In India , it doesn't matter what the capacity of a bike is , people adjust themselves anyhow 😂😂😂

The one riding bike never walked 😂😂he Is real genius

Bob is going to win Gold medal in next Olympics by walking 86 km at 15 km/hr. 🥇

Everyone : Calculate how many hours Me : who the hell able to walk 15km/hr

this is how i carry my two friends on a bike while in blue zone in pubg...

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.

Legends say: They never reached their uncle's house as they are still trying to solve the complex equation. 😂

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.

Ultra optimised solution: Call their uncle to the brothers house Then it would take only 5hrs😂😂😂

***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

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.

Meanwhile in *INDIA* : ALEX SITTING ON TANKER OF BIKE BOB DRIVING CARL SITTING IN BACK

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

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?

“Simple calculation” Maybe when I was in year 10. As an engineer 20 years on. No chance

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

First i get the ans - 11 hrs and then i know the correct approach to solve this. Thanks man👍

What if uncle isn't at home😂.They all will waste time

Imagine the calculations goes wrong and Bob arrives a month later

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

We must also add the time taken to solve the problem in final answer😂

More than 5 hours and half "to walk" at a speed of 15 km/h ... WOW! Google employs supermen.

"All brothers are brilliant logicians" then let them solve their own problem themselves lol

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*

Everything in your solution went way above my head after 3:17. Let me give a simpler solution. 1. Distance travelled by Alex & Bob on bike till Bob gets off the bike is ‘x’ kms 2. So, distance travelled by Carl during this time is 0.25x kms 3. Now, the distance between Alex & Carl is 0.75x kms. Both of them are now moving towards each other. Using 1:4 ratio, Alex only needs to travel 4/5th of 0.75x while Carl travels remaining 1/5th. So, Alex travels 4/5 * 0.75x i.e., 0.6x kms. 4. Meanwhile, Bob has travelled 1/4th of 0.6x i.e., 0.15x kms towards destination. Now the distance between Alex & Bob is 0.75x kms (0.6x + 0.15x). 5. Now, all Alex has to do is travel x kms once again because we saw in Point No. 1 to 3 that if the biker travels x kms, then the walker will travel 0.25x kms and the distance between them will be 0.75x kms which is exactly what we want now. 6. After adding all the three bike trips, we get: x kms (-) 0.6x kms (+) x kms = 300 kms i.e., 1.4x kms = 300 kms So, x = 214.29 kms. By substituting this, we get: 1. Alex & Bob travel 214.29 kms (x) on bike while Carl travels 53.57kms (0.25x) on foot. 2. Alex drops Bob at this point & returns to pickup Carl. He travels back 128.57 kms (0.6x) while Carl has travelled another 32.14 kms (0.15x) and both of them meet at this point. 3. Meanwhile, Bob has travelled 32.14 kms (0.15x) towards destination and now, the distance between Alex & Bob is 160.71 kms (0.75x) 4. Alex travels another 214.29 kms (x) with Carl on bike while Bob walks 53.57 kms (0.25x) and all thee of them reach the destination at the same time. Total distance travelled is: 214.29 kms (-) 128.57 kms (+) 214.29 kms = 300 kms. Total time taken for travelling: (60 km/hr = 1 km per minute) 214.29 min + 128.57 min + 214.29 min = 557.14 min i.e., 9 Hours 17 Minutes & 8.4 Seconds. Of course, this whole equation assumes that there is no time wasted in bike acceleration, braking, dropping a brother, taking a U-Turn to pick up another & most importantly, there is no traffic which may potentially slow down the biker. 😂 Such an ideal situation. 😅

That animation and your explanation is very awesome 🥰 Thanks man for the Video 😘

If their walking speed is 15km/hr , literally they are sprinting!

This is by far the most impractical thing i have seen in a while

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 ;)

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!

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.

More precisely, It's 9 hours, 16 minutes, 8 seconds. Awesome video, brother. loved the explanation & animations.

What an amazing explanation 🔥

Amazing! Amazing! Awesome puzzle 🤯🤯🤯first I thought I cracekd it but when I saw the sloution I just say 'Mind blowing' !!

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~~!!

My solution gave 11 hours and it took me 2 mins. I am still happy with the accuracy/ efficiency ratio 😁😁

This riddle is mind-blowing. I figured it out years ago. Nice explanation bro.

15km/h that's not walking, that's running, and at a pretty decent pace

This actually made me laugh. I realized where you were going with the optimized solution before you explicitly stated it and busted out laughing. Clever.

Lol better take the uncle to there (Three brothers home)😂😂

Now, how can one figure this out over there in front of the interviewer?😓

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 beautifully proven. I did the passive approach and got 11 hours, I was happy thinking I solved it, lol. I love your channel.

Took me exactly 13 minutes to figure it out. Very good riddle!

Loved this puzzle, made you think harder. Thanks

One of the genuine puzzle channel. Keep it up✌️✌️

I am your great fan. Nice problem, equally well explained.

>walking at 15km/h Damn, those brothers are pretty fast.

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 😂😂😂

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.

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 I were Bob, I'd be mad at Alex for dropping me off half way and making me walk 85+ km

In Asia you will find that upto 4 persons can travel easily on long journey on a BIKE

Alex so lucky just drop and receive Bob and Carl😂😂😂

With this one question I liked , shared the video and also subscribed it .

They stopped at a restaurant and took rest for 5 hours Wasted their money and come back to their home 🤣🤣

I would love to see anyone "walking" 85km at 15km/h.

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.

I’m just impressed that they walked 15 km/h

Absolutely brilliant ❤️❤️

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

Thank you so much for the nice puzzle, solution, animations and easy way of explanation. Thank you

Kudos to those 2 who can walk at 15 km/h. I don't even know if I can run that fast

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😇

Amazing puzzle and the way of solving is also awesome.

Khrtnak sandar lajwab mn khus ho gya iska solution dekh ke.

Logic dictates they found a solution but are looking for way to make it shorter therefore never reaching their uncle's house

Wait! Uncle is wasting time call the uncle and fix a piont in the middle of root save time😀

Carl and Bob be like : hatja aaj bike tera bhai chalayega 😂😂😂

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).

Wow..I loved the optimization thing

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."

I got 9 hrs but missed in minutes over all good attempt I guess haha

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.

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!

Bob's legs after doing 86km at 15km/h (53mi at 9mph) 💪🦵

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.

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

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.

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 :)

How bob will feel when you drop him 50km before destination

Great video Also 4Q too buddy.

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

What If the Uncle also start walking at the same time from opposite direction

I got an answer 65/7 without this much complex, but I don't know my method was right!

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.

Jaw dropping solution 🔥🔥

It's really interesting and tough one for me.

Thanks a lot, sir. Solving this puzzle lifted my spirits up. Keep posting regularly.

Uber driver: ok you walk from here Me: why? Uber driver: I don't have time to explain

Great puzzle. Great explanation. Thanks!

When I did by my own I got 11 hours...🤪 But now I know the correct approach 😃 Thank you AMAR sir💖😊💖

The 'c' person is first reach that place why he was wasting time he was able to use mobile phone 😂

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...

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.

I totally enjoyed his type of explanation and I remained at long time

The exact answer is 9hours,17 minutes, and 60/7 seconds

Got the answer on the first try by solving for x where 300-x was the first drop off point. Algebra was simple but was a pain to sub x back into the equations for t1 + t2 + t3. t1 = (300-x)/60 x1 = t1 * 15 = (300-x)/4 x2 will be the point where the walker and the bike meet up and since the bike is 4 times faster than the walker, they meet at 1/5 the distance between their initial positions. So x2 = x1 + 1/5 * [(300-x) - x1] = x1 + (3/20)(300-x) t2 = (3/20)(300-x)/15 t3 = (300 - x2)/60 Since everyone arrives at same time, 15(t2 + t3) = x Substitute t2 and t3 into that and x = 1800/21 Then substitute x to find the total time = t1 + t2 + t3

Did anyone notice the Poor Old Brother (White Hairs) Gets to Walk First and A Young Energetic (Long Hair) Brother Never Walks the Entire Way. So Sad 😂😂😂

Answer to this question In India :- 5 hrs. In abroad :- 9 hrs. 17 mins.