Sure, let's break down the problem step-by-step.
1. Identify the initial volumes:
The mixture is originally 3 cups in total. This mixture is divided into flow and milk, with [tex]\(\frac{2}{5}\)[/tex] of it being flow and [tex]\(\frac{3}{5}\)[/tex] of it being milk.
2. Calculate the volume of flow and milk in the original mixture:
- Volume of flow = [tex]\(\frac{2}{5}\)[/tex] of 3 cups:
[tex]\[
\text{Volume of flow} = 3 \times \frac{2}{5} = 1.2 \text{ cups}
\][/tex]
- Volume of milk = [tex]\(\frac{3}{5}\)[/tex] of 3 cups:
[tex]\[
\text{Volume of milk} = 3 \times \frac{3}{5} = 1.8 \text{ cups}
\][/tex]
3. Add 1 cup of milk to the mixture:
With the addition of 1 cup of milk:
[tex]\[
\text{New volume of milk} = 1.8 + 1 = 2.8 \text{ cups}
\][/tex]
4. Calculate the new total volume of the mixture:
The original mixture was 3 cups, and we added 1 more cup of milk. So:
[tex]\[
\text{New total volume} = 3 + 1 = 4 \text{ cups}
\][/tex]
5. Determine the percentage of milk in the new mixture:
The volume of milk in the new mixture is 2.8 cups out of the total 4 cups.
[tex]\[
\text{Percentage of milk} = \left( \frac{2.8}{4} \right) \times 100\% = 70\%
\][/tex]
Hence, when 1 cup of milk is added to a 3-cup mixture that is [tex]\(\frac{2}{5}\)[/tex] flow and [tex]\(\frac{3}{5}\)[/tex] milk, the percentage of the 4-cup mixture that is milk is [tex]\(70\%\)[/tex].
