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Two dogsled teams raced across a 300-mile course in Wyoming. Team A

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Two dogsled teams raced across a 300-mile course in Wyoming. Team A [#permalink]

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Two dogsled teams raced across a 300-mile course in Wyoming. Team A finished the course in 3 fewer hours than did team B. If team A’s average speed was 5 miles per hour greater than that of team B, what was team B’s average speed, in miles per hour?

A. 12
B. 15
C. 18
D. 20
E. 25
[Reveal] Spoiler: OA

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Two dogsled teams raced across a 300-mile course in Wyoming. Team A [#permalink]

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New post 29 Aug 2016, 09:50
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let t=B's time
300/(t-3)-300/t=5
t=15
300/15=20 mph

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Re: Two dogsled teams raced across a 300-mile course in Wyoming. Team A [#permalink]

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New post 29 Aug 2016, 09:51
Bunuel wrote:
Two dogsled teams raced across a 300-mile course in Wyoming. Team A finished the course in 3 fewer hours than did team B. If team A’s average speed was 5 miles per hour greater than that of team B, what was team B’s average speed, in miles per hour?

A. 12
B. 15
C. 18
D. 20
E. 25


For team A time taken, \(T = \frac{300}{A}\) , where A is the avg speed of Team A
For team B, \(T + 3 = \frac{300}{(A-5)}\)

Substituting the values from choices, D, satisfies the condition. \(T = \frac{300}{25} = 12
T + 3 = \frac{300}{20} = 15 = 12 + 3\)

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Re: Two dogsled teams raced across a 300-mile course in Wyoming. Team A [#permalink]

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New post 29 Aug 2016, 12:24
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Bunuel wrote:
Two dogsled teams raced across a 300-mile course in Wyoming. Team A finished the course in 3 fewer hours than did team B. If team A’s average speed was 5 miles per hour greater than that of team B, what was team B’s average speed, in miles per hour?

A. 12
B. 15
C. 18
D. 20
E. 25



300/T- 300/T+3 = 5

300( 1/T - 1/T+3) = 5

Can form the quadratic T^2 +3T -180 = 0

or can go for option choice substitution fro here.

Answer D
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Re: Two dogsled teams raced across a 300-mile course in Wyoming. Team A [#permalink]

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New post 29 Aug 2016, 21:39
Bunuel wrote:
Two dogsled teams raced across a 300-mile course in Wyoming. Team A finished the course in 3 fewer hours than did team B. If team A’s average speed was 5 miles per hour greater than that of team B, what was team B’s average speed, in miles per hour?

A. 12
B. 15
C. 18
D. 20
E. 25


Speed of B= S
Speed of A= S+5

Time taken by B=300/S Time Taken by A= 300/(S+5)
Difference in their time is 3 hrs.
300/S - 3 = 300/(S+5)
Plug in the values from the option
No need of choosing 12 & 18 it won't satisfy
a) if we take 15 then
T(B) = 300/15 =20
T(A)= 300/20 =15. Hence from this option difference comes out to be 5 which is not correct
b) if we take 20 then
T(B) = 300/20 = 15
T(A)= 300/25 = 12
Difference in timing is 3 hence Option D is correct
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Re: Two dogsled teams raced across a 300-mile course in Wyoming. Team A [#permalink]

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New post 04 Oct 2017, 10:39
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Re: Two dogsled teams raced across a 300-mile course in Wyoming. Team A   [#permalink] 04 Oct 2017, 10:39
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