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# Last weekend, Sid visited his friend Mia who lives at 90 miles from hi

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Joined: 10 Oct 2018
Posts: 129
Location: United States
GPA: 4
WE: Human Resources (Human Resources)
Last weekend, Sid visited his friend Mia who lives at 90 miles from hi  [#permalink]

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Updated on: 18 Jan 2019, 08:19
1
00:00

Difficulty:

55% (hard)

Question Stats:

69% (03:22) correct 31% (03:31) wrong based on 15 sessions

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Last weekend, Sid visited his friend Mia who lives at 90 miles from his place. He divided his onward and return journey in different manner. He reached Veronica’s place by dividing the total distance in two equal part, at a speed of 15 mph and 30 mph respectively. He stayed there for 2 hours and took the return journey via the same path by dividing the total journey time in three equal parts, at a speed of 5 mph, 10 mph, and 15 mph respectively. If x, y, and z denote the average speed of his onward, return, and overall journey respectively, what is the value of x + y + z?

A) 10
B) 11.6
C) 16.25
D) 20
E) 41.6

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Originally posted by topper97 on 18 Jan 2019, 07:17.
Last edited by topper97 on 18 Jan 2019, 08:19, edited 1 time in total.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8891
Location: Pune, India
Re: Last weekend, Sid visited his friend Mia who lives at 90 miles from hi  [#permalink]

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18 Jan 2019, 08:15
topper97 wrote:
Last weekend, Sid visited his friend Mia who lives at 90 miles from his place. He divided his onward and return journey in different manner. He reached Veronica’s place by dividing the total distance in two equal part, at a speed of 15 mph and 30 mph respectively. He stayed there for 2 hours and took the return journey via the same path by dividing the total journey time in three equal parts, at a speed of 5 mph, 10 mph, and 15 mph respectively. If x, y, and z denote the average speed of his onward, return, and overall journey respectively, what is the value of x + y + z?

A) 10
B) 11.6
C) 16.25
D) 20
E) 41.6

Bunuel pls help. Thank-you in advance!

Average speed when equal distances are travelled for onward journey = 2ab/(a + b) = 2*15*30/(15 + 30) = 20 mph

Average speed when time taken is same for return journey = (5 + 10 + 15)/3 = 10 mph

Overall average speed when travelling for same distance (onward and return) = 2ab/(a + b) = 2*20*10/(20 + 10) = 40/3 = 13.33 mph

Sum of all speeds = 20 + 10 + 13.33 = 43.33 mph

Check this post for a discussion of this formula: https://www.veritasprep.com/blog/2015/0 ... -the-gmat/
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Karishma
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Joined: 10 Oct 2018
Posts: 129
Location: United States
GPA: 4
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Last weekend, Sid visited his friend Mia who lives at 90 miles from hi  [#permalink]

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18 Jan 2019, 08:36

Average speed when equal distances are travelled for onward journey = 2ab/(a + b) = 2*15*30/(15 + 30) = 20 mph

Average speed when time taken is same for return journey = (5 + 10 + 15)/3 = 10 mph

Overall average speed when travelling for same distance (onward and return) = 2ab/(a + b) = 2*20*10/(20 + 10) = 40/3 = 13.33 mph

Sum of all speeds = 20 + 10 + 13.33 = 43.33 mph

Check this post for a discussion of this formula: https://www.veritasprep.com/blog/2015/0 ... -the-gmat/

I had doubt from the highlighted part. Generally, outward and return speeds can be different then why did you take the average speed of both to find overall avg speed? Maybe I'm missing something

Also, for return journey, why didn't you use 3abc/(ab + bc + ca) as the total journey time is in three equal parts.
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Last weekend, Sid visited his friend Mia who lives at 90 miles from hi   [#permalink] 18 Jan 2019, 08:36
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