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2015 AMC 10B Problem 7

2015 AMC 10B Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

Functions · difficulty 1 of 5 · about 2 min · position P1-10

Consider the operation "minus the reciprocal of," defined by $a\diamond b=a-\frac{1}{b}$ . What is $((1\diamond2)\diamond3)-(1\diamond(2\diamond3))$ ?

A)
$-\dfrac{7}{30}$
B)
$-\dfrac{1}{6}$
C)
$0$
D)
$\dfrac{1}{6}$
E)
$\dfrac{7}{30}$

Answer

Key insight

Evaluate inside out: (1 diamond 2) diamond 3 = 1/2 - 1/3 = 1/6, and 1 diamond (2 diamond 3) = 1 - 3/5 = 2/5; the difference is -7/30.

Solution

Work from the innermost parentheses outward.

Left expression: $1 \diamond 2 = 1 - \tfrac12 = \tfrac12$ , then
$\tfrac12 \diamond 3 = \tfrac12 - \tfrac13 = \tfrac16.$

Right expression: $2 \diamond 3 = 2 - \tfrac13 = \tfrac53$ , then
$1 \diamond \tfrac53 = 1 - \frac{1}{5/3} = 1 - \tfrac35 = \tfrac25.$

Difference:
$\tfrac16 - \tfrac25 = \tfrac{5}{30} - \tfrac{12}{30} = -\tfrac{7}{30}.$

The answer is $\boxed{\textbf{(A)}\ -\frac{7}{30}}$ .

Why this works

A custom operation is just a formula: substitute carefully, one layer at a time. The point of the problem is that $\diamond$ is not associative, so the two groupings differ. Keep every intermediate value as an exact fraction; the reciprocal of $\tfrac53$ is $\tfrac35$ , not $\tfrac53$ .

The trap

Computing 1 diamond (5/3) as 1 - 5/3 instead of 1 minus the reciprocal 3/5.

Common mistakes

Computing $1 \diamond \tfrac53$ as $1 - \tfrac53$ instead of $1$ minus the reciprocal $\tfrac35$ .
Subtracting in the wrong order and reporting $+\tfrac{7}{30}$ , choice (E).

Techniques

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