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Juan bikes halfway to school at 9 miles per hour, and walks the rest

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Bunuel
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Juan bikes halfway to school at 9 miles per hour, and walks the rest [#permalink] New post   01 Jul 2019, 01:50
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Juan bikes halfway to school at 9 miles per hour, and walks the rest of the distance at 3 miles per hour. What is Juan's average speed for the whole trip?

A. 4 mph
B. 4.5 mph
C. 5 mph
D. 5.5 mph
E. 6 mph
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B
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CareerGeek
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Juan bikes halfway to school at 9 miles per hour, and walks the rest [#permalink] New post   Updated on: 01 Jul 2019, 01:56
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Bunuel wrote:
Juan bikes halfway to school at 9 miles per hour, and walks the rest of the distance at 3 miles per hour. What is Juan's average speed for the whole trip?

A. 4 mph
B. 4.5 mph
C. 5 mph
D. 5.5 mph
E. 6 mph


Let the distance between school and home = 2x
He travels x miles @9 mph speed and remaining x miles @ 3 mph speed

Total time = \(t_1\) + \(t_2\)
= x/9 + x/3
= (x + 3x)/9
= 4x/9

\(Average Speed = \frac{Total Distance}{Total Time}\)
= 2x/(4x/9)
= 2*9/4
= 4.5 mph

IMO Option B

Pls Hit kudos if you like the solution

Originally posted by CareerGeek on 01 Jul 2019, 01:56.
Last edited by CareerGeek on 01 Jul 2019, 01:56, edited 1 time in total.
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Re: Juan bikes halfway to school at 9 miles per hour, and walks the rest [#permalink] New post   01 Jul 2019, 01:56
Bunuel wrote:
Juan bikes halfway to school at 9 miles per hour, and walks the rest of the distance at 3 miles per hour. What is Juan's average speed for the whole trip?

A. 4 mph
B. 4.5 mph
C. 5 mph
D. 5.5 mph
E. 6 mph


Let us assume distance = 2d
Total time taken = \(\frac{d}{9} + \frac{d}{3} = \frac{4d}{9}\)
Average speed = \(2d/\frac{4d}{9} = \frac{9}{2} = 4.5 mph\)

IMO B
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Re: Juan bikes halfway to school at 9 miles per hour, and walks the rest [#permalink] New post   01 Jul 2019, 09:45
Bunuel wrote:
Juan bikes halfway to school at 9 miles per hour, and walks the rest of the distance at 3 miles per hour. What is Juan's average speed for the whole trip?

A. 4 mph
B. 4.5 mph
C. 5 mph
D. 5.5 mph
E. 6 mph


time for first half distance ; x/2*9; x/18 and other half = x/2*3 ; x/6

total avg speed ; x/x/18 + x/6 ; 18x /4x; 4.5
IMO B
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Re: Juan bikes halfway to school at 9 miles per hour, and walks the rest [#permalink] New post   02 Jul 2019, 18:51
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Bunuel wrote:
Juan bikes halfway to school at 9 miles per hour, and walks the rest of the distance at 3 miles per hour. What is Juan's average speed for the whole trip?

A. 4 mph
B. 4.5 mph
C. 5 mph
D. 5.5 mph
E. 6 mph

We can let each distance = 9, and can use the formula average = (total distance)/(total time):

average = 18/(9/9 + 9/3)

average = 18/4

Average= 4.5

Answer: B
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Re: Juan bikes halfway to school at 9 miles per hour, and walks the rest [#permalink] New post   02 Jul 2019, 21:27
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Bunuel wrote:
Juan bikes halfway to school at 9 miles per hour, and walks the rest of the distance at 3 miles per hour. What is Juan's average speed for the whole trip?

A. 4 mph
B. 4.5 mph
C. 5 mph
D. 5.5 mph
E. 6 mph


When distance travelled is the same,

Avg Speed = 2ab/(a + b) = 2*9*3/(9 + 3) = 4.5 mph

Answer (B)

Check this post for more: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/0 ... -the-gmat/
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Re: Juan bikes halfway to school at 9 miles per hour, and walks the rest [#permalink] New post   03 Jul 2019, 03:20
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Bunuel wrote:
Juan bikes halfway to school at 9 miles per hour, and walks the rest of the distance at 3 miles per hour. What is Juan's average speed for the whole trip?

A. 4 mph
B. 4.5 mph
C. 5 mph
D. 5.5 mph
E. 6 mph


The same distance is traveled at each speed.
For every hour traveled at 9 mph, 3 hours must be traveled at 3 mph to cover the same distance.
Average speed per 4 hours \(=\frac{9+(3*3)}{4} = \frac{18}{4} = 4.5\)

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Re: Juan bikes halfway to school at 9 miles per hour, and walks the rest [#permalink]
03 Jul 2019, 03:20
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