2017 AMC 10B Problem 7

Below is the professionally curated solution for Problem 7 of the 2017 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 10B solutions, or check the answer key.

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Concepts:distance rate and timelinear equation

Difficulty rating: 1220

7.

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?

2.02.0

2.22.2

2.82.8

3.43.4

4.44.4

Solution:

Let the distance she walked be d.d. Since this is the same as the amount she biked, represented as s,s, we know that d=s.d=s.

Furthermore, let the time she walked (in hours) be t.t. Therefore, the amount of time she biked is 4460t.\dfrac{44}{60}-t.

Now, using the definition of speed, we can see that 5=st5 = \frac st17=s4460t 17 = \dfrac s{\frac{44}{60}-t} This implies that: s=5t=17(4460t)s=5t = 17\left(\frac{44}{60}-t\right) so 22t=17446022t = \frac{17\cdot 44}{60} Therefore, t=1730t = \frac{17}{30} Since s=5t,s=5t, we have s=51730=176s = 5\cdot \frac{17}{30} = \frac{17}{6} which approximates to 2.8.2.8.

Thus, the correct answer is C .

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