AMC 10 · 2013 · #1
Grade 5 arithmeticWhat is 1+3+52+4+6−2+4+61+3+5?
Pick an answer.
AMC 10 2013 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: Add the numbers in each group, form the two fractions $\dfrac{2+4+6}{1+3+5}$ and $\dfrac{1+3+5}{2+4+6}$, then subtract the second from the first.
Givens: The expression $\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6}$; Top group: $2+4+6$. Bottom group: $1+3+5$; Answer choices: (A) $-1$, (B) $\dfrac{5}{36}$, (C) $\dfrac{7}{12}$, (D) $\dfrac{49}{20}$, (E) $\dfrac{43}{3}$
Unknowns: The single number the whole expression equals
Understand
Restated: Add the numbers in each group, form the two fractions $\dfrac{2+4+6}{1+3+5}$ and $\dfrac{1+3+5}{2+4+6}$, then subtract the second from the first.
Givens: The expression $\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6}$; Top group: $2+4+6$. Bottom group: $1+3+5$; Answer choices: (A) $-1$, (B) $\dfrac{5}{36}$, (C) $\dfrac{7}{12}$, (D) $\dfrac{49}{20}$, (E) $\dfrac{43}{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
The expression is built from small pieces, so Tool #7 (Identify Subproblems) breaks it into three easy jobs: add each group of numbers, reduce the two fractions, then subtract. Tool #3 (Eliminate Possibilities) uses the shape of the problem as a guardrail — the first fraction is bigger than 1 and the second is smaller than 1, so their difference is a small positive number, which throws out the negative choice (A) and the large choices (D) and (E) before any careful arithmetic.
Execute — Answer: C
2.NBT.B.5 Step 1 Add the numbers in each group
- Work out the two sums that keep repeating.
- The top group is $2+4+6=12$ and the bottom group is $1+3+5=9$.
- So the expression becomes $\dfrac{12}{9}-\dfrac{9}{12}$.
- Notice the same two numbers, $12$ and $9$, just trade places between the fractions.
💡 Doing the repeated addition once turns a scary-looking expression into two plain fractions.
4.NF.A.1 Step 2 Reduce each fraction
- Simplify to make the subtraction easier.
- Divide top and bottom of $\dfrac{12}{9}$ by $3$ to get $\dfrac{4}{3}$, and divide top and bottom of $\dfrac{9}{12}$ by $3$ to get $\dfrac{3}{4}$.
- The problem is now $\dfrac{4}{3}-\dfrac{3}{4}$, and the two fractions are flips of each other.
💡 Dividing the top and bottom by the same number keeps a fraction's value but makes the digits smaller.
5.NF.A.1 Step 3 Subtract using a common denominator
- To subtract $\dfrac{4}{3}-\dfrac{3}{4}$, rewrite both over the shared denominator $12$: $\dfrac{4}{3}=\dfrac{16}{12}$ and $\dfrac{3}{4}=\dfrac{9}{12}$.
- Now subtract the tops: $\dfrac{16}{12}-\dfrac{9}{12}=\dfrac{7}{12}$.
- This is already in lowest terms and matches choice (C); the earlier size check had already ruled out (A), (D), and (E).
💡 Fractions can only be subtracted once they are cut into the same-size pieces, so match denominators first.
2.NBT.B.5 Work out the two sums that keep repeating. The top group is $2+4+6=12$ and the b 4.NF.A.1 Simplify to make the subtraction easier. Divide top and bottom of $\dfrac{12}{9} 5.NF.A.1 To subtract $\dfrac{4}{3}-\dfrac{3}{4}$, rewrite both over the shared denominato Review
Reasonableness: The first fraction $\tfrac43\approx1.33$ is a bit above $1$ and the second $\tfrac34=0.75$ is a bit below $1$, so their difference should be a small positive number near $0.5$. The result $\tfrac{7}{12}\approx0.58$ fits exactly. A negative answer like (A) is impossible because the bigger fraction is in front, and the large choices (D) $2.45$ and (E) $\approx14.3$ are far too big.
Alternative: Use decimals as a check. $\tfrac{12}{9}=1.\overline{3}$ and $\tfrac{9}{12}=0.75$, so $1.333\ldots-0.75=0.583\ldots$. Comparing the choices, $\tfrac{7}{12}=0.58\overline{3}$ matches, while $\tfrac{5}{36}\approx0.14$ and $\tfrac{49}{20}=2.45$ do not, confirming (C).
CCSS standards used (min grade 5)
2.NBT.B.5Fluently add and subtract within 100 (Adding each group of numbers: $2+4+6=12$ and $1+3+5=9$.)4.NF.A.1Explain why a fraction is equivalent to another fraction (Reducing $\dfrac{12}{9}$ to $\dfrac{4}{3}$ and $\dfrac{9}{12}$ to $\dfrac{3}{4}$ by dividing top and bottom by $3$.)5.NF.A.1Add and subtract fractions with unlike denominators (Subtracting $\dfrac{4}{3}-\dfrac{3}{4}=\dfrac{16}{12}-\dfrac{9}{12}=\dfrac{7}{12}$ over the common denominator $12$.)
⭐ Add the repeated groups first, then subtract the two fractions over a common denominator — a bigger-than-one fraction minus a smaller-than-one fraction leaves a small positive number.
⭐ Add the repeated groups first, then subtract the two fractions over a common denominator — a bigger-than-one fraction minus a smaller-than-one fraction leaves a small positive number.
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