Pre-algebra math worksheet with fraction division problems and a joke puzzle.

To find the answer to the riddle, we need to solve each division problem and match the result to the code at the bottom of the page.

Here is the step-by-step solution for each letter:

1. Solve the math problems:

* (S) $-\frac{3}{7} \div \frac{1}{2} = -\frac{3}{7} \times \frac{2}{1} = \mathbf{-\frac{6}{7}}$
* (T) $\frac{25}{3} \div \frac{15}{6} = \frac{25}{3} \times \frac{6}{15}$. Simplify before multiplying: $\frac{5}{1} \times \frac{2}{3} = \mathbf{\frac{10}{3}}$
* (N) $\frac{-7}{10} \div 7 = \frac{-7}{10} \times \frac{1}{7} = \mathbf{-\frac{1}{10}}$
* (C) $\frac{-4}{5} \div \frac{-4}{3} = \frac{-4}{5} \times \frac{3}{-4}$. The negatives cancel out and the 4s cancel out: $\mathbf{\frac{3}{5}}$
* (G) $\frac{-7}{9} \div \frac{-21}{6} = \frac{-7}{9} \times \frac{6}{-21}$. Negatives become positive. Simplify: $\frac{1}{3} \times \frac{2}{3} = \mathbf{\frac{2}{9}}$
* (L) $\frac{11}{12} \div \frac{-33}{8} = \frac{11}{12} \times \frac{8}{-33}$. Simplify: $\frac{1}{3} \times \frac{2}{-3} = \mathbf{-\frac{2}{9}}$
* (Y) $\frac{5}{8} \div \frac{-7}{12} = \frac{5}{8} \times \frac{12}{-7}$. Simplify (divide 8 and 12 by 4): $\frac{5}{2} \times \frac{3}{-7} = \mathbf{-\frac{15}{14}}$
* (I) $\frac{-11}{4} \div \frac{2}{3} = \frac{-11}{4} \times \frac{3}{2} = \mathbf{-\frac{33}{8}}$
* (P) $\frac{-45}{4} \div \frac{-15}{16} = \frac{-45}{4} \times \frac{16}{-15}$. Negatives cancel. Simplify ($45/15=3$ and $16/4=4$): $3 \times 4 = \mathbf{12}$
* (A) $\frac{-15}{13} \div \frac{-10}{11} = \frac{-15}{13} \times \frac{11}{-10}$. Negatives cancel. Simplify ($15/10 = 3/2$): $\frac{3}{13} \times \frac{11}{2} = \mathbf{\frac{33}{26}}$
* (V) $\frac{-9}{10} \div \frac{-12}{5} = \frac{-9}{10} \times \frac{5}{-12}$. Negatives cancel. Simplify ($5/10=1/2$ and $9/12=3/4$): $\frac{3}{2} \times \frac{1}{4} = \mathbf{\frac{3}{8}}$
* (H) $\frac{6}{20} \div \frac{7}{10} = \frac{6}{20} \times \frac{10}{7}$. Simplify ($10/20=1/2$): $\frac{6}{2} \times \frac{1}{7} = \frac{3}{1} \times \frac{1}{7} = \mathbf{\frac{3}{7}}$
* (E) $\frac{5}{6} \div \frac{-7}{8} = \frac{5}{6} \times \frac{8}{-7}$. Simplify ($8/6 = 4/3$): $\frac{5}{3} \times \frac{4}{-7} = \mathbf{-\frac{20}{21}}$
* (O) $3 \div \frac{-2}{3} = \frac{3}{1} \times \frac{3}{-2} = \mathbf{-\frac{9}{2}}$
* (R) $-10 \div \frac{-4}{7} = \frac{-10}{1} \times \frac{7}{-4}$. Negatives cancel. Simplify ($10/4 = 5/2$): $\frac{5}{1} \times \frac{7}{2} = \mathbf{\frac{35}{2}}$
* (D) $\frac{-8}{9} \div 2 = \frac{-8}{9} \times \frac{1}{2} = \mathbf{-\frac{4}{9}}$
* (U) $\frac{4}{15} \div \frac{-14}{5} = \frac{4}{15} \times \frac{5}{-14}$. Simplify ($5/15=1/3$ and $4/14=2/7$): $\frac{2}{3} \times \frac{1}{-7} = \mathbf{-\frac{2}{21}}$
* (F) $\frac{-48}{9} \div \frac{16}{21} = \frac{-48}{9} \times \frac{21}{16}$. Simplify ($48/16=3$ and $21/9=7/3$): $\frac{-3}{1} \times \frac{7}{3} = \mathbf{-7}$

2. Decode the message:

Now we look at the numbers in the code boxes at the bottom and replace them with their corresponding letters.

* First Line:
* $\frac{3}{7}$ → H
* $\frac{-2}{21}$ → U
* $\frac{2}{9}$ → G
* $\frac{3}{7}$ → H
* $\frac{33}{26}$ → A
* $\frac{-1}{10}$ → N
* $\frac{-4}{9}$ → D
* $\frac{-9}{2}$ → O
* $\frac{-1}{10}$ → N
* $\frac{-2}{9}$ → L
* $\frac{-15}{14}$ → Y
* *(Word: HUGH AND ONLY)*

* Second Line:
* $\frac{3}{7}$ → H
* $\frac{-2}{21}$ → U
* $\frac{2}{9}$ → G
* $\frac{3}{7}$ → H
* $\frac{3}{5}$ → C
* $\frac{33}{26}$ → A
* $\frac{-1}{10}$ → N
* $12$ → P
* $\frac{35}{2}$ → R
* $\frac{-20}{21}$ → E
* $\frac{3}{8}$ → V
* $\frac{-20}{21}$ → E
* $\frac{-1}{10}$ → N
* $\frac{10}{3}$ → T
* *(Word: HUGH CAN PREVENT)*

* Third Line:
* $-7$ → F
* $\frac{-2}{9}$ → L
* $\frac{-9}{2}$ → O
* $\frac{35}{2}$ → R
* $\frac{-33}{8}$ → I
* $\frac{-6}{7}$ → S
* $\frac{10}{3}$ → T
* $-7$ → F
* $\frac{35}{2}$ → R
* $\frac{-33}{8}$ → I
* $\frac{33}{26}$ → A
* $\frac{35}{2}$ → R
* $\frac{-6}{7}$ → S
* *(Word: FLORIST FRIARS)*

Final Answer:
HUGH AND ONLY HUGH CAN PREVENT FLORIST FRIARS