AMC 8 · 2010 · #3

Grade 6 rate-ratio
percentagegraph-readingfraction-arithmetic identify-subproblems ↑ Prerequisites: percentagemulti-digit-arithmetic
📏 Medium solution 💡 3 insights 📊 Diagram
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Problem

The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?

Pick an answer.

(A)
50
(B)
62
(C)
70
(D)
89
(E)
100
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Toolkit + CCSS Solution

Understand

Restated: A bar graph shows the price of $5$ gallons of gasoline for each of the first $10$ months of the year. Reading the bars, the highest price is $\$17$ (Month $1$) and the lowest price is $\$10$ (Month $3$). By what percent does the highest price exceed the lowest price?

Givens: Bar graph of prices for $5$ gallons of gasoline, Months $1$ through $10$; Vertical axis is price in dollars, with gridlines at $0, 5, 10, 15, 20$; Tallest bar (Month $1$) reaches $\$17$; Shortest bar (Month $3$) reaches $\$10$; Answer choices: (A) $50$, (B) $62$, (C) $70$, (D) $89$, (E) $100$ (percent)

Unknowns: The percent by which the highest price exceeds the lowest price

Understand

Restated: A bar graph shows the price of $5$ gallons of gasoline for each of the first $10$ months of the year. Reading the bars, the highest price is $\$17$ (Month $1$) and the lowest price is $\$10$ (Month $3$). By what percent does the highest price exceed the lowest price?

Givens: Bar graph of prices for $5$ gallons of gasoline, Months $1$ through $10$; Vertical axis is price in dollars, with gridlines at $0, 5, 10, 15, 20$; Tallest bar (Month $1$) reaches $\$17$; Shortest bar (Month $3$) reaches $\$10$; Answer choices: (A) $50$, (B) $62$, (C) $70$, (D) $89$, (E) $100$ (percent)

Plan

Primary tool: #5 Look for a Pattern

Secondary: #7 Identify Subproblems

The problem has a simple shape — read two numbers off the graph, then compute a percent increase — so the work splits cleanly into Tool #7 subproblems: (1) extract the maximum and minimum from the bar chart, (2) compute the percent increase using $(\text{high} - \text{low}) / \text{low} \times 100\%$. Tool #5 (Look for a Pattern) is what makes the second step quick: $\$7$ on a base of $\$10$ is the familiar pattern $\tfrac{7}{10} = 0.70 = 70\%$, so no long division is needed.

Execute — Answer: C

#7 Identify Subproblems 3.MD.B.3 Step 1
  • Read the extremes off the bar graph.
  • Scanning all $10$ bars, Month $1$ is the tallest and reaches the $\$17$ level; Month $3$ is the shortest and reaches exactly the $\$10$ gridline.
$\text{highest} = \$17, \quad \text{lowest} = \$10$

💡 Reading a maximum and minimum off a scaled bar graph is the Grade 3 "draw and interpret a scaled bar graph" skill.

#7 Identify Subproblems 4.OA.A.3 Step 2
  • Find the difference between the two prices.
  • This is the dollar amount by which the high price exceeds the low price.
$\$17 - \$10 = \$7$

💡 Subtraction in a multi-step word problem is a Grade 4 operations standard.

#5 Look for a Pattern 6.RP.A.3 Step 3
  • Set up the percent-increase ratio.
  • "$A$ is more than $B$ by what percent" means we divide the excess by $B$ (the lowest price, our base) and convert to a percentage.
$\text{percent more} = \dfrac{\$17 - \$10}{\$10} \times 100\% = \dfrac{\$7}{\$10} \times 100\%$

💡 Choosing the lowest price as the base of the percent — not the highest — is the key Grade 6 ratio-reasoning move.

#5 Look for a Pattern 6.RP.A.3 Step 4
  • Compute the ratio and convert to a percent.
  • The fraction $\tfrac{7}{10}$ is exactly $0.7$, and $0.7 \times 100\% = 70\%$.
$$\dfrac{7}{10} \times 100\% = 70\% \;\Rightarrow\; \textbf{(C)}$$

💡 Recognizing $\tfrac{7}{10} = 70\%$ without long division is the percent-pattern shortcut from Grade 6.

[1] #7 3.MD.B.3 Read the extremes off the bar graph. Scanning all $10$ bars, Month $1$ is the ta
[2] #7 4.OA.A.3 Find the difference between the two prices. This is the dollar amount by which t
[3] #5 6.RP.A.3 Set up the percent-increase ratio. "$A$ is more than $B$ by what percent" means
[4] #5 6.RP.A.3 Compute the ratio and convert to a percent. The fraction $\tfrac{7}{10}$ is exac

Review

Reasonableness: If the high price were double the low price, the increase would be $100\%$; doubling $\$10$ would mean a high of $\$20$. Since $\$17$ is below $\$20$ but well above $\$10$, the answer must be less than $100\%$ but more than $50\%$. That rules out (A) and (E) and leaves (B) $62$, (C) $70$, (D) $89$. Our calculation $\tfrac{7}{10} = 70\%$ lands on (C), which sits in the expected window.

Alternative: Tool #6 (Guess and Check) on the choices: for each choice $p$, the high price would be $\$10 \times (1 + p/100)$. (A) $50\%$ gives $\$15$, (B) $62\%$ gives $\$16.20$, (C) $70\%$ gives $\$17$, (D) $89\%$ gives $\$18.90$, (E) $100\%$ gives $\$20$. Only (C) reproduces the chart value $\$17$.

CCSS standards used (min grade 6)

  • 3.MD.B.3 Draw and interpret scaled picture graphs and bar graphs (Reading the tallest bar ($\$17$ in Month $1$) and the shortest bar ($\$10$ in Month $3$) off the bar graph.)
  • 4.OA.A.3 Solve multistep word problems using the four operations (Subtracting the two prices to find the $\$7$ excess used in the percent-increase formula.)
  • 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including percent problems (Setting up and computing the percent increase $(\$17 - \$10) / \$10 \times 100\% = 70\%$, with the lowest price as the base.)

⭐ This AMC 8 problem only needs Grade 6 percent reasoning — divide the gap by the smaller number — that you already know!

⭐ This AMC 8 problem only needs Grade 6 percent reasoning — divide the gap by the smaller number — that you already know!