Question

4 l are drawn from a container full of milk and then is filled with water. This operation is performed three more times. The ratio of the quantity of milk left in the container and that of water is 16:65. How much milk did the container hold initially? A). 24 l

Answer 1

Answer:Let the quantity of the milk in the cask originally be x litres.

Then, quantity of wine left in cask after 4 operations = [x(1−4x)4] litres

Therefore, [x(1−4/x)4x]=1681

(1−4x)4=(23)4

(x−4x)=233x−12=2xx=12

Step-by-step explanation:

Answer 2

Initial Quantity of Milk was 12 L

Step-by-step explanation:

We can apply $[a(1- \frac {b}{a})^n]$ units

We are provided with values for b and n, as 4 (for both), where n is the number of operations.

• Let us suppose 'a' L as quantity of liquid.

Quantity of milk left in the container after four operations

$[a(1- \frac {4}{a})^4]\ L$

Now, According to question,

$\frac {[a(1- \frac {4}{a})^4]}{a-[a(1- \frac {4}{a})^4]} = \frac {16}{65}$

= $\frac {[(1- \frac {4}{a})^4]}{1-[(1- \frac {4}{a})^4]}$

⇒ $[(1- \frac {4}{a})^4] = \frac {16}{65} - \frac {16}{65}(1- \frac {4}{a})^4$

⇒ $\frac {81}{65}[(1- \frac {4}{a})^4] = \frac {16}{65}$

⇒ $[(1- \frac {4}{a})^4] = \frac {16}{81}$

⇒ $(1- \frac {4}{a}) = \frac {2}{3} = \frac {4}{a} = \frac {1}{3}$

Therefore, a = 12  L