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Martha biked 18 2/3 miles in 2 hours and 40 minutes. What was her aver

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Martha biked 18 2/3 miles in 2 hours and 40 minutes. What was her aver  [#permalink]

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New post 02 Aug 2018, 00:16
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Martha biked \(18 \frac{2}{3}\) miles in 2 hours and 40 minutes. What was her average rate of speed in miles per hour?


A. 7

B. \(7 \frac{2}{3}\)

C. \(8 \frac{1}{3}\)

D. 9

E. \(9 \frac{1}{3}\)

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Re: Martha biked 18 2/3 miles in 2 hours and 40 minutes. What was her aver  [#permalink]

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New post 02 Aug 2018, 00:33
Ans: A
Miles = 56/3 miles Time taken: 2 +2/3 hrs = 8/3 hrs
Velocity = Miles/Time = 7 miles/hrs

Bunuel wrote:
Martha biked \(18 \frac{2}{3}\) miles in 2 hours and 40 minutes. What was her average rate of speed in miles per hour?

A. 7

B. \(7 \frac{2}{3}\)

C. \(8 \frac{1}{3}\)

D. 9

E. \(9 \frac{1}{3}\)

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Re: Martha biked 18 2/3 miles in 2 hours and 40 minutes. What was her aver  [#permalink]

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New post 02 Aug 2018, 03:11
Bunuel wrote:
Martha biked \(18 \frac{2}{3}\) miles in 2 hours and 40 minutes. What was her average rate of speed in miles per hour?

Distance = \(18 \frac{2}{3}\) = 56 / 3 miles
Time = 2 hours + 40 minutes = 2 hrs + 2 / 3 hrs = 8 / 3 hrs

Speed = Distance / Time = 56/3 / 8/3 = 7 miles per hour

Hence, A.
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Martha biked 18 2/3 miles in 2 hours and 40 minutes. What was her aver  [#permalink]

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New post 03 Aug 2018, 12:20
\(18 \frac{2}{3} = \frac{56}{3}\)
2 hours and 40 minutes equals to 160 minutes
Lets say that R is the average rate

\(\frac{56}{3R} = \frac{160}{60}\)
\(\frac{56}{3R} = \frac{8}{3}\)
\(\frac{56}{R} = 8\)
Cross multiplication:
\(56 = R*8\)
\(R = \frac{56}{8}\)
\(R = 7\)

Option A
Martha biked 18 2/3 miles in 2 hours and 40 minutes. What was her aver &nbs [#permalink] 03 Aug 2018, 12:20
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