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A man covers the first 5 miles of the journey at the rate of 25 miles

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Bunuel
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A man covers the first 5 miles of the journey at the rate of 25 miles [#permalink] New post   12 Jun 2019, 02:11
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A man covers the first 5 miles of the journey at the rate of 25 miles per hour, the next 10 miles at the rate of 30 miles per hour and finally the remaining distance in 30 minutes, thereby averaging 25 miles per hour for the entire journey. Approximately how much is the distance travelled in the last phase for the journey?

A. 9
B. 10
C. 11
D. 15
E. 20
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Re: A man covers the first 5 miles of the journey at the rate of 25 miles [#permalink] New post   12 Jun 2019, 02:20
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Bunuel wrote:
A man covers the first 5 miles of the journey at the rate of 25 miles per hour, the next 10 miles at the rate of 30 miles per hour and finally the remaining distance in 30 minutes, thereby averaging 25 miles per hour for the entire journey. Approximately how much is the distance travelled in the last phase for the journey?

A. 9
B. 10
C. 11
D. 15
E. 20


Average speed=Total distance/Total time; (15+x)/(5/25+10/30+1/2)=25; 15+x/(1.03)=25; x=10.8 IMO C
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Re: A man covers the first 5 miles of the journey at the rate of 25 miles [#permalink] New post   12 Jun 2019, 02:22
5 miles = 25mph = 5/25*60 = 12 min
10 miles = 30mph = 10/30*60 = 20min
X miles = 30min.

Average speed = 25 = (5+10+x)/(12+20+30) = 15/62/60 ( to convert 62 min into hour, divide by 60)
Cross multiply, we get 31*25 = 30(15+x)
x = 325/30 = approx 11 miles.

C is the answer
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Re: A man covers the first 5 miles of the journey at the rate of 25 miles [#permalink] New post   12 Jun 2019, 02:30
Bunuel wrote:
A man covers the first 5 miles of the journey at the rate of 25 miles per hour, the next 10 miles at the rate of 30 miles per hour and finally the remaining distance in 30 minutes, thereby averaging 25 miles per hour for the entire journey. Approximately how much is the distance travelled in the last phase for the journey?

A. 9
B. 10
C. 11
D. 15
E. 20


Let the distance covered in 30 minutes = x.

Total distance = 5 + 10 + x
Total time = 5/25 + 10/30 + 30/60 (minutes to hour)
= 1/5 + 1/3 + 1/2 = (6 + 10 + 15)/30 = 31/30

Average speed = total distance/total time
—> 25 = (15 + x)/(31/30)
—> 25*31/30 = 15 + x
—> x = 155/6 - 15
—> x = 65/6 = 11 (approximately)

IMO Option C

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Re: A man covers the first 5 miles of the journey at the rate of 25 miles [#permalink] New post   17 Jun 2019, 17:50
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Bunuel wrote:
A man covers the first 5 miles of the journey at the rate of 25 miles per hour, the next 10 miles at the rate of 30 miles per hour and finally the remaining distance in 30 minutes, thereby averaging 25 miles per hour for the entire journey. Approximately how much is the distance travelled in the last phase for the journey?

A. 9
B. 10
C. 11
D. 15
E. 20


We can let d = the distance of the last phase of the journey and create the equation for the time traveled as:

5/25 + 10/30 + 0.5 = (5 + 10 + d)/25

Multiplying the equation by 150, we have:

30 + 50 + 75 = 6(15 + d)

155 = 90 + 6d

6d = 65

d = 10 ⅚ ≈ 11

Answer: C
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Re: A man covers the first 5 miles of the journey at the rate of 25 miles [#permalink] New post   24 Oct 2021, 04:35
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Re: A man covers the first 5 miles of the journey at the rate of 25 miles [#permalink]
24 Oct 2021, 04:35
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