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2017 AMC 10B Problem 7

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2017 AMC 10B Problem 7AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 2 of 5 · about 3 min · position P1-10

Samia set off on her bicycle to visit her friend, traveling at an average speed of 1717 kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at 55 kilometers per hour. In all it took her 4444 minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?

  1. A)

    2.02.0

  2. B)

    2.22.2

  3. C)

    2.82.8

  4. D)

    3.43.4

  5. E)

    4.44.4

Answer

C

Key insight

Both halves have the same length d, so d/17 + d/5 = 44/60 hours; solve for d, which is exactly the walking distance.

Solution

Let dd be the length of each half of the trip in kilometers; the walking distance is exactly dd. Biking dd km at 1717 km/h takes d17\frac{d}{17} hours and walking dd km at 55 km/h takes d5\frac{d}{5} hours. The total is 4444 minutes =4460=1115= \frac{44}{60} = \frac{11}{15} hours:
d17+d5=1115.\frac{d}{17} + \frac{d}{5} = \frac{11}{15}.
Combine the left side: 5d+17d85=22d85\frac{5d + 17d}{85} = \frac{22d}{85}. So
d=1115⋅8522=8530=176≈2.83.d = \frac{11}{15} \cdot \frac{85}{22} = \frac{85}{30} = \frac{17}{6} \approx 2.83.
Rounded to the nearest tenth, Samia walked 2.82.8 km.

The answer is (C) 2.8\boxed{\textbf{(C)}\ 2.8}.

Why this works

Two legs of equal distance at different speeds give a one-variable time equation. Choosing the unknown to be the quantity asked for (the half distance, not the whole trip) avoids a final conversion step. Keep the time in hours since the speeds are per hour.

Alternative approach

Walking is the slow part: at 55 km/h it takes 17/5=3.417/5 = 3.4 times as long as biking the same distance. Of the 4444 minutes, walking takes 1722⋅44=34\frac{17}{22} \cdot 44 = 34 minutes, and 3460⋅5=176≈2.8\frac{34}{60} \cdot 5 = \frac{17}{6} \approx 2.8 km.

The trap

Solving for the total distance 2d = 5.7 km or forgetting to convert 44 minutes to hours before setting up the equation.

Common mistakes

  • Solving for the total distance 2d = 5.7 km or forgetting to convert 44 minutes to hours before setting up the equation.
  • Averaging the speeds to 1111 km/h and using 11⋅446011 \cdot \frac{44}{60} for the trip length.

Techniques

Set up the equation/formula and compute; no special trick needed