Question

vincent had a pitcher containing $6\frac{2}{3}$ cups of lemonade, but he accidentally spilled $4\frac{1}{3}$ cups while carrying the pitcher outside. luckily, he had an extra 3 cups of lemonade in another container, so he poured that into the pitcher. how much lemonade is in the pitcher now? \\(\frac{\square}{\square}\\) cups

Explanation:

Step1: Calculate remaining after spill

$6\frac{2}{3} - 4\frac{1}{3} = \frac{20}{3} - \frac{13}{3} = \frac{7}{3}$

Step2: Add extra lemonade

$\frac{7}{3} + 3 = \frac{7}{3} + \frac{9}{3} = \frac{16}{3}$? No, wait: $6\frac{2}{3}$ is $\frac{20}{3}$, $4\frac{1}{3}$ is $\frac{13}{3}$, so $\frac{20}{3}-\frac{13}{3}=\frac{7}{3}$. Then add 3 cups: $\frac{7}{3}+3=\frac{7}{3}+\frac{9}{3}=\frac{16}{3}$? Wait no, original calculation: $6\frac{2}{3}-4\frac{1}{3}=2\frac{1}{3}$, then $2\frac{1}{3}+3=5\frac{1}{3}=\frac{16}{3}$. Oh right, $\frac{16}{3}$ is the correct answer. I made a typo earlier. So final answer is $\frac{16}{3}$ cups.

Corrected:

Step1: Subtract spilled amount

$6\frac{2}{3} - 4\frac{1}{3} = \frac{20}{3} - \frac{13}{3} = \frac{7}{3}$

Step2: Add extra lemonade

$\frac{7}{3} + 3 = \frac{7}{3} + \frac{9}{3} = \frac{16}{3}$

Answer:

$\frac{14}{3}$ cups