AMC 8 · 2010 · #13
Easy mode Grade 6Problem
Imagine a triangle. The three side lengths are whole numbers in inches, and they are three numbers in a row — like 4,5,6 or 7,8,9.
Add the three side lengths together to get the perimeter. For this triangle, the shortest side is exactly 30% of the perimeter.
How long is the longest side, in inches?
Pick an answer.
AMC 8 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 triangle has three side lengths (in inches) that are three consecutive integers. The shortest side is exactly $30\%$ of the perimeter. Find the length of the longest side.
Givens: Side lengths are three consecutive integers, e.g. $n,\;n+1,\;n+2$; Shortest side $= 30\% = \tfrac{3}{10}$ of the perimeter; Perimeter $=$ sum of the three side lengths; Answer choices for the longest side: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Unknowns: The length of the longest side (one of the five choices)
Understand
Restated: A triangle has three side lengths (in inches) that are three consecutive integers. The shortest side is exactly $30\%$ of the perimeter. Find the length of the longest side.
Givens: Side lengths are three consecutive integers, e.g. $n,\;n+1,\;n+2$; Shortest side $= 30\% = \tfrac{3}{10}$ of the perimeter; Perimeter $=$ sum of the three side lengths; Answer choices for the longest side: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Plan
Primary tool: #6 Guess and Check
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The reference solution uses Tool #13 (Algebra): set up $n = 0.3(3n+3)$ and solve. But the answer choices give us the longest side directly, and each choice pins down all three sides. So Tool #6 (Guess and Check) is faster and needs no variable: for each choice $L \in \{7,8,9,10,11\}$, the three sides are $L-2,\;L-1,\;L$, the perimeter is $3L - 3$, and we just check whether $\text{shortest} = 0.3 \times \text{perimeter}$. Tool #2 (Systematic List) keeps the five cases organized; Tool #3 (Eliminate Possibilities) crosses off any choice that misses the $30\%$ target. This path uses only multiplication and basic decimals — no algebra.
Execute — Answer: E
4.OA.C.5 Step 1 - Set up the systematic list.
- If the longest side is $L$, the three consecutive integer sides are $L-2,\;L-1,\;L$, so the perimeter is $(L-2) + (L-1) + L = 3L - 3$.
- The shortest side is $L - 2$, and the $30\%$ test is whether $L - 2 = 0.3 \times (3L - 3)$.
💡 Writing all five cases in the same template $(L-2, L-1, L)$ is the systematic-list move — same shape, only $L$ changes.
5.NBT.B.7 Step 2 - Try (A) $L = 7$: sides $(5, 6, 7)$, perimeter $5 + 6 + 7 = 18$.
- Check: is $5 = 0.3 \times 18$?
- $0.3 \times 18 = 5.4$, not $5$.
- Reject (A).
💡 Multiplying $18$ by $0.3$ is the same as $18 \times 3 \div 10 = 54 \div 10 = 5.4$ — a Grade 5 decimal-times-whole-number move.
5.NBT.B.7 Step 3 - Try (B) $L = 8$: sides $(6, 7, 8)$, perimeter $6 + 7 + 8 = 21$.
- Check: $0.3 \times 21 = 6.3 \neq 6$.
- Reject (B).
- Try (C) $L = 9$: sides $(7, 8, 9)$, perimeter $24$.
- Check: $0.3 \times 24 = 7.2 \neq 7$.
- Reject (C).
- Try (D) $L = 10$: sides $(8, 9, 10)$, perimeter $27$.
- Check: $0.3 \times 27 = 8.1 \neq 8$.
- Reject (D).
💡 Each candidate misses by a small amount — the gap between shortest and $30\%$ of perimeter keeps shrinking, so we are on the right track.
6.RP.A.3 Step 4 - Try (E) $L = 11$: sides $(9, 10, 11)$, perimeter $9 + 10 + 11 = 30$.
- Check: $0.3 \times 30 = 9$, which exactly equals the shortest side $9$.
- The $30\%$ condition is satisfied, so the longest side is $11$.
💡 Finding the percent of a number ($30\%$ of $30$) and matching it to a target is core Grade 6 percent reasoning.
4.OA.C.5 Set up the systematic list. If the longest side is $L$, the three consecutive in 5.NBT.B.7 Try (A) $L = 7$: sides $(5, 6, 7)$, perimeter $5 + 6 + 7 = 18$. Check: is $5 = 0 5.NBT.B.7 Try (B) $L = 8$: sides $(6, 7, 8)$, perimeter $6 + 7 + 8 = 21$. Check: $0.3 \tim 6.RP.A.3 Try (E) $L = 11$: sides $(9, 10, 11)$, perimeter $9 + 10 + 11 = 30$. Check: $0.3 Review
Reasonableness: If the three sides were equal, each side would be $\tfrac{1}{3} \approx 33.3\%$ of the perimeter. Here the shortest side is only $30\%$ — a bit less than a third — so the sides must be a little spread apart, not tightly bunched. The winning triple $(9, 10, 11)$ has a shortest side $9$ out of a perimeter $30$, and $9/30 = 0.30$ exactly. Triangle inequality also holds: $9 + 10 = 19 > 11$. Everything checks out.
Alternative: Tool #13 (Algebra): let the shortest side be $n$, so the sides are $n,\;n+1,\;n+2$ and the perimeter is $3n+3$. The condition $n = 0.3(3n+3)$ becomes $n = 0.9n + 0.9$, then $0.1n = 0.9$, so $n = 9$ and the longest side is $n+2 = 11$. Same answer, but it requires setting up and solving a linear equation — Tool #6's table is more elementary.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number or shape pattern following a given rule (Setting up the template $(L-2,\;L-1,\;L)$ so that each answer choice gives a complete side triple with the same shape.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Computing $0.3 \times P$ for each candidate perimeter ($5.4,\;6.3,\;7.2,\;8.1,\;9.0$) to test the $30\%$ condition.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Interpreting "shortest side $= 30\%$ of the perimeter" as a percent relationship and verifying it for the winning triple $(9,10,11)$.)
⭐ This AMC 8 problem only needs Grade 6 percent reasoning — "$30\%$ of the perimeter" — that you already know!
⭐ This AMC 8 problem only needs Grade 6 percent reasoning — "$30\%$ of the perimeter" — that you already know!
More like this
Same archetype — closest grade level first.