A milkman ordered his assistant to first remove 3 litres out of the 8

Question

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

A 1:1
B 21:52
C 55:64
D 3:4
E 6:11

yoliwu · Accepted Answer

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

KarishmaB · Answer

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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... -mixtures/

Kem12 · Answer

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.

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chetan2u · Answer

Hi..

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

UB001 · Answer

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..

jfranciscocuencag · Answer

Hello!

Could someone explain to me where does the 5/16 come from?

Regards!

khorwei01 · Answer

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

Kritisood · Answer

@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?

ScottK · Answer

we get 8/11 because

(11-3) <-- 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.

effatara · Answer

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 ,  Kritisood 
Hope it is clear now.

Sreeragc · Answer

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.

Namig · Answer

let assume there is m liters of milk. So initial ratio of milk in the mixture is m/8.

1) expected ratio: Cf = Ci x Vi/Vf m/8 x 5/8 =  5m/64 
2) actual ratio: first 3 liters water added ratio of milk became m/11 then 3 liters of mixture removed ratio of milk still remain  m/11 
3) expected / actual = 5m/64 : m/11 = 55/64.
Answer:C

Kinshook · Answer

@cfc198Given: 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.
Asked: Find the ratio of the expected concentration of the milk to the actual concentration of the milk.
 As per intructions of milkman 
Let the ratio of milk and water be k.
In 8 liters: - 
Milk = 8k/(k+1)
Water = 8/(k+1)

After removing 3 liters out and replacing with water: -
Milk = 5k/(k+1)
Water = 8 - 5k/(k+1)

Expected Concentration of Milk = 5k/8(k+1)

As done by Lab Assistant

In 8 liters: - 
Milk = 8k/(k+1)
Water = 8/(k+1)

After adding 3 liters of water: -
Milk = 8k/(k+1)
Water = 8/(k+1) + 3 = (11 + 3k)/(k+1)

After removing 3 liters of solution :- 
Milk = 8*8k/11(k+1)

Actual Concentration of Milk = 8k/11(k+1)

Ratio of the expected concentration of the milk to the actual concentration of the milk = (5/8)/(8/11) = 55/64

IMO C

AbhiS101 · Answer

Let the qty of milk be m & qty of water be w
so, m+w= 8, given

Expected: Milkmnan had ordered to remove first 3 ltrs out of 8, since its a mixture so 3 ltrs will be remove in the ratio of milk in mixture to water in mixture
Example: If 100 is in ratio of 4&6, 40 + 60 = 100, if I have to take 20 out then both 40 & 60 will reduced in their ratio to make 80
(40/100)*80 + (60/100)*80 = 80

similarly, (m/8)*5 + (w/8)*5 = 5
now , 3 ltrs water is added, it will only increase the overall qty & qty of water in mixture, milk qty will remains same
so, ratio of milk in new 8 ltrs mixture can be defined as
(5m/8/8) = 5m/64

Actual:  Lab assistant first added 3 ltrs of water then removed 3 ltrs from mixture
so, new qty of liquid is 11, it can be defined as
m/11, now 3 ltrs is taken from 11 ltrs of mixture to make it 8
 8m/11 is the new qty of milk in mixture & ratio of milk would be
8m/88

now, expected concentration/actual concentration
= 5m/64/8m/88 = 5m*88/64*8m = 5*11/64

IMO C

pradeepbhuyan · Answer

Your explanation blows my mind 😔 😔 
Please explain it in more simpler form 
I find what is water that is milk in your explanation

mcrea · Answer

you have X liters of milk in the solution out of 8 liters, you reomve 3 liters of total solution and in doing so you remove 3*x/8 since there are x liters of milk out of 8 total liters --> x - 3*x/8 = 5x/8 (expected)

you have X liters of milk in the solution out of 8 liters, you add 3 liters of water and now you have X liters of milk out of 8+3 total liters, so 11 liters, then you remove 3 liters of total solution and in doing so you remove 3*x/1 since there are x liters of milk out of 11 total liters --> x - 3*x/1 = 8x/11 (actual)

expected / actual = (5x/8) / (8x/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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