Question

A barrel contains 300 litres of pure wine. 30 litres was withdrawn from the barrel and replaced with water. This process is repeated two more times. Find the concentration of the wine in resultant solution.

Answer 1

Given:

A barrel contains 300 litres of pure wine.

30 litres was withdrawn from the barrel and replaced with water.

This process is repeated two more times.

To find:

The concentration of the wine in resultant solution.

Solution:

To solve the above-given problem we will use the following formula:

$\boxed{\bold{Final\:Quantity = x [1 - \frac{y}{x} ]^n}}$

where,

x = Initial quantity of pure liquid (wine, milk etc) in the vessel

y = quantity taken out at one time

n = no. of times the process is repeated

Here we have,

The initial quantity of pure wine in the barrel, x = 300 litres

The quantity of wine withdrawn and replaced with water, y = 30 litres

The no. of time this process is repeated, n = 3

Now, on substituting the values of x, y & n in the above formula, we get

The quantity of wine left in the barrel after 3 operations are,

=  $300 [1 - \frac{30}{300} ]^3$

= $300 [\frac{300 - 30}{300} ]^3$

= $300 [\frac{270}{300} ]^3$

= $300 \times [\frac{9}{10} ]^3$

= $\frac{3 \times 9^3}{10}$

= $\bold{218.7\:litres }$

Thus, the concentration of the wine in the resultant solution is → 218.7 litres.

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Answer 2

Given:

A barrel contains 300 litres of pure wine.

30 litres was withdrawn from the barrel and replaced with water.

This process is repeated two more times.

To find:

The concentration of the wine in resultant solution.

Solution:

To solve the above-given problem we will use the following formula:

$\boxed{\bold{Final\:Quantity = x [1 - \frac{y}{x} ]^n}}$

where,

x = Initial quantity of pure liquid (wine, milk etc) in the vessel

y = quantity taken out at one time

n = no. of times the process is repeated

Here we have,

The initial quantity of pure wine in the barrel, x = 300 litres

The quantity of wine withdrawn and replaced with water, y = 30 litres

The no. of time this process is repeated, n = 3

Now, on substituting the values of x, y & n in the above formula, we get

The quantity of wine left in the barrel after 3 operations are,

=  $300 [1 - \frac{30}{300} ]^3$

= $300 [\frac{300 - 30}{300} ]^3$

= $300 [\frac{270}{300} ]^3$

= $300 \times [\frac{9}{10} ]^3$

= $\frac{3 \times 9^3}{10}$

= $\bold{218.7\:litres }$

Thus, the concentration of the wine in the resultant solution is → 218.7 litres.

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