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???