He could walk, but he would have to get off real fast (such as within a second) and let the other brother ride (then he walks) such as if his first stop is at the 71.428 km mark to return to pick up the other brother behind. In this case, all three would meet up at the 100km mark. If the bike stops at the 71.428 km mark when Alex was the rider (Alex and Bob on the bike) then that would take 1.19 hours (71.428 divided by 60). At the same, Carl would be at the 17.857 km mark( 1.19 at 15 miles per hour). So after 1.19 hours, they are just 53.571 km apart ( 71.428 - 17.857). Bob could return at 60 km/hr to get Carl who is walking at 15 km/hr while Alex continues to also walk at 15 km/hr. Bob and Carl would meet in 0.71428 seconds (53.571 divided by 75 since traveling in OPPOSITE DIRECTION add 60 +15) at the 28.57 km mark (17.857 + 10.7142) while Alex is at the 82.1422 km mark (71.428 + 10.7142, the 10.714 comes from multiplying 15 by 0.71428 hours). Now they are still 53.571 km apart, but this time traveling in the SAME DIRECTION and will meet up after traveling for 71.428 km (53.571 x 4/3, the '4' is the 60 miles per hour and '3' is 45 miles from substracting 60-15, so the 4/3 ratio)). But since Bob and Carl are at the 28.57 km mark and will meet up with Alex after 71.428 km of traveling they will meet up at the 100 km mark (71.428 + 28.57). Note the 3 times (1.19 then 0.71428 then 1.19) or 1.19 + 0.71428 + 1.19 = 3.09428 hrs just to reach the 100km but since the trip is 300 km, just multiply 3.09428 by 3 to get 9.28284 hours or about 9 hours 17 minutes. How did I get 71.428? I came up with a formula and reduce it to just dividing 100 km where all three would meet after the bike return by 1.4. So in this case, since I want them to meet at 100km I divide the 100 km mark by 1.4 to get 71.428. If you want all three to meet up at 50 km just divide 50 by 1.4. In other words, the time it took them to meet up at the 50 km mark, they would just need to multiply that time by 6 since 6 times 50=300km
You got me laughning very well. Also Carl too - same distance for same amount of time. Like i said: this is not logical because nobody can walk 15 km/hr
@@motijewelsghatkoparw8516 That's his top speed, right. But it is for a short period of time (sprinting). Usain Bolt sprint was 100m in 9.58s. Now if we want to see him do this for 53.57km... 53.57km = 53570m 53570m / 100m * 9.58s = 5132s = 85.53min = 1.425h That would be 1 hours and 25.5 minutes where he needs to sprint at his top speed.
I took a slightly modified approach that of course yields the same results but is, I think, less confusing as it requires only one unknown and is perhaps a little more intuitive. The basic idea is very similar to the one of the video: the "biking brother" carries one of the "walking brothers" some distance, drops him off, and returns to the point which the other "walking brother" has reached, picks him up and drives again towards the uncle's house where all of them arrive simultaneously. The "trick" is to realize that the first walking brother walks a distance x before he's carried the remaining way to the uncle, while the second walker is first carried on motorbike and then walks the rest - but since they must both spend the same time walking and going on motorbike respectively in order to arrive at the uncle's house, the distance where the second walker is dropped off is the same distance x that the first walker goes before being picked up. So, for both walkers the following equation is true: t = x / v₁ + (s-x) / v₂; x being the unknown partial distance, s being the total distance (i.e. the 300 km), and v₁ and v₂ the velocities of the walking and riding the motorbike respectively I habitually approach problems with the general case and plug in the specific numbers later. The downside, I know, is that plugging in the numbers first might speed up the solution but the upside is that one gets a solution that is valid for *all* cases - different numbers? --> you just plug in different values at the very end! ;-) Back to the problem. Let's now focus on the lucky bastard who doesn't have to walk at all. He will first cover the total distance s minus the partial distance x where he drops off the first brother. When he goes back, the other brother's already walked partial distance x and the go the remainder all the way to the uncle, i.e. also (s - x). What about the bit when he's going the other way after having dropped of the first brother to pick up the second? This is the total distance s with the partial distance x missing on both ends, i.e. (s-2x). In total, the biker goes (s-x)+(s-2x)+(s-x) = (3s-4x). Going with motorbike speed, he'll need the following time for this: t = (3s-4x) / v₂ These two expressions must be equivalent and therefore: x / v₁ + (s-x) / v₂ = (3s-4x) / v₂ We perform, I quote, "simple calculation" and plug in the numbers to learn that x is 600/7 km and t is 9 ²/₇ h. OK, that last line was a bit facetious. It is easier to plug in the numbers first but if one continues with the general case one finds that the partial distance x is: x = 2sv₁ / (v₂ + 3v₁) and the time to reach the uncle's house is: t = s/v₂ * [3 - 8 / (v₂/v₁ + 3)] Like I said above, the beauty of these general solutions is that they will work for any value of the given parameters.
***Alternative solution*** We know optimal solution involves a drop off and a pick up. Let x, y, and z be time to first drop off, time between drop off and pick up, and time until finish. Eqn1: 15(x+y) + 60z = 300 (distance of initial walker) Eqn2: 60x + 15(y+z) = 300 (distance of initial passenger) Eqn3: 60(x - y + z) = 300 (distance traveled by motorcycle, has to backtrack so y is negative) 3 equations, 3 unknowns, no problem. Eqn1 + Eqn2 + 2*Eqn3 -> x + z = 50/7 Sub that solution back into Eqn3 and find y = 15/7 Total time elapsed x + y + z = 65/7 = 9.28
Uncle :- Why you guys are late? 3 Fellows :- Actually we are wasting time in equating a way to reach in less time. Edit :- I think I have a good humour that I make 1k+ people laugh.
The basic solution comes out to 11 hours, so since it is unlikely it would take more than 1h43m to come up with a more optimized solution, it would be worth considering. That’s a little post hoc though, since you wouldn’t know it’s worth it until after calculating it, and it doesn’t account for making two people walk for ~2.5 hours each. Still, it’s the same principle as sharpening the axe for 5 minutes to chop a tree in 1 minute versus chopping for 10 minutes.
I did it this way in my head: Once I figured out that to maximize efficiency, all 3 brothers needed to arrive at the same time, and that meant the "walking" brothers each had to walk (and ride) the same distance, I called the walking distance "X". The dropoff point is X from the destination, and when the biker returns to the pickup point, it will be X from the origin. The distance in between the pickup and dropoff is "Y". The entire distance can be split in three sections: X -> Y -> X Now the question is: how far is X and Y? To figure it out, we can compare how far the brothers go in the same time period; from the start until the original walker is picked up by the rider. One brother walks distance X to the pickup in the time that the biker goes X+Y to the dropoff, then Y again back to the pickup, or X+2Y. We know that riding is 4x as fast as walking (60km/h) / (15km/h), so if the walker went X, he would have gone 4X, had he been riding. We know that the distance the rider went, and the distance the walker would have went if he were riding, are equal Therefore the equation is X+2Y=4X reduce to Y=1.5X The distance from origin to destination can now be rewritten as a ratio: 1:1.5:1 (from X:Y:X) To use integers for ease, double it to 2:3:2 This splits the total distance into sevenths, X is 2/7 of the total distance, and Y is 3/7 of the total distance So the "walking" brothers walk 2/7 of the way and ride 5/7 of the way The rider rides 5/7 of the way (X+Y), drops off his brother, rides back 3/7 of the way (Y), picks up the other brother, and rides 5/7 of the way to the origin, for a total of: 5/7 + 3/7 + 5/7 = 13/7 The rider rode 13/7 of the total distance, which was 300km, or 300*13/7=557 & 1/7km 557 1/7km @ 60km/h = 9 2/7 hours No complex equations or calculations required. I'm curious, how easy was it to follow my solution?
I also thought about that and a more fair solution would be like this They will solve the problem as it is 100km and and in each 100km someone else will go with bike You can calculate it is the same result
Bob doesn't wasting time. He would be cleaning their uncle's house while waiting for the other 2. Their uncle would gives bob some money to buy fuel for their motorcycle. Everyone is happy now
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I took the easy route. I figured out what would happen if Alex dropped Bob off at the 300km mark in 5 hours and let him keep walking. Carl moved 75km and is 225km from the 300km mark. As Alex moves back to pick up Carl they will be moving towards each other at 75km/h and will meet in 3 hours at the 120km mark, and Bob moved another 45km so he's at 345km. Now Alex and Carl have to catch up to Bob at 45 km/h relative speed and he is 225 km away so that's another 5 hours. They all end up at the 420km mark in 13 hours so no matter the distance using this arrangement their average speed will be 420/13 km/h. 420km/300km is 7/5, so they could get to 300km in (13*5)/7 hours, or ~9.29 hours
I lived this puzzle way back during 2001 to 2003; where I used to drop my Mom-Dad to bus-stop from home. And Always used to follow this approach. During this commute, I suddenly thought of finding the exact point where I can drop the first one and then pick the second one. First I worked on this lot and found the answer and then identified the formula as well. Then, I shared that puzzle with many friends many times; but none appeared with correct answer so far. Then I decided to make a video on that, but that time never came and you got my thoughts ;)
@@mageshm3997 It was just situation that I shared, and out of which I imagine the format of puzzle. Also the distance was not that short, more than 10 km (i guess)
From information I could find it is actually normal speed for professional ultramaraphon runners: current record for 5-7hr run seems to be in 100km distance: just above 6hr (meaning 16-17km/hr) Although it seems like for amateur 6hr runner 10km/hr is more realistic (but still would require significant prior training)
@@kriparane6370 Well the world record in sports walking on distance 50km is 3h32m so it's pretty close to 15km/h (more than 14km/h). Yet I doubt if Carl or Bob are olimpic medalists in that discipline.
They can start off immediately. That gives Bob (they're all brilliant logicians, so any of them could figure it out) a few hours while he's the passenger to figure out where he should get dropped off. Though if I were Bob I'd be tempted to adjust the numbers so that I wouldn't have to walk/sprint quite so fast. To make it a little more fair, Alex and Bob should trade places once they catch up to Carl again.
I am not really sure but when I think of the solution, I realise "Alex" was always on the bike during the entire course of the event. Usually (though not always), Optimal solutions are symmetric too. And here, given the fact each of the 3 brothers are identical in every aspect, there seems to be a pressing need of symmetry in the solution. That implies, each one of them should get a ride on the bike for the same amount of distance (though in parts) and should get to walk on the road the same amount of distance (though in parts). This in-turn implies, we need to have at least 2 "stops" (or 5, 8, 11, ....) in between so that we can break the course of the event into 3 parts wherein each brother gets to walk for exactly one time (and gets to ride for one time). Makes sense ? Great video though @LogicallyYours PS: Will take out some time to see if the above actually gives a more optimal solution.
Thanks for sharing the riddle and explaining the results so that we can all understand it clearly. I understand these numbers in the question are just out of the blue, but would it be worth mentioning to the interviewers that 15 km/hr is not walking speed. As a running speed, these three brothers are solid ultra marathoners. They ran a little over two marathon's worth of distance under 6 hours. After watching many your videos, one thing I learned a lot from you is to really pay attention to every detail and consider every information given. Again, I understand that your videos are meant to share and solve riddles, but it has definitely helped me think outside of the box. Thanks!
@@LOGICALLYYOURS Hello bro just one question, how much time do you think it will take to solve these type of questions in the interview. I did it in 20-30 mins or so in my home, how long should a person will take it during an interview.
@@МихаилВолков-л2я But here they have to cover almost twice that distance. Absolute world record for 50km walk is 3:32 and its below 15km/h (close to but still below).
I did it simpler with less variables. Actually just 1. 1. We understand, that all 3 must move all the time. 2. We understand, that bike must be running all the time. 3. Alex is a biker and will be always driving. 4. Bob and Carl will need to take part of way by bike and part by feet. 5. Since Bob's and Carl's walking speed is the same we can state, that proportion of walking and driving for them is the same. 6. X is the point, where Alex will drop Bob. 7. Bob will be (300-x)km traveling by bike with speed 60km/h and (x)km traveling be feet with speed 15km/h 8. Carl will start from traveling (x)km by feet with speed 15km/h and then (300-x)km by bike. 9. Alex will travel (300-x) to Bob drop point, then (300-2x) to Carl pickup point and again (300-x) 10. Time for Alex=((300-x)+(300-2x)+(300-x))/60= *(900-4x)/60* 11. The same time for Bob/Carl = (300-x)/60+x/15= *(300+3x)/60* 12. Now T=T and 900-4x=300+3x, so x=600/7= *85.7 km* , so *T=9.28 h*
Actually, the shortest time can be found, without calculating any distances, explicitly. Lets say: T0: time, when the first guy is dropped off the bike, T1: time, which this guy needs to get to the destination, T2: time, for the bike to get back to the third guy, who is already walking toward the destination, T3: time, for the two guys on bike to get to the destination. As you said, they all arrive at the same time. We have the following equations: a) 60 T0 +15 T1 = 300 (the dropped-off guy) b) 15 T0 +15 T2 + 60 T3 = 300 (the picked-up guy) c) 60 T0 - 60 T2 + 60 T3 = 300 (the bike-guy), d) T1 = T2 + T3 (simultaneous arrival) 4 equations, can be solved for T0..3, to obtain the total shortest time: T = T0 + T1 ~ 9,28 h
If they give true 10 minutes to think then it's absolutely possible. And I believe all the interviewers must understand this point... that these type of puzzles are not supposed to be answered quickly.... like an instant coffee.... rather they should provide sufficient time to think.
By having the right intuition. I think the explanation on the video was unnecessarily complex. After following this channel for many years and solving similar puzzles elsewhere, the idea of all three brothers arriving at the same time came instantly. Furthermore, it was obvious that there would be at least one drop-off and turning back. Then I realized that with one drop-off there would be two walkers, and in order to arrive at the same time, the walking distances should be equal. Considering the symmetry, the drop-off point should be at 300-x and the pick-up point at x. All this didn't take me more than half a minute. The rest is just algebra. On the left side you have the distance travelled by the bike after drop-off. First it drives back 300-2x to pick up the second passenger. Then it has the distance of 300-x to destination. The dropped-off walker has distance of x to destination. From there you have a simple proportion equation. Distance travelled by bike divided by bike's speed is equal to distance travelled by walker divided by walking speed. Getting to this point should satisfy the interviewer.
I did it by another way- Let the remaining distance which B has to travel after getting off the bike be x Let the distance travelled by the bike after dropping B and picking up C be q So 4x=2q+x Then q=1.5x The other side will be symmetrical so it is also x 3.5x=300 Total distance traveled by the bike is 300+2q=300+3x =557.142857 Time=557.142857÷60=9.28 hours
Wonderful video, and the logic is beautifully explained. However, for countries where this scenario is a common occurrence, I propose another solution. The third person hangs precariously from the motorcycle, and all three riders hang on and pray for 5 hours until they get to the uncle's house. I would also predict that the driver would never let off the motorcycle's horn the entire trip. :) Your videos are always awesome. Thanks again.
I haven't commented on this channels after videos for many times ,because those last videos are simple ,but not gonna lie this one is pretty unique. Keep loving your videos
So basically you'll take 10 hrs 2 mins Better anyone take bike, reach your uncle's house, pick him and bring him to your home, it will take 5+5=10hrs...2 min less than your time🤣
@@ravi-k-gupta or, even the optimal way would be the uncle starts walking (with the same speed, 15kpm) towards the opposite direction, so that the one with the bike would get him at exact 4 hours and return at exact 4 hours, maybe they could even grab some lunch on their way back and would still reach back quicker🤣
@@ravi-k-gupta They assume two guys walking 85km constantly with the speed of 15kmph, so why not the uncle walks with the same speed, for a little shorter distance ?🤣
I found this much easier: Brother 1 and 2 drive 1 h: 60km Brother 3 walks: 15 km Brother 2 drives back: meets Brother 3 after 36m - brother 1 walked 9 further kilometer There is a gap between the brothers of 45 km now. At a catch up speed of 45km/h it takes them another hour to all meet again. All brothers are together with their bikes at Kilometer 84 after 2,6h. They moved with 32,3 Km/h with that speed it takes them 9h 17 minutes and about 48 seconds to reach the goal.
Thing is, if this riddle was there at the time of my Jee advance preparation i surely had done it, now 4 years later my fate brought me here.. Thank you for the explanation
That was what I thought. The correct solution is that you can't walk at 15km/h therefore the person who wrote the question should be fired and the QC of the questions as well. I just saved Google about $300k p.a
Let's assume they are traveling 1.4 km instead and divide by the constant 1.4/1.4 = 1km So Alex/Carl will ride to the 1km mark, drop off Carl, and turn around to pick up Bob. Once he picks up Bob, he goes straight to the 1.4 km destination, where Alex/Bob will reach home at the same time as Carl. The beauty of the 1.4 constant is that WHEN Alex picks up Bob, they will be at a distance of 4 times away from home as the distance Carl is away from 'home.' And since Alex/Bob is traveling four times the speed of Carl, they all will reach at the same time. It will take Alex /Carl [ 0.01667 ]hours to reach the 1km traveling at 60km per hour. Hence, Bob will be at the 0.25km mark since he travels 15 km per hour or 1/4 the bike's speed. Alex drops off Carl (recall at the 1km mark) to turn around to get Bob. Since Alex is at the 1km mark and Bob is at the 0.25 km mark, they are 0.75 km apart . So, how long before they meet? It will take them the 0.75km divided by 75 km per hour or 0.01 hours So, at what distance will Bob travel in [0.01] hour? Distance = speed * time = 15km per hour * 0.01 = 0.15km But since he has already traveled 0.25km, when Alex meets him, he will be at the 0.40 km mark (0.25 + 0.15) So where is Carl? Since he walks at the same speed as Bob, he travels for 0.15km. So, now Carl is at the 1.15km mark (1 km + 0.15km) Notice that Carl is now just 0.25km away from home ( 1.4 km - 1.15 km = 0.25km), and notice that Alex/Bob is 1km from home (1.4 km - 0.40km). Notice that 1km is 4 times the distance of 0. 25 km, Hence, since Alex/Bob is traveling four times, Carl, all three will reach 'home' at the same time. So, how long before they reach home? 1km divided by 60 miles per hour or 0.25km divided by 15 miles per hour. Either one will give you the exact time 1/60 = [0.01667hour] So let's add up all the hours travel so far 0.01667 0.01 0.01667 ------------- 0.04334 Total hours to reach 1.4 km It takes 0.04334 hours to reach 1.4 km (say they do this every week) But the total distance is not 1.4 km, it is 300 km So, let's divide 300km by 1.4 = 214. 2857 Then let's multiply this by 0.04334 Hence, 0.04334 hours * 214.2857 =9.287 hours Answer
Bob could ride Alex to the 214.285 km Mark. This would take 3.5714 hours (214.285/60) while Carl would be at the 53.57 km mark (214.285/4) since he is walking 1/4 the bike's speed (15/60). The distance between Bob/Alex and Carl is now 160.713 km. Bob drops off Alex to return for Carl, and it would take Bob 2.1428 hours (160.713/75) to reach Carl. Carl then would walk just 32.142 km (2.1428 x 15) while Bob would ride for 128.57 km (2.1428 x 60 or 32.142 x 4) before they met. Bob and Carl are now at the 85.714 km mark (214.285- 128.57 or 32.142 + 53.57), while Alex is at the 246.427 km mark (214.285 + 32.142), AND they have so far completed 5.7142 hours (3.5714 + 2.1428). Bob and Carl now have 214.285 km to reach the destination, while Alex has 53.57 km to reach the destination. But since Bob and Carl are traveling at 4 times the speed Alex is walking, and since 214.285 is four times 53.57, they would reach the same time or in 3.5714 hours ( 214.285/60 or 53.47/15). Total hours to reach hence = 5.7142 + 3.5714 or 9.28 hours or 9 hours and 17 minutes (Answer) (8:13) (8:26) PS So Bob rode 557.14 km in total (214.285 + 128.57 + 214.285), 257.14 km MORE THAN the 300 km journey, while Carl and Alex walked 85.714 km and rode with Bob for 214.285 km (notice that 85.714 + 214.285 =300km). Since all three reached the same time, by just dividing Bob's 557.14 km on the bike by 60 miles per hour, you still get 9.285 hours (the Answer). And if you divide Carl's (or Alex's) 85.714 km walking by 15, you get 5.714 hours, then divide their 214.285 km (riding with Bob) by 60, you get 3.5714. And 5.174 + 3.5715 also = 9.28 hours. The 214.285 km (the drop-off) came from dividing 300 by 1.4 from a formula I created. PS The key is once all three started the journey (riding and walking), no one stops (except when Bob drops off Alex to return for Carl to pick him up, yet Alex has to begin to walk immediately, and Carl also has to continue to walk towards Bob, though Bob is riding to get him), and they must reach their uncle's house EXACTLY the same time. What if Bob drops off Alex at the 250 km mark (as opposed to the 214.285 km mark)? Then Alex would reach first (before 9.28 hours), but the time would be greater than 9.28 hours or 9 hours 17 minutes for all three, which they don't want. PS. The difference between Alex and Clark during the 300-km journey was that while Alex walked 85.714 km at the start of the trip and rode 214.285 at the end of the trip, Clark walked the same distance but at the end of the trip and rode the same distance but at the start of the trip (214.285 + 85.714 = 300km).
Well, my approach was in separation of the whole journey into the three segments: 1. 1st&2nd pal ride on a bike, while third one walk X time afoot. 2. 1st pal and 3rd pal walk afoot, while 2nd pal rides back to 3d pal Y time. 3. 1st pal still walks afoot while 2nd and 3rd ones ride to him on a bike Z time. After such division, we'll have the system of three linear equations: 60*x + 15*y + 15*z = 300 60*x - 60*y + 60*z = 300 | "-60*y" means that 2nd pal moved backward comparatively to general movement direction 15*x + 15*y + 60*z = 300 where, in the end, we got the next values: x = 25/7, z = 25/7, y = 15/7; and our lowest time is equal to (x+y+z) = 65/7 -> 9 2/7 hrs.
I got the same answer a different way: First, the way they travel is the same as what you suggest. Second, I just picked an arbitrary first leg of 1 hour T = 1 h, A = 60 km, B = 60 km, C = 15 km Third, I calculated where Alex would meet Carl if he backtracked now. Carl will cross 1/5 and Bob will cross 4/5 of the 45 km gap between then when they meet. So carl will walk 9 miles, in 0.6 hours, when Bob finds him. T = 1.6 h, A = 69 km, B = 24 km, C = 24 km Fourth, I calculated where they'd all meet up. Since it's symmetrical, I bet it takes 1 hour. T= 2.6 h, A = 84 km, B = 84 km, C = 84 km Using this method they travel 84 km in 2.6 hours, which is 32.3 km/h. Their speed won't change when we scale it to 300 km. At 32.3 km/h, it takes 9.28 hours to go 300 km.
My working (i have not watched the solution): The concept would be 2 people riding while 1 walk. Then, 1 from the bike would walk to the house while the person ride the bike to pick up the person walking from the start. Then they would travel to their destination. Setting up variables: We will have the time from the start to where one person would drop off the bike to variable x. Then the remaining time would be variable y. Then, we will create a distance variable named a. Working: a/15=x (This shows how to calculate x based on a and the walking speed) Here, we will start to define variable a If that's for walking distance, the bike can travel 4a. 300-4a would be the remaining distance before the bike travels back. The time would be y=(300-4a)/15 To find the location of where the person walking from the start to the bike would be a+3a/5. The additional distance for the walking person would be 3a/5. The bike distance to meet would be 12a/5. Then to reach the house, they would travel 300-8a/5. So another way to calculate y is (12a/5+300-8a/5)/60=y and the simplified one would be (300+4a/5)/60=y. Now we could compare (300+4a/5)/60=(300-4a)/15. To balance them, we have 300+4a/5=1200-16a. We can calculate a by using balancing method so it would be 16a+4a/5=1200-300 to 84a/5=900 to a=5/84×900 to a=375/7. Keep it improper fraction for easier use. Back to a/15=x, we can make it to x=(375/7)×(1/15). Simplify and you have x as 25/7. Back to (300-4a)/15=y, we can make it into (2100/7-1500/7)/15=y. Here, we have y=(600/7)×(1×15). Simplify and you have y as 40/7. Now we have values for both x and y, we add them up to find the total time so the result would be 25/7+40/7=65/7. Answer: 9+2/7 hours
I literally spent almost an hour thinking this through. It's been way too long since I quit studying math ... :( I'm just relieved that I got the right answer :)
haha it took me only 3min to get the wrong answer, because as a practical person 2 dudes went to final destination with the bike and then he turned to get his mate :P thats how we did it when we were young and didnt have a car, that makes the equation way easier :P hey it only took them 11h :P
@@sankarnathan6787 He is going to have the other two (so all three will be on the bike) on the bike (though legally only two persons at most can ride on it), and since 300 km at 60km/hr then it would take him 300/60 =5 hours.
Without all these maths, I came up with the answer of a little less than 10 hours. Logic is simple, Alex drops Bob after 4 hours to walk the remaining 60 km on foot for next 4 hours. Then come back to pick the 3rd bro who's resting a bit after walking 90 km in 6 hours. Alex would be back at 90km mark in 2:30 mins and then take him to uncle house 210 km away in 3:hours 30 mins. Bob would have already reached by the time. 4+2.5+3.5=10 hrs. Less than 10 hours if Carl doesnt wait for Alex after 6 hours and walks a bit further till Alex reaches him.
Wow! Can't believe Google still does that!? The only thing you can find out by interviewing engineers like this is a guy who can solve riddles very well. But it has no correlation with how good a person can actually write code!
My approach: I tried to find the overall optimal speed (km/hr) they can achieve as a group of 3, by using an example of a + b on the motorcycle for an hour, then a meets with c 0.6 hours after when backtracking to pick him up, and then them meeting up again at 2.6 hours at a distance of 84km from home. this gives us a net optimal speed of 84km/2.6hr, or (84/2.6)km/hr, so the time it takes to travel 300 km would be 300*2.6/84 hours, or if you must simplify, 75*2.6/21 hours, returning 9.28571
Thank you for sharing, love the question. And the awesome part is showing step by step how to solve the question. The explanation along with the video is perfect. The ownage teaching technique~~!!
Am I the only one that did not actually explicitly calculate for distance in any way? It seemed unnecessary. At the start, I approached it the same way: view the simple method of dropping him at goal, go back and pick up, and then drive rest of the way. The wasted time of the second person not walking means you need to drop them off earlier so they can walk and arrive at the same time as the other two on the bike. And then you just set up three equations, where x represents the time spent driving the first person, y represents the time driving back, and z represents the time driving the third person: Person 1: 60x - 60y + 60z = 300 Person 2: 60x + 15y + 15z = 300 Person 3: 15x + 15y + 60z = 300 Solve for x, y, and z, and add them up for your total time: 25/7 + 15/7 + 25/7 = 65/7, or around 9.3. You have to do a little bit of algebra to solve the system of equations, but conceptually it seems simpler.
lemme tell you , " They don't ask these type of questions.These are easy . They usually tell you to solve complex math problems with your preferred programming language. Stay in school , don't trust this."
Alex forgot about the new road works which meant he had to divert his route half way and add an extra 300km to the distance. Bob still ends up wasting time. Carl is very tired as a result and doesn't enjoy the visit to their uncle's house.
@@indianbot0077 Two time intervals, the first one is until the motorcycle turns, and the second one is the time to pick up the last brother + drive to the finish line.
I calculated it a bit differently. a = walking speed = 15 b = bike speed = 60 h1 = the time it took for Bob to get to x on bike h2 = the time it took for Alex to get to z after dropping Bob off at x on bike OR the time it took Carl to get to z by walking h3 = the time it took for Alex and Carl to get to their uncle's house on bike from z OR the time it took for Bob to get to his uncle's house by walking after h2 We can describe the brothers' journey like this Bob: h1b + a(h2+h3) = 300 Alex: b(h1-h2+h3) = 300 Carl: a(h1+h2) + h3b = 300 (Bob & Carl) h1b + h2a + h3a = h1a + h2a + h3b h1b + h3a = h1a + h3b h1b - h3b = h1a - h3a b(h1 - h3) = a(h1 - h3) Canceling out the h terms would entail that a and b are equal, but this is not true. This implies that we're somewhere dividing by zero. Therefore the only way for this equality to make sense is for (h1-h3) to be 0, that is: h1=h3 (Alex & Bob) b(h1-h2+h3) = h1b + a(h2+h3) b(2(h1) - h2) = h1b + h2a + h1a b(2(h1) - h2) - h1b = a(h2+h1) b(2(h1) - h2 - h1) = a(h2+h1) b(h1-h2) = a(h2+h1) b = 4a -> 4a(h1-h2) = a(h2+h1) 4h1 - 4h2 = h2 + h1 3h1 = 5h2 h2 = (3/5)h1 Now we can solve the puzzle. h1 + h2 + h3 = (13/5)h1 = ? (Bob) h1b + a(h2+h3) = 300 h1(4a) + h2a + h3a = 300 a(4h1 + h2 + h3) = 300 (15)(4h1 + h2 + h3) = 300 4h1 + h2 + h3 = 20 4h1 + (3/5)h1 + h1 = 20 (28/5)h1 = 20 13/28( (28/5)h1 = 20 ) (13/5)h1 = 9.28...
Hey, i really loved solving the problem! Great video! However, you never proved that your solution was the fastest. Here a way to do it: If the bike covered more distance, it could potentially make people walk less but as here they reach the house at the same time than the bike, the total time would be higher because of the bike Time. If the bike covered less distance the opposite would happen. I know you tell it in the video but you instantly exclude others solutions like ride every people of 1km, or whatever other.
YOU are totally correct, and everybody else is WRONG!!! Because you CAN be more efficient if you use the turn around multiple times! I calculated the time for 2 drop offs, and I got the total time of
The interviewer failed totally by forgetting it's dangerous to walk alone on the road at night. It's safer for Bob to reach and stay inside the uncle's house, while Carl to wait inside their own house.
This answer in this video is so convoluted... Here is how I solved it: The biker goes away from the walker at 45 and towards at 75... so to swing around and pick up the slower walker, it takes 45/75 = 0.6 times as long as the biker spent going forwards. Once the bike picks up the walker, its just gonna go forwards again for the same time as it did when it had the carried the first passenger, so thats a total of 2.6 time units forwards, of which it spent 0.6 going backwards(which counts as negative), so its as if it only went forwards by 1.4 out of the full 2.6. Traveling at regular speed, the bike takes 5 hours, so 5 * 2.6/1.4 = 9.285.
![Quando Rondo - Grow Up [Official Music Video]](http://i.ytimg.com/vi/8zFnCg3BO4Q/mqdefault.jpg)
In India , it doesn't matter what the capacity of a bike is , people adjust themselves anyhow 😂😂😂
😂😂😂😂
Poora khandaan chalta hai ek bike par 😎😎😎😎
Superb😂
😂😂🔥
Indians are always in a hurry but never on time😂
The one riding bike never walked 😂😂he Is real genius
You are not understood the question
@@muthukannur7149 have*
He's the interviewer 😂
He could walk, but he would have to get off real fast (such as within a second) and let the other brother ride (then he walks) such as if his first stop is at the 71.428 km mark to return to pick up the other brother behind. In this case, all three would meet up at the 100km mark.
If the bike stops at the 71.428 km mark when Alex was the rider (Alex and Bob on the bike) then that would take 1.19 hours (71.428 divided by 60). At the same, Carl would be at the 17.857 km mark( 1.19 at 15 miles per hour). So after 1.19 hours, they are just 53.571 km apart ( 71.428 - 17.857). Bob could return at 60 km/hr to get Carl who is walking at 15 km/hr while Alex continues to also walk at 15 km/hr. Bob and Carl would meet in 0.71428 seconds (53.571 divided by 75 since traveling in OPPOSITE DIRECTION add 60 +15) at the 28.57 km mark (17.857 + 10.7142) while Alex is at the 82.1422 km mark (71.428 + 10.7142, the 10.714 comes from multiplying 15 by 0.71428 hours). Now
they are still 53.571 km apart, but this time traveling in the SAME DIRECTION and will meet up after traveling for 71.428 km (53.571 x 4/3, the '4' is the 60 miles per hour and '3' is 45 miles from substracting 60-15, so the 4/3 ratio)). But since Bob and Carl are at the 28.57 km mark
and will meet up with Alex after 71.428 km of traveling they will meet up at the 100 km mark (71.428 + 28.57). Note the 3 times (1.19 then 0.71428 then 1.19) or 1.19 + 0.71428 + 1.19 = 3.09428 hrs just to reach the 100km but since the trip is 300 km, just multiply 3.09428 by 3 to get 9.28284 hours or about 9 hours 17 minutes.
How did I get 71.428? I came up with a formula and reduce it to just dividing 100 km where all three would meet after the bike return by 1.4. So in this case, since I want them to meet at 100km I divide the 100 km mark by 1.4 to get 71.428. If you want all three to meet up at 50 km just divide 50 by 1.4. In other words, the time it took them to meet up at the 50 km mark, they would just need to multiply that time by 6 since 6 times 50=300km
@@devondevon4366 you have some patience
Bob is going to win Gold medal in next Olympics by walking 86 km at 15 km/hr. 🥇
You got me laughning very well.
Also Carl too - same distance for same amount of time.
Like i said: this is not logical because nobody can walk 15 km/hr
Meanwhile Usain bolt -37km/hr world record
@@motijewelsghatkoparw8516 That's his top speed, right. But it is for a short period of time (sprinting).
Usain Bolt sprint was 100m in 9.58s.
Now if we want to see him do this for 53.57km...
53.57km = 53570m
53570m / 100m * 9.58s = 5132s = 85.53min = 1.425h
That would be 1 hours and 25.5 minutes where he needs to sprint at his top speed.
😂😂😂 we better reduce the distanc from 300 on to 30 km 😉
And he won
I took a slightly modified approach that of course yields the same results but is, I think, less confusing as it requires only one unknown and is perhaps a little more intuitive.
The basic idea is very similar to the one of the video: the "biking brother" carries one of the "walking brothers" some distance, drops him off, and returns to the point which the other "walking brother" has reached, picks him up and drives again towards the uncle's house where all of them arrive simultaneously.
The "trick" is to realize that the first walking brother walks a distance x before he's carried the remaining way to the uncle, while the second walker is first carried on motorbike and then walks the rest - but since they must both spend the same time walking and going on motorbike respectively in order to arrive at the uncle's house, the distance where the second walker is dropped off is the same distance x that the first walker goes before being picked up.
So, for both walkers the following equation is true:
t = x / v₁ + (s-x) / v₂; x being the unknown partial distance, s being the total distance (i.e. the 300 km), and v₁ and v₂ the velocities of the walking and riding the motorbike respectively
I habitually approach problems with the general case and plug in the specific numbers later. The downside, I know, is that plugging in the numbers first might speed up the solution but the upside is that one gets a solution that is valid for *all* cases - different numbers? --> you just plug in different values at the very end! ;-)
Back to the problem. Let's now focus on the lucky bastard who doesn't have to walk at all.
He will first cover the total distance s minus the partial distance x where he drops off the first brother. When he goes back, the other brother's already walked partial distance x and the go the remainder all the way to the uncle, i.e. also (s - x).
What about the bit when he's going the other way after having dropped of the first brother to pick up the second? This is the total distance s with the partial distance x missing on both ends, i.e. (s-2x).
In total, the biker goes (s-x)+(s-2x)+(s-x) = (3s-4x).
Going with motorbike speed, he'll need the following time for this:
t = (3s-4x) / v₂
These two expressions must be equivalent and therefore:
x / v₁ + (s-x) / v₂ = (3s-4x) / v₂
We perform, I quote, "simple calculation" and plug in the numbers to learn that x is 600/7 km and t is 9 ²/₇ h.
OK, that last line was a bit facetious. It is easier to plug in the numbers first but if one continues with the general case one finds that the partial distance x is:
x = 2sv₁ / (v₂ + 3v₁)
and the time to reach the uncle's house is:
t = s/v₂ * [3 - 8 / (v₂/v₁ + 3)]
Like I said above, the beauty of these general solutions is that they will work for any value of the given parameters.
Everyone : Calculate how many hours
Me : who the hell able to walk 15km/hr
Same haha
@@eestitulevik-et-eestitegij5436 and for so many hours
it's just a brisky walk/stroll
@@erishkigal1 Could you please film yourself strolling at 15km/hr?
@@b99andla damn, you're right - I was sure it was pace: 15min/km
***Alternative solution***
We know optimal solution involves a drop off and a pick up. Let x, y, and z be time to first drop off, time between drop off and pick up, and time until finish.
Eqn1: 15(x+y) + 60z = 300 (distance of initial walker)
Eqn2: 60x + 15(y+z) = 300 (distance of initial passenger)
Eqn3: 60(x - y + z) = 300 (distance traveled by motorcycle, has to backtrack so y is negative)
3 equations, 3 unknowns, no problem.
Eqn1 + Eqn2 + 2*Eqn3 -> x + z = 50/7
Sub that solution back into Eqn3 and find y = 15/7
Total time elapsed
x + y + z = 65/7 = 9.28
Gr8 solution simple and fast
A bit hard
nothing extraordinary he used distance as variable and u used times as that and yeah which surely gives u the right ans
@@palashagrawal2343 sorry but i am a lower classes to calcualte this eqn
@@Thelegendarian- what? This is basic stuff. How can this be hard?
Uncle :- Why you guys are late?
3 Fellows :- Actually we are wasting time in equating a way to reach in less time.
Edit :- I think I have a good humour that I make 1k+ people laugh.
There no way the waste 90 min doing that
Lol😂
😂
The basic solution comes out to 11 hours, so since it is unlikely it would take more than 1h43m to come up with a more optimized solution, it would be worth considering. That’s a little post hoc though, since you wouldn’t know it’s worth it until after calculating it, and it doesn’t account for making two people walk for ~2.5 hours each. Still, it’s the same principle as sharpening the axe for 5 minutes to chop a tree in 1 minute versus chopping for 10 minutes.
🤣
this is how i carry my two friends on a bike while in blue zone in pubg...
Same with a buggy😂😂
😂😂😂
😂
Hahahahhahahahahahhahaahhhahahahahahhahahahahahhahahahahahhahaah
😑
I did it this way in my head:
Once I figured out that to maximize efficiency, all 3 brothers needed to arrive at the same time, and that meant the "walking" brothers each had to walk (and ride) the same distance, I called the walking distance "X". The dropoff point is X from the destination, and when the biker returns to the pickup point, it will be X from the origin. The distance in between the pickup and dropoff is "Y".
The entire distance can be split in three sections: X -> Y -> X
Now the question is: how far is X and Y? To figure it out, we can compare how far the brothers go in the same time period; from the start until the original walker is picked up by the rider.
One brother walks distance X to the pickup in the time that the biker goes X+Y to the dropoff, then Y again back to the pickup, or X+2Y.
We know that riding is 4x as fast as walking (60km/h) / (15km/h), so if the walker went X, he would have gone 4X, had he been riding.
We know that the distance the rider went, and the distance the walker would have went if he were riding, are equal
Therefore the equation is X+2Y=4X
reduce to Y=1.5X
The distance from origin to destination can now be rewritten as a ratio: 1:1.5:1 (from X:Y:X)
To use integers for ease, double it to 2:3:2
This splits the total distance into sevenths, X is 2/7 of the total distance, and Y is 3/7 of the total distance
So the "walking" brothers walk 2/7 of the way and ride 5/7 of the way
The rider rides 5/7 of the way (X+Y), drops off his brother, rides back 3/7 of the way (Y), picks up the other brother, and rides 5/7 of the way to the origin, for a total of:
5/7 + 3/7 + 5/7 = 13/7
The rider rode 13/7 of the total distance, which was 300km, or 300*13/7=557 & 1/7km
557 1/7km @ 60km/h = 9 2/7 hours
No complex equations or calculations required. I'm curious, how easy was it to follow my solution?
If they are brothers in their teens, they will fight for who will go on bike first and end up all three walking dragging their bike along.
And then arises the problem as to who should drag the bike.
Would they still be able to keep their speed of 15kph if they do that? If not, shouldn't they just leave the bike behind?
Or Carl gets stuck on the handlebars.
@@lohithashwa6653 The youngest brother - that's easy.
I also thought about that and a more fair solution would be like this
They will solve the problem as it is 100km and and in each 100km someone else will go with bike
You can calculate it is the same result
Ultra optimised solution:
Call their uncle to the brothers house
Then it would take only 5hrs😂😂😂
No man..then time will be indefinite...
@@subashsahu8925 no if uncle use 100kmh car
@Chor Narendra Modi 😂😂
Yeah🤣🤣🤣 but uncle is very very busy
Your hired with promotion..were fireing this guy that's wasting everyone's time with over complications
Bob doesn't wasting time. He would be cleaning their uncle's house while waiting for the other 2. Their uncle would gives bob some money to buy fuel for their motorcycle. Everyone is happy now
😅😅
😂😂😂
😂😂😂😂😂
lmao 😂
Nah.. I think the same LOL 😂
Meanwhile in *INDIA* :
ALEX SITTING ON TANKER OF BIKE
BOB DRIVING
CARL SITTING IN BACK
Sure 🤣🤣
Yhi hota h😂🤣
And they are taking one hitchhiker with big backpack on their way.
And sining "ye dosti...."
Abe agar car hi tha to bike ki jarurat hi kya, sab milkar car me ja sakte the. This not wastage of time, this is wastage of fuel LOL
Legends say:
They never reached their uncle's house as they are still trying to solve the complex equation.
😂
🤣oh boi
Because Bob only went 299.99 km ... Bob couldn't finish the last 10 meters with his math.
Xeno's other Dilemma.
🤣
“Simple calculation”
Maybe when I was in year 10.
As an engineer 20 years on. No chance
He made it overly complicated
First i get the ans - 11 hrs and then i know the correct approach to solve this. Thanks man👍
Well done!
Me too
Same here....
ARE BHAIYA AGAR TUM IS QUESTION KO SOLVE KARLETE TO ISKA COMMENT NAHI KAR RAHE HOTE BALKI TUM 20 LACK SALARY KE SAATH MAST GOOGLE ME KAM KAR RAHE HOTE SAMJHE.
@@sbxsurya3438 😂😂😂
I took the easy route. I figured out what would happen if Alex dropped Bob off at the 300km mark in 5 hours and let him keep walking. Carl moved 75km and is 225km from the 300km mark. As Alex moves back to pick up Carl they will be moving towards each other at 75km/h and will meet in 3 hours at the 120km mark, and Bob moved another 45km so he's at 345km. Now Alex and Carl have to catch up to Bob at 45 km/h relative speed and he is 225 km away so that's another 5 hours. They all end up at the 420km mark in 13 hours so no matter the distance using this arrangement their average speed will be 420/13 km/h. 420km/300km is 7/5, so they could get to 300km in (13*5)/7 hours, or ~9.29 hours
5Head
this is really good
I lived this puzzle way back during 2001 to 2003; where I used to drop my Mom-Dad to bus-stop from home. And Always used to follow this approach. During this commute, I suddenly thought of finding the exact point where I can drop the first one and then pick the second one.
First I worked on this lot and found the answer and then identified the formula as well.
Then, I shared that puzzle with many friends many times; but none appeared with correct answer so far. Then I decided to make a video on that, but that time never came and you got my thoughts ;)
Lovely !!!! Solving a real life problem gives you much higher confidence and satisfaction than solving a puzzle in theory.
@@LOGICALLYYOURS
Yes... Nice
Ur story is incredible 😁
But your mom and dad probably had different walking speed and its useless if the distance to cover is short.
@@mageshm3997 It was just situation that I shared, and out of which I imagine the format of puzzle. Also the distance was not that short, more than 10 km (i guess)
So we basically have 3 walking Usain Bolts that never get tired, and they have a weird motorcycle that reach 60km/h instantly, but can't go faster
Walking Usain Bolt? They can accelerate to 15kmph in no time! Usain Bolt couldn't do that!
50 cc scooter
@@wiono Good point.
With no firctional costs for dismounting, changing directions or changing passengers
Those who are not interested in maths, comment like this
Imagine the calculations goes wrong and Bob arrives a month later
😂😂😂
He went out to get milk
😂🤣
More than 5 hours and half "to walk" at a speed of 15 km/h ... WOW! Google employs supermen.
From information I could find it is actually normal speed for professional ultramaraphon runners: current record for 5-7hr run seems to be in 100km distance: just above 6hr (meaning 16-17km/hr)
Although it seems like for amateur 6hr runner 10km/hr is more realistic (but still would require significant prior training)
@@aleksandertrubin4869 "walking" bruh.
Walking speed couldn't be 15 kmph it would running speed
If I could walk that fast I would have never bought a bike with 60km/h limit.
@@kriparane6370 Well the world record in sports walking on distance 50km is 3h32m so it's pretty close to 15km/h (more than 14km/h). Yet I doubt if Carl or Bob are olimpic medalists in that discipline.
What if uncle isn't at home😂.They all will waste time
😂😂😂😂
Aur effort bhi 😂😂😂
@@asadshams1681 Yeah
🤣🤣
😂
That animation and your explanation is very awesome 🥰 Thanks man for the Video 😘
Thanks bro for the appreciation :)
We must also add the time taken to solve the problem in final answer😂
They can start off immediately. That gives Bob (they're all brilliant logicians, so any of them could figure it out) a few hours while he's the passenger to figure out where he should get dropped off. Though if I were Bob I'd be tempted to adjust the numbers so that I wouldn't have to walk/sprint quite so fast. To make it a little more fair, Alex and Bob should trade places once they catch up to Carl again.
I am not really sure but when I think of the solution, I realise "Alex" was always on the bike during the entire course of the event. Usually (though not always), Optimal solutions are symmetric too. And here, given the fact each of the 3 brothers are identical in every aspect, there seems to be a pressing need of symmetry in the solution.
That implies, each one of them should get a ride on the bike for the same amount of distance (though in parts) and should get to walk on the road the same amount of distance (though in parts). This in-turn implies, we need to have at least 2 "stops" (or 5, 8, 11, ....) in between so that we can break the course of the event into 3 parts wherein each brother gets to walk for exactly one time (and gets to ride for one time). Makes sense ?
Great video though @LogicallyYours
PS: Will take out some time to see if the above actually gives a more optimal solution.
This riddle is mind-blowing. I figured it out years ago. Nice explanation bro.
Glad you liked it
Thanks for sharing the riddle and explaining the results so that we can all understand it clearly.
I understand these numbers in the question are just out of the blue, but would it be worth mentioning to the interviewers that 15 km/hr is not walking speed. As a running speed, these three brothers are solid ultra marathoners. They ran a little over two marathon's worth of distance under 6 hours.
After watching many your videos, one thing I learned a lot from you is to really pay attention to every detail and consider every information given. Again, I understand that your videos are meant to share and solve riddles, but it has definitely helped me think outside of the box. Thanks!
Many thanks Ryo for your kind words :) it certainly made me feel confident about my work.
@@LOGICALLYYOURS Hello bro just one question, how much time do you think it will take to solve these type of questions in the interview. I did it in 20-30 mins or so in my home, how long should a person will take it during an interview.
15km/h that's not walking, that's running, and at a pretty decent pace
It almost break world record of 21 km hour 😅
Speed of professional race walk athlete on 20 km
@@МихаилВолков-л2я But here they have to cover almost twice that distance. Absolute world record for 50km walk is 3:32 and its below 15km/h (close to but still below).
I did it simpler with less variables. Actually just 1.
1. We understand, that all 3 must move all the time.
2. We understand, that bike must be running all the time.
3. Alex is a biker and will be always driving.
4. Bob and Carl will need to take part of way by bike and part by feet.
5. Since Bob's and Carl's walking speed is the same we can state, that proportion of walking and driving for them is the same.
6. X is the point, where Alex will drop Bob.
7. Bob will be (300-x)km traveling by bike with speed 60km/h and (x)km traveling be feet with speed 15km/h
8. Carl will start from traveling (x)km by feet with speed 15km/h and then (300-x)km by bike.
9. Alex will travel (300-x) to Bob drop point, then (300-2x) to Carl pickup point and again (300-x)
10. Time for Alex=((300-x)+(300-2x)+(300-x))/60= *(900-4x)/60*
11. The same time for Bob/Carl = (300-x)/60+x/15= *(300+3x)/60*
12. Now T=T and 900-4x=300+3x, so x=600/7= *85.7 km* , so *T=9.28 h*
More precisely, It's 9 hours, 16 minutes, 8 seconds.
Awesome video, brother.
loved the explanation & animations.
No
More precisely it‘s 9h, 16m and 48s.
Because 0,28 x 60 is 16,8.
0,8 minutes are 48 seconds.
@@Atook774 Very correct; thanks for pointing it out bro
Amazing! Amazing! Awesome puzzle 🤯🤯🤯first I thought I cracekd it but when I saw the sloution I just say 'Mind blowing' !!
Thanks Subham! Cheers!
@@LOGICALLYYOURS My Pleasure☺
OMG I am feeling sad for Bob that he was being left alone by the biker on a way to his uncle..... Just imagine his feelings 😂😂😂
lmao 😂
That's the same feeling what Carl was having at the start of their ride.
No man Bob is the clever he is resting in uncle house and eating all the snacks alone when Alex going back to Carl
@@secretunknown2782 lmao but he was walking when Alex and carl riding on their bikes .
@@yogeshbhati1839 example 1
In Asia you will find that upto 4 persons can travel easily on long journey on a BIKE
So true!
I have seen that in the Netherlands too
Yesterday 4 boys gone on one bike. 16wheel lorry ran over his head . Time is precious but life is priceless.
@@nijuuu6630 very true that's why I dislike bikes especially in Asia
@@nijuuu6630 Thala.... Namma country full ah apdi than iruku...
Actually, the shortest time can be found, without calculating any distances, explicitly. Lets say:
T0: time, when the first guy is dropped off the bike,
T1: time, which this guy needs to get to the destination,
T2: time, for the bike to get back to the third guy, who is already walking toward the destination,
T3: time, for the two guys on bike to get to the destination.
As you said, they all arrive at the same time. We have the following equations:
a) 60 T0 +15 T1 = 300 (the dropped-off guy)
b) 15 T0 +15 T2 + 60 T3 = 300 (the picked-up guy)
c) 60 T0 - 60 T2 + 60 T3 = 300 (the bike-guy),
d) T1 = T2 + T3 (simultaneous arrival)
4 equations, can be solved for T0..3, to obtain the total shortest time: T = T0 + T1 ~ 9,28 h
"All brothers are brilliant logicians" then let them solve their own problem themselves lol
That was my thought ... why do they need my help. I'll optimize my time by helping someone else.
Now, how can one figure this out over there in front of the interviewer?😓
If they give true 10 minutes to think then it's absolutely possible. And I believe all the interviewers must understand this point... that these type of puzzles are not supposed to be answered quickly.... like an instant coffee.... rather they should provide sufficient time to think.
By having the right intuition. I think the explanation on the video was unnecessarily complex. After following this channel for many years and solving similar puzzles elsewhere, the idea of all three brothers arriving at the same time came instantly. Furthermore, it was obvious that there would be at least one drop-off and turning back. Then I realized that with one drop-off there would be two walkers, and in order to arrive at the same time, the walking distances should be equal. Considering the symmetry, the drop-off point should be at 300-x and the pick-up point at x. All this didn't take me more than half a minute.
The rest is just algebra. On the left side you have the distance travelled by the bike after drop-off. First it drives back 300-2x to pick up the second passenger. Then it has the distance of 300-x to destination. The dropped-off walker has distance of x to destination. From there you have a simple proportion equation. Distance travelled by bike divided by bike's speed is equal to distance travelled by walker divided by walking speed. Getting to this point should satisfy the interviewer.
I did it by another way-
Let the remaining distance which B has to travel after getting off the bike be x
Let the distance travelled by the bike after dropping B and picking up C be q
So 4x=2q+x
Then q=1.5x
The other side will be symmetrical so it is also x
3.5x=300
Total distance traveled by the bike is 300+2q=300+3x
=557.142857
Time=557.142857÷60=9.28 hours
@@LOGICALLYYOURS ya may be the interviewer just only want the approach of the candidate to solve this puzzle.
And then there are geniuses like me who proudly calculated 11hrs😂😂
Wonderful video, and the logic is beautifully explained.
However, for countries where this scenario is a common occurrence, I propose another solution. The third person hangs precariously from the motorcycle, and all three riders hang on and pray for 5 hours until they get to the uncle's house.
I would also predict that the driver would never let off the motorcycle's horn the entire trip. :)
Your videos are always awesome. Thanks again.
Haha.. solution accepted :D
What an amazing explanation 🔥
This is a very very beautiful puzzle....
Lovely.... Thanks.... 👍💐🙏❤👌👍👍
So beautifully proven. I did the passive approach and got 11 hours, I was happy thinking I solved it, lol. I love your channel.
Me too! We missed by several levels of logic 🤪
@@ringing7 we sure did.
I actually skipped to the end just to realize I was wrong 😂
I haven't commented on this channels after videos for many times ,because those last videos are simple ,but not gonna lie this one is pretty unique. Keep loving your videos
Many thanks buddy :)
Sir, myself BAPAN BISWAS,19 years old, I paused the video and tried to solve it. It consumed my 45 minutes. But luckily I solved it. I am very happy😇
So basically you'll take 10 hrs 2 mins
Better anyone take bike, reach your uncle's house, pick him and bring him to your home, it will take 5+5=10hrs...2 min less than your time🤣
@@ravi-k-gupta or, even the optimal way would be the uncle starts walking (with the same speed, 15kpm) towards the opposite direction, so that the one with the bike would get him at exact 4 hours and return at exact 4 hours, maybe they could even grab some lunch on their way back and would still reach back quicker🤣
@@bibekaryaal yeah but please consider his uncle is a senior citizen, let's not make him walk and that too at 15kmph😂
@@ravi-k-gupta They assume two guys walking 85km constantly with the speed of 15kmph, so why not the uncle walks with the same speed, for a little shorter distance ?🤣
They stopped at a restaurant and took rest for 5 hours
Wasted their money and come back to their home
🤣🤣
Meanwhile Carl was walking on expecting to see Alex on bike pretty soon 🤣😂.
If their walking speed is 15km/hr , literally they are sprinting!
I found this much easier:
Brother 1 and 2 drive 1 h: 60km
Brother 3 walks: 15 km
Brother 2 drives back: meets Brother 3 after 36m - brother 1 walked 9 further kilometer
There is a gap between the brothers of 45 km now.
At a catch up speed of 45km/h it takes them another hour to all meet again.
All brothers are together with their bikes at Kilometer 84 after 2,6h. They moved with 32,3 Km/h with that speed it takes them 9h 17 minutes and about 48 seconds to reach the goal.
Thing is, if this riddle was there at the time of my Jee advance preparation i surely had done it, now 4 years later my fate brought me here.. Thank you for the explanation
Bruh your from which IIT ❤️
Now are you not in iit??????
@@the_dopeyone6428 if he would've, he would've written that he's from 'X' IIT
How to know if someone is from IIT?
They'll tell you.
>walking at 15km/h
Damn, those brothers are pretty fast.
That was what I thought. The correct solution is that you can't walk at 15km/h therefore the person who wrote the question should be fired and the QC of the questions as well. I just saved Google about $300k p.a
Probably was written by an American who had no idea what the hell a kilometer is. :)
My solution gave 11 hours and it took me 2 mins. I am still happy with the accuracy/ efficiency ratio 😁😁
Same
Let's assume they are traveling 1.4 km instead and divide by the constant
1.4/1.4 = 1km
So Alex/Carl will ride to the 1km mark, drop off Carl, and turn around to pick up Bob. Once he picks up Bob, he goes straight to the 1.4 km destination, where Alex/Bob will reach home at the same time as Carl. The
beauty of the 1.4 constant is that WHEN Alex picks up Bob, they will be at a distance of 4 times away from home as the distance Carl is away from 'home.' And since Alex/Bob is traveling four times the speed of Carl,
they all will reach at the same time.
It will take Alex /Carl [ 0.01667 ]hours to reach the 1km
traveling at 60km per hour.
Hence, Bob will be at the 0.25km mark since he travels 15 km per hour or 1/4 the bike's speed.
Alex drops off Carl (recall at the 1km mark) to turn
around to get Bob.
Since Alex is at the 1km mark and Bob is at the 0.25 km
mark, they are 0.75 km apart . So, how long before
they meet? It will take them the 0.75km divided by
75 km per hour or 0.01 hours
So, at what distance will Bob travel in [0.01] hour?
Distance = speed * time = 15km per hour * 0.01
= 0.15km
But since he has already traveled 0.25km, when Alex
meets him, he will be at the 0.40 km mark (0.25 + 0.15)
So where is Carl? Since he walks at the same speed
as Bob, he travels for 0.15km. So, now Carl is
at the 1.15km mark (1 km + 0.15km)
Notice that Carl is now just 0.25km away from home (
1.4 km - 1.15 km = 0.25km),
and notice that Alex/Bob is 1km from home (1.4 km - 0.40km).
Notice that 1km is 4 times the distance of 0. 25 km,
Hence, since Alex/Bob is traveling four times, Carl,
all three will reach 'home' at the same time.
So, how long before they reach home?
1km divided by 60 miles per hour
or 0.25km divided by 15 miles per hour. Either one will
give you the exact time 1/60 = [0.01667hour]
So let's add up all the hours travel so far
0.01667
0.01
0.01667
-------------
0.04334 Total hours to reach 1.4 km
It takes 0.04334 hours to reach 1.4 km (say they do this every week)
But the total distance is not 1.4 km, it is 300 km
So, let's divide 300km by 1.4 = 214. 2857
Then let's multiply this by 0.04334
Hence, 0.04334 hours * 214.2857 =9.287 hours
Answer
Bob could ride Alex to the 214.285 km Mark. This would take 3.5714 hours (214.285/60) while Carl would be at the 53.57 km mark (214.285/4) since he is walking 1/4 the bike's speed (15/60). The distance between Bob/Alex and Carl is now 160.713 km. Bob drops off Alex to return for Carl, and it would take Bob 2.1428 hours (160.713/75) to reach Carl. Carl then would walk just 32.142 km (2.1428 x 15) while Bob would ride for 128.57 km (2.1428 x 60 or 32.142 x 4) before they met. Bob and Carl are now at the 85.714 km mark (214.285- 128.57 or 32.142 + 53.57), while Alex is at the 246.427 km mark (214.285 + 32.142), AND they have so far completed 5.7142 hours (3.5714 + 2.1428).
Bob and Carl now have 214.285 km to reach the destination, while Alex has 53.57 km to reach the destination. But since Bob and Carl are traveling at 4 times the speed Alex is walking, and since 214.285 is four times 53.57, they would reach the same time or in 3.5714 hours ( 214.285/60 or 53.47/15). Total hours to reach hence = 5.7142 + 3.5714 or 9.28 hours or 9 hours and 17 minutes (Answer) (8:13) (8:26)
PS
So Bob rode 557.14 km in total (214.285 + 128.57 + 214.285), 257.14 km MORE THAN the 300 km journey,
while Carl and Alex walked 85.714 km and rode with Bob for 214.285 km (notice that 85.714 + 214.285 =300km). Since all three reached the same time, by just dividing Bob's 557.14 km on the bike by 60 miles per hour, you still get 9.285 hours (the Answer). And if you divide Carl's (or Alex's) 85.714 km walking by 15, you get 5.714 hours, then divide their 214.285 km (riding with Bob) by 60, you get 3.5714. And 5.174 + 3.5715 also = 9.28 hours.
The 214.285 km (the drop-off) came from dividing 300 by 1.4 from a formula I created.
PS
The key is once all three started the journey (riding and walking), no one stops (except when Bob drops off Alex to return for Carl to pick him up, yet Alex has to begin to walk immediately, and Carl also has to continue to walk towards Bob, though Bob is riding to get him), and they must reach their uncle's house EXACTLY the same time. What if Bob drops off Alex at the 250 km mark (as opposed to the 214.285 km mark)? Then Alex would reach first (before 9.28 hours), but the time would be greater than 9.28 hours or 9 hours 17 minutes for all three, which they don't want.
PS.
The difference between Alex and Clark during the 300-km journey was that while Alex walked 85.714 km at the start of the trip and rode 214.285 at the end of the trip, Clark walked the same distance but at the end of the trip and rode the same distance but at the start of the trip (214.285 + 85.714 = 300km).
This is by far the most impractical thing i have seen in a while
Loved this puzzle, made you think harder. Thanks
I would love to see anyone "walking" 85km at 15km/h.
Well, my approach was in separation of the whole journey into the three segments:
1. 1st&2nd pal ride on a bike, while third one walk X time afoot.
2. 1st pal and 3rd pal walk afoot, while 2nd pal rides back to 3d pal Y time.
3. 1st pal still walks afoot while 2nd and 3rd ones ride to him on a bike Z time.
After such division, we'll have the system of three linear equations:
60*x + 15*y + 15*z = 300
60*x - 60*y + 60*z = 300 | "-60*y" means that 2nd pal moved backward comparatively to general movement direction
15*x + 15*y + 60*z = 300
where, in the end, we got the next values: x = 25/7, z = 25/7, y = 15/7;
and our lowest time is equal to (x+y+z) = 65/7 -> 9 2/7 hrs.
I got the same answer a different way:
First, the way they travel is the same as what you suggest.
Second, I just picked an arbitrary first leg of 1 hour
T = 1 h, A = 60 km, B = 60 km, C = 15 km
Third, I calculated where Alex would meet Carl if he backtracked now.
Carl will cross 1/5 and Bob will cross 4/5 of the 45 km gap between then when they meet. So carl will walk 9 miles, in 0.6 hours, when Bob finds him.
T = 1.6 h, A = 69 km, B = 24 km, C = 24 km
Fourth, I calculated where they'd all meet up. Since it's symmetrical, I bet it takes 1 hour.
T= 2.6 h, A = 84 km, B = 84 km, C = 84 km
Using this method they travel 84 km in 2.6 hours, which is 32.3 km/h. Their speed won't change when we scale it to 300 km.
At 32.3 km/h, it takes 9.28 hours to go 300 km.
Alex so lucky just drop and receive Bob and Carl😂😂😂
Amazing puzzle and the way of solving is also awesome.
Thanks Dishant bhai :)
I’m just impressed that they walked 15 km/h
Yes, especially Carl, at his age.
My working (i have not watched the solution):
The concept would be 2 people riding while 1 walk. Then, 1 from the bike would walk to the house while the person ride the bike to pick up the person walking from the start. Then they would travel to their destination.
Setting up variables: We will have the time from the start to where one person would drop off the bike to variable x. Then the remaining time would be variable y. Then, we will create a distance variable named a.
Working:
a/15=x (This shows how to calculate x based on a and the walking speed)
Here, we will start to define variable a
If that's for walking distance, the bike can travel 4a.
300-4a would be the remaining distance before the bike travels back.
The time would be y=(300-4a)/15
To find the location of where the person walking from the start to the bike would be a+3a/5. The additional distance for the walking person would be 3a/5. The bike distance to meet would be 12a/5. Then to reach the house, they would travel 300-8a/5.
So another way to calculate y is (12a/5+300-8a/5)/60=y and the simplified one would be (300+4a/5)/60=y.
Now we could compare (300+4a/5)/60=(300-4a)/15. To balance them, we have 300+4a/5=1200-16a. We can calculate a by using balancing method so it would be 16a+4a/5=1200-300 to 84a/5=900 to a=5/84×900 to a=375/7. Keep it improper fraction for easier use.
Back to a/15=x, we can make it to x=(375/7)×(1/15). Simplify and you have x as 25/7.
Back to (300-4a)/15=y, we can make it into (2100/7-1500/7)/15=y. Here, we have y=(600/7)×(1×15). Simplify and you have y as 40/7.
Now we have values for both x and y, we add them up to find the total time so the result would be 25/7+40/7=65/7.
Answer: 9+2/7 hours
Took me exactly 13 minutes to figure it out. Very good riddle!
When I did by my own I got 11 hours...🤪
But now I know the correct approach 😃
Thank you AMAR sir💖😊💖
I literally spent almost an hour thinking this through. It's been way too long since I quit studying math ... :( I'm just relieved that I got the right answer :)
haha it took me only 3min to get the wrong answer, because as a practical person 2 dudes went to final destination with the bike and then he turned to get his mate :P thats how we did it when we were young and didnt have a car, that makes the equation way easier :P hey it only took them 11h :P
@@grasgruen84Relative motion😅
Answer to this question
In India :- 5 hrs.
In abroad :- 9 hrs. 17 mins.
How is it so?
@@navinpnk526 ?
@@sankarnathan6787 3 people ride the motorbike at the same time.
@@sankarnathan6787 He is going to have the other two (so all three will be on the bike) on the bike (though legally only two persons at most can ride on it), and since 300 km at 60km/hr then it would take him 300/60 =5 hours.
@@devondevon4366 You got it 👍🏻✌
Khrtnak sandar lajwab mn khus ho gya iska solution dekh ke.
One of the genuine puzzle channel. Keep it up✌️✌️
Wait! Uncle is wasting time call the uncle and fix a piont in the middle of root save time😀
Bob's legs after doing 86km at 15km/h (53mi at 9mph) 💪🦵
Bob looks pretty young and healthy. Think about Carl!
Without all these maths, I came up with the answer of a little less than 10 hours. Logic is simple, Alex drops Bob after 4 hours to walk the remaining 60 km on foot for next 4 hours. Then come back to pick the 3rd bro who's resting a bit after walking 90 km in 6 hours. Alex would be back at 90km mark in 2:30 mins and then take him to uncle house 210 km away in 3:hours 30 mins. Bob would have already reached by the time.
4+2.5+3.5=10 hrs. Less than 10 hours if Carl doesnt wait for Alex after 6 hours and walks a bit further till Alex reaches him.
Wow! Can't believe Google still does that!? The only thing you can find out by interviewing engineers like this is a guy who can solve riddles very well. But it has no correlation with how good a person can actually write code!
It's really interesting and tough one for me.
Carl and Bob be like : hatja aaj bike tera bhai chalayega 😂😂😂
Lol better take the uncle to there (Three brothers home)😂😂
With this one question
I liked , shared the video and also subscribed it .
My approach:
I tried to find the overall optimal speed (km/hr) they can achieve as a group of 3, by using an example of a + b on the motorcycle for an hour, then a meets with c 0.6 hours after when backtracking to pick him up, and then them meeting up again at 2.6 hours at a distance of 84km from home. this gives us a net optimal speed of 84km/2.6hr, or (84/2.6)km/hr, so the time it takes to travel 300 km would be 300*2.6/84 hours, or if you must simplify, 75*2.6/21 hours, returning 9.28571
Thanks a lot, sir. Solving this puzzle lifted my spirits up. Keep posting regularly.
I don't think this question is hard enough to be an interview question of google. I solved this in 2 minutes and the solution is much cleaner.
Best solution is to use public transport
Absolutely brilliant ❤️❤️
Thank you for sharing, love the question.
And the awesome part is showing step by step how to solve the question.
The explanation along with the video is perfect.
The ownage teaching technique~~!!
Am I the only one that did not actually explicitly calculate for distance in any way? It seemed unnecessary. At the start, I approached it the same way: view the simple method of dropping him at goal, go back and pick up, and then drive rest of the way. The wasted time of the second person not walking means you need to drop them off earlier so they can walk and arrive at the same time as the other two on the bike. And then you just set up three equations, where x represents the time spent driving the first person, y represents the time driving back, and z represents the time driving the third person:
Person 1: 60x - 60y + 60z = 300
Person 2: 60x + 15y + 15z = 300
Person 3: 15x + 15y + 60z = 300
Solve for x, y, and z, and add them up for your total time: 25/7 + 15/7 + 25/7 = 65/7, or around 9.3. You have to do a little bit of algebra to solve the system of equations, but conceptually it seems simpler.
How bob will feel when you drop him 50km before destination
Ikr😂😂😂😂🤣🤣🤣!!!!
😂😂😂🤣
Not 50km bro it's 85.71km 😄
@@sureshganga3780 sure thats gonna hurt his legs 😂
lemme tell you , " They don't ask these type of questions.These are easy . They usually tell you to solve complex math problems with your preferred programming language. Stay in school , don't trust this."
How do you know? Utube is full of these kind of videos ,famous math utuber 3blue1brown also upload these kind of problems.
Alex forgot about the new road works which meant he had to divert his route half way and add an extra 300km to the distance. Bob still ends up wasting time. Carl is very tired as a result and doesn't enjoy the visit to their uncle's house.
Wow..I loved the optimization thing
I am your great fan. Nice problem, equally well explained.
The exact answer is 9hours,17 minutes, and 60/7 seconds
And a bit more ;)
you sure?
I got an answer 65/7 without this much complex, but I don't know my method was right!
Ya bro near 9.2 hrs
Did u solved using three variables and considering the as the three time intervals??? Plz tell.
@@indianbot0077
Two time intervals, the first one is until the motorcycle turns, and the second one is the time to pick up the last brother + drive to the finish line.
Logic dictates they found a solution but are looking for way to make it shorter therefore never reaching their uncle's house
I calculated it a bit differently.
a = walking speed = 15
b = bike speed = 60
h1 = the time it took for Bob to get to x on bike
h2 = the time it took for Alex to get to z after dropping Bob off at x on bike OR the time it took Carl to get to z by walking
h3 = the time it took for Alex and Carl to get to their uncle's house on bike from z OR the time it took for Bob to get to his uncle's house by walking after h2
We can describe the brothers' journey like this
Bob: h1b + a(h2+h3) = 300
Alex: b(h1-h2+h3) = 300
Carl: a(h1+h2) + h3b = 300
(Bob & Carl)
h1b + h2a + h3a = h1a + h2a + h3b
h1b + h3a = h1a + h3b
h1b - h3b = h1a - h3a
b(h1 - h3) = a(h1 - h3)
Canceling out the h terms would entail that a and b are equal, but this is not true. This implies that we're somewhere dividing by zero. Therefore the only way for this equality to make sense is for (h1-h3) to be 0, that is: h1=h3
(Alex & Bob)
b(h1-h2+h3) = h1b + a(h2+h3)
b(2(h1) - h2) = h1b + h2a + h1a
b(2(h1) - h2) - h1b = a(h2+h1)
b(2(h1) - h2 - h1) = a(h2+h1)
b(h1-h2) = a(h2+h1)
b = 4a -> 4a(h1-h2) = a(h2+h1)
4h1 - 4h2 = h2 + h1
3h1 = 5h2
h2 = (3/5)h1
Now we can solve the puzzle.
h1 + h2 + h3 = (13/5)h1 = ?
(Bob)
h1b + a(h2+h3) = 300
h1(4a) + h2a + h3a = 300
a(4h1 + h2 + h3) = 300
(15)(4h1 + h2 + h3) = 300
4h1 + h2 + h3 = 20
4h1 + (3/5)h1 + h1 = 20
(28/5)h1 = 20
13/28( (28/5)h1 = 20 )
(13/5)h1 = 9.28...
Hey, i really loved solving the problem! Great video!
However, you never proved that your solution was the fastest.
Here a way to do it:
If the bike covered more distance, it could potentially make people walk less but as here they reach the house at the same time than the bike, the total time would be higher because of the bike Time.
If the bike covered less distance the opposite would happen.
I know you tell it in the video but you instantly exclude others solutions like ride every people of 1km, or whatever other.
YOU are totally correct, and everybody else is WRONG!!! Because you CAN be more efficient if you use the turn around multiple times! I calculated the time for 2 drop offs, and I got the total time of
What If the Uncle also start walking at the same time from opposite direction
with his house in the backpacket.
"A" person is also wasting her time by giving lift to B and C
The interviewer failed totally by forgetting it's dangerous to walk alone on the road at night. It's safer for Bob to reach and stay inside the uncle's house, while Carl to wait inside their own house.
Calm down. Interview has taken place in Europe. Here during the night is as safe as in bright light day.
Great puzzle. Great explanation. Thanks!
Very nice logical question and explanation is very clear. Thanks for the vedio 😊😀
I would need the acceleration and the breaking time of the bike 🤓🥸🤷♂️
Lol nice, infinite power here I come
How can walk at speed of 15 km/h 😂😂😂😂
Usain bolt with genius
It doesnt matter if they have a car. Problem solved. And everyone is happy.
This answer in this video is so convoluted... Here is how I solved it: The biker goes away from the walker at 45 and towards at 75... so to swing around and pick up the slower walker, it takes 45/75 = 0.6 times as long as the biker spent going forwards. Once the bike picks up the walker, its just gonna go forwards again for the same time as it did when it had the carried the first passenger, so thats a total of 2.6 time units forwards, of which it spent 0.6 going backwards(which counts as negative), so its as if it only went forwards by 1.4 out of the full 2.6. Traveling at regular speed, the bike takes 5 hours, so 5 * 2.6/1.4 = 9.285.
Great video
Also 4Q too buddy.
The 'c' person is first reach that place why he was wasting time he was able to use mobile phone 😂