Question

a bag contains 5 cups of rice. teagan removes \\(\frac{2}{3}\\) cup of rice for a meal. how much rice is left in the bag? write your answer as a mixed number. 4 1 3 cups teagan removes another \\(\frac{2}{3}\\) cup of rice from the bag. how much rice is left now? write your answer as a mixed number.? ? ? cups

Explanation:

Step1: Analyze the first removal

The bag initially has 5 cups of rice. Teagan first removes $\frac{2}{3}$ cup. So we calculate $5 - \frac{2}{3}$. Convert 5 to a fraction with denominator 3: $5=\frac{15}{3}$. Then $\frac{15}{3}-\frac{2}{3}=\frac{13}{3}=4\frac{1}{3}$ cups left after first removal.

Step2: Analyze the second removal

Then Teagan removes another $\frac{2}{3}$ cup. Now we take the remaining $4\frac{1}{3}$ (which is $\frac{13}{3}$) and subtract $\frac{2}{3}$. So $\frac{13}{3}-\frac{2}{3}=\frac{11}{3}=3\frac{2}{3}$ cups left after second removal. Wait, but maybe the first part: initial 5 cups, first remove $\frac{2}{3}$, so $5 - \frac{2}{3}=\frac{15 - 2}{3}=\frac{13}{3}=4\frac{1}{3}$ (first "how much left" after first removal). Then second removal: $4\frac{1}{3}-\frac{2}{3}=\frac{13}{3}-\frac{2}{3}=\frac{11}{3}=3\frac{2}{3}$. But let's check the first problem: "A bag contains 5 cups of rice. Teagan removes $\frac{2}{3}$ cup for a meal. How much rice is left in the bag?". So that's $5-\frac{2}{3}=\frac{15}{3}-\frac{2}{3}=\frac{13}{3}=4\frac{1}{3}$. Then the next: "Teagan removes another $\frac{2}{3}$ cup... How much left now?" So $4\frac{1}{3}-\frac{2}{3}=3\frac{2}{3}$.

Answer:

First remaining: $4\frac{1}{3}$ cups. Second remaining: $3\frac{2}{3}$ cups. (Assuming two sub - questions: first after one removal, second after two removals. If only one, clarify, but based on text: first "How much rice is left..." after first removal: $4\frac{1}{3}$; then after second removal: $3\frac{2}{3}$)