Question

we have 100 litres of a mixture of milk and water which is 10% water. How much more pure milk should be added so that the new mixture has only 5% water?






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Answer 1

Answer:

100 liters pure milk to 100+100 = 200 (milk+water) has 10 liters water. So the percentage of water in the mixture = 10 x 100/200 = 5%. Answer. Add 100 liters of pure milk.

Answer 2

Given:

• Total Quantity mixture of milk and water = 100
• Percentage (quantity) of water = 10 %

To find:

• How much more pure milk should be added so that the new mixture has only 5% water?

Solutions:

Total Quantity of mixture of milk and water = 100 l

In mixture,Quantity of water = 10 %

Quantity of Pure milk = ( 100 - 10 ) l = 90 l

Let 'x' be the milk added to the mixture to make water 5 % and pure milk 95 % .

As per the question,

$(90 + x)l = 95\% \times( 100 + x)l$

$= > 90 + x = 0.95(100 + x)$

$= > 90 + x = 95 + 0.95x$

$= > x - 0.95x = 95 - 90$

$= > 0.05x = 5$

$= > x = \frac{5}{0.05} = 100l$

So, 100l of pure milk should be added so that the new mixture has only 5 % water.