Choose some smart number for the distance from Mayville to Rome: let the distance be 120 miles (120 is LCM of 30 and 40 and this will make further calculations easier).

\(Average \ speed=\frac{total \ distance}{total \ time}\) --> so the nominator, total distance would be \(2*120=240\);

Total time equals time from Mayville to Rome plus time from Rome to Mayville --> \(\frac{120}{30}+\frac{120}{40}=7\);

\(Average \ speed=\frac{total \ distance}{total \ time}=\frac{240}{7}\approx{34.3}\).

Answer: C.
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IMO C



Solution:

Average speed for the entire trip = 2uv/(u+v),
where u = speed of going
v = speed of returning

Thus, average speed = 2*40*30/(70) = 34.3 mi/h

Hence, C

It is pretty involuntary to come up with 35 miles per hour as the average speed in this case. But, that is wrong. Since, the car has taken more time to travel at 30 mph and less time to travel at 40 mph, the avg speed of the car would be somewhere between 30 and 35(mean) and closer to 35. If I had no time on my hands and had to make a calculated guess, I would have chosen C.

The correct way to find the avg speed is to calculate the total distance and the total time and divide the total distance by total time.

Let the one way distance in this case be D. So the total distance is 2D(travelled over the same route twice)

Total Time = Time taken to travel distance D at 30 mph + time taken to travel distance D at 40 mph
= \((D/30) + (D/40)\)
= \(7D/120\)
Thus avg speed = Total distance/total time
= \((2D)/(7D/120)\)
= 34.285

Hope it helps,
meshtrap

We can pick any number for distance that is divisible by both 30 and 40. Since we have rate and distance, by plugging in the rate formula (r=d/t) we can solve for the average rate/speed

City``` Rate--- Time ---Distance
M-R.... 30--- 4 ---- 120
R-M ... 40 --- 3 ---- 120
Total 34.28 --- 7 ---- 240

34.28 ~ 34.3
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This is just an approximation and round trip problem- we simply pick a value- which is usually the LCM

Total Distance= Avgspeed * Total Time

120/30= 4
120/40= 3

240= average speed * 7

240/7= 34.3

APPROACH #1: Apply the average speed formula
Average speed formula = (total distance travelled)/(total travel time)
If we let d = the distance from Mayville to Rome, then total distance travelled = 2d (since the car traveled from Mayville to Rome and then back)

time = distance/rate
So, the travel time from Mayville to Rome = d/30
Similarly, the travel time from Rome to Mayville = d/40
So, the total travel time = d/30 + d/40 = 4d/120 + 3d/120 = 7d/120

The average speed for the round trip= (2d)/(7d/120)
= (2d)(120/7d)
= 240d/7d
= 240/7 = 34 2/7 ≈ 34.3
Answer: C

APPROACH #2: Assign a "nice" distance between Mayville and Rome
So, let's say the distance between Mayville and Rome = 120 miles - since this distance that works well with 30 miles per hour AND 40 miles per hour

Time = distance/speed
So, time from Mayville and Rome (at 30 mph) = 120/30 = 4 hours
Time from Rome to Mayville (at 40 mph) = 120/40 = 3 hours

Average speed = (total distance traveled)/(total time)
= (120 + 120)/(4 + 3)
= 240/7 = 34 2/7 ≈ 34.3
Answer: C

Cheers.
Brent
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We can use the average rate formula: average = total distance/total time, and let the distance each way = 120 miles. (Note that we could use any number for the distance, but we have chosen the convenient number 120 because it is evenly divisible by both 30 and 40, making the arithmetic easier.)
average = 240/(120/30 + 120/40)

average = 240/(4 + 3) = 240/7 = 34.3

Answer: C
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Total distance = D + D = 2D

Total time = D/30 +D/40

Avg. Speed = Total distance / Total time

Avg. Speed =\(\frac{(2D)}{(D/30 + D/40)}\)

Avg. Speed = \(\frac{240D}{70D}\)

Avg. Speed = 34.2 approx

Hence (C)
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For Avg speed if two speeds are given formula is Avg speed=2*S1*S2/S1+S2=34.2

can you please explain how you reached to Avg. Speed = \(\frac{240D}{70D}\) ?

If the distance is equal in both trips, then
Average speed =2xy/(x+y)
=2*30*40/70
=34.3 miles/hr
x=speed in first trip
y= speed in second trip

Straight formula : If distance is same and different speeds.

Av speed: 2* 1/ (1/30 +1/40)
LCM of 30 and 40 is 120

2* 1/(4/120 +3/120)
2*120 /7 = 240/7 : 34.3

Answer :C

Hi VeritasKarishma,

I tried to solve this question using your weighted average method (below) and got a different answer. Could you please help me figure out why the method did not work in this case?

V1/V2 = 40-x/x-30 ----------- where x = average speed

30/40 = 40-x/x-30

on solving further, we get

x = 250/7 = 35.7 (this is in fact none of the options given)

Average speed (when distance is same / constant) = \(\frac{{2 * a * b} }{ {a + b}}\)

=> Average speed = \(\frac{{2 * 30 * 40} }{ {30 + 40}}\)

=> Average speed = \(\frac{{2 * 30 * 40} }{ {70}}\)

=> Average speed = 34.28 ≈ 34.3

Answer C
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What are V1 and V2 in your equation?
The weight when averaging speeds is time taken. Since the distance is the same in the two cases, the time taken would be in the ratio of inverse of speeds i.e. 4:3.

4/3 = 40-x/x-30
x = 34.3 mph

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2014/1 ... -averages/
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Given: A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour.
Asked: Of the following, which is closest to the average speed, in miles per hour, for the round trip?

Total distance = 2x
Total time = x/30 + x/40 = 7x/120

Average speed = 2x*120/7x = 240/7 = 34.3

IMO C
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