AMC 8 · 2002 · #7
Grade 6 rate-ratioProblem
The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?
Pick an answer.
AMC 8 2002 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: Mrs. Sawyer's class voted between five kinds of candy ($A$ through $E$), and the results are shown on a bar graph. Reading the bars: $A = 6$, $B = 8$, $C = 4$, $D = 2$, $E = 5$ students. What percent of the class chose candy $E$?
Givens: Five candies labeled $A, B, C, D, E$, each chosen by some students; Bar heights on the graph: $A = 6$, $B = 8$, $C = 4$, $D = 2$, $E = 5$; Every student picks exactly one candy, so the bars together account for the whole class; Answer choices: (A) $5$, (B) $12$, (C) $15$, (D) $16$, (E) $20$ (percent)
Unknowns: The percent of the class that chose candy $E$
Understand
Restated: Mrs. Sawyer's class voted between five kinds of candy ($A$ through $E$), and the results are shown on a bar graph. Reading the bars: $A = 6$, $B = 8$, $C = 4$, $D = 2$, $E = 5$ students. What percent of the class chose candy $E$?
Givens: Five candies labeled $A, B, C, D, E$, each chosen by some students; Bar heights on the graph: $A = 6$, $B = 8$, $C = 4$, $D = 2$, $E = 5$; Every student picks exactly one candy, so the bars together account for the whole class; Answer choices: (A) $5$, (B) $12$, (C) $15$, (D) $16$, (E) $20$ (percent)
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern
The work splits cleanly into Tool #7 subproblems: (1) read the five bar heights, (2) add them to get the class total, (3) form the fraction $\tfrac{E}{\text{total}}$ and convert to a percent. Tool #5 (Look for a Pattern) finishes the last step quickly — $5$ out of $25$ is the familiar $\tfrac{1}{5} = 20\%$ pattern, so no long division is needed.
Execute — Answer: E
3.MD.B.3 Step 1 - Read the five bar heights off the graph.
- Each bar's top reaches a horizontal gridline that tells how many students picked that candy.
💡 Reading values off a scaled bar graph is the Grade 3 "draw and interpret a scaled bar graph" skill.
4.OA.A.3 Step 2 - Add the five bar heights to get the total number of students.
- Each student is counted exactly once, so this sum is the size of Mrs.
- Sawyer's class.
💡 Combining several known quantities to get a total is the Grade 4 multi-step word-problem move.
6.RP.A.3 Step 3 Write the percent of the class that chose $E$ as a fraction over the class total, then convert to a percent.
💡 Choosing the class total as the base of the percent (the "whole") is the key Grade 6 ratio move.
6.RP.A.3 Step 4 - Simplify the fraction and convert.
- $\tfrac{5}{25} = \tfrac{1}{5}$, and $\tfrac{1}{5}$ is the well-known $20\%$ pattern.
💡 Recognizing $\tfrac{1}{5} = 20\%$ without long division is the Grade 6 percent-pattern shortcut.
3.MD.B.3 Read the five bar heights off the graph. Each bar's top reaches a horizontal gri 4.OA.A.3 Add the five bar heights to get the total number of students. Each student is co 6.RP.A.3 Write the percent of the class that chose $E$ as a fraction over the class total 6.RP.A.3 Simplify the fraction and convert. $\tfrac{5}{25} = \tfrac{1}{5}$, and $\tfrac{1 Review
Reasonableness: Sanity check by comparing $E$ to the rest. $E = 5$ students out of $25$ is one-fifth, and one-fifth of $100\%$ is $20\%$, which lands on (E). Also, $E$'s bar is shorter than $A$ and $B$ but taller than $D$, so $E$ should account for a middling slice of the class — well below $A$'s $\tfrac{6}{25} = 24\%$ and well above $D$'s $\tfrac{2}{25} = 8\%$. Twenty percent sits exactly in that window. Finally, the five percentages $24, 32, 16, 8, 20$ add to $100\%$, confirming the class total of $25$ is correct.
Alternative: Tool #6 (Guess and Check) on the choices: for each candidate $p$, the number of $E$-voters would be $25 \times p/100$. (A) $5\% \to 1.25$, (B) $12\% \to 3$, (C) $15\% \to 3.75$, (D) $16\% \to 4$, (E) $20\% \to 5$. Only (E) produces the bar height of $5$ students, so (E) is the answer.
CCSS standards used (min grade 6)
3.MD.B.3Draw and interpret scaled picture graphs and bar graphs (Reading the five bar heights ($A = 6, B = 8, C = 4, D = 2, E = 5$) off the scaled bar graph.)4.OA.A.3Solve multistep word problems using the four operations (Adding the five bar heights $6 + 8 + 4 + 2 + 5 = 25$ to find the class total.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems, including percent problems (Forming the part-to-whole ratio $\tfrac{5}{25}$ and converting it to the percent $20\%$.)
⭐ Read each bar, add them for the class total, then write the $E$-share as a fraction and turn it into a percent — a Grade 6 part-to-whole move that turns an AMC 8 problem into one step of arithmetic.
⭐ Read each bar, add them for the class total, then write the $E$-share as a fraction and turn it into a percent — a Grade 6 part-to-whole move that turns an AMC 8 problem into one step of arithmetic.
More like this
Same archetype — closest grade level first.