159 replies to this topic

[Naniki]

Ok, i have read a bunch of replies, but not all but, isnt this question not even valid, you are missing info. How much time do you have left to complete the trip, a lot are assuming it was just the hour. but wouldn't the answer be any kph over 90km/h (ex. 90.0000000000000001 km/h) in the end average out. Given at the example and just adjusting it for time spend traveling, the unknown variable? the km/h is dependent, so this answer should be undefined? Please dont yell at me if this is just all wrong, but time remaining is unknown, unless their is some rule I am not aware of.

[Yoshia]

Yeah, here is the part your not quite getting: the average is not calculated by speed over distance. Your model produces an average of 60(km/h)/km. The average we're looking for is calculated by distance over time(km/h). Because 90km/h is faster then 30km/h it takes less time to cover the distance remaining. So you are going slow for a long time, and fast for a short time. So you don't get to spend 30 minutes at 90mph to bring the average up if you are traveling in a straight line. To spend 30 minutes at 90km/h you must overshoot your mark by 15km. I hope that helps clear things up a little.

You've brought up the point I was going to make. The wording of this question led me to believe it was asking for the average speed over the distance. Here was the question:

"If I go halfway to the town (which is 60 km away) at the speed of 30 km/hour, how fast do I have to go for the rest of the way to have the average speed of the entire way 60 km/hour?

Edit: "rest of the way" means to the town and not an inch farther and the total distance traveled has to be exactly 60 km (this is just to explain how I meant the riddle to be understood)"

It certainly sounds like the speed is a possession of the "entire way."

If half the distance was traveled at 30 km/hr and the other half of the distance traveled at 90 km/hr... then the average speed over that distance is 60km/hr. This satisfies the question. The question should have been worded differently if it meant to only imply that all it wanted was [ (30km first half) + (30km second half)] / [(60 min first half + mystery time second half)] to equal 60km/60min.... or in simpler terms, the question didn't say "How long should it take you to travel the remaining distance in order to have traveled at a speed of 60km/hr (average) the whole way?"

So 90km/hr is a valid and correct answer.

[rookie1ja]

If half the distance was traveled at 30 km/hr and the other half of the distance traveled at 90 km/hr... then the average speed over that distance is 60km/hr.

I am missing your logic somehow ... let's say the distance from the starting point to the city is 60 km, then you say:
If half the distance (30 km) was traveled at 30 km/hr (so 1 hour) and the other half (remaining 30 km) of the distance traveled at 90 km/hr (so 20 minutes) ... then the average speed over that distance (60 km) is 60km/hr (60 km in 80 minutes ... so it's not 60 km in 60 minutes)

[Yoshia]

I am missing your logic somehow ... let's say the distance from the starting point to the city is 60 km, then you say:
If half the distance (30 km) was traveled at 30 km/hr (so 1 hour) and the other half (remaining 30 km) of the distance traveled at 90 km/hr (so 20 minutes) ... then the average speed over that distance (60 km) is 60km/hr (60 km in 80 minutes ... so it's not 60 km in 60 minutes)

Imagine the distance as a 60km line of infinite points. At half of those points, your speed is 30 km/hr (regardless of the time you spent doing it) and at the other half of those points your speed is 90 km/hr (also regardless of time spent doing it). The question was worded in a way that would mean it was asking for the (average speed) of (the entire way)... meaning:

sum of speeds at every point/ every point = 60km/hr (the average speed OF THE DISTANCE) and a satisfied question with the answer of 90.

Edited by Yoshia, 31 January 2008 - 08:29 PM.

[rookie1ja]

Imagine the distance as a 60km line of infinite points. At half of those points, your speed is 30 km/hr (regardless of the time you spent doing it) and at the other half of those points your speed is 90 km/hr (also regardless of time spent doing it). The question was worded in a way that would mean it was asking for the (average speed) of (the entire way)... meaning:

sum of speeds at every point/ every point = 60km/hr (the average speed OF THE DISTANCE) and a satisfied question with the answer of 90.

I get your points theory and as mentioned - you would really have to exclude the time factor (which is the most important - it's all about the time)

the whole puzzle seemed so clear and simple to me ... until I posted it in this forum

[superbc]

It is possible since it's an average. Your first leg was 30kph the second leg would have to be 90kph. Your AVERAGE would be 60kph, it doesn't state it has to be completed in the hour stated. When I drive to town at 50kph but I'm only in the car for 20 minutes I don't adjust my figures--it's an average.

[rookie1ja]

It is possible since it's an average. Your first leg was 30kph the second leg would have to be 90kph. Your AVERAGE would be 60kph, it doesn't state it has to be completed in the hour stated. When I drive to town at 50kph but I'm only in the car for 20 minutes I don't adjust my figures--it's an average.

pls show me how you get the average speed 60 kph when you go half way at the speed of 30 kph ... choose any distance (some specific number in km), any total time (some specific number in hours - it does not have to be 1 hour - choose any) ... let's see if the average of 60 kph is possible

[Emapher]

pls show me how you get the average speed 60 kph when you go half way at the speed of 30 kph ... choose any distance (some specific number in km), any total time (some specific number in hours - it does not have to be 1 hour - choose any) ... let's see if the average of 60 kph is possible

I................/\.................I
/\This will rep. half way
which we can agree has 30km on each side.

I................/\.................I

On the first half, the driver goes
30kph which we can agree will take 1hr.

I................/\.................I

On the second half the driver moves at a speed of 90kph which i hope we can agree is 20 min.



Now had to origianl question asked - If I go halfway to the town (which is 60 km away) at the speed of 30 km/hour, how fast do I have to go for the rest of the way to have the average speed of the entire way 60 km/hour and an average time of 1 hour?
Then yes you would most assuredly be correct. However, the question is " how fast will she have to travel the rest of the way to have an average speed of 60kph?"

She will have to travel 90kph, as 90 and 30 average 60!

[rookie1ja]

to Emapher:
in your example you would travel (the whole way) 60 km in 80 minutes (the whole time) which is not the desired average speed of 60 km/hour
try using any other distance (which would not make the approximately 1 hour you mentioned) and see what happens

[Xspyder]

I stopped reading all the pages of up 4 when complete idiocy took the field over, much like a brawl at a soccer game,
or maybe an election, none are listening to the issues and most have lost the point. "Elduderino2085" has/had it correct
perhaps though he too was infected with the chaotic spell cast amongst the hurried crowd. Read ja's lips, exactly 60 miles, and not an inch more, thus defining parameters of the statement and the outcome of the answer. NO TIME LIMIT is mentioned and therefore there is exactly no time limit, not one second more nor less. Several had the answer there
within the first several pages of complete mayhem with the squares and roots and times all agagle over nought. It is 30 for the first half and then 90 for the second half distance equalling an average of 60 Km / hour, and not 60 km/ within an hour. "How I love thee, how I love thee.....Brain Den Maestros. Hey, I haven't had my Martini yet. What a beauuutifulll thing is the round table, oh and sweet the medecine that salves the spirit, yes it is so to think not in parrallel
with horse blinders on, though many could have been bridled with such, and perhaps some wax for the ears also. I'm sure more than several stable horse won the race laughing as most went careening into and off of the quide rails. RSVP