when I came up with the answer in 20 secs. Mind you, you need to go through the link provided below to understand this theory. Else the 20 sec solution will probably not make sense to you.
We are substituting milk so we should work with water.
Final concentration of water = 9/49 There were two iterations. So, 9/49 = (100%)*(Vi/Vf)^2 Vi/Vf = 3/7 Since we are putting 8 liters of water but difference between Vi and Vf is 4, final volume (which is also equal to volume of the vessel) must be twice too i.e. 7*2 = 14 liters.
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48% (03:00) correct 
