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# Two containers contain milk and water solutions of volume x liters and

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Intern
Joined: 30 Sep 2017
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Location: India
Concentration: Entrepreneurship, General Management
Schools: IIM Udaipur '17
GMAT 1: 700 Q50 V37
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WE: Engineering (Energy and Utilities)
Two containers contain milk and water solutions of volume x liters and  [#permalink]

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08 Dec 2017, 07:35
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75% (hard)

Question Stats:

59% (02:23) correct 41% (02:25) wrong based on 190 sessions

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Two containers contain milk and water solutions of volume x liters and y liters, respectively. What would be the minimum concentration of milk in either container so that when the entire contents of both containers are mixed, 30 liters of 80 percent milk solution is obtained?

(1) x = 2y
(2) x = y +10

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Two containers contain milk and water solutions of volume x liters and  [#permalink]

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08 Dec 2017, 08:03
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My understanding is that we are being asked what is the minimum concentration of milk in either X or Y in order to achieve a 30-liters mixture of 80% milk and 20% water when both X and Y are combined.

Elaborating the question stem, we get:
30liters -> 24 milk, 6 water.

By doing some quick math with X+Y=30 and the two answer options, we easily find out that they basically tell us the same thing, i.e. X=20, Y=10. We can therefore boil down our answers to either D) or E).

We now need to find the scenario in which the milk concentration in either X or Y is lowest. Since our "target" milk amount is 24, every percentage point of concentration in X yields a higher amount of milk, therefore we can assume that X has a 100% concentration of milk, leaving us with 20 liters of milk out of the 24 we need. Finally, we proceed to fill in the remaining 4 liters with Y, with a concentration of 4/10=40%.

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Re: Two containers contain milk and water solutions of volume x liters and  [#permalink]

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26 Jan 2019, 15:50
1
Aim800score wrote:
Two containers contain milk and water solutions of volume x liters and y liters, respectively. What would be the minimum concentration of milk in either container so that when the entire contents of both containers are mixed, 30 liters of 80 percent milk solution is obtained?

(1) x = 2y
(2) x = y +10

This is a Data Sufficiency word problem. It's usually best to put these problems into mathematical terms before we try to either solve, or test cases. Otherwise it might be easy to miss a constraint or a subtle detail in the question.

The question itself gives us some constraints. We need to mix x liters and y liters, and end up with 30 liters of 80% milk. The 30 liters part corresponds to this equation:

x + y = 30

The 80% part corresponds to this equation:

x(percent of milk in x) + y(percent of milk in y) = 80%(30) = 24

So, we have two equations and four variables. We're trying to find the minimum percent of milk in x or y. On to the statements!

Statement 1: x = 2y. In this case, we can solve for x and y, using this equation and the first equation from the question stem.

x + y = 30
2y + y = 30
3y = 30
y = 10
x = 20

Can we now find the minimum percent of milk required? 20(percent of milk in x) + 10(percent of milk in y) = 24. To minimize one value in an equation, you generally want to maximize everything else. We'd want to make the percent of milk in x as large as possible (100%), then find the resulting percent of milk in y, or vice versa. We shouldn't actually do that work, though - it's data sufficiency, so it's enough to know that we could figure out the minimum. Sufficient.

Statement 2: This will follow the exact same solution path as the above statement, starting with finding the values of x and y. So, it's also sufficient.
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Re: Two containers contain milk and water solutions of volume x liters and  [#permalink]

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24 Jan 2019, 22:49
Aim800score wrote:
Two containers contain milk and water solutions of volume x liters and y liters, respectively. What would be the minimum concentration of milk in either container so that when the entire contents of both containers are mixed, 30 liters of 80 percent milk solution is obtained?

(1) x = 2y
(2) x = y +10

So, this is a question of weighted average, you mix the contents of 2 containers to get a container which will give you 30 liters of 80% milk solution.
Since its a DS question, if you are able to realize the inline you should be good.

If i get a relationship between x and y i should be good here, I can know what will be the ratio of milk and water in the new milk solution.

Statement 1) x = 2y, 3y = 30, y =10 and x =20, From Allegation we can get the value of minimum milk concentration.

Statement 2) x = y +10, 2y = 20, y =10 and x =20, From Allegation again we can get the value of minimum milk concentration.

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Re: Two containers contain milk and water solutions of volume x liters and  [#permalink]

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26 Jan 2019, 23:49
Aim800score wrote:
Two containers contain milk and water solutions of volume x liters and y liters, respectively. What would be the minimum concentration of milk in either container so that when the entire contents of both containers are mixed, 30 liters of 80 percent milk solution is obtained?

(1) x = 2y
(2) x = y +10

By the Rule of Alligations

$$\frac{x}{y}$$ = $$\frac{(80-b)}{(a-80)}$$ ..... Eqn 1

Where a and b are the concentrations of milk in the first and second container respectively

To minimize b , We maximize a and make it 100%

Each Statement alone gives x/y = 2 ( Since, x + y = 30)

Now, there is only one unknown b in the Eqn 1 which can be calculated

Each Statement alone is Sufficient

Choice D
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Re: Two containers contain milk and water solutions of volume x liters and  [#permalink]

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24 Sep 2019, 06:02
Aim800score wrote:
Two containers contain milk and water solutions of volume x liters and y liters, respectively. What would be the minimum concentration of milk in either container so that when the entire contents of both containers are mixed, 30 liters of 80 percent milk solution is obtained?

(1) x = 2y
(2) x = y +10

x+y=30
x/y=0.8-a/b-0.8… maximize b=1 to find minimum a
x/y=0.8-a/1-0.8…x/y=0.8-a/0.2…0.2x=0.8y-ay…ay=0.8y-0.2x

(1) x=2y: x+y=30…2y+y=30…y=10; a=0.8y-0.2x…ay=0.8y-0.2(2y)…a(10)=0.4(10)…a=0.4, sufic.
(1) x=y+10: x+y=30…(y+10)+y=30…y=10; a=0.8y-0.2x…ay=0.8y-0.2(2y)…a(10)=0.4(10)…a=0.4, sufic.

Re: Two containers contain milk and water solutions of volume x liters and   [#permalink] 24 Sep 2019, 06:02
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