A dishonest milkman sells a 40 liter mixture of milk and water that co

Hi maxk297,

While many ratio questions can be solved by setting up an algebraic equation, this really isn't one of those questions. It's pure Arithmetic - we're given the starting numbers, we're told what to add and subtract, and then we're asked for an end result. In that way, this is somewhat rare among the typical ratio questions that you'll see on Test Day. THOSE questions can normally be solved by doing Algebra, TESTing THE ANSWERS or by doing 'brute force' Arithmetic.

GMAT assassins aren't born, they're made,
Rich

VERITAS PREP OFFICIAL SOLUTION:

In this word problem, the key to success is staying organized and following the changes in value. The mixture begins with a 3:2 ratio of milk to water, so out of 40 liters you have:

24 milk, 16 water

The first change is that the milkman removes half of that mixture, so cut those totals in half:

12 milk, 8 water

Then he adds 20 liters of milk, so add 20 to the milk total, making the running tally:

32 milk, 8 water

The next step is to take out half of that mixture, so you'll cut those totals in half:

16 milk, 4 water

And then add 20 liters of water to get the final mixture:

16 milk, 24 water

Since that's the final mixture, take that as a ratio. 16 and 24 are each multiples of 8, so you have (2 * 8) milk and (3 * 8) water, leading to a ratio of 2:3.

ans C 2:3...
initial 40 ltrs ... when 20 ltr left m:w=3:2=12:8..
when 20l milk added... m:w=32:8..
now 20l removed, means half of each left=16:4..
now add 20 l water m:w=16:24=2:3

Actually, there is a formula for such replacement questions....

\(Final_Concentration = Initial_Concentration * (1 - \frac{Replace}{Total})^{Number_of_replacements}\)

Although nopt tried the formula on this Q, I think the formulae will stand if same item is replaced or added.. here u have two different items being added on both occassions

Hi UNSTOPPABLE12,

When dealing with an established mixture - and you remove 50% from the TOTAL (in this case, 40L of liquid), you are removing liquid that matches the original ratio (here, that 20L of liquid is 3:2 - milk:water, so we're removing 12L of milk and 8L of water), NOT 50% of each liquid. In other versions of this type of question, you might be able to remove "part" of the mixture (for example, if you were mixing types of nuts, then you could conceivably remove only one type of nut) - but the prompt would explicitly state that that is what you were doing. If you don't see that type of language, then you have to assume that when you remove some of the OVERALL mixture, then what you are removing matches the original ratio of ingredients.

GMAT assassins aren't born, they're made,
Rich

C: 2:3

I took the long route.. i would like to solve this algebraically.. I am pondering over that now...

The original mixture has a ratio of milk to water = 3x : 2x, and thus:

40 = 5x

x = 8

So, originally there were 3(8) = 24 liters of milk and 2(8) = 16 liters of water.

Since 20 liters of the mixture (i.e., half of the mixture) are removed, half of the milk (12 liters) and half of the water (8 liters) are removed. Thus, after 20 liters are removed, there are:

12 liters of milk and 8 liters of water in the mixture.

Since 20 liters of milk are added back in, we now have:

32 liters of milk and 8 liters of water in the mixture.

Since another 20 liters of the mixture (i.e., half of the mixture) are removed, half of the milk (16 liters) and half of the water (4 liters) are removed. Thus, after 20 liters are removed, there are:

16 liters of milk and 4 liters of water in the mixture.

Since 20 liters of water are added back in, we now have:

16 liters of milk and 24 liters of water in the mixture.

Thus, the final ratio of milk to water is 16/24 = 2/3.

Answer: C

took some time but did it this way:
1) 40 = 3x + 2x --> 5x = 40, x = 8. so 24 milk, 16 water

2) removed 20 L of mix but added 20 milk --> 20=5x --> 12 milk, 8 water and 20 more of milk. New amounts: 32 milk and 8 water.
3)He then takes out 20 liters of this new mixture and replaces it with an equal amount of water to create his final mixture..
20 water + (20=5x, x =4--> 16 milk, 4 water and 20 water).
Thus 24 water /16 milk = 2:3
Ans C

Kudos if helpful!

Hi Using Allegation Concept

Key here is to remember that conc of Mixture does not change if we remove Liquid from it .

So final conc of M: W=2:3

Intially Conc of M:W is 60:40

When 20 Lts is removed Conc is same = 60:40

Now 20 Ltrs Milk is added
The conc of mixture is 80:20

When 20 Lts is removed Conc is same = 80:20

Now 20 Ltrs Water is added
The conc of final mixture is 40:60

so we have 2:3

Hello guys , Bunuel ScottTargetTestPrep

I have a general question regarding this kind of problems, so in the beginning the ratio of milk to water is 3:2 so there is more milk than water , hence when we remove 20litres (reduce by 50%) , how do we know that there will be a uniform decrease(50% each) , and not a proportionate one that will be correlated with the ratios the two elements have.

A dishonest milkman sells a 40 liter mixture of milk and water that contains milk and water in the ratio of 3:2. He takes out 20 liters of the mixture and replaces it with an equal amount of milk. He then takes out 20 liters of this new mixture and replaces it with an equal amount of water to create his final mixture. What is the ratio of milk and water in the final mixture?

A. 2:5
B. 3:5
C. 2:3
D. 3:2
E. 5:3

\(Final/Initial\) = (1 - \(b/a\))^n

Final = The final quantity of that component who's concentration is being reduced
Initial = The 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 since milk is being added, the concentration of water is reduced in the first repacement iteration
\(F/16\) = (1 - 20/40)^N
16 = \(2/5 \)* 40 (Since the initial mixture is in the ratio of milk 3 parts : water 2 parts) & N =1
Final water concentration = \(F/16\) = \(1/2 \)F = 8 Final milk concentration = 32
In the second iteration , the concentration of milk is being reduced
\(F/32\) = (1 - \(20/40\))^N
Final concentration of milk = 16
Final concentration of water = 40 - 16 = 24
Final ratio of milk : water = 16:24 = 2:3

C

Use weighted average
Total= 40
M/W=3/2 --> so, M/T= 3/5=6/10,--> Solution have a concetration of Milk of 0.6

1st scenario : add 20 of 100%milk
******************************
Solut.----|---------100%Milk
20----------|------------20
|------------|------------|
0.6---------|------------1 --> Concentration

--> weighted average: the ratio of distances is the same as the ratio of concentration,
because the ratio of weights is 1/1, so the ratio of concetration is the same
Solut.----|---------100%Milk
20----------|------------20
|------------|------------|
0.6-------- 0.8------------1 --> Concentration
--> New solution has a concentration of milk of 0.8

2st scenario : add 20 of Water (0%Milk)
******************************
Solut.----|---------100%Water
20----------|------------20
|------------|------------|
0.8---------|------------0 --> Concentration

--> weighted average: the ratio of distances is the same as ratio of concetration ,
because the ratio of weights is 1/1, so the ratio of concetration is the same
Solut.----|---------100%Water
20----------|------------ 20
|------------|------------|
0.8--------- 0.4------------0 --> Concentration
--> New solution has a concentration of milk of 0.4=2/5=Milk/Total
-->Milk=2k, Total=5k and Water=3k
--> Therefore the ratio of Milk out of Water is 2k/3k=2/3
-->Right answer: C