This problem involves finding the initial quantity of a mixture based on initial and final ratios after adding specific amounts of components.
We are given the initial ratio of milk to water in a pot and the ratio after adding equal amounts of milk and water. We need to determine the total initial quantity.
Let the initial quantities of milk and water be $3x$ litres and $4x$ litres, respectively, based on the given ratio of 3:4.
The initial total quantity of the mixture is the sum of milk and water quantities:
$ \text{Initial Total Quantity} = 3x + 4x = 7x \text{ litres} $According to the problem, 10 litres of milk and 10 litres of water are added to the mixture.
The new ratio of milk to water is given as 4:5.
Therefore, we can set up the equation:
$ \frac{3x + 10}{4x + 10} = \frac{4}{5} $Cross-multiply the terms to solve for $x$:
$ 5(3x + 10) = 4(4x + 10) $Expand both sides:
$ 15x + 50 = 16x + 40 $Rearrange the terms to isolate $x$:
$ 50 - 40 = 16x - 15x $ $ 10 = x $So, the value of $x$ is 10.
Substitute the value of $x = 10$ back into the expression for the initial total quantity ($7x$):
$ \text{Initial Total Quantity} = 7x = 7 \times 10 = 70 \text{ litres} $The initial quantity of the total mixture was 70 litres.
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