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???