Dracula is painting pictures. He has \( 4 \frac{3}{4} \) containers of paint remaining. Each picture requires \( 2 \frac{1}{12} \) of a container of paint. How many pictures can Dracula complete with the remaining paint?

To find out how many pictures Dracula can complete, we first need to convert the mixed numbers into improper fractions. For the paint he has: \[ 4 \frac{3}{4} = 4 \times 4 + 3 = 16 + 3 = 19 \rightarrow \frac{19}{4} \] For the paint needed for each picture: \[ 2 \frac{1}{12} = 2 \times 12 + 1 = 24 + 1 = 25 \rightarrow \frac{25}{12} \] Now, we divide the total amount of paint by the amount needed for one picture: \[ \frac{19}{4} \div \frac{25}{12} = \frac{19}{4} \times \frac{12}{25} = \frac{19 \times 12}{4 \times 25} = \frac{228}{100} = \frac{57}{25} \] To convert \( \frac{57}{25} \) back to a mixed number: - 25 goes into 57 two times (because \( 2 \times 25 = 50 \)). - The remainder is \( 57 - 50 = 7 \). So, \( \frac{57}{25} = 2 \frac{7}{25} \). Dracula can complete **2 pictures** fully, with some paint left over!