Question

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

Explanation:

Step1: Analyze the first removal

The bag initially has \( 3 \) cups of flour. Noah first removes \( \frac{1}{3} \) cup. So we calculate the remaining flour after the first removal: \( 3-\frac{1}{3}=\frac{9}{3}-\frac{1}{3}=\frac{8}{3}=2\frac{2}{3} \) cups.

Step2: Analyze the second removal

Then Noah removes another \( 3\frac{4}{3} \) (wait, that seems like a typo, probably \( \frac{4}{3} \)) cups. Wait, maybe the second removal is \( \frac{4}{3} \) cups. Wait, the original problem: first removal \( \frac{1}{3} \), then second removal \( \frac{4}{3} \). Wait, let's re - check. The initial amount is \( 3=\frac{9}{3} \). First removal \( \frac{1}{3} \), remaining \( \frac{9 - 1}{3}=\frac{8}{3} \). Then second removal \( \frac{4}{3} \), so remaining \( \frac{8}{3}-\frac{4}{3}=\frac{4}{3}=1\frac{1}{3} \)? Wait, maybe the second removal is \( 1\frac{4}{3} \)? No, the problem says "Noah removes another \( 3\frac{4}{3} \)" which is incorrect, \( 3\frac{4}{3}=4\frac{1}{3} \), which is more than the initial \( 3 \). So maybe it's a typo, and it's \( \frac{4}{3} \). Let's assume the second removal is \( \frac{4}{3} \).

Wait, maybe the first part: initial \( 3 \) cups. First removal \( \frac{1}{3} \), so \( 3-\frac{1}{3}=\frac{9 - 1}{3}=\frac{8}{3}=2\frac{2}{3} \). Then second removal \( \frac{4}{3} \), so \( \frac{8}{3}-\frac{4}{3}=\frac{4}{3}=1\frac{1}{3} \). But maybe the second removal is \( 1\frac{1}{3} \) (since \( 3\frac{4}{3} \) is wrong). Alternatively, maybe the first question: after first removal ( \( \frac{1}{3} \) cup), remaining is \( 3-\frac{1}{3}=\frac{9 - 1}{3}=\frac{8}{3}=2\frac{2}{3} \) cups. Then after second removal ( \( \frac{4}{3} \) cups), remaining is \( 2\frac{2}{3}-\frac{4}{3}=\frac{8 + 2}{3}-\frac{4}{3} \)? No, \( 2\frac{2}{3}=\frac{8}{3} \), \( \frac{8}{3}-\frac{4}{3}=\frac{4}{3}=1\frac{1}{3} \).

Wait, maybe the problem has a typo, and the second removal is \( \frac{4}{3} \) cups. Let's proceed with the first part: when Noah removes \( \frac{1}{3} \) cup from \( 3 \) cups.

\( 3-\frac{1}{3}=\frac{9}{3}-\frac{1}{3}=\frac{8}{3}=2\frac{2}{3} \) cups left after first removal.

Then, if we consider the second removal: if the second removal is \( \frac{4}{3} \) cups, then \( 2\frac{2}{3}-\frac{4}{3}=\frac{8}{3}-\frac{4}{3}=\frac{4}{3}=1\frac{1}{3} \) cups.

But maybe the first question (how much after first removal) is \( 2\frac{2}{3} \) and after second removal is \( 1\frac{1}{3} \).

Wait, let's start over.

First scenario: Initial flour = \( 3 \) cups.

First removal: \( \frac{1}{3} \) cup.

Step 1: Calculate remaining after first removal.

We know that to subtract a fraction from a whole number, we convert the whole number to a fraction with the same denominator. \( 3=\frac{3\times3}{3}=\frac{9}{3} \).

So \( 3-\frac{1}{3}=\frac{9}{3}-\frac{1}{3}=\frac{9 - 1}{3}=\frac{8}{3} \).

Convert \( \frac{8}{3} \) to a mixed number: \( \frac{8}{3}=2\frac{2}{3} \) (since \( 8\div3 = 2\) with a remainder of \( 2 \)).

Second removal: Let's assume the second removal is \( \frac{4}{3} \) cups (since \( 3\frac{4}{3} \) is an improper mixed number, probably a typo).

Step 2: Calculate remaining after second removal.

We have \( \frac{8}{3}-\frac{4}{3}=\frac{8 - 4}{3}=\frac{4}{3} \).

Convert \( \frac{4}{3} \) to a mixed number: \( \frac{4}{3}=1\frac{1}{3} \) (since \( 4\div3 = 1\) with a remainder of \( 1 \)).

Answer:

After first removal: \( 2\frac{2}{3} \) cups; After second removal: \( 1\frac{1}{3} \) cups (assuming the second removal is \( \frac{4}{3} \) cups, correcting the likely typo in \( 3\frac{4}{3} \)).