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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]

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08 Aug 2012, 06:33
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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

Source: 4gmat
[Reveal] Spoiler: OA

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

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08 Aug 2012, 07:00
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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 $$average \ speed=\frac{total \ distance}{total \ time}=\frac{100}{\frac{40}{80}+\frac{60}{40}}=50$$.

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

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08 Aug 2012, 07:05
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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.

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

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08 Aug 2012, 07:09
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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]

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08 Aug 2012, 07:31
Expert's post
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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]

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20 Jan 2013, 20:45
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]

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20 Jan 2013, 21:03
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 03 Oct 2009 Posts: 62 Followers: 0 Kudos [?]: 115 [0], given: 8 Re: Bill travels first 40% of the distance to his destination at [#permalink] ### Show Tags 21 Jan 2013, 00:07 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: 80 Followers: 0 Kudos [?]: 11 [0], given: 1 Re: Bill travels first 40% of the distance to his destination at [#permalink] ### Show Tags 21 Jan 2013, 04:20 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: 39061 Followers: 7755 Kudos [?]: 106562 [0], given: 11628 Re: Bill travels first 40% of the distance to his destination at [#permalink] ### Show Tags 21 Jan 2013, 04:34 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: 62 Followers: 0 Kudos [?]: 115 [0], given: 8 Re: Bill travels first 40% of the distance to his destination at [#permalink] ### Show Tags 22 Jan 2013, 22:19 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: 7380 Location: Pune, India Followers: 2292 Kudos [?]: 15164 [0], given: 224 Re: Bill travels first 40% of the distance to his destination at [#permalink] ### Show Tags 22 Jan 2013, 22:28 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 Get started with Veritas Prep GMAT On Demand for$199

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

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15 Sep 2014, 12:46
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Re: Bill travels first 40% of the distance to his destination at [#permalink]

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04 Oct 2014, 13:50
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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?

When we have these Average Rate for the whole trip problems, the key is to understand that:
- We only need the Rates of both part of the trips.
- We can plug in for Distance and infer for Time
- Then, you just have to solve the simple formula (Distance 1 + Distance 2) / (Time 1 + Time 2)

I´ll tell you how I did this in around 75 seconds with the RTD chart

Question Stem
R * T = D
1st Part: 80 * ? =(0.4) d
2nd Part: 40 * ? =(0.6) d

Note that Rates, Times, and Distances ARE ALL DIFFERENT in each part of the trip.

Next, come up with some Distances to plug in (400 and 600 are a safe bet here)
R * T = D
1st Part: 80 * ? = 400
2nd Part: 40 * ? =600

Now, deduce the Times for each part of the trip
R * T = D
1st Part: 80 * 5 = 400
2nd Part: 40 * 15 =600

Now, solve the Average Rate Formula
(400 + 600) / (5 + 15) = 1,000/20 = 50

I know it looks time consuming but this is just because I broke every step down. You obviously don´t need to do 3 charts: it is just the same chart in each step of the resolution.

Hope it helps!
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Re: Bill travels first 40% of the distance to his destination at [#permalink]

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13 Dec 2015, 06:27
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Re: Bill travels first 40% of the distance to his destination at   [#permalink] 13 Dec 2015, 06:27
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