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A mixture of 40 liters of milk and water contains 10% water. [#permalink]
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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]
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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]
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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: howtoimprovetheforumsearchfunctionforothers99451.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 nonwater to decrease from 90% to 80%. Note that when we add \(x\) liters of water the amount of nonwater 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]
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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: howtoimprovetheforumsearchfunctionforothers99451.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 nonwater to decrease from 90% to 80%. Note that when we add \(x\) liters of water the amount of nonwater 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]
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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: howtoimprovetheforumsearchfunctionforothers99451.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 nonwater to decrease from 90% to 80%. Note that when we add \(x\) liters of water the amount of nonwater 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]
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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 .

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Re: A mixture of 40 liters of milk and water contains 10% water. [#permalink]
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19 Apr 2014, 00:34
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]
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27 Sep 2014, 00:40
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. 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]
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29 Sep 2014, 01:18
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Initial water = 4 Initial Solution = 40 Say Quantity "x" water added 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]
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Re: A mixture of 40 liters of milk and water contains 10% water. [#permalink]
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Re: A mixture of 40 liters of milk and water contains 10% water. [#permalink]
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16 Apr 2017, 13:43
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. let w=water to be added .10*40+w=.20(40+w) w=5 liters

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