AMC 8 · 2023 · #9
Easy mode Grade 5Problem
Picture Malaika skiing down a hill. As she skis, her height above the bottom of the hill keeps changing.
The graph below shows her height in meters at each moment in time. The bottom line is time in seconds. The side line is her height in meters.
Sometimes she is high up, sometimes lower. We want to look at the moments when her height is somewhere between 4 meters and 7 meters.
Add up all the seconds she spends in that height range. How many seconds is it in total?
Diagram
Malaika's elevation versus time; the yellow band marks heights between 4 and 7 meters.
Pick an answer.
AMC 8 2023 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 graph shows Malaika's elevation (in meters) above the base as time (in seconds) passes while she skis. We need the total number of seconds her elevation $E$ stays inside the band $4 \le E \le 7$ meters.
Givens: The horizontal axis is time in seconds; the vertical axis is elevation in meters; A smooth curve traces her elevation as a function of time; From the curve: $E=7$ at $t = 2, 6, 10, 14$ seconds; From the curve: $E=4$ at $t = 4$ and $t = 12$ seconds; Between $t=6$ and $t=10$ the curve sits above $7$ meters; Answer choices: (A) 6, (B) 8, (C) 10, (D) 12, (E) 14
Unknowns: The total time (in seconds) during which $4 \le E \le 7$
Understand
Restated: A graph shows Malaika's elevation (in meters) above the base as time (in seconds) passes while she skis. We need the total number of seconds her elevation $E$ stays inside the band $4 \le E \le 7$ meters.
Givens: The horizontal axis is time in seconds; the vertical axis is elevation in meters; A smooth curve traces her elevation as a function of time; From the curve: $E=7$ at $t = 2, 6, 10, 14$ seconds; From the curve: $E=4$ at $t = 4$ and $t = 12$ seconds; Between $t=6$ and $t=10$ the curve sits above $7$ meters; Answer choices: (A) 6, (B) 8, (C) 10, (D) 12, (E) 14
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The problem already gives a graph, so the heavy lifting is *reading* it carefully — Tool #1 (Draw/use a diagram) applies directly: we add two horizontal reference lines $y=4$ and $y=7$ on top of the given curve and read off where they cross. Once those crossings split the timeline into pieces, Tool #7 (Subproblems) lets us treat each piece independently — duration of piece $1$, plus duration of piece $2$, and so on — instead of trying to handle the whole curve at once. Finally, since this is multiple-choice, Tool #3 confirms the total against the five listed options.
Execute — Answer: B
5.G.A.2 Step 1 - Mark the band on the diagram.
- Draw two horizontal lines on the graph: one at elevation $y = 4$ and one at $y = 7$.
- The region between them is the "target band" — any time the curve is inside this strip counts toward our total.
💡 Plotting a horizontal line $y = c$ on a coordinate graph is exactly the Grade 5 "graph points / lines on a coordinate plane" skill.
5.G.A.2 Step 2 - Read off the crossing times.
- Following the curve from left to right, it crosses $y=7$ at $t = 2, 6, 10, 14$ seconds (four times — coming down, going back up, coming down again, going back up) and crosses $y=4$ at $t = 4$ and $t = 12$ seconds.
- Mark these six time stamps on the time axis.
💡 Each crossing point is just an ordered pair $(t, y)$ read off the coordinate plane — Grade 5 graphing.
5.G.A.2 Step 3 - Break the timeline into pieces (Tool #7).
- Using the six crossings, the relevant part of the time axis splits into four sub-intervals where the curve is inside the band: $[2,4]$ (going from $7$ down to $4$), $[4,6]$ (going from $4$ up to $7$), $[10,12]$ (going from $7$ down to $4$), and $[12,14]$ (going from $4$ up to $7$).
- Between $t=6$ and $t=10$ the curve is above $7$, so it is OUTSIDE the band and contributes nothing.
💡 Deciding "is the curve between $y=4$ and $y=7$ here?" for each strip is Grade 5 coordinate-plane reading.
2.OA.A.1 Step 4 - Find each piece's length, then add them up.
- Each sub-interval has length (right endpoint) $-$ (left endpoint), and they are all $2$ seconds long.
- Adding the four pieces gives the total in-band time.
💡 Subtracting small whole numbers and adding the four results is a Grade 2 multi-step word-problem skill.
2.OA.A.1 Step 5 - Match to the answer choices.
- The total is $8$ seconds, which is choice (B).
- The other choices fail: $6$ would mean only three $2$-second pieces; $10, 12, 14$ would require pieces between $t=6$ and $t=10$, but the curve is above $7$ there, so those are excluded.
💡 Checking the total against the five listed values is straightforward Grade 2 arithmetic comparison.
5.G.A.2 Mark the band on the diagram. Draw two horizontal lines on the graph: one at ele 5.G.A.2 Read off the crossing times. Following the curve from left to right, it crosses 5.G.A.2 Break the timeline into pieces (Tool #7). Using the six crossings, the relevant 2.OA.A.1 Find each piece's length, then add them up. Each sub-interval has length (right 2.OA.A.1 Match to the answer choices. The total is $8$ seconds, which is choice (B). The Review
Reasonableness: Does $8$ seconds make sense? The curve enters the $[4,7]$ band four separate times and each crossing-to-crossing trip takes the same $2$ seconds, so $4 \times 2 = 8$. The middle hump (where she is highest, above $7$ m, from $t=6$ to $t=10$) and the deepest dip (where she is below $4$ m, between $t=4$ and $t=12$ on the $y=4$ side except the band-band pieces) are correctly EXCLUDED. The answer (B) $8$ is consistent.
Alternative: Tool #3 (Eliminate Possibilities) can be used as the main path: every in-band sub-interval has the same length (because the cubic curve is symmetric in its dips and the band is hit symmetrically), so the total must be a multiple of $2$. Visual scanning shows clearly more than $1$ but fewer than $5$ such intervals, narrowing the answer to $4$ or $8$ seconds. Only $8$ appears among the choices, giving (B).
CCSS standards used (min grade 5)
5.G.A.2Represent real-world and mathematical problems by graphing points (Reading the elevation-vs-time graph: locating the horizontal lines $y=4$ and $y=7$, identifying where the curve crosses them, and deciding which time intervals lie inside the $[4,7]$ band.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Computing each interval's length by subtraction ($4-2$, $6-4$, $12-10$, $14-12$) and adding the four $2$-second pieces to get $8$ seconds.)
⭐ This AMC 8 problem only needs Grade 5 reading points on a coordinate graph you already know!
⭐ This AMC 8 problem only needs Grade 5 reading points on a coordinate graph you already know!
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