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Joined: 29 Nov 2018
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A milkman ordered his assistant to first remove 3 litres out of the 8
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19 Dec 2018, 09:55
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61% (02:48) correct 39% (02:37) wrong based on 134 sessions
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A milkman ordered his assistant to first remove 3 liters out of the 8 liters of a solution of milk and water and then add 3 liters of water to it. Thinking that it would not make any difference, the lab assistant first added 3 liters of water and then removed 3 liters of the solution. Find the ratio of the expected concentration of the milk to the actual concentration of the milk. A 1:1 B 21:52 C 55:64 D 3:4 E 6:11
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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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19 Dec 2018, 10:56
Imagine 8L of the solution includes 4L water and 4L milk. reduce 3L of the solution, then has 2.5L water and 2.5L milk Then add 3L of water, now in the bottle, there are 5.5L water and 2.5L milk, so the expected concentration of the milk is 5/16
The way that the assistant did makes the milk in the bottle before remove is still 4L, and the water is 7L. No matter it's removed or not,the concentration of the milk in the bottle is 4/11
So, answer: (5/16)/(4/11)=C




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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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28 Dec 2018, 14:11
Hi, chetan2u, Bunuel, please help me out with this. I don't understand a part of this explanation. The question doesn't explicitly state the ratio of milk to water in the original solution. Are we just goingbto assume that it's 1:1 as stated in the above explanation or am I missing something stated in the question that implies the ratio in the explanation. Happy Holidays btw. Posted from my mobile device



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A milkman ordered his assistant to first remove 3 litres out of the 8
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28 Dec 2018, 19:35
Kem12 wrote: Hi, chetan2u, Bunuel, please help me out with this. I don't understand a part of this explanation. The question doesn't explicitly state the ratio of milk to water in the original solution. Are we just goingbto assume that it's 1:1 as stated in the above explanation or am I missing something stated in the question that implies the ratio in the explanation. Happy Holidays btw. Posted from my mobile deviceHi.. It will not make a difference what you take except when you take initial milk to be 0. Because then the concentration will be 0 in both times. Let the entire solution be water, so 8 litres of water.. (I) expected milk 5...after taking out 3 litres of milk and replaced with water. (II) actual ..After addition of 3 litres of water total milk is 8 out of 11 litres of solution. therefore if you take out 3 litres, milk taken out 3*8/11=24/11.. Milk left is 8(24/11)=64/11. Ratio of concentration.. (5)/8: (64/11)/8 =5:64/11 =55:64
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A milkman ordered his assistant to first remove 3 litres out of the 8
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29 Dec 2018, 02:25
Hi chetan2u, thanks for the explanation. Much clearer now. Posted from my mobile device



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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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30 Dec 2018, 11:25
Hi chetan2u, I didn't get some part of your reply, which I've mentioned below. Can you please explain it. (I) expected milk 5. (II) actual .. Total milk 8/11..
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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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02 Jan 2019, 20:53
yoliwu wrote: Imagine 8L of the solution includes 4L water and 4L milk. reduce 3L of the solution, then has 2.5L water and 2.5L milk Then add 3L of water, now in the bottle, there are 5.5L water and 2.5L milk, so the expected concentration of the milk is 5/16
The way that the assistant did makes the milk in the bottle before remove is still 4L, and the water is 7L. No matter it's removed or not,the concentration of the milk in the bottle is 4/11
So, answer: (5/16)/(4/11)=C Hello! Could someone explain to me where does the 5/16 come from? Regards!



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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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09 Jan 2019, 03:25
jfranciscocuencag wrote: yoliwu wrote: Imagine 8L of the solution includes 4L water and 4L milk. reduce 3L of the solution, then has 2.5L water and 2.5L milk Then add 3L of water, now in the bottle, there are 5.5L water and 2.5L milk, so the expected concentration of the milk is 5/16
The way that the assistant did makes the milk in the bottle before remove is still 4L, and the water is 7L. No matter it's removed or not,the concentration of the milk in the bottle is 4/11
So, answer: (5/16)/(4/11)=C Hello! Could someone explain to me where does the 5/16 come from? Regards! You don’t have to follow the solution provided. First, it is about concentration ratio It is up to you to define the initiative ratio. So let us make it 8L of milk instead. You reduce 3L milk and add 3L water Concentration ratio: 5/8 You add 3L water and reduce 3L of the mixture Concentration ratio :8/11 5/8:8/11 55/88:64/88 55:64 Posted from my mobile device



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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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16 Jan 2019, 23:03
@khorwie "You add 3L water and reduce 3L of the mixture Concentration ratio :8/11" if we remove 3 L of the mixture the remaining mixture should be 8L not 11 L right? so how is the ratio coming out to be 8/11?



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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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08 Aug 2019, 08:10
Kritisood wrote: @khorwie "You add 3L water and reduce 3L of the mixture Concentration ratio :8/11" if we remove 3 L of the mixture the remaining mixture should be 8L not 11 L right? so how is the ratio coming out to be 8/11? we get 8/11 because (113) < 11 is the amount of liquid after the water is added, 3 is the liquid removed  11 < the amount of liquid we started with before removing anything. This is necessary if we want to find out the percentage. The change over the original.



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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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08 Aug 2019, 23:58
Let the qty of milk in the original 8L solution be 'x'L. (A) If the assistant had done as instructed, he would have FIRST removed 3L from the 8L solution (thereby reducing the qty of solution to 5L) Qty of milk removed in this process = (x/8)*3 = (3x/8)L Qty of milk remaining = x  (3x/8) = (5x/8)L. (Since he then adds 3L of water, qty of solution remains unchanged at 8L). Resultant concentration of milk = (5x/8)/8 = 5x/64.......(a) (B) But what he did was add 3L of water (thus increasing the volume of the solution to 11L) and THEN remove 3L of the solution. Qty of milk removed in this process = (3x/11)L Qty of milk remaining = x  (3x/11) = (8x/11) (He added 3L of water and removed 3L of solution so final volume of solution remains unchanged at 8L) Resultant concentration of milk = (8x/11)/8 = x/11.......(b) a/b = (5x/64)/(x/11) = 55/64. ANS: C UB001, jfranciscocuencag, KritisoodHope it is clear now.



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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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28 Aug 2019, 07:42
cfc198 wrote: A milkman ordered his assistant to first remove 3 liters out of the 8 liters of a solution of milk and water and then add 3 liters of water to it. Thinking that it would not make any difference, the lab assistant first added 3 liters of water and then removed 3 liters of the solution. Find the ratio of the expected concentration of the milk to the actual concentration of the milk.
A 1:1 B 21:52 C 55:64 D 3:4 E 6:11 VeritasKarishma : Cf = \(\frac{Vi*Ci}{Vf}\) Taking expected Cf, Vi= 5, Vf=8, Ci=x Cfe = \(\frac{x*5}{8}\) Taking actual Cf, Vi=11, Vf=8, Ci=x Cfa = \(\frac{11*x}{8}\) So the ratio Cfe : Cfa = \(\frac{5x}{8}\) * \(\frac{8}{11x}\) = 5/11 So what exactly went wrong? Can you help.



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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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01 Sep 2019, 00:12
Sreeragc wrote: cfc198 wrote: A milkman ordered his assistant to first remove 3 liters out of the 8 liters of a solution of milk and water and then add 3 liters of water to it. Thinking that it would not make any difference, the lab assistant first added 3 liters of water and then removed 3 liters of the solution. Find the ratio of the expected concentration of the milk to the actual concentration of the milk.
A 1:1 B 21:52 C 55:64 D 3:4 E 6:11 VeritasKarishma : Cf = \(\frac{Vi*Ci}{Vf}\) Taking expected Cf, Vi= 5, Vf=8, Ci=x Cfe = \(\frac{x*5}{8}\) Taking actual Cf, Vi=11, Vf=8, Ci=x Cfa = \(\frac{11*x}{8}\) So the ratio Cfe : Cfa = \(\frac{5x}{8}\) * \(\frac{8}{11x}\) = 5/11 So what exactly went wrong? Can you help. Note that Ci is concentration and Vi is volume before addition step and Cf (or C2) is concentration and Vf is volume after addition step. Removal does not change the concentration so is ignored. "A milkman ordered his assistant to first remove 3 liters out of the 8 liters of a solution of milk and water and then add 3 liters of water to it. "Ci * Vi = C2 * V2 Ci * 5 = C2 * 8 C2 = Ci*(5/8) "the lab assistant first added 3 liters of water and then removed 3 liters of the solution."Ci * Vi = C3 * V3 Ci * 8 = C3 * 11 C3 = Ci*(8/11) C2/C3 = (5/8)*(11/8) = 55/64 For more, check: https://www.veritasprep.com/blog/2012/0 ... mixtures/
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Re: A milkman ordered his assistant to first remove 3 litres out of the 8
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