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Waugh jogged to a restaurant at x miles per hour, and jogged back home

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Waugh jogged to a restaurant at x miles per hour, and jogged back home  [#permalink]

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New post 14 Apr 2019, 04:27
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E

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  45% (medium)

Question Stats:

61% (01:31) correct 39% (02:01) wrong based on 46 sessions

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Waugh jogged to a restaurant at x miles per hour, and jogged back home along the same route at y miles per hour. He took 30 minutes for the whole trip. If the restaurant is 2 miles from home, what is the average speed in miles per hour at which he jogged for the whole trip?

(A) 0.13
(B) 0.5
(C) 2
(D) 4
(E) 8

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Re: Waugh jogged to a restaurant at x miles per hour, and jogged back home  [#permalink]

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New post 14 Apr 2019, 04:38
Bunuel wrote:
Waugh jogged to a restaurant at x miles per hour, and jogged back home along the same route at y miles per hour. He took 30 minutes for the whole trip. If the restaurant is 2 miles from home, what is the average speed in miles per hour at which he jogged for the whole trip?

(A) 0.13
(B) 0.5
(C) 2
(D) 4
(E) 8


total distance = 4 miles
and total time = 1/2 hrs ; 30 mins
so total speed = 2*4 ; 8 mph

IMO E
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Re: Waugh jogged to a restaurant at x miles per hour, and jogged back home  [#permalink]

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New post 14 Apr 2019, 04:38
Bunuel wrote:
Waugh jogged to a restaurant at x miles per hour, and jogged back home along the same route at y miles per hour. He took 30 minutes for the whole trip. If the restaurant is 2 miles from home, what is the average speed in miles per hour at which he jogged for the whole trip?

(A) 0.13
(B) 0.5
(C) 2
(D) 4
(E) 8


Waugh walks 2 miles from the water to restaurant and back(total of 4 miles) in 30 minutes.

Therefore, the average speed at which Waugh completed this whole trip is 8 miles/hour(Option E)
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Re: Waugh jogged to a restaurant at x miles per hour, and jogged back home  [#permalink]

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New post 14 Apr 2019, 04:40
\(Speed_{avg} = \frac{Total distance}{Total time taken}\)

Total distance = 4 miles

Total time took = 30 minutes = \(\frac{1}{2}\) hours

\(Speed_{avg} = \frac{4}{\frac{1}{2}}\)

\(Speed_{avg}\) = 8 miles per hour

OPTION: E
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Re: Waugh jogged to a restaurant at x miles per hour, and jogged back home  [#permalink]

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New post 14 Apr 2019, 05:47
1
IMO E ,


avg speed = total distance/total time

total distance = 4

total time 1/2 hr

so avg speed = 4/(1/2)
= 4*2 = 8

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Re: Waugh jogged to a restaurant at x miles per hour, and jogged back home   [#permalink] 14 Apr 2019, 05:47
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