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

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!

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 \(average \ speed=\frac{total \ distance}{total \ time}=\frac{m+n}{\frac{m}{r_1}+\frac{n}{r_2}}\).

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

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

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?

@Bunuel,is it true that we always solve average speed problems by plugin method? :O

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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.

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.

Average speed =
total distance / total time
Let distance =100
Therefore
100/(40/80 + 60/40)

=100/(0.5+1.5)
= 50
Option C is the answer

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