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# Are there many ways to solve this?

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For travel from Washington DC to Baltimore by car, a constant speed of 60 mph was maintained.  For the return trip, there was a constant speed of 50 mph. What was the average speed over the whole two-way trip?

May 28, 2018

#1
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$$\boxed{\text{Note: } \\\text{speed} = \dfrac{\text{distance}}{\text{time}}\\\text{distance} = \text{speed}\cdot\text{time}\\\text{time}=\dfrac{\text{distance}}{\text{speed}}}$$

$$\text{Let }d\text{ miles be the distance between Washington DC and Baltimore.}\\\quad\;\;\;\!\;\! t\text{ hours be the time of return trip.}$$

(magic: we can solve that without knowing the actual distance and time)

$$t = \dfrac{d}{50}$$

Time taken by Washington DC -> Baltimore = d/60

Total time used = $$d(\dfrac{1}{60}+\dfrac{1}{50})$$

$$\quad\text{Average speed} \\= \dfrac{d}{d\left(\dfrac{1}{60}+\dfrac{1}{50}\right)}\\=\dfrac{1}{\dfrac{1}{60}+\dfrac{1}{50}}\\=\dfrac{300}{11}\text{mph}$$

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May 28, 2018
#2
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i feel like you have to mulitple 300 by 2 before dividing by 11

May 28, 2018
#4
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It printed it twice!!!

May 28, 2018
edited by Guest  May 28, 2018
edited by Guest  May 28, 2018
edited by Guest  May 28, 2018
#4
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Let the average speed = S

[1/(60) + 1/(50)]S =2 [2 being twice the distance covered]

Solve for S:

(11 S)/300 = 2

Multiply both sides of (11 S)/300 = 2 by 300/11:

(300×11 S)/(11×300) = 300/11×2

300/11×11/300 = (300×11)/(11×300):

(300×11)/(11×300) S = 300/11×2

300/11×2 = (300×2)/11:

(300×11 S)/(11×300) = (300×2)/11

(300×11 S)/(11×300) = (11×300)/(11×300)×S = S:

S = (300×2)/11

300×2 = 600:

S = 600/11- Average speed over two-way trip.

Guest May 28, 2018