1. EASY

\(\frac{1}{2}\div\frac{1}{2}=\)

  1. \(\frac{1}{4}\)
  2. 1
  3. \(\frac{2}{4}\)
  4. \(\frac{0}{0}\)
  5. None of the above
 
2. EASY

\(\frac{1}{2}\div\frac{1}{4}=\)

  1. \(\frac{1}{8}\)
  2. \(\frac{1}{4}\)
  3. \(\frac{1}{2}\)
  4. 1
  5. None of the above
 
3. EASY

\(\frac{4}{3}\div\frac{1}{3}=\)

 
4. MEDIUM

\(\frac{1}{8}\div\frac{1}{4}=\)

  1. \(\frac{1}{2}\)
  2. \(\frac{1}{8}\)
  3. 2
  4. \(\frac{1}{4}\)
  5. None of the above
 
5. MEDIUM

\(\frac{1}{3}\div\frac{5}{6}=\)

  1. \(\frac{8}{15}\)
  2. \(\frac{2}{5}\)
  3. \(\frac{5}{16}\)
  4. \(\frac{7}{15}\)
  5. None of the above
 
6. MEDIUM

\(\frac{6}{7}\div\frac{3}{2}=\)

 
7. MEDIUM

\(\frac{7}{4}\div\frac{4}{7}=\)

 
8. HARD

\(\frac{5}{7}\div\frac{4}{9}=\)

  1. \(\frac{20}{63}\)
  2. \(\frac{28}{45}\)
  3. \(\frac{63}{20}\)
  4. \(\frac{45}{28}\)
  5. None of the above
 
9. HARD

\(\frac{3}{13}\div\frac{5}{8}=\)

 
10. HARD

Show that in the diagram below, \(|AX|\cdot|XB|=|CX|\cdot|XD|\). This is known as the Power of a Point Theorem.

  • foo
  • bar
  • baz
  • qux