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# Bill travels first 40% of the distance to his destination at

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Bill travels first 40% of the distance to his destination at [#permalink]  08 Aug 2012, 05:33
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25% (low)

Question Stats:

74% (02:32) correct 25% (01:26) wrong based on 147 sessions
Bill travels first 40% of the distance to his destination at 80 mph and covered the balance distance at 40 mph. What is the average speed of his travel?

A) 45
B) 38
C) 50
D) 52
E) 55

Source: 4gmat
[Reveal] Spoiler: OA

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Re: Bill travels first 40% of the distance to his destination at [#permalink]  08 Aug 2012, 06:00
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Expert's post
SOURH7WK wrote:
Bill travels first 40% of the distance to his destination at 80 mph and covered the balance distance at 40 mph. What is the average speed of his travel?

A) 45
B) 38
C) 50
D) 52
E) 55

Source: 4gmat

Say total distance is 100 miles (first part 40 miles and second part 60 miles), then average \ speed=\frac{total \ distance}{total \ time}=\frac{100}{\frac{40}{80}+\frac{60}{40}}=50.

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Re: Bill travels first 40% of the distance to his destination at [#permalink]  08 Aug 2012, 06:05
2
KUDOS
SOURH7WK wrote:
Bill travels first 40% of the distance to his destination at 80 mph and covered the balance distance at 40 mph. What is the average speed of his travel?

A) 45
B) 38
C) 50
D) 52
E) 55

Source: 4gmat

Average speed can be computed as (total distance)/(total time).
In our case, if we denote by D the distance, then the average speed is given by:

S_a=\frac{D}{\frac{0.4D}{80}+\frac{0.6D}{40}}=...=50
The distance cancels out in the above computations.

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Senior Manager
Joined: 15 Jun 2010
Posts: 363
Schools: IE'14, ISB'14, Kellogg'15
WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)
Followers: 5

Kudos [?]: 106 [0], given: 50

Re: Bill travels first 40% of the distance to his destination at [#permalink]  08 Aug 2012, 06:09
Bunuel wrote:

Say total distance is 100 miles (first part 40 miles and second part 60 miles), then average \ speed=\frac{total \ distance}{total \ time}=\frac{100}{\frac{40}{80}+\frac{60}{40}}=50.

Can we put it into any kind of formula to solve this kind of problem Quickly???
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Re: Bill travels first 40% of the distance to his destination at [#permalink]  08 Aug 2012, 06:31
Expert's post
SOURH7WK wrote:
Bunuel wrote:

Say total distance is 100 miles (first part 40 miles and second part 60 miles), then average \ speed=\frac{total \ distance}{total \ time}=\frac{100}{\frac{40}{80}+\frac{60}{40}}=50.

Can we put it into any kind of formula to solve this kind of problem Quickly???

Well, formula can be derived, though I don't think that memorizing such kind of formulas is more time efficient than understating the logic behind it.

If the total distance is d=m+n kilometers, and m kilometers is covered at r_1 kilometers per hour while n kilometers is covered at r_2 kilometers per hour, then average \ speed=\frac{total \ distance}{total \ time}=\frac{m+n}{\frac{m}{r_1}+\frac{n}{r_2}}.
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Re: Bill travels first 40% of the distance to his destination at [#permalink]  20 Jan 2013, 19:45
Why doesn't weighted average formula work in this case?

(80 * 0.4) + (40 * 0.6) = 32 + 24 = 56

or

let average speed be s.

(80-s)/(s-40) = 3/2
s = 56
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Re: Bill travels first 40% of the distance to his destination at [#permalink]  20 Jan 2013, 20:03
Expert's post
Apex231 wrote:
Why doesn't weighted average formula work in this case?

(80 * 0.4) + (40 * 0.6) = 32 + 24 = 56

or

let average speed be s.

(80-s)/(s-40) = 3/2
s = 56

Because in the weighted average formula, when you are trying to find the average speed, the weights will be the time taken, not the distance traveled. Here 0.4 and 0.6 is the distance traveled.
You have to be careful when you are choosing the weights.

Check out a related discussion here:
a-certain-car-traveled-twice-as-many-miles-from-town-a-108368.html#p859543
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Manager Joined: 03 Oct 2009 Posts: 64 Followers: 0 Kudos [?]: 13 [0], given: 8 Re: Bill travels first 40% of the distance to his destination at [#permalink] 20 Jan 2013, 23:07 I am still not sure of how to correctly identify whether time or distance be considered as value/weight in weighted avg formula. Is there a generic rule to identify this? Manager Joined: 04 Jan 2013 Posts: 82 Followers: 0 Kudos [?]: 4 [0], given: 1 Re: Bill travels first 40% of the distance to his destination at [#permalink] 21 Jan 2013, 03:20 Bunuel wrote: SOURH7WK wrote: Bill travels first 40% of the distance to his destination at 80 mph and covered the balance distance at 40 mph. What is the average speed of his travel? A) 45 B) 38 C) 50 D) 52 E) 55 Source: 4gmat Say total distance is 100 miles (first part 40 miles and second part 60 miles), then average \ speed=\frac{total \ distance}{total \ time}=\frac{100}{\frac{40}{80}+\frac{60}{40}}=50. Answer; C. @Bunuel,is it true that we always solve average speed problems by plugin method? :O Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 16852 Followers: 2774 Kudos [?]: 17607 [0], given: 2226 Re: Bill travels first 40% of the distance to his destination at [#permalink] 21 Jan 2013, 03:34 Expert's post chiccufrazer1 wrote: Bunuel wrote: SOURH7WK wrote: Bill travels first 40% of the distance to his destination at 80 mph and covered the balance distance at 40 mph. What is the average speed of his travel? A) 45 B) 38 C) 50 D) 52 E) 55 Source: 4gmat Say total distance is 100 miles (first part 40 miles and second part 60 miles), then average \ speed=\frac{total \ distance}{total \ time}=\frac{100}{\frac{40}{80}+\frac{60}{40}}=50. Answer; C. @Bunuel,is it true that we always solve average speed problems by plugin method? :O Posted from my mobile device No, definitely not ALWAYS. But when fractions and/or percentages are involved to define distance for example, then we can assign some number to it and proceed this way. _________________ Manager Joined: 03 Oct 2009 Posts: 64 Followers: 0 Kudos [?]: 13 [0], given: 8 Re: Bill travels first 40% of the distance to his destination at [#permalink] 22 Jan 2013, 21:19 Apex231 wrote: I am still not sure of how to correctly identify whether time or distance be considered as value/weight in weighted avg formula. Is there a generic rule to identify this? Can someone explain why time and not distance is considered in weighted avg formula for speed?; just trying to understand concepts. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4028 Location: Pune, India Followers: 858 Kudos [?]: 3613 [0], given: 144 Re: Bill travels first 40% of the distance to his destination at [#permalink] 22 Jan 2013, 21:28 Expert's post Apex231 wrote: Apex231 wrote: I am still not sure of how to correctly identify whether time or distance be considered as value/weight in weighted avg formula. Is there a generic rule to identify this? Can someone explain why time and not distance is considered in weighted avg formula for speed?; just trying to understand concepts. Speed is measured in miles/hr. The weights will be in 'hr' Look at the weighted average formula: Cavg = (C1*w1 + C2*w2)/(w1 + w2) = miles/hr = (miles/hr * hr + miles/hr * hr) / (hr + hr) = (miles + miles)/(hr + hr) i.e. total distance/total time weights cannot be distance: miles/hr = (miles/hr * miles + miles/hr * miles) / (miles + miles) -> miles^2/hr is no physical quantity. Did you check out the link I gave you above? It discusses this concept with miles/gallon. Weights should be in gallons here. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Bill travels first 40% of the distance to his destination at   [#permalink] 22 Jan 2013, 21:28
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