AMC 10 · 2009 · #3
Easy mode Grade 4On a number line, you walk from 41 to 43. You stop once you have covered one third of that walk. What number are you standing on?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: On the number line, start at $\frac{1}{4}$ and travel toward $\frac{3}{4}$. Find the number you land on once you have covered one third of that trip.
Givens: The trip starts at $\frac{1}{4}$ and finishes at $\frac{3}{4}$; The share of the trip to be covered is $\frac{1}{3}$, counted from the starting end; Answer choices: (A) $\frac{1}{3}$, (B) $\frac{5}{12}$, (C) $\frac{1}{2}$, (D) $\frac{7}{12}$, (E) $\frac{2}{3}$
Unknowns: The single number sitting one third of the way along the trip
Understand
Restated: On the number line, start at $\frac{1}{4}$ and travel toward $\frac{3}{4}$. Find the number you land on once you have covered one third of that trip.
Givens: The trip starts at $\frac{1}{4}$ and finishes at $\frac{3}{4}$; The share of the trip to be covered is $\frac{1}{3}$, counted from the starting end; Answer choices: (A) $\frac{1}{3}$, (B) $\frac{5}{12}$, (C) $\frac{1}{2}$, (D) $\frac{7}{12}$, (E) $\frac{2}{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #15 Organize Information in More Ways, #7 Identify Subproblems, #11 Work Backwards
Everything hard about this problem is in reading the phrase, not in the arithmetic, so Tool #1 (Draw a Diagram) comes first: a number line makes the start, the finish, and the direction visible all at once, and it is the direction that four of the five answer choices attack. Tool #15 (Organize Information in More Ways) then does the real work — rewrite both ends in twelfths instead of fourths, chosen so that a trip which must be cut into three equal parts cuts on whole marks. Tool #7 (Identify Subproblems) splits the job into measure-the-trip and then cut-it-into-thirds, and it forces the trisection to be shown rather than eyeballed: the marks are third-marks only because the gaps between them come out equal. Finally Tool #11 (Work Backwards) checks the candidate against the definition from the other end — measure the two pieces it creates and confirm they land in a $1:2$ ratio.
Execute — Answer: B
3.NF.A.2 Step 1 Draw the trip and its direction
- Mark $\frac{1}{4}$ and $\frac{3}{4}$ on a number line.
- The words "from $\frac{1}{4}$ to $\frac{3}{4}$" say which end is the start: you leave $\frac{1}{4}$ and move right toward $\frac{3}{4}$.
- So the number wanted lies between the two ends and closer to $\frac{1}{4}$, and it is the point reached after covering one third of the distance — it is not the number $\frac{1}{3}$, which is a value rather than a share of a distance.
💡 A fraction of a trip is a share of the distance, so the trip needs a starting end before the share means anything.
4.NF.A.1 Step 2 Switch the ruler to twelfths
- The trip must be cut into three equal pieces, so choose units that survive being cut into thirds.
- Fourths are too coarse: the trip is $\frac{1}{2}$ long, and a third of $\frac{1}{2}$ is not a whole number of fourths.
- Tripling the top and bottom of each end turns fourths into twelfths, and $3$ is now a factor of the counts, so thirds will come out exact.
💡 Pick the ruler so that the length you have to split three ways lands on whole marks.
4.NF.B.3 Step 3 Cut the trip into three equal pieces
- From $\frac{3}{12}$ to $\frac{9}{12}$ is $6$ twelfths of distance.
- Those $6$ twelfths break into three equal pieces of $2$ twelfths each, and the break is exact because $6 = 2 + 2 + 2$ in whole twelfths.
- Walking the pieces from the start puts marks at $\frac{3}{12}$, $\frac{5}{12}$, $\frac{7}{12}$, $\frac{9}{12}$, with every consecutive gap equal to $\frac{2}{12}$.
- Those equal gaps are what earns these marks the name "third-marks"; it is not enough that two numbers happen to sit somewhere inside the interval.
💡 "Thirds" means three jumps of the same size, so show the jumps are the same size instead of trusting the picture.
4.NF.B.3 Step 4 Take one jump from the start
- One third of the way is one of those three equal jumps, taken from the end the problem starts at.
- Adding a single jump to the start gives $\frac{3}{12} + \frac{2}{12} = \frac{5}{12}$.
- Two jumps would give $\frac{7}{12}$, which is two thirds of the way — and it is also exactly what you get if you slip and run the trip backwards, starting at $\frac{3}{4}$ and walking one jump toward $\frac{1}{4}$.
💡 One third of the way is one jump, not two, and the jumps are counted from the end the problem names first.
4.NF.A.2 Step 5 Check the 1-to-2 split
- Test the candidate against the definition by measuring instead of building.
- The piece behind it is $\frac{5}{12} - \frac{3}{12} = \frac{2}{12}$ and the piece ahead is $\frac{9}{12} - \frac{5}{12} = \frac{4}{12}$, so behind-to-ahead is $2 : 4 = 1 : 2$.
- That is precisely "one of the three equal parts walked, two still to go", which is what one third of the way means.
- It also explains the shortcut of averaging the ends with weights $2$ and $1$: the near end carries the heavier weight because two of the three parts still lie ahead of the point, not because of a rule to memorise.
- And no other number passes — sliding right makes the piece behind grow while the piece ahead shrinks, so the $1:2$ split occurs at exactly one place.
- The answer is $\textbf{(B)}\ \frac{5}{12}$.
💡 If one third of the trip is behind you, two thirds are still ahead, so the two pieces have to measure $1$ to $2$.
3.NF.A.2 Mark $\frac{1}{4}$ and $\frac{3}{4}$ on a number line. The words "from $\frac{1} 4.NF.A.1 The trip must be cut into three equal pieces, so choose units that survive being 4.NF.B.3 From $\frac{3}{12}$ to $\frac{9}{12}$ is $6$ twelfths of distance. Those $6$ twe 4.NF.B.3 One third of the way is one of those three equal jumps, taken from the end the p 4.NF.A.2 Test the candidate against the definition by measuring instead of building. The Review
Reasonableness: Put every choice on the same twelfths ruler: (A) $\frac{4}{12}$, (B) $\frac{5}{12}$, (C) $\frac{6}{12}$, (D) $\frac{7}{12}$, (E) $\frac{8}{12}$, against a start of $\frac{3}{12}$ and a finish of $\frac{9}{12}$. Since $\frac{1}{3} < \frac{1}{2}$, the point must sit strictly between the start and the midpoint $\frac{6}{12}$, which already kills (C), (D) and (E). Of the two survivors, (A) sits only $\frac{1}{12}$ past the start — one sixth of the $\frac{6}{12}$ trip, not one third — while (B) sits $\frac{2}{12}$ past, exactly one third. Each wrong choice is a named misreading: (A) treats "one third" as the number itself, (C) is halfway rather than a third of the way, (D) is two thirds of the way (equivalently, one third of the way if the trip is run backwards from $\frac{3}{4}$), and (E) is the number two thirds.
Alternative: Do it with algebra from the ratio instead of from the distance. Let $x$ be the number. One third of the trip lies behind $x$ and two thirds ahead, so the piece ahead is twice the piece behind: $2\left(x - \frac{1}{4}\right) = \frac{3}{4} - x$. Then $2x - \frac{1}{2} = \frac{3}{4} - x$, so $3x = \frac{5}{4}$ and $x = \frac{5}{12}$. The coefficient $3$ on $x$ is nonzero, so this equation has exactly one solution — uniqueness comes free. This route also derives, rather than assumes, the popular weighted-average trick $x = \frac{2 \cdot \frac{1}{4} + 1 \cdot \frac{3}{4}}{3}$: the weights $2$ and $1$ are just the two pieces measured in thirds of the trip. In general the point $t$ of the way from $a$ to $b$ is $a + t(b - a) = (1 - t)a + tb$, so the near end always carries weight $1 - t$; at $t = \frac{1}{3}$ that is $\frac{2}{3} \cdot \frac{1}{4} + \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{6} + \frac{1}{4} = \frac{5}{12}$, matching the number-line count.
CCSS standards used (min grade 4)
3.NF.A.2Understand a fraction as a number on the number line (Placing $\frac{1}{4}$ and $\frac{3}{4}$ on a number line and reading "one third of the way from" as a share of the distance travelled in a fixed direction.)4.NF.A.1Explain why a fraction is equivalent to another fraction (Rewriting $\frac{1}{4} = \frac{3}{12}$ and $\frac{3}{4} = \frac{9}{12}$ so the trip length becomes a whole number of twelfths that splits into thirds exactly.)4.NF.B.3Understand a fraction with numerator greater than one as sum of unit fractions (Measuring the trip as $\frac{9}{12} - \frac{3}{12} = \frac{6}{12}$, decomposing it as $\frac{2}{12} + \frac{2}{12} + \frac{2}{12}$ to exhibit three equal pieces, and adding one piece to the start.)4.NF.A.2Compare two fractions with different numerators and different denominators (Ordering the marks along the trip, comparing the $\frac{2}{12}$ behind against the $\frac{4}{12}$ ahead, and ranking the five answer choices once they are all written in twelfths.)
⭐ Rewrite both ends in twelfths so the trip cuts evenly into three, then count one jump from the end the problem starts at — that lands you on $\frac{5}{12}$.
⭐ Rewrite both ends in twelfths so the trip cuts evenly into three, then count one jump from the end the problem starts at — that lands you on $\frac{5}{12}$.
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