AMC 8 Daily Practice Round 6

Complete problem set with solutions and individual problem pages

Problem 8 Easy

What is the value of \frac{3}{4} \times \frac{8}{9} \times \frac{15}{16} \times \frac{24}{25} \times \frac{35}{36} \times \frac{48}{49}?

  • A.

    \frac{3}{4}

  • B.

    \frac{8}{25}

  • C.

    \frac{1}{2}

  • D.

    \frac{4}{7}

  • E.

    \frac{48}{49}

Answer:D

By observation, we notice that all denominators are perfect squares.

We can attempt the following transformation: \frac{3}{4} \times \frac{8}{9} \times \frac{15}{16} \times \frac{24}{25} \times \frac{35}{36} \times \frac{48}{49}=\frac{1 \times 3}{2 \times 2} \times \frac{2 \times 4}{3 \times 3} \times \frac{3 \times 5}{4 \times 4} \times \frac{4 \times 6}{5 \times 5} \times \frac{5 \times 7}{6 \times 6} \times \frac{6 \times 8}{7 \times 7} = \frac{1}{2} \times \frac{3}{2} \times \frac{2}{3} \times \frac{4}{3} \times \frac{3}{4} \times \frac{5}{4} \times \frac{4}{5} \times \frac{6}{5} \times \frac{5}{6} \times \frac{7}{6} \times \frac{6}{7} \times \frac{8}{7}

Now we group terms to make the cancellation explicit: = \frac{1}{2} \times \left(\frac{3}{2} \times \frac{2}{3}\right) \times \left(\frac{4}{3} \times \frac{3}{4}\right) \times \left(\frac{5}{4} \times \frac{4}{5}\right) \times \left(\frac{6}{5} \times \frac{5}{6}\right) \times \left(\frac{7}{6} \times \frac{6}{7}\right) \times \frac{8}{7}

Each pair in parentheses cancels out completely: = \frac{1}{2} \times 1 \times 1 \times 1 \times 1 \times 1 \times \frac{8}{7}

After all cancellations, we are left with: = \frac{1}{2} \times \frac{8}{7} = \frac{4}{7}

Final result: \boxed{\frac{4}{7}}