AMC 10 · 2014 · #3
Grade 6 arithmeticRandy drove the first third of his trip on a gravel road, the next 20 miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip?
Pick an answer.
AMC 10 2014 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: Randy's trip is split into three parts. The first one-third of the whole trip is on gravel, the next $20$ miles are on pavement, and the last one-fifth of the whole trip is on dirt. Find the total length of the trip in miles.
Givens: The gravel part is one-third of the whole trip; The pavement part is exactly $20$ miles; The dirt part is one-fifth of the whole trip; The three parts together make up the whole trip; Answer choices: (A) $30$, (B) $\frac{400}{11}$, (C) $\frac{75}{2}$, (D) $40$, (E) $\frac{300}{7}$
Unknowns: The total length of the trip in miles
Understand
Restated: Randy's trip is split into three parts. The first one-third of the whole trip is on gravel, the next $20$ miles are on pavement, and the last one-fifth of the whole trip is on dirt. Find the total length of the trip in miles.
Givens: The gravel part is one-third of the whole trip; The pavement part is exactly $20$ miles; The dirt part is one-fifth of the whole trip; The three parts together make up the whole trip; Answer choices: (A) $30$, (B) $\frac{400}{11}$, (C) $\frac{75}{2}$, (D) $40$, (E) $\frac{300}{7}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #13 Convert to Algebra
The whole trip is the one number everything else is measured against, so Tool #4 (Introduce a Variable) names it $x$ and turns "one-third" and "one-fifth" into $\frac{x}{3}$ and $\frac{x}{5}$. Tool #7 (Identify Subproblems) splits the job cleanly: first combine the two fraction parts, then see what fraction the pavement must fill. Tool #13 (Convert to Algebra) writes the sentence "the parts add to the whole" as one equation, so the $20$ miles pins down $x$ instead of floating free.
Execute — Answer: E
6.EE.B.6 Step 1 Name the whole trip
- Let $x$ be the total length of the trip in miles.
- Then the gravel part is $\frac{x}{3}$, the dirt part is $\frac{x}{5}$, and the pavement part is the given $20$ miles.
💡 Giving the unknown whole a name lets every "fraction of the trip" become a real expression you can add up.
5.NF.A.1 Step 2 Combine the two fraction parts
- Add the gravel and dirt fractions to see how much of the trip they cover together.
- Using the common denominator $15$, $\frac{1}{3} = \frac{5}{15}$ and $\frac{1}{5} = \frac{3}{15}$, so together they are $\frac{8}{15}$ of the trip.
💡 Rewriting both fractions over the same denominator lets you add them into a single share of the trip.
6.EE.B.7 Step 3 Find the pavement's share
- The whole trip is $1$, or $\frac{15}{15}$.
- Since gravel and dirt take $\frac{8}{15}$, the pavement must fill what is left: $\frac{15}{15} - \frac{8}{15} = \frac{7}{15}$ of the trip.
- That leftover share is the $20$ miles of pavement.
💡 Whatever the two fractions do not cover, the pavement covers, so its fraction of the trip equals $20$ miles.
6.NS.A.1 Step 4 Solve for the total
- Undo the multiplication by $\frac{7}{15}$ by multiplying both sides by its reciprocal $\frac{15}{7}$.
- This gives $x = 20 \cdot \frac{15}{7} = \frac{300}{7}$ miles, which matches choice (E).
💡 Multiplying by the reciprocal strips off the $\frac{7}{15}$ and leaves the whole trip $x$ alone.
6.EE.B.6 Let $x$ be the total length of the trip in miles. Then the gravel part is $\frac 5.NF.A.1 Add the gravel and dirt fractions to see how much of the trip they cover togethe 6.EE.B.7 The whole trip is $1$, or $\frac{15}{15}$. Since gravel and dirt take $\frac{8}{ 6.NS.A.1 Undo the multiplication by $\frac{7}{15}$ by multiplying both sides by its recip Review
Reasonableness: Plug $x = \frac{300}{7}$ back in: gravel $= \frac{1}{3}\cdot\frac{300}{7} = \frac{100}{7}$ and dirt $= \frac{1}{5}\cdot\frac{300}{7} = \frac{60}{7}$. Their sum is $\frac{160}{7}$, and adding the $20 = \frac{140}{7}$ miles of pavement gives $\frac{300}{7}$, exactly the whole trip. The value $\frac{300}{7} \approx 42.9$ miles is a bit larger than $40$, which fits: the pavement is only $\frac{7}{15}$ (under half) of the trip, so the whole trip must be more than twice $20$.
Alternative: Skip naming a variable and reason in shares directly (Tool #15, Organize Information in More Ways). Gravel plus dirt is $\frac{1}{3}+\frac{1}{5}=\frac{8}{15}$ of the trip, so pavement is the remaining $\frac{7}{15}$. If $\frac{7}{15}$ of the trip is $20$ miles, then $\frac{1}{15}$ of the trip is $\frac{20}{7}$ miles, and the full $\frac{15}{15}$ is $15\cdot\frac{20}{7}=\frac{300}{7}$ miles.
CCSS standards used (min grade 6)
5.NF.A.1Add and subtract fractions with unlike denominators (Adding $\frac{1}{3}+\frac{1}{5}=\frac{8}{15}$ over the common denominator $15$ and subtracting from $1$ to get the pavement's share $\frac{7}{15}$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting $x$ stand for the total trip and writing the gravel and dirt parts as $\frac{x}{3}$ and $\frac{x}{5}$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Writing the equation $\frac{7}{15}x = 20$ from the fact that the leftover share of the trip is the $20$-mile pavement.)6.NS.A.1Interpret and compute quotients of fractions and solve word problems (Solving $\frac{7}{15}x = 20$ by multiplying by the reciprocal $\frac{15}{7}$ to get $x = \frac{300}{7}$.)
⭐ The $20$ miles is whatever fraction of the trip the two fractions leave over, so find that leftover fraction first.
⭐ The $20$ miles is whatever fraction of the trip the two fractions leave over, so find that leftover fraction first.
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