AMC 8 Daily Practice Round 11

Complete problem set with solutions and individual problem pages

Problem 9 Medium

A new operation is defined as:

a \Theta b = \frac{a+b}{2}

Given the equation:

\frac{3}{4} \Theta \left( \frac{1}{6} \Theta \square \right) = \frac{1}{2}

What is the value that should be placed in \square?

  • A.

    3

  • B.

    \frac{1}{3}

  • C.

    \frac{13}{24}

  • D.

    4

  • E.

    \frac{1}{4}

Answer:B

Observing the new operation, it calculates the average of two numbers.

First, treating the parentheses as a whole, the average of the number inside the parentheses and \frac{3}{4} equals \frac{1}{2}, meaning:

\frac{1}{6} \Theta \square = 1 - \frac{3}{4} = \frac{1}{4}.

Thus, the average of \square and \frac{1}{6} equals \frac{1}{4}, leading to:

\square = \frac{1}{4} \times 2 - \frac{1}{6} = \frac{1}{3}.

Thus, the correct answer is B.

Related Problems
Problem 7 AMC 8 Daily Practice Round 11
Problem 8 AMC 8 Daily Practice Round 11
Problem 10 AMC 8 Daily Practice Round 11
Problem 11 AMC 8 Daily Practice Round 11
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