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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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29 Aug 2016, 04:08
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Difficulty:

45% (medium)

Question Stats:

70% (02:57) correct 30% (02:16) wrong based on 134 sessions

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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

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

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29 Aug 2016, 10:50
3
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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29 Aug 2016, 10:51
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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Joined: 15 Feb 2016
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GMAT 1: 710 Q48 V40
Re: Two dogsled teams raced across a 300-mile course in Wyoming. Team A  [#permalink]

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29 Aug 2016, 13:24
1
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.

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

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29 Aug 2016, 22:39
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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04 Oct 2017, 11:39
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Re: Two dogsled teams raced across a 300-mile course in Wyoming. Team A &nbs [#permalink] 04 Oct 2017, 11:39
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