A can contains $200$ liters of pure milk. From this can, $20$ liters of milk is removed and replaced with water. THis process is repeated is repeated two more times. Find the concentration $(i n %)$ of milk in the resultant solution.

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Solution

1. $200 L - 20 L = 180 L$
Now, We have 180L Milk, 20L Water

2. $180 L - 20 L = 160 L$
Now, We have 160L Milk, 40L Water

3. $160 L - 20 L = 140 L$
Now, We have 140L Milk, 60L Water

Hence, Resultant Solution will have
140L Milk, 60L Water

Now, % Concentration of Milk
$= \frac{140}{200} \times 100 = 70 %$

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A can contains 200 liters of pure milk. From this can, 20liters of milk is removed and replaced with water. THis process is repeated is repeated two more times. Find the concentration (in%) of milk in the resultant solution.

1. 200L−20L=180LNow, We have 180L Milk, 20L Water2. 180L−20L=160LNow, We have 160L Milk, 40L Water3.160L−20L=140LNow, We have 140L Milk, 60L WaterHence, Resultant Solution will have 140L Milk, 60L WaterNow, % Concentration of Milk=140200×100=70%

A can contains $200$ liters of pure milk. From this can, $20$ liters of milk is removed and replaced with water. THis process is repeated is repeated two more times. Find the concentration $(i n %)$ of milk in the resultant solution.