During a trip Michael traveled x% of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine´s average speed for the entire trip?
A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12000/(x+200)
Average speed question
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I'll suggest two approaches you could take here. We can certainly solve algebraically:jfranco23 wrote:During a trip Michael traveled x% of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine´s average speed for the entire trip?
A. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12000/(x+200)
I'd normally make a table to solve these problems -- with three columns: first part of the trip; second part of the trip; total for the trip -- and three rows: distance, time and speed. I'm not sure how to display that here, so I'll do each step one at a time:
Let D be the total distance.
For the first part of the trip we have:
distance = (x/100)D
speed = 40
time = d/s = xD/4000
For the second part of the trip we have:
distance = [(100 - x)*D]/100
speed = 60
time = d/s = [(100 - x)*D]/6000
For the total trip we have:
distance = D
time = xD/4000 + [(100 - x)*D]/6000 = [3xD + 200D -2xD]/12000 = (xD + 200D)/12000
speed = d/t = D/[(xD + 200D)/12000] = 12000D/[D*(x + 200)] = 12000/(x + 200)
Alternatively, if one of the answers is right, it must be right for any reasonable value of x (between 0 and 100). You could, for example, solve the problem for some specific value of x, then plug that value into each answer choice to see which gives you the right answer. The lone problem there is that two answer choices might give you the right answer, and you might need to do this all over again to see which is right. For that reason, it's best to use very simple choices for x to rule out as many answers as possible.
Luckily we don't even need to solve the problem for any particular number here. We can see immediately that if x = 0, the answer must be 60 mph. Plugging x = 0 into each answer choice, only C and E give the right answer (60). We now need to decide which of those is correct, and unfortunately, plugging in x = 100 doesn't help to decide. However, if x = 50, then half the distance is driven at 40 mph, and half at 60 mph. The average speed must then be less than 50 mph, because more time will be spent at the slower speed. Plugging in x = 50 rules out answer choice C, leaving us with E.
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Great solution, Ian!
On the same note take x=50 (no restriction to not to do so)
When equal distances are covered at diffferent speeds/same speeds the shortcut to find average speed is
2xy/ x+y where x,y are the speeds
If we take x=50
We get average speed = 48
When u substitute x=50 only E gives us 48
If u can think like Ian on the GMAT then QUANT is a cakewalk
Regards,
CR
On the same note take x=50 (no restriction to not to do so)
When equal distances are covered at diffferent speeds/same speeds the shortcut to find average speed is
2xy/ x+y where x,y are the speeds
If we take x=50
We get average speed = 48
When u substitute x=50 only E gives us 48
If u can think like Ian on the GMAT then QUANT is a cakewalk
![Very Happy :D](./images/smilies/grin.png)
Regards,
CR