To determine how many cups of chili were left over from the second batch of chili, we start with the given equation:
[tex]\[ 3 \frac{1}{3} + x = 7 \frac{3}{4} \][/tex]
First, we need to convert the mixed numbers to improper fractions.
1. Convert [tex]\( 3 \frac{1}{3} \)[/tex] to an improper fraction:
- The whole number part is 3.
- The fractional part is [tex]\( \frac{1}{3} \)[/tex].
- [tex]\( 3 \frac{1}{3} = 3 + \frac{1}{3} = \frac{3 \cdot 3}{3} + \frac{1}{3} = \frac{9}{3} + \frac{1}{3} = \frac{10}{3} \)[/tex].
2. Convert [tex]\( 7 \frac{3}{4} \)[/tex] to an improper fraction:
- The whole number part is 7.
- The fractional part is [tex]\( \frac{3}{4} \)[/tex].
- [tex]\( 7 \frac{3}{4} = 7 + \frac{3}{4} = \frac{7 \cdot 4}{4} + \frac{3}{4} = \frac{28}{4} + \frac{3}{4} = \frac{31}{4} \)[/tex].
Now, we rewrite the original equation with the improper fractions:
[tex]\[ \frac{10}{3} + x = \frac{31}{4} \][/tex]
Next, solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{31}{4} - \frac{10}{3} \][/tex]
To subtract these fractions, we find a common denominator. The least common multiple of 3 and 4 is 12.
3. Express [tex]\(\frac{10}{3}\)[/tex] with a denominator of 12:
- [tex]\(\frac{10}{3} = \frac{10 \cdot 4}{3 \cdot 4} = \frac{40}{12} \)[/tex]
4. Express [tex]\(\frac{31}{4}\)[/tex] with a denominator of 12:
- [tex]\(\frac{31}{4} = \frac{31 \cdot 3}{4 \cdot 3} = \frac{93}{12} \)[/tex]
Now, subtract the fractions:
[tex]\[ x = \frac{93}{12} - \frac{40}{12} = \frac{93 - 40}{12} = \frac{53}{12} \][/tex]
Finally, convert the improper fraction back to a mixed number:
[tex]\[ \frac{53}{12} \][/tex]
5. Divide the numerator by the denominator:
- 53 divided by 12 is 4 with a remainder of 5.
- This gives us the mixed number [tex]\( 4 \frac{5}{12} \)[/tex].
Therefore, the number of cups of chili left over from the second batch of chili is:
[tex]\[ \boxed{4 \frac{5}{12}} \][/tex]
So the correct answer is [tex]\( B \)[/tex].
