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AMC 8 · 2023 · #9

Grade 5 rate-ratio
Archetype Coordinate Plane Figure Computation
graph-readingcoordinate-geometryinterval-arithmetic graph-readingidentify-subproblems ↑ Prerequisites: coordinate-geometrymulti-digit-arithmetic
📏 Medium solution 💡 2 insights 📊 Diagram
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Problem

Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between 44 and 77 meters?

Pick an answer.

(A)
6
(B)
8
(C)
10
(D)
12
(E)
14
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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

#1 Draw a Diagram 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.
$$\text{band:}\ 4 \le y \le 7$$

💡 Plotting a horizontal line $y = c$ on a coordinate graph is exactly the Grade 5 "graph points / lines on a coordinate plane" skill.

#1 Draw a Diagram 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.
$$y=7\colon\ t = 2, 6, 10, 14;\quad y=4\colon\ t = 4, 12$$

💡 Each crossing point is just an ordered pair $(t, y)$ read off the coordinate plane — Grade 5 graphing.

#7 Identify Subproblems 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.
$$\text{in-band intervals: }[2,4],\ [4,6],\ [10,12],\ [12,14]$$

💡 Deciding "is the curve between $y=4$ and $y=7$ here?" for each strip is Grade 5 coordinate-plane reading.

#7 Identify Subproblems 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.
$$(4-2) + (6-4) + (12-10) + (14-12) = 2+2+2+2 = 8$$

💡 Subtracting small whole numbers and adding the four results is a Grade 2 multi-step word-problem skill.

#3 Eliminate Possibilities 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.
$$8\ \text{seconds} \;\Rightarrow\; \textbf{(B)}$$

💡 Checking the total against the five listed values is straightforward Grade 2 arithmetic comparison.

[1] #1 5.G.A.2 Mark the band on the diagram. Draw two horizontal lines on the graph: one at ele
[2] #1 5.G.A.2 Read off the crossing times. Following the curve from left to right, it crosses
[3] #7 5.G.A.2 Break the timeline into pieces (Tool #7). Using the six crossings, the relevant
[4] #7 2.OA.A.1 Find each piece's length, then add them up. Each sub-interval has length (right
[5] #3 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.2 Represent 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.1 Solve 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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