Guest Posted April 17, 2008 Report Share Posted April 17, 2008 I don't know how you possibly took my question as derogatory. I premised the question by saying I didn't mean it to be IN ANY WAY OFFENSIVE. Sorry I didn't add a smiley face our add lol. So if Martini found asking "who the hell he was" offensive I apologize to him/or her. Anybody else is obviously LOOKING for something to get uppity about and I'm not real concerned. Scraff didn't say he found your question derogatory; he asked why you would ask "but who the hell is MARTINI." I'm not offended and I don't find that necessarily derogatory, but I'm also puzzled as to why you would ask that. Was it something I said? Link to comment Share on other sites More sharing options...
Guest Posted April 19, 2008 Report Share Posted April 19, 2008 Scraff didn't say he found your question derogatory; he asked why you would ask "but who the hell is MARTINI." I'm not offended and I don't find that necessarily derogatory, but I'm also puzzled as to why you would ask that. Was it something I said? I only asked because your presence every where and uber-intelligence scared me. Its like knowing some ones there but not sure if its an omnipotent god or a blood crazed ax murderer. I asked the question because as I was reading posts I was wondering just that. "Who the hell is Martini". Who is this person that knows all the answers and can make me look so stupid and, unlike other posters, do it so tactfully. I've read numerous remarks by what I know now are other moderators that I thought were snippy and arrogant. You've never been that way and I admire that. Obviously I don't have that same tact. At the time I didnt realize you were a moderator. I have to sroll over on my screen to see that edge of the page. Link to comment Share on other sites More sharing options...
Guest Posted April 19, 2008 Report Share Posted April 19, 2008 Aww, schucks. I just got a warm fuzzy. Thanks for the compliments, tawanna. Alright folks, move along. Nothing to see here. Link to comment Share on other sites More sharing options...
Guest Posted May 21, 2008 Report Share Posted May 21, 2008 Sorry to post to a puzzle that's already been beaten to death. The intended solution ("it's impossible") has been questioned & explained enough; I don't have a thing to add there. I want to go back to the alternate solution ("detour") and look at a different angle. If the town is 60km away, and I go half way, it's assumed that I've gone 30km. The detour solution then suggests that I would take a 90km route (for example) to get the rest of the way to town, thereby averaging 60km/h. However, the original post does not state that the first portion of the journey is 30km, it only says "halfway". If I take the suggested detour, my entire trip is 120km, so halfway is 60km. At 30km, this has taken me 2 hours. If I then go 90km/h for the remaining 60km, I'll have driven a total of 120km in 2.667 hours... which doesn't average 60km/h. Same math problem we started with, which I didn't mean to bring up My point is not about the math. I just wanted to note that anyone suggesting the detour logic should use that logic from start to finish. One shouldn't (in my opinion) start with straight-route logic (halfway = 30km) and then suddenly change the base assumptions for the remainder of the explanation. I'm curious whether others would read "halfway" as a variable as I have, or am I nitpicking the wording of the question? Link to comment Share on other sites More sharing options...
Guest Posted June 5, 2008 Report Share Posted June 5, 2008 what if you go at 90mph back to your house and then to the town? 30miles + 60miles = 90miles @ 90mph = 1 hour Link to comment Share on other sites More sharing options...
Guest Posted June 5, 2008 Report Share Posted June 5, 2008 what if you go at 90mph back to your house and then to the town? 30miles + 60miles = 90miles @ 90mph = 1 hour Welcome to the boards, Gnome. I assume 'back to your house" means back to the starting point. If you're halfway to town, you could go go back to the starting point and then from there to the finish point all at an average speed of 90 mph and make the trip in an hour. There's only one problem: You're discounting the time it took to get to the half-way point. Link to comment Share on other sites More sharing options...
Guest Posted June 12, 2008 Report Share Posted June 12, 2008 3 x 10^8 m/s^2 Link to comment Share on other sites More sharing options...
Guest Posted December 23, 2008 Report Share Posted December 23, 2008 3 x 10^8 m/s^2 nice... but not possible. Damn relativistic mass increase!!!! grrr Link to comment Share on other sites More sharing options...
Guest Posted April 11, 2009 Report Share Posted April 11, 2009 I read this riddle and thought is was fairly easy one I solved it. But what shocked me was when I saw people debating whether or not there is a solution or not. And even the person who posted this riddle said there was no solution and went on showing off that he is familiar with Einsein's theory of relativity which doesn't apply to the question. First let me say that one does NOT have to travel for an hour to average a speed. I can go from my house to work which doesn't take me an hour to drive and still average a MP/H or KM/H. Now if the riddle stated that you had a time limit of one hour to average a certain speed then yes it is impossible since you already used up your hour. Now I read all previous posts and agree with all the equations about measuring time and distance and speed. However, the problem still has a solution and everyone is going about it the wrong way because they think you need to be traveling for one hour to average a speed which is, I'm sorry to say retarded Total distance to cover is 60km. You already spent 60 min. covering a distance of 30km going at a rate of 30km/h. CORRECT Remaining distance to cover is 30 KM. If you SPEED UP the second half of the trip to 90 KM/H you can cover the remaining 30KM in 15 min. It is 90 KM/H because you have to factor in 30 KM/H you spent driving the first 30KM and divide by 2 which will give you the AVERAGE SPEED over the ENTIRE distance. 90 + 30 = 120; 120/2 = 60KM/H problem solved! Time spent: 60 minutes the first half, 15 minutes the second half of the way. The time it took to get to your destination is 1 hour and 15 minutes. At an averaging speed of 60KM/H You don't have to drive backwards or take a detour or have rocket boosters, you just need to speed up. People get it through your heads that you don't need to be traveling for an hour to average a speed. And if the person who posted this riddle intended there to be a time limit of hour then he should learn to write better. Man can average running a top speed of around 25 MILES PER HOUR. It took trials of sprinting to get this average speed. No one made a man run as fast as he can for ONE HOUR. Plus if you have the endurance to run 25 MPH for one hour then marothons would be won in an hour and not the usual three hours. Link to comment Share on other sites More sharing options...
Guest Posted April 11, 2009 Report Share Posted April 11, 2009 Welcome to the boards, JAY10. If you haven't done so already, please read the sticky in the red box titled, "Important: READ BEFORE POSTING", especially the part about not resurrecting old threads. The rationale for there being no solution has been explained many times and so has the correct way to calculate average speed. Now I read all previous posts and agree with all the equations about measuring time and distance and speed. Some formulas were correct and lead to the correct answer and some were incorrect and lead to incorrect answers. How can you agree with all of them? The formula that should be used was given in the third response to this thread and multiple times later and you obviously don't agree with them. Remaining distance to cover is 30 KM. If you SPEED UP the second half of the trip to 90 KM/H you can cover the remaining 30KM in 15 min. No, it would take 20 minutes to travel 30 km at 90 km/hour. It is 90 KM/H because you have to factor in 30 KM/H you spent driving the first 30KM and divide by 2 which will give you the AVERAGE SPEED over the ENTIRE distance. 90 + 30 = 120; 120/2 = 60KM/H problem solved! That is not how average speed is calculated. The very first response in this thread was by someone asking if driving the second half of the trip at 90 km/hour would solve the problem and the next two posts (and several others in this thread) explained why this doesn't give an average speed of 60 km/hour for the trip. I also explained this in post #104. Travel 30 km/hour for 30 km, and you've taken an hour. Travel 90 km/hour for 30 km, and you've taken 20 minutes. 60 km in 80 minutes is an average of 45 km/hour. or average speed = distance traveled / time taken x = 60 km / 1.33.... hours (80 mins.) x = 45 km/hour The only way to average a speed of 60 km/hour in a trip that totals 60km is to finish the trip in one hour. If you've already traveled for an hour at the half way point... And if the person who posted this riddle intended there to be a time limit of hour then he should learn to write better. No, the riddle is written just fine; it's your understanding of the solution given that's lacking. You can, however, learn to read better. I brought your attention the the sticky in the red box above. Read it and pay attention to posting here with respect. Link to comment Share on other sites More sharing options...
Guest Posted April 11, 2009 Report Share Posted April 11, 2009 I really meant no disrespect and read all the posts, but I still felt this issue needed handled because there is a solution. First, your math is wrong your calculating just one part of the trip thats why you're answer is wrong. By the way the only part of my math that was wrong was the time for 90kmh over a distance of 30km, sorry I was in a hurry trying to get my point across. You have to do the equation TWICE because you traveled at TWO different speeds as I proved earlier. Maybe you should learn to read and learn how to calculate an average. LETS USE THE FORMULA: time = speed / distance 1st half of trip you traveled: 1 hour since 30kmh / 30 km = 1 The remaining distance of the trip you traveled: 20 min. since 90kmh / 30 km = 20 Total distance traveled: 30km + 30km = 60 km Total time traveled: 1 hour + 20 min. = 1 hr and 20 min. or 80 min. AVERAGE SPEED: hmmm lets see.... 90kmh + 30 kmh = 120 kmh ; 120kmh / 2 = 60 KMH! WOW! Im sorry you felt that you can only use the formula one time. So think before you reply you only proved that I got the time wrong not the average speed but thanks for catching that little error. Can you prove my point wrong about a human's average speed running. I thought not. Who is this Martini clown? Link to comment Share on other sites More sharing options...
rookie1ja Posted April 11, 2009 Author Report Share Posted April 11, 2009 Total distance traveled: 30km + 30km = 60 km Total time traveled: 1 hour + 20 min. = 1 hr and 20 min. or 80 min. AVERAGE SPEED: hmmm lets see.... 90kmh + 30 kmh = 120 kmh ; 120kmh / 2 = 60 KMH! WOW! Im sorry you felt that you can only use the formula one time. So think before you reply you only proved that I got the time wrong not the average speed but thanks for catching that little error. Can you prove my point wrong about a human's average speed running. I thought not. Who is this Martini clown? JAY10, maybe I could help. In your example, you traveled 60 km in 80 minutes. And that is not 60 km in 60 minutes (60 kmh). I am admin and Martini is moderator of this forum. Calling someone "clown" and messing up basic math is not a great start but I'm sure you will enjoy some other brain teasers (perhaps word riddles by Shakeepuddn). Link to comment Share on other sites More sharing options...
Guest Posted April 11, 2009 Report Share Posted April 11, 2009 I really meant no disrespect In your first post: "even the person who posted this riddle said there was no solution and went on showing off" "However, the problem still has a solution and everyone is going about it the wrong way because they think you need to be traveling for one hour to average a speed which is, I'm sorry to say retarded" "People get it through your heads " "And if the person who posted this riddle intended there to be a time limit of hour then he should learn to write better." In your second post- the very post you claim you meant no disrespect: "Maybe you should learn to read and learn how to calculate an average." "Who is this Martini clown?" It's clear you're not familiar with what having respect entails. Keep it up and your time on this board won't be long. LETS USE THE FORMULA: time = speed / distance 1st half of trip you traveled: 1 hour since 30kmh / 30 km = 1 The remaining distance of the trip you traveled: 20 min. since 90kmh / 30 km = 20 Total distance traveled: 30km + 30km = 60 km Total time traveled: 1 hour + 20 min. = 1 hr and 20 min. or 80 min. AVERAGE SPEED: hmmm lets see.... 90kmh + 30 kmh = 120 kmh ; 120kmh / 2 = 60 KMH! WOW! You didn't use the formula you claimed you would. You didn't even include time or distance in your calculation; you just added together two speeds and divided it by 2! The formula does not include averaging together two speeds. How could it since the information you provided above was total distance and total time? It's speed that is unknown in the information you gave above. You also gave the wrong formula. It should be: time = distance / speed not time = speed / distance (not that it mattered, since you didn't use that formula, although you said, "LETS USE THE FORMULA: time = speed / distance") If you would have used that (incorrect) formula, the answer for average speed should have been 80 km/hour. Let's use the correct formula. You gave this information: "Total distance traveled: 30km + 30km = 60 km Total time traveled: 1 hour + 20 min. = 1 hr and 20 min. or 80 min." Since you gave both both time and distance: 1.33.... hours = 60 km / x We divide 60 by what number to get 1.333...? Im sorry you felt that you can only use the formula one time. It could be used as many times as you like; it just needs to be used correctly. Link to comment Share on other sites More sharing options...
Guest Posted April 11, 2009 Report Share Posted April 11, 2009 I apologize for the disrespect. Lets focus on the problem and help each other out if we can look past the previous postings. Yes I said the total time took 80 min. And traveling a total distance of 60 km in 80 min. gave you the speed 45 kmh. However 45 kmh isn't an average, you just found exact speed for the entire trip without speeding up. So if you traveled 60 km at a constant speed of 45 km/h the time it took you to complete your travel would be 80 min. Correct? To get an average you need to add all the factors and divide by how many factors there are. agree? So think of the trip as being two trips because you changed speeds, even though it is still one trip, because you want to get and "average" speed. Kinda like an order of operations. You have to find the numbers for the first half of the trip (which are given), then the numbers for the second half of the trip. Then you can average the two speeds together. 1st half of trip took: 30kmh for 30 km took 60 min. 2nd half of trip took: 90 kmh for 30 km took 20 min. To find Average add both speeds and divide by 2. 30kmh + 90kmh = 120kmh ; 120/2 = 60 kmh (Average Speed) Again sorry for the dispute Martini. Link to comment Share on other sites More sharing options...
rookie1ja Posted April 11, 2009 Author Report Share Posted April 11, 2009 1st half of trip took: 30kmh for 30 km took 60 min. 2nd half of trip took: 90 kmh for 30 km took 20 min. To find Average add both speeds and divide by 2. 30kmh + 90kmh = 120kmh ; 120/2 = 60 kmh (Average Speed) I advise to study weighted average and give importance (weight) to time of each part of travel (instead of distance). I think you don't agree that total average speed is total distance divided by total time. So you don't agree that if you travel 60 km in 80 minutes you will have average speed of 45 kmh (no matter if you speed up, slow down, stop etc.). If you went to a pub 2 km away and the first km you travel by car in 1 minute and the second km you step out of the car and walk for 1 hour (you enjoy the nature). 1km = driven for 1 minute ... speed = 60 kmh 1km = walked for 1 hour ... speed = 1 kmh You spend the whole hour walking and drive the car just 1 minute. Do you really want to give both parts the same weight? Do you think that the total average speed is 30.5 kmh? Wouldn't it be fair to give more importance to the speed that you go for 1 whole hour and give less importance to the speed that you go just 1 short minute? Think about it Link to comment Share on other sites More sharing options...
Guest Posted April 11, 2009 Report Share Posted April 11, 2009 I apologize for the disrespect. Lets focus on the problem and help each other out if we can look past the previous postings. It would help if you acknowledged the points I raised in my last post. You didn't respond to me bringing up that you are not following the formula for speed you agreed with, and just added two speeds together and divided by two. This discounts the formula entirely as it doesn't take into account distance or time. Yes I said the total time took 80 min. And traveling a total distance of 60 km in 80 min. gave you the speed 45 kmh. However 45 kmh isn't an average, you just found exact speed for the entire trip without speeding up. The exact speed? The exact speed of what? It's the exact average speed for the trip! What do you mean "without speeding up"? The 45 km/hour figure I arrived at includes speeding up to 90 km/hour for the second half of the trip. So if you traveled 60 km at a constant speed of 45 km/h the time it took you to complete your travel would be 80 min. Correct? Correct. Do you agree that traveling at 60 km at a constant speed of 60 km/hour, then the total time it takes to complete the trip is 60 minutes? Then how can the following also be true?: Total distance traveled: 30km + 30km = 60 km Total time traveled: 1 hour + 20 min. = 1 hr and 20 min. or 80 min. AVERAGE SPEED: hmmm lets see.... 90kmh + 30 kmh = 120 kmh ; 120kmh / 2 = 60 KMH! WOW! To get an average you need to add all the factors and divide by how many factors there are. agree? No, to get an average, we follow the correct formula. I brought up in my last post how you didn't follow the formula for finding speed and you didn't acknowledge this. And btw, you did not "add all the factors and divide by how many factors there are", you are just adding two speeds and dividing by two. You are ignoring time and distance. So think of the trip as being two trips because you changed speeds, even though it is still one trip, because you want to get and "average" speed. Kinda like an order of operations. You have to find the numbers for the first half of the trip (which are given), then the numbers for the second half of the trip. Then you can average the two speeds together. 1st half of trip took: 30kmh for 30 km took 60 min. 2nd half of trip took: 90 kmh for 30 km took 20 min. To find Average add both speeds and divide by 2. 30kmh + 90kmh = 120kmh ; 120/2 = 60 kmh (Average Speed) That's not how average speed is calculated. You are totally ignoring time and distance. As has been mentioned, the formula for speed is: speed = distance / time You even seem to agree with this in your last post, although you transposed 'distance' and 'speed' (and then totally ignored using the formula in your calculation). Look at it this way: Since you think all you must do is add the speeds together and divide by two, you must think the two trips below are traveled at the same average speed. I travel for 100 km. I go 99 km at 99 km/hour. The last km I travel at 1 km/hour. Using you calculation: "To find Average add both speeds and divide by 2." 99 km/hour + 1 km/hour = 100 km/hour ; 100/2 = 50 km/hour (Average Speed) I travel for 100 km. I go 99 km at 1 km/hour. The last km I travel at 99 km/hour. Using you calculation: "To find Average add both speeds and divide by 2." 1kmh + 99kmh = 100 km/hour ; 100/2 = 50 km/hour (Average Speed) Do you think I averaged the same speed for both trips? I didn't. My average speed was much higher for the first trip. We don't just add the two speeds together and divide by two; we have to take into account how long I drove at each speed, hence the formula: time = distance / speed or distance = time * speed or speed = distance / time The fist trip: Distance: 100 km I drive 99 km at 99 km/hour. Time = 1 hour I drive 1 km at 1 km/hour. Time = 1 hour Total time: 2 hour We're looking for speed and we know speed = distance / time. x = 100 km / 2 hours Average speed- 50 km/hour Hey, that works out just like your formula. Why is that? It's because we traveled at different speeds for the same amount of time (one hour each). In a situation like that, we can just add the two speeds together and divide by two. But what if we travel at different speeds for different periods of time?: The second trip: Distance: 100 km I drive 99 km at 1 km/hour. Time = 99 hours I drive 1 km at 99 km/hour. Time = 0.010101_ hour Total time: 99.010101_ hours We're looking for speed and we know speed = distance / time. x = 100 km / 99.010101_ hours Average speed- 1.009... km/hour On preview, I see that rookie1ja gave a similar anecdote. If you still don't agree, respond to all my points and questions as I have responded to yours. Link to comment Share on other sites More sharing options...
Guest Posted April 13, 2009 Report Share Posted April 13, 2009 Cheers! This is my first post. I don't really enjoy the lack of formatting in forums, especially when it comes to anything that must be expressed mathematically. So, I just typed my solution out. The image shouldn't be too big! (Hopefully). To be clear, I'm solving this riddle: "If I went halfway to a town 60 km away at the speed of 30 km/hour, how fast do I have to go the rest of the way to have an average speed of 60 km/hour over the entire trip?" Since we're finding the average speed or, velocity, there should be no problems in getting an answer. Also, there is nothing particularly tricky in the wording that would suggest this problem should be unsolvable. The only way to make this problem impossible is if there are speed restrictions, time restrictions, and any other extraneous factors (such as getting pulled over by the police). For example, you'd have to travel the speed of light AND make sure that it takes you at least thirty minutes to get to the final destination. Even if we could theoretically go that fast, we'd be way past our target in 20 minutes. To those who insist this problem is impossible to solve, think about real life examples. (Probably more meaningful to those that actually drive, or those that normally do these calculations while trying to be on time.) For example, say you get stuck behind some slow moving vehicle for half of your trip, which takes about three hours to complete on average. Then for the second half, you find yourself on an open stretch of road leading to your final destination. There are no cops, obstacles, objecting love ones around to prevent you from going as fast as you want. Theoretically, if you go fast enough, you can still finish your trip in the three hours it usually takes you. It's also possible to go so fast, you lower your average time, but we're all driving safely, right, JAY10? Link to comment Share on other sites More sharing options...
andromeda Posted April 13, 2009 Report Share Posted April 13, 2009 So there you have it. If you drive 30km/hr for the first leg of your trip, then you drive 90 km/hr for the second leg and the total distance you want to travel is 60km, then you will average 60 km/hr by the end of your trip. Not really! As the mod and the admin had already said in one of the previous posts SPEED = DISTANCE / TIME SPEED = (30km + 30km)/(60min. + 20min.) SPEED = 60km/1.33hr SPEED = 45km/hr That's the average speed! Same goes for solutions for example. If you have two solutions with known concentrations (in %) and you mixed them up you won't get their average concentration by adding those two together and dividing them by two. You have to add the masses of the solutions together and you have to add the masses of the solutes together and then you set up a proportion in order to get the average concentration. Link to comment Share on other sites More sharing options...
Guest Posted April 13, 2009 Report Share Posted April 13, 2009 Not really! As the mod and the admin had already said in one of the previous posts SPEED = DISTANCE / TIME SPEED = (30km + 30km)/(60min. + 20min.) SPEED = 60km/1.33hr SPEED = 45km/hr That's the average speed! Same goes for solutions for example. If you have two solutions with known concentrations (in %) and you mixed them up you won't get their average concentration by adding those two together and dividing them by two. You have to add the masses of the solutions together and you have to add the masses of the solutes together and then you set up a proportion in order to get the average concentration. You're still looking for AVERAGE SPEED Which means you still have to divide by two since you're calculating for TWO speeds. You're just finding speed overall. Not the average speed. First off, if what you're saying is correct, how did they get 30 km/20 min? What I think is happening is that someone is applying the math incorrectly. a+b/A+B is different from a/A + b/B (prove it to yourself by setting a=1, b=2, A=3, B=4) Since we're given DISCRETE speeds, namely 30km/hr, then we CAN'T use the form a+b/A+B. YOU WILL GET THE WRONG ANSWER ALL THE TIME. Think about speed in this manner. Speed = d(1)/t(1) + d(2)/t(2) +d(n)/t(n).... But we're not done. If you recall, in order to find the average of something, you have to ADD all the items, then DIVIDE by the number of items involved. So, we would have something like: AVERAGE SPEED = (d(1)/t(1) + d(2)/t(2)) / 2 So you MUST calculate the speed in this way: 30/60 + 90/60 (<This figure comes from 30/20, scaled up to the proper unit) Ultimately we get 120/60, we convert the 60 to hours, and get 120km/hr! Oh boy! We got the speed, but not the average speed. 120km/hr / 2 = 60 km/hr. Fractions are tricky, but don't let them fool you! Link to comment Share on other sites More sharing options...
andromeda Posted April 13, 2009 Report Share Posted April 13, 2009 Fractions are tricky, but don't let them fool you! This link was provided by the admin several posts earlier! It's a bit tricky, but don't let it arithmetically fool you Link to comment Share on other sites More sharing options...
Guest Posted April 13, 2009 Report Share Posted April 13, 2009 Hello again to all who think that this riddle still has no solution. As for Martini and rookie1ja you two, I apologize in advance, are giving wrong examples and submitting wrong anecdotes. You still don't get the concept of what AVERAGE speed means and what it entails. MARTINI said if it takes you 80 min. to travel 60 km. then your AVERAGE speed is 45km/h. NO it is not AVERAGE speed. 45kmh is just the exact speed that you traveled the 60km and it took a duration of 80 min., as I said before. Also those numbers do not factor in both speeds, you think it does but look carefully because it doesn't. The reason why is because you have to do the formula twice (since there are two different speeds) to get the average. EXAMPLE: If I traveled 30 km at 30kmh. Then traveled 30 km again at 90 kmh. What do you think my AVERAGE speed will be for the ENTIRE trip????? 30 kmh + 90 kmh = 120 kmh : 120 kmh / 2 (since two different speeds where used) = 60 kmh FOR THE ENTIRE DISTANCE of 60km. These formulas Calculate SPEED, TIME and DISTANCE : SPEED = Distance / Time TIME = Distance / Speed DISTANCE = Speed x Time They DO NOT calculate an AVERAGE. You see you are misinterpreting what AVERAGE is. To get the average speed for the riddle you would have to calculate it this way. IT'S LIKE FOLLOWING AN ORDER OF OPERATION. First Half of trip you traveled 30 km at a rate of 30kmh. Second half of the trip you traveled 30 km at a rate of 90 kmh. The average speed between 90kmh and 30kmh is 60kmh! So your total distance traveled is 60km. Total time is 80 min. IF YOU WANTED TO SEE WHAT YOUR CONSTANT SPEED IS FROM START TO FINISH WITHOUT SLOWING DOWN OR SPEEDING UP, THEN YOU WOULD USE THE FORMULA ABOVE AND PLUG IN THE NUMBERS. WE ESTABLISHED THAT IT IS 45KMH. BUT THAT IS NOT THE AVERAGE SPEED FOR THE ENTIRE TRIP. YOU JUST MERELY FOUND OUT HOW FAST YOU WOULD HAVE TO TRAVEL TO COVER A DISTANCE OF 60KM IN 80 MIN. WHICH IS 45KMH. In other words IF YOU DROVE 60KM AT A CONSTANT SPEED OF 45KMH THEN THE TIME IT TAKES TO TRAVEL IS 80 MIN. IN OTHER WORDS IF I DROVE FOR 80 MIN. AT A SPEED OF 45KMH, HOW FAR HAVE I GONE? ANSWER: 60KM I'VE ANSWERED AND PROVEN ALL MY POINTS. IF YOU NEED FURTHER EXPLANATION AFTER THIS, THEN I JUST FEEL SORRY FOR THE WAY PEOPLE THINK. SO KICK ME OFF THE BOARDS IF YOU WANT. I'VE ASKED TEN PEOPLE THIS RIDDLE SO FAR, TEACHERS AND PEERS AND ALL HAVE PROVIDED AN ANSWER. AND TO SEIAI, I DON'T KNOW WHO YOU ARE BUT WOW, YOUR MATHEMATICAL SKILLS AND PROBLEM SOLVING METHODS ARE AMAZING. Link to comment Share on other sites More sharing options...
rookie1ja Posted April 13, 2009 Author Report Share Posted April 13, 2009 JAY10 and Seiai, pls write that you are younger than 13 years just to be sure that it's not a joke. Seiai, look at my short comments in spoiler below. First, give some time to study weighted average, my and Martini's post above and then write if it is more fair to give each speed the same importance (as you say) or to give more importance to speed at which you travel for longer time (as we say) when calculating average. JAY10, I see your point. The question is do you see ours? In particular, pls address Martini's examples and then we could move further. I travel for 100 km. I go 99 km at 99 km/hour. The last km I travel at 1 km/hour. Using you calculation: "To find Average add both speeds and divide by 2." YOUR CALCULATION: 99 km/hour + 1 km/hour = 100 km/hour ; 100/2 = 50 km/hour (Average Speed) MY CALCULATION: Distance: 100 km I drive 99 km at 99 km/hour. Time = 1 hour I drive 1 km at 1 km/hour. Time = 1 hour Total time: 2 hour We're looking for speed and we know speed = distance / time. x = 100 km / 2 hours Average speed = 50 km/hour I travel for 100 km. I go 99 km at 1 km/hour. The last km I travel at 99 km/hour. Using you calculation: "To find Average add both speeds and divide by 2." YOUR CALCULATION: 1kmh + 99kmh = 100 km/hour ; 100/2 = 50 km/hour (Average Speed) MY CALCULATION: Distance: 100 km I drive 99 km at 1 km/hour. Time = 99 hours I drive 1 km at 99 km/hour. Time = 0.010101_ hour Total time: 99.010101_ hours We're looking for speed and we know speed = distance / time. x = 100 km / 99.010101_ hours Average speed = 1.009... km/hour Correct me if I am wrong: YOU say: Average speed is when you add all speeds and divide them by how many different speeds there are. I say: Average speed is when you add all speeds, give them their importance/weight and then divide them by how many different speeds there are. So I say that when you travel one speed for majority of time then the average speed should be closer to that speed since we went that speed longer (and not just simply fifty-fifty). Which approach seems more fair? First think about posted above and then give me solutions for 2 tasks below: I have 2 apples, my friend has 2 apples and you have 4 apples. What is the average number of apples that one has? How do you calculate it? I travel at 30 kmh for 20 minutes, then I travel at 30 kmh for another 20 minutes, then I travel 90 kmh for another 20 minutes and then I travel at 30 kmh for another 20 minutes, What is the average speed? How do you calculate it? Link to comment Share on other sites More sharing options...
Guest Posted April 13, 2009 Report Share Posted April 13, 2009 JAY10 and Seiai, pls write that you are younger than 13 years just to be sure that it's not a joke. Seiai, look at my short comments in spoiler below. First, give some time to study weighted average, my and Martini's post above and then write if it is more fair to give each speed the same importance (as you say) or to give more importance to speed at which you travel for longer time (as we say) when calculating average. JAY10, I see your point. The question is do you see ours? In particular, pls address Martini's examples and then we could move further. I travel for 100 km. I go 99 km at 99 km/hour. The last km I travel at 1 km/hour. Using you calculation: "To find Average add both speeds and divide by 2." YOUR CALCULATION: 99 km/hour + 1 km/hour = 100 km/hour ; 100/2 = 50 km/hour (Average Speed) MY CALCULATION: Distance: 100 km I drive 99 km at 99 km/hour. Time = 1 hour I drive 1 km at 1 km/hour. Time = 1 hour Total time: 2 hour We're looking for speed and we know speed = distance / time. x = 100 km / 2 hours Average speed = 50 km/hour I travel for 100 km. I go 99 km at 1 km/hour. The last km I travel at 99 km/hour. Using you calculation: "To find Average add both speeds and divide by 2." YOUR CALCULATION: 1kmh + 99kmh = 100 km/hour ; 100/2 = 50 km/hour (Average Speed) MY CALCULATION: Distance: 100 km I drive 99 km at 1 km/hour. Time = 99 hours I drive 1 km at 99 km/hour. Time = 0.010101_ hour Total time: 99.010101_ hours We're looking for speed and we know speed = distance / time. x = 100 km / 99.010101_ hours Average speed = 1.009... km/hour Correct me if I am wrong: YOU say: Average speed is when you add all speeds and divide them by how many different speeds there are. I say: Average speed is when you add all speeds, give them their importance/weight and then divide them by how many different speeds there are. So I say that when you travel one speed for majority of time then the average speed should be closer to that speed since we went that speed longer (and not just simply fifty-fifty). Which approach seems more fair? First think about posted above and then give me solutions for 2 tasks below: I have 2 apples, my friend has 2 apples and you have 4 apples. What is the average number of apples that one has? How do you calculate it? I travel at 30 kmh for 20 minutes, then I travel at 30 kmh for another 20 minutes, then I travel 90 kmh for another 20 minutes and then I travel at 30 kmh for another 20 minutes, What is the average speed? How do you calculate it? Apples: If you want to find the average number of apples a person has, then you add up how many apples everyone has, and divide by the number of people. 2+2+4= 8 is the total number of apples then you divide 8 by 3, 2.6666. If it helps you see it this way: (2/1) + (2/1) + (4/1) = (8/1) is similar to how speed is being used in my example. (50/1) + (30/1) + (40/1) = (90/1) there are THREE different speeds so you must divide (90/1) by 3. In your question: I travel at 30 kmh for 20 minutes, then I travel at 30 kmh for another 20 minutes, then I travel 90 kmh for another 20 minutes and then I travel at 30 kmh for another 20 minutes, What is the average speed? How do you calculate it? 30+30+90+30 = 120 km/hr 120/4 = 60 km/hr I can also tell you how far you traveled. 10+10+30+10 = 60 km Overall, your trip took you 1 hr and 20 mins, with an average speed of 60 km/hr. If you drove at a constant rate of 30 km/hr, the whole trip would take you two hours to complete. If you drove at rate of 60 km/hr, the whole trip would take you one hour to complete. If you drove at a rate of 90 km/hr, the whole trip would take you thirty. As your riddle is written, there is no limitation on when we have to make it into town, merely that we need to make it in while averaging 60 km/hr. If you said something like, "I drove 30 km/hr in one hour, then my mum called me up and told me to get my a** home in ten minutes" or "I drove 30 km/hr in one hour, then it started to hail, so I could go no faster than x km/hr..." etc. The correct formula for speed is indeed S = d/t each incidence of km/hr accounts for a speed. If you want the average speed, you still have to do what we did with the apples: S(1) + S(2) + S(3)+ .....S(n), which is to take the sum of your speeds (in this case S) then divide by the number of instances (in this case (n)) Also, did you realize that (a+b)/(A+B) is different from (a/A) + (b/B)? You're doing in your example is actually altering the speed, not accounting for it. So you can't say 30km/hr, it would be something else. Since we agree on the formula, you have to realize that S = d/t, so if you already have S, the d/t bit is embedded in the number. It is still necessary to use Speed as a discrete entity. If you look at my paper, I showed (pretty much) all the steps. Your way finds SPEED, just not AVERAGE SPEED. If you're upset about that, you should re-word your riddle correctly. Otherwise, people will keep solving it. And In martini's case: If you traveled 99km/hr, for one hour, you've driven 99 km. Sure, you can drive at 1km/hr, and make the 100km trip in two hours (averaging 50 km/hr), but you could drive 2km/hr and make the trip in 1.5 hours (averaging 50.5 km/hr), if you decide to speed up a bit more your average speed will keep increasing. Why would he suddenly want to drive 1 km/hr after driving 99 km/hr beats me. Sure you want to use the formula for speed, but don't you want to take the average after that to get the average speed? (You stopped working too soon, and you applied the formula incorrectly.) At least if I were <13yrs, there would be some hope for maths education today. Link to comment Share on other sites More sharing options...
andromeda Posted April 13, 2009 Report Share Posted April 13, 2009 At least if I were <13yrs, there would be some hope for maths education today. If you are not 13 then click on the link that the admin has provided several post back and share your opinion on that!! I'm REALLY interested to what you have to say on that! Link to comment Share on other sites More sharing options...
rookie1ja Posted April 13, 2009 Author Report Share Posted April 13, 2009 Seiai, I am glad that you are interested in this brain teaser. We might probably set a few steps/questions/issues and address each. 1st issue 30+30+90+30 = 120 km/hr How exactly did you calculate that? I am sure that you know that "30 + 30" is "60" and "90 + 30" is "120" so the total should be "60 + 120" so "180". And "180 / 4 = 45" Once we get this 1st issue straigt, we might continue. Actually, there won't be any continuation needed if you admit calculating mistake. But let's see Link to comment Share on other sites More sharing options...
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