Guest Report post Posted September 15, 2007 I will still be disagreeing with this and submit that "halfway to town" could be 30 minutes if the answer that is being sought is the average speed for one hour. Therefore, the answer would be 90km/hr. Share this post Link to post Share on other sites

Guest Report post Posted September 15, 2007 I will still be disagreeing with this and submit that "halfway to town" could be 30 minutes if the answer that is being sought is the average speed for one hour. Therefore, the answer would be 90km/hr. No. If you go halfway to town (30 km) at 30 km/hr, it will take 60 minutes. The riddle states that that's what's done; there is no going half way to town in 30 minutes. Share this post Link to post Share on other sites

Guest Report post Posted September 15, 2007 I understand what the riddle states about halfway. Maybe, the breakdown starts in the line the rest of the way to have the average speed of the entire way 60 km/hour? I understand rest of the way... 30km I understand average speed... 60km/hr "of the entire way" what does that mean. AND... I still don't get the mysterious hour in which the trip seems to need to take. Share this post Link to post Share on other sites

Guest Report post Posted September 15, 2007 "of the entire way" what does that mean. That just means to average 60km/hr over the entire 60 km. AND... I still don't get the mysterious hour in which the trip seems to need to take. If the trip is 60 km and you are to average 60/km an hour, the trip must take exactly one hour. Share this post Link to post Share on other sites

Guest Report post Posted October 1, 2007 With the question as originally stated, I personally believe the answer is infinity, since the question was "how fast". However, if you want to ignore the "implied" requirement of a fixed distance of 60km and answer the alternate question of "how far, and at what speed" (the detour alternative), then you have an infinite set of ordered rate/distance pairs which solve the problem. Where: the rate > 60km/hr and distance > 30km. While it is an interesting and un-arguable fact that there are an infinite number of solutions to the alternate question, I find the original question to be more interesting as it has only one solution. Share this post Link to post Share on other sites

Guest Report post Posted October 4, 2007 u dont have that much distance to cover, and make the average 60, got it? (its just like -> day by day[day/day=1 and its !0, so u are not zero's ] u ppl are getting increasingly brainy[pun intended]) Share this post Link to post Share on other sites

Guest Report post Posted October 10, 2007 "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) " Average speed = Distance Traveled / Time of Travel Average speed = 30km + D / 1hr + T D=30KM of remaining distance T=0 Basically you would have to travel 30km in 0 amount of time... so the answer is not currently possible. To go exactly 60km at an average speed of exactly 60km/h you would have to travel at exactly 60km/h the entire way.... Share this post Link to post Share on other sites

Guest Report post Posted October 15, 2007 I just had to respond to this teaser. For those who need to create visual supports to understand, imagine two people, car 1 & car 2, leaving point A side-by-side. Point A and point B are 60km apart. Car 1 travels at 60km/h and arrives at point B in one hour. Car 2 was driving at 30km/hr, so he is only halfway to point B when car 1 arrives. Car 1 was traveling at 60km/h, there is no way for car 2 to catch-up to arrive at point B at the same time as car 1. Even if car 2 traveled at 90km/h the second half of the way, he still would be slower than an average of 60km/h. Another possible solution: The car crosses a time line half way. So the car travels 30kmh the 1st hr, crosses the time line, then maintains 30kmh the second hr. According to the clocks at point A and point B, it would have taken 1 hr to travel 60km. This may be bending the rules, thinking outside the box rather, but it is more logical than a detour solution. Share this post Link to post Share on other sites

Guest Report post Posted November 10, 2007 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? but it says how fast do you have to go for the rest of the way to have the average speed of the entire way 60 km/r. so you have to make the average 120, thus going 90 km/h. Share this post Link to post Share on other sites

Guest Report post Posted November 12, 2007 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? u can't..u already use the 1hr in going 30kms.... Share this post Link to post Share on other sites

Guest Report post Posted December 3, 2007 Well, I didn't read all 9 pages of replies so far, so I may have missed this solution a few times. If you travel the remaining distance at a speed of 30km/second, you will add 1 second to your time, which, when averaging, will round nicely to 60km/hour. Ahem. Now, where to get a vehicle with rocket engines... Share this post Link to post Share on other sites

Guest Report post Posted December 6, 2007 Consider this posibble solution. Given, my algebra is a little rusty. Is the solution to travel at 60km ph for the last 30km? If you go 30km ph for 60 minutes, you cover 30km, or .5km per minute. If you go at 60km ph for 30 minutes, you cover 30km, or 2km per minute. You traveled a total of 60km in 1.5 hours, or 90 minutes, at an added 90km ph. Which gives you an average of 1km per minute, which is 60km ph. Share this post Link to post Share on other sites

Guest Report post Posted December 11, 2007 Hmm... you know, this looks possible. Lets see..... For the first part of the journey, he travelled 30km at a speed of 30 kmh. Time in the inertial frame of reference: 1 hour unknown time = Time with respect to observer in inertial frame of reference sqrt[1-(30kmh^2/c^2)] unknown time: 59999999999.999699584484865522271 nanoseconds Which gives him 0.00030041551513447773 nanoseconds to cover 30 km =0.0000000000000050069252522412955 hours to cover 30 km =5991701191578769.8403966508288831 kmh He only needs to travel at 5991701191578769.8403966508288831 kmh to average 60kmh. =) fuzzywallaby is on the right track to giving the only mathematically teneble solution if we do not allow detours or turn-arounds. For the person in the stationary reference frame, the problem is indeed impossible. But for the (hypothetically indestructible) person making the trip (with instantaneous acceleration and arbitrarily large amounts of energy), the situation is different. Specifically, less time has passed for the moving person, and thus s/he can still accelerate to some finite speed that will allow him/her to traverse the remaining distance in an arbitrarily short period. v = 30 km/h = 8 1/3 m/s (exact) c = 299792458 m/s (exact) (v/c)^2 = 7.7267365003723502234311803114445 e -16 (approximate) 1 - (v/c)^2 = 0.99999999999999922732634996276498 (appx) sqrt(1-(v/c)^2) = 0.999999999999999613663174981382 (appx) 3600 seconds (= 1 hour) times above factor: 3599.9999999999986091874299329767 seconds Subtracting this from one hour (3600 seconds) leaves 1.39081257 e -12 second, approximately, to finish the trip. Let's call this t(remaining). Now of course, this is far too short a time to travel said distance even at the speed of light. But remember, the distance is foreshortened for someone traveling at v very close to c. So what we want is to travel at some velocity v such that t(relativistic) = t(remaining). So t(remaining) (= 1.39 e -12) = sqrt(1 - (v/c)^2) (1.39 e-12)^2 = 1 - (v/c)^2 v = c * sqrt(1 - (1.39 e-12)^2) = 0.99999999999999999999999903282 c = 299792457.9999999999999997100468 m/s Share this post Link to post Share on other sites

Guest Report post Posted December 21, 2007 ok speed up to the intended time frame 60mph,then runaround the town tell you hours almost up then run in at presisly 1 hour...i win hahaha Share this post Link to post Share on other sites

Guest Report post Posted December 21, 2007 v = c * sqrt(1 - (1.39 e-12)^2) = 0.99999999999999999999999903282 c More conveniently represented as (1 - 9.6718 e-25)c, or about (1 - 10^-24)c. Very, very fast, but less than c. Share this post Link to post Share on other sites

Guest Report post Posted January 18, 2008 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. Share this post Link to post Share on other sites

Guest Report post Posted January 30, 2008 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. Share this post Link to post Share on other sites

rookie1ja 11 Report post Posted January 30, 2008 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) Share this post Link to post Share on other sites

Guest Report post Posted January 31, 2008 (edited) 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 January 31, 2008 by Yoshia Share this post Link to post Share on other sites

rookie1ja 11 Report post Posted January 31, 2008 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 Share this post Link to post Share on other sites

Guest Report post Posted February 6, 2008 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. Share this post Link to post Share on other sites

rookie1ja 11 Report post Posted February 6, 2008 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 Share this post Link to post Share on other sites

Guest Report post Posted February 7, 2008 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! Share this post Link to post Share on other sites

rookie1ja 11 Report post Posted February 7, 2008 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 Share this post Link to post Share on other sites

Guest Report post Posted February 7, 2008 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 1 Share this post Link to post Share on other sites