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2010 AMC 10A Problem 6

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2010 AMC 10A Problem 6AMC 10 Step by Step
Name:                             Date: September 13, 2026
Functions · difficulty 1 of 5 · about 1 min · position P1-10

For positive numbers xx and yy the operation x♠yx\spadesuit y is defined as

x♠y=x−1yx\spadesuit y = x -\dfrac{1}{y}

What is 2♠(2♠2)2\spadesuit (2\spadesuit 2) ?

  1. A)

    23\dfrac{2}{3}

  2. B)

    11

  3. C)

    43\dfrac{4}{3}

  4. D)

    53\dfrac{5}{3}

  5. E)

    22

Answer

C

Key insight

Evaluate the inner operation first: 2 spade 2 = 3/2, then 2 spade (3/2) = 2 - 2/3 = 4/3.

Solution

Work from the inside out. The inner expression is
2♠2=2−12=32.2\spadesuit 2 = 2 - \frac{1}{2} = \frac{3}{2}.

Now substitute that as the second input:
2♠32=2−13/2=2−23=43.2\spadesuit \frac{3}{2} = 2 - \frac{1}{3/2} = 2 - \frac{2}{3} = \frac{4}{3}.

The answer is (C) 43\boxed{\textbf{(C)}\ \dfrac{4}{3}}.

Why this works

A made-up operation is just a two-variable function; parentheses tell you which evaluation happens first, and the result feeds into the outer call as the yy input. The only real hazard is the reciprocal of a fraction, so flip carefully.

The trap

Computing 1/(3/2) as 3/2 instead of 2/3, which gives 1/2, or evaluating the operations left to right.

Common mistakes

  • Computing 1/(3/2) as 3/2 instead of 2/3, which gives 1/2, or evaluating the operations left to right.
  • Plugging 32\tfrac32 into the xx slot instead of the yy slot, giving 32−12=1\tfrac32 - \tfrac12 = 1, choice (B).

Techniques

Set up the equation/formula and compute; no special trick needed