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A mixture of 40 liters of milk and water contains 10% water.

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A mixture of 40 liters of milk and water contains 10% water. [#permalink]  27 Oct 2010, 11:57
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a mixture of 40 liters of milk and water contains 10% water. how much water should be added to this so that water may be 20% in the new mixture?

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Re: another math problem [#permalink]  27 Oct 2010, 12:33
40 liters of mixture contains 4 liters of water (10%). So, it has 36 liters of milk.

if 36 liters <-> 80%,
100% <-> (36/80)x100 = 45 liters

So, the total water in the mixture should be 9 liters. Since there is already 4 liters in the mixture, 5 liters need to be added.
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Re: another math problem [#permalink]  27 Oct 2010, 16:44
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munia123 wrote:

a mixture of 40 liters of milk and water contains 10% water. how much water should be added to this so that water may be 20% in the new mixture?

Provide answer choices for PS questions.

Let the amount of water to be added be $$x$$ liters.

We want the percentage of water to increase from 10% to 20% or percentage of non-water to decrease from 90% to 80%. Note that when we add $$x$$ liters of water the amount of non-water in mixture in liters remains the same, so:

$$0.9*40=0.8(40+x)$$ --> $$x=5$$.

Or, the amount of water after adding $$x$$ liters of water ($$0.2*(40+x)$$) should be equal to initial amount of water ($$0.1*40$$) plus the amount of water we add ($$x$$)

$$0.2*(40+x)=0.1*40+x$$ --> $$x=5$$.

Hope it helps.
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Re: another math problem [#permalink]  28 Oct 2010, 01:42
Bunuel wrote:
munia123 wrote:

a mixture of 40 liters of milk and water contains 10% water. how much water should be added to this so that water may be 20% in the new mixture?

Provide answer choices for PS questions.

Let the amount of water to be added be $$x$$ liters.

We want the percentage of water to increase from 10% to 20% or percentage of non-water to decrease from 90% to 80%. Note that when we add $$x$$ liters of water the amount of non-water in mixture in liters remains the same, so:

$$0.9*40=0.8(40+x)$$ --> $$x=5$$.

Or, the amount of water after adding $$x$$ liters of water ($$0.2*(40+x)$$) should be equal to initial amount of water ($$0.1*40$$) plus the amount of water we add ($$x$$)

$$0.2*(40+x)=0.1*40+x$$ --> $$x=5$$.

Hope it helps.

Hi Bunuel, i got the different ans.. let me know where i m wrong....

Problem again :
a mixture of 40 liters of milk and water contains 10% water. how much water should be added to this so that water may be 20% in the new mixture?

a mixture of 40 ltrs water + milk contains 10% of water so 4 ltrs are water rest 36 ltrs are milk.

if i want to make water 20% in the mixture
40*20% = (4+ x) here x is the water to be added to the mixture

so, i got x as 4

please suggest me if i gone wrong???

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Re: another math problem [#permalink]  28 Oct 2010, 01:56
Expert's post
vitamingmat wrote:
Bunuel wrote:
munia123 wrote:

a mixture of 40 liters of milk and water contains 10% water. how much water should be added to this so that water may be 20% in the new mixture?

Provide answer choices for PS questions.

Let the amount of water to be added be $$x$$ liters.

We want the percentage of water to increase from 10% to 20% or percentage of non-water to decrease from 90% to 80%. Note that when we add $$x$$ liters of water the amount of non-water in mixture in liters remains the same, so:

$$0.9*40=0.8(40+x)$$ --> $$x=5$$.

Or, the amount of water after adding $$x$$ liters of water ($$0.2*(40+x)$$) should be equal to initial amount of water ($$0.1*40$$) plus the amount of water we add ($$x$$)

$$0.2*(40+x)=0.1*40+x$$ --> $$x=5$$.

Hope it helps.

Hi Bunuel, i got the different ans.. let me know where i m wrong....

Problem again :
a mixture of 40 liters of milk and water contains 10% water. how much water should be added to this so that water may be 20% in the new mixture?

a mixture of 40 ltrs water + milk contains 10% of water so 4 ltrs are water rest 36 ltrs are milk.

if i want to make water 20% in the mixture
40*20% = (4+ x) here x is the water to be added to the mixture

so, i got x as 4

please suggest me if i gone wrong???

4+x liters of water is 20% of 40+x liters of mixture not 40.
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Re: another math problem [#permalink]  28 Oct 2010, 03:06
thanks to bunuel and also thanks to that person who asked the question to bunnel ( i also thought that 40*20% = 40+x). now i have understood. thanks .
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Re: A mixture of 40 liters of milk and water contains 10% water. [#permalink]  18 Apr 2014, 23:34
No need to complicate much.

4 ltr - Water
36 ltr - milk

Let X water be added then the new mixture must have 20% water

Then

4 + X = 20/100 (40+X)

X=5
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Re: A mixture of 40 liters of milk and water contains 10% water. [#permalink]  26 Sep 2014, 23:40
munia123 wrote:

a mixture of 40 liters of milk and water contains 10% water. how much water should be added to this so that water may be 20% in the new mixture?

hi how would u solve this using the allegation method ?
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Re: A mixture of 40 liters of milk and water contains 10% water. [#permalink]  29 Sep 2014, 00:18
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Initial water = 4

Initial Solution = 40

Final water = 4+x

Final Solution = 40+x

$$\frac{4+x}{40+x} = \frac{20}{100} = \frac{1}{5}$$

x = 5

Answer = 5 (No OA for this question)
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Re: A mixture of 40 liters of milk and water contains 10% water.   [#permalink] 29 Sep 2014, 00:18
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