AMC 10 · 2010 · #6
Grade 6 arithmeticFor positive numbers x and y the operation ♠(x,y) is defined as
♠(x,y)=x−y1
What is ♠(2,♠(2,2))?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A made-up operation $\spadesuit(x,y)$ means "take $x$ and subtract $\dfrac{1}{y}$." We need the value of $\spadesuit(2,\spadesuit(2,2))$, where one $\spadesuit$ is sitting inside another.
Givens: The rule $\spadesuit(x,y) = x - \dfrac{1}{y}$ for positive numbers $x$ and $y$.; The expression to evaluate is $\spadesuit(2,\spadesuit(2,2))$.
Unknowns: The single number that $\spadesuit(2,\spadesuit(2,2))$ equals.
Understand
Restated: A made-up operation $\spadesuit(x,y)$ means "take $x$ and subtract $\dfrac{1}{y}$." We need the value of $\spadesuit(2,\spadesuit(2,2))$, where one $\spadesuit$ is sitting inside another.
Givens: The rule $\spadesuit(x,y) = x - \dfrac{1}{y}$ for positive numbers $x$ and $y$.; The expression to evaluate is $\spadesuit(2,\spadesuit(2,2))$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
The operation is nested, so the outer $\spadesuit$ depends on a number we do not have yet. Split it into the inner subproblem $\spadesuit(2,2)$ first, plug that result into the outer $\spadesuit$, and match the final value to one of the five choices.
Execute — Answer: C
6.EE.A.2 Step 1 Read the rule, spot the inner piece
The operation is just a recipe: $\spadesuit(x,y)$ says "first number minus one over the second number." In $\spadesuit(2,\spadesuit(2,2))$ the second slot is itself a $\spadesuit$, so the inner $\spadesuit(2,2)$ has to be computed before anything else.
💡 You cannot use a number you have not found yet, so the inside comes first.
5.NF.A.1 Step 2 Compute the inner spade
- Put $x=2$ and $y=2$ into the rule.
- This gives $2 - \dfrac{1}{2}$.
- Since $2 = \dfrac{4}{2}$, subtracting $\dfrac{1}{2}$ leaves $\dfrac{3}{2}$.
💡 Rewrite the whole number as halves so both parts share the same denominator.
6.NS.A.1 Step 3 Set up the outer spade
- Now the expression is $\spadesuit\left(2,\dfrac{3}{2}\right)$.
- Apply the rule with $x=2$ and $y=\dfrac{3}{2}$: this is $2 - \dfrac{1}{3/2}$.
- Dividing $1$ by $\dfrac{3}{2}$ means multiplying by its reciprocal, so $\dfrac{1}{3/2} = \dfrac{2}{3}$.
💡 One divided by a fraction just flips the fraction over.
5.NF.A.1 Step 4 Finish the subtraction
- Compute $2 - \dfrac{2}{3}$.
- Write $2 = \dfrac{6}{3}$, then $\dfrac{6}{3} - \dfrac{2}{3} = \dfrac{4}{3}$.
- That matches choice (C).
💡 Same trick as before: give both terms thirds, then subtract the tops.
6.EE.A.2 The operation is just a recipe: $\spadesuit(x,y)$ says "first number minus one o 5.NF.A.1 Put $x=2$ and $y=2$ into the rule. This gives $2 - \dfrac{1}{2}$. Since $2 = \df 6.NS.A.1 Now the expression is $\spadesuit\left(2,\dfrac{3}{2}\right)$. Apply the rule wi 5.NF.A.1 Compute $2 - \dfrac{2}{3}$. Write $2 = \dfrac{6}{3}$, then $\dfrac{6}{3} - \dfra Review
Reasonableness: The final value $\dfrac{4}{3}$ is a little more than $1$, which fits: each $\spadesuit$ starts at $2$ and subtracts a positive amount less than $2$, so the results should stay between $0$ and $2$. Both $\dfrac{3}{2}$ and $\dfrac{4}{3}$ land in that range, and $\dfrac{4}{3}$ is exactly choice (C).
Alternative: You could keep everything as one fraction: $\spadesuit(2,2)=\dfrac{3}{2}$, then the outer becomes $2 - \dfrac{2}{3}$ directly. Either way the reciprocal step $\dfrac{1}{3/2}=\dfrac{2}{3}$ is the key move; skipping it (writing $\dfrac{1}{3/2}=\dfrac{3}{2}$) would wrongly give $\dfrac{1}{2}$, which is not even a choice.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Reading the defined operation $\spadesuit(x,y)=x-\dfrac{1}{y}$ as a formula and substituting numbers into its slots.)5.NF.A.1Add and subtract fractions with unlike denominators (Computing $2-\dfrac{1}{2}=\dfrac{3}{2}$ and $2-\dfrac{2}{3}=\dfrac{4}{3}$ by rewriting whole numbers with a common denominator.)6.NS.A.1Interpret and compute quotients of fractions and solve word problems (Evaluating $\dfrac{1}{3/2}$ as $1$ divided by $\dfrac{3}{2}$, i.e. multiplying by the reciprocal to get $\dfrac{2}{3}$.)
⭐ When one operation sits inside another, solve the inside first, and remember that one over a fraction just flips it.
⭐ When one operation sits inside another, solve the inside first, and remember that one over a fraction just flips it.
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