Question

A bucket contains 20 litres of milk 2 liters of the milk were takrn out from this bucket and replace with the same amount of water

Answer 1

Answer:

14.58 litres of milk are left in the bucket now.

Step-by-step explanation:

Given information:

The quantity of milk in the container = 20 litres

Amount of litres taken out and replaced by water = 2 litres

The process was repeated another two times.

Concept Used:

Suppose a container contains $x$ of liquid from which $y$ units are taken out and replaced by water.

After $n$ operations, the quantity of pure liquid = $x\left(1-\dfrac{y}{x}\right)^n \,\,\text{units}$

Step 1:

Here, $x=20$ and $y=2$.

After 1+2=3 operations, the quantity of pure liquid = $20\left(1-\dfrac{2}{20}\right)^3 \,\,\text{units}$

$20\left(1-\dfrac{2}{20}\right)^3 \,\,\text{units}=20\left(\dfrac{10-1}{10}\right)^3\\=20\times \dfrac{729}{1000}\\=14.58 \,\, \text{litres}$

Hence, 14.58 litres of milk are left in the bucket now.

Answer 2

Answer:

The quantity of milk left in the bucket is 14.58 litres.

Assuming the complete question is:

A bucket contains 20 litres of milk 2 liters of the milk were taken out from this bucket and replace with the same amount of water. If this operation is done further two more times, then how much milk is left in the container?

Step-by-step explanation:

Given:

The quantity of milk in the container = 20 litres

Amount of milk replaced with water = 2 litres

Number of times the operation repeated =1+2=3

The quantity of pure liquid remaining after replacing the liquid with other substances  $=x\Big(1-\frac{y}{x} \Big)^{n}$

Where

x is quantity of liquid

y is quantity of other substance added

n is the number of times, the operation is carried.

Substituting the given values,

$x\Big(1-\frac{y}{x} \Big)^{n}=20\Big(1-\frac{2}{20} \Big)^{3}$

$=20\Big(\frac{9}{10} \Big)^{3}$

$=20\times \frac{729}{1000}$

$= \frac{729}{50}=14.58litres$

Hence, the quantity of milk left in the bucket is 14.58 litres.