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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] New post 27 Oct 2010, 12:57
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hey guys, i have got another problem. please help me to solve it.the problem says,

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?

my answer is 4 liter but, i am not sure. please help me.
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Re: another math problem [#permalink] New post 27 Oct 2010, 13:33
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] New post 27 Oct 2010, 17:44
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munia123 wrote:
hey guys, i have got another problem. please help me to solve it.the problem says,

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?

my answer is 4 liter but, i am not sure. please help me.


Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please:
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] New post 28 Oct 2010, 02:42
Bunuel wrote:
munia123 wrote:
hey guys, i have got another problem. please help me to solve it.the problem says,

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?

my answer is 4 liter but, i am not sure. please help me.


Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please:
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] New post 28 Oct 2010, 02:56
vitamingmat wrote:
Bunuel wrote:
munia123 wrote:
hey guys, i have got another problem. please help me to solve it.the problem says,

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?

my answer is 4 liter but, i am not sure. please help me.


Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please:
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.
_________________

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


What are GMAT Club Tests?
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Find out what's new at GMAT Club - latest features and updates

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Re: another math problem [#permalink] New post 28 Oct 2010, 04:06
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 .
Re: another math problem   [#permalink] 28 Oct 2010, 04:06
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