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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] New post 08 Aug 2012, 05:33
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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

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[Reveal] Spoiler: OA

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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 08 Aug 2012, 06:00
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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.
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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 08 Aug 2012, 06:05
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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.

Answer C
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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 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.

Answer; C.


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] New post 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.

Answer; C.


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] New post 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] New post 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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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 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?
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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 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

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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 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 Image


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.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 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.
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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 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.
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Re: Bill travels first 40% of the distance to his destination at [#permalink] New post 15 Sep 2014, 11:46
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Re: Bill travels first 40% of the distance to his destination at   [#permalink] 15 Sep 2014, 11:46
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