Seemingly Impossible Walking To College Riddle

[@MindYourDecisions]

I agree that "he leaves at the same time" should have been stated. In my meager defense, the problem made sense in my head, and after you solve enough of these questions you just assume details like that. Here is the verbatim of the original problem, which tacitly assumes but does not explicitly say he is leaving at the same exact time. "Narayan walking at a speed of 20 km/h reaches his college 10 minutes late. Next time he increases his speed by 5 km/h, but finds that he is still late by 4 minutes. What is the distance of his college from his house?"

[@mohitrawat5225]

Presh, you have lost a subscriber. Please leave the world of maths. If you teach maths. Then I am very sad for your students. I bet they make fun of you behind your back😂😂😂😂😂😂

[@JavedAlam24]

Ignore the guy above me. I agree that you shouldn't have put that in the question, it means you have to use your brain to figure out that the only way to solve the problem would be to assume that he left at the same time both days. If you can't figure that out then you aren't smart enough to solve it. It's better that way and it is a measure of intelligence.

[@justjuniorjaw]

​​@@mohitrawat5225Such a meaningless reason to leave. In class, these kinds of incomplete questions exist a lot. Back in my days before computer/online exams were a thing, teachers had to save paper by reducing the amount of words. Once you start printing exam papers for thousands of students, saving a single piece of paper goes a long way. This causes a lot of questions to be very short and simple. We as the students are expected to solve a math question, and some information are implied. Sure adding the information IS useful, but in this case the added information isn't necessary in an exam setting. Edit : As for a YT video, he solved it by putting it in a pinned comment.

[@mohitrawat5225]

Presh if half knowledge is dangerous than you've made a nuclear Bomb here😂😂😂😂😂

[@gumbitoicic9977]

​@@mohitrawat5225 *then

[@nagarajabhat4807]

@3:25, instead of writing the equations for distance, write the equations for the time -- i.e., d/20 = t+1/6 and d/25=t+1/15. Subtract the one from the other to eliminate 't' without ever solving for t. You'll directly get d/100 =1/10 or d = 10km

[@Jo-yr2fl]

The correct way to solve the problem is to sleep earlier, especially when you need to run 10km at the pace of world record next morning.

[@user-zw4sg9yd7s]

and still be late

[@guylamoureux880]

Take a bicycle ;-)

[@smallfieldlo4658]

Just assume a distance of 100km (LCM of 20and 25) If he goes on 20km/h then it will be 5 hours, and 4 hours for 25km/h, so the difference in time is 1 hour(60 mins). now the actual time difference is 10-4=6 mins, so the real distance will be 100km x 6/60=10km.

[@michaelmahoney6782]

Ooh that's a nice solution!

[@bobross7473]

@@michaelmahoney6782it’s elegant but hard to think of on the spot. I prefer the long way

[@chriskennedy2846]

I assumed a distance of 25 km, since that was the faster rate (25 km/h). I was able to quickly determine that that distance would produce a 15 minute gap between the two different rates. Since I was seeking a distance that would produce a 6 minute gap, I saw that 6 is 40% of 15, so the desired distance must be 10 Km, since 10 is 40% of 25.

[@philliberatore4265]

That's actually very much like the Goal Seek function demonstrated with Excel.

[@Deldee117]

I was able to figure it out but I had to make the assumption he left at the same time each time. It is critical in order to find a distance.

[@ChristopherFoxglove]

Exactly. The question just says he wakes up “late” each day, but it does not confirm it’s the same time. Perhaps he woke up a little less late on the second day. No big deal. I assumed, as you did, and came up with the answer.

[@templarknight7]

you don't have to make that assumption. you only have to assume he traveled the same distance.

[@Deldee117]

@@templarknight7 they live at distance 0, wake up 10 min late day 1, 4 min day 2. They live both at 0 and 10 km away [contradiction]

[@turtrooper8310]

Exactly, and it’s only solvable if you assume that.

[@templarknight7]

@@Deldee117 if they live at distance 0, they can't travel 20/25 km/hr. Look, imagine he leaves at 1pm one day, and 2 pm the next day. it doesn't change the answer, but he didn't leave at the same time in the scenarios. Just face the fact that you are wrong.

[@talkingscribe8898]

With "this information" we can't actually solve the problem. We only know that he woke up late each day. We don't know how late, specifically, if he was delayed the same amount each day.

[@MACADEMIAC]

Correct, we have to assume the same start time from home in both cases!

He has to travel the same amount of distance so the time he is late determines everything we need to calculate

[@conorslater8295]

We're not actually told the distance is the same on both days. He could've taken a different, longer route with less traffic (hence the faster average speed). I think it's fair to assume it's implied the distance is the same though.

[@quantumbuddha777]

The time he leaves each day will affect the answer. We don't actually know how long it took him each day, just that he arrived 6 minutes earlier the second day. If he also left 2 minutes later than the first day, this will change the answer.

[@georgelaing2578]

It doesn't matter WHEN he started, only WHERE he started!😐😐😐

[@maxlevel4425]

So on the first day he runs as fast as the fastest women competing in the 10km run. On the second day he runs fast enough to obliterate even the men´s world record. The question is: Why does he still study instead of starting a sports career? :D

[@yurenchu]

Because he's always late for training, and sports coaches don't like that.

[@dr.johnslab7502]

Presh never said that he walked or ran. But I agree with you that he couldn't do it on foot! So maybe he skateboarded, biked, drove, or even Segwayed! 😅

[@verkuilb]

@@dr.johnslab7502 He also never said he left at the same time each day-you need to assume that in order for the puzzle to be solvable. And if you make that assumption, then it’s also safe to assume that the graphic of a runner which accompanies the puzzle indicates that the person is, indeed, running the 10k. Which means the answer to the question of why he isn’t pursuing a sports career is: He keeps sleeping through all the qualifying events. 😂 Or, he’s taking ridiculously high amounts of illegal performance enhancers.

[@Meshamu]

I guess he's in college on an athletic scholarship? 🤷

[@yurenchu]

@@verkuilb You may be onto something: not "illegal", but "experimental" -- he's possibly a _chemistry_ student! :-)

[@joaogonzalez7672]

6 minutes is the difference in travel time from 20 to 25 Km/h. 6 minutes is equal to 0.1 hours. Considering "t" is the time it takes travelling at 25Km/h we have tx25 = (t+0.1)x20. We can calculate t which comes to 2/5 of an hour (24 minutes). t+0.1=30 minutes, at 20Km/h gives us the 10Km distance.

[@HackedPC]

If you don't mind What job do you do? A teacher?

[@joaogonzalez7672]

@@HackedPC 😄no, not a teacher. I am retired now but I worked for tech companis all my professional career: pre-sales, sales and sales management. however, for Presh's problems, it is my electronics engineering degree that kicks in!

[@shelleyweiss9920]

I used this same approach. It does not matter how late he arrives, rather it only matters the *Difference* in the times it takes at different speeds. This approach sets up the equation based on a baseline time and the fater (or slower) time, rather than creating a third "hypothetical" trip where he arrives on time. I think this technique is cleaner than the one presented by Presh

[@updating.....]

this type of questions is asked in govt exams ,just apply the formula d= (S1*S2/(S1~S2)) * time difference

[@samradhakrishnan]

This seems to be a cleaner method. You could simplify this even more In case one time taken is a d/20 and in the second case it's d/25 We know that the difference between the times is 6 minutes or 1/10 hours Hence d/20 - d/25 = 1/10, which simplifies to d = 10

[@bpark10001]

First, one must assume (it is not stated in the problem) that student leaves at the same time each day before the class is to begin. The algebra can be short cut by looking at only the 2 "late" cases, & that one is 6 minutes later in arrival time than the other. That gives equation d/20 = d/25 + 1/10. Multiplying through by 100 gives 5d = 4d + 10, d = 10.

[@Hokiebird428]

I read this problem and immediately realized that “10 minutes late” and “4 minutes late” were deliberately meant to mislead and confuse. So I reframed it: Imagine if he travels at 20 km/h and arrives on time, and the next day he travels at 25 km/h and arrives 6 minutes early. Same question: speed increases by 5km/h, time decreases by 6 mins. But much MUCH easier to understand and solve than whatever silliness was going on in the video with fractions of 1/15 and 1/6. v1 * t1 = v2 * t2 t2 = t1 - 0.1 20t1 = 25(t1 - 0.1) 20t1 = 25t1 - 2.5 5t1 = 2.5 t1 = 0.5 hrs t2 = 0.4 hrs d = 10 km

[@Ramon314]

I just immediately went 6/15*25 in my head. Then I was amazed by the explanation in the video

[@toddarmstrong7038]

Yep, that’s how I did it too.

[@corpsie666]

You call Presh's method silly, but you did the same thing except without showing the simplification steps being written out in full.

[@randomxnp]

I used the same technique but set up for mental arithmetic because I could not be bothered to find a piece of paper: At T+4 (4 minutes late) on day 1 he still has 6 minutes to go at 20 km/h which is 2 km. Day 2 achieved that extra 2 km by T+4 by walking an extra 5 km/h (this is a technique I teach pilots for calculating effect of wind on distance) which would take 24 minutes. Thus the total distance is 24 minutes at 25 km/h or 10 km. To check, 24 minutes at 20 km/h is 8 km, so day 1 was still 2 km short at T+4. At 20 km/h he would achieve 10 km by T+10. This is all easy in your head as every time is a multiple of 6 minutes, 0.1 hours so speeds are 2.5 km/6 minutes and 2 km/6 minutes. Another trick I teach pilots, except in knots (NM/hr).

[@ChristofAbsolution]

I punched these numbers into my calculator and it keeps coming back "he needs to wake up sooner".

[@vanchinamas9582]

If we use the formula speed=dostance/time, we get the following equation: (d/20)-(d/25)=6/60, where d is the distance b/n his home and college and 6 min is the time difference. Solving this directly is much easier

[@Skippy3rd]

I'm disappointed that you didn't state the assumption that the departure time is the same each day explicitly. That's the "Impossible" part, and it was just swept under the rug.

[@lam_xyz]

The same with the time school starts, this could also vary.

[@douglastaylor7947]

Do not complain a just enjoy and admire this beautiful math problem.

[@VeryGoodDeals]

@@douglastaylor7947 there is nothing beautiful about untold assumptions

[@JonathanKahan-q2g]

The departure time does not need to be assumed to be a the exact same time, as long as the school class starts at the same time every day which he does state.

[@Yusso]

@@JonathanKahan-q2g Yes it matters. Let's say the second day he woke up 5 minutes later than the day before. That will either shorten the distance or he will be 9 minutes late at school instead of 4.

[@Chris-hf2sl]

Presh's approach is not the most efficient way to solve the problem. You don't need to take into account the lateness in both journeys - one will suffice. When travelling at 25 km/h, he gets there in time t. (That's not the same t as Presh's.) When travelling at 20 km/h, he needs an extra 10-4 = 6 minutes (ie. 0.1 hours). So, now write the two simultaneous equations and solve - the maths is far simpler: d = 25 x t and d = 20 x (t+0.1). So 25t = 20t + 2. Therefore 5t = 2 and the distance = 25t = 5 x 5t = 5 x 2 = 10 km.

[@awfuldynne]

I just wrote an algebraic equation d/20kph - d/25kph = 0.1 hours "The difference in the time it takes to travel this unknown distance at 20 kph vs 25 kph is six minutes" but we write those six minutes in terms of hours because it's using km per hour and I want to cancel units if applicable. It ended up being _really_ efficient because my first step was to multiply both sides by the LCD of 100 kph to "clear the fractions", which resulted in 5d - 4d = 10 km

[@adipy8912]

Yet another fantastic question. Also, how did I never know about the excel trick? This can come in handy in so many situations.

[@douglastaylor7947]

Wow, I really enjoyed solving this problem!

[@georgemathiellis5486]

How do we know that he woke up the same time ? Waking up "late" does not mean that he woke up the same time. We are facing a problem with t1 and t2 and that is 2 equations and 3 unknowns....

[@georgemathiellis5486]

...I see I am not the only one who spend time trying to figure out the impossible riddle ...that is actually impossible unless you take t1=t2 .....

[@mubeenplayz5246]

With sums like these there are assumptions that must made

[@georgemathiellis5486]

@@mubeenplayz5246 I agree if you are dealing with a mathematical problem, but this is supposed to be an "impossible riddle".... So I was seeking an out of the box clue....

[@jacklinders7495]

We even dont know college start on the samen time

[@ADITYA-u5v1v]

Simple use v = d/t so d(diatance) = velocity × time d = 20[t + (10/60)] = 25[t + (4/60)] on solving equation t =1/3 put it in any equation d comes out to be 10km

[@brunogrieco5146]

BTW, according to Wikipedia for the Chicago Marathon, the record is around 2 hours for 42km i.e 21km/h. This kid may register for an athletic scholarship and or just quit school. 🙂

[@hippophile]

Assuming he left at the same time each day and college lectures started at the same time, I get the same equations at 3:18. Slightly quicker you can multiply the first eqn by 5 and the second by 4 to give s without calculating t. He really needs to be sprinting the 10km at 30 km/h. But as he already beat the world 10km record (26:24 minutes) with the 24 minute run, that is asking a lot!

[@raileite5994]

Important assumption: He leaves the house at the exact same time both days. So that's how we can compare the difference in time taken.

[@davidsorensen2116]

Which is a poor assumption, given the setup is that he wakes up late. Waking up on time can be attributed to an alarm and assuming he left the house at the same time would be reasonable. Waking up late is much more likely to result in variance in the starting time.

[@kenmore01]

I was wondering how it would work out if he left his house three minutes late already.

[@raileite5994]

@@davidsorensen2116 I agree, just pointing out exactly because of that.

[@kurzackd]

That's not an *"assumption"* because it's implicitly asserted in the way that the conditions for the problem are stated. .

[@gavindeane3670]

​@@davidsorensen2116It's not a poor assumption. It's an obvious assumption, necessary for the puzzle to be meaningful. But still, a surprising omission from the question.

[@dividebyzero1000]

This bring back memories of math being fun. Algebra just made so much sense to me. Problems like these were just fun logic problems to figure out. Higher level math, like multi-variable calculus with 6 dimensions- not so much. That was more like a magic class where you followed rules and got an answer that could not be justified or validated by any sort of logic or reasoning- you just made sure you followed all the rules and blindly trusted result.

[@quenchize]

If he was late you need to add how much late he was on each day. He could have woken up late the second day ran at 20km/h and been 4 minutes late. If he woke up 20 min late the first day and 14 min late the second day.

[@michaelwinpisinger9458]

d = 20t and d = 25(t-6) is even easier to solve. Gets rid of ugly fractions to work with. Just call the first day's time t and the second day his time was (t-6).

[@russellblake9850]

you're assuming he started at the same time both days.

[@gavindeane3670]

That's what I did (except I wanted the time in hours so I had t-0.1 rather than t-6). I haven't watched the video solution. I'm fascinated to see how the video finds a more complicated solution.

[@lupifa4395]

A tiny little issue : t-6*1/60, minutes vs hours...

[@RacingAnt]

​@russellblake9850 that's implied.

[@danmerget]

@@lupifa4395 Assuming that the OP solved the equation by setting 20t = 25(t-6) and solving for t, then no, you don't need to convert minutes to hours when solving for t. Both 20t and 25(t-6) are distances measured in units of km*min/hour. That's a very weird unit of distance; you would need to multiply it by a scale factor of 1/60 (the number of hours per minute) to convert the distance to km. But once you set 20t = 25(t-6), the scale factor cancels out, and solving for t gives you the time in minutes. It's pretty brilliant IMHO.

[@Stefan-xm9qb]

I remember when I was younger solving these puzzles. At some point I stopped solving them and don’t really know why. Now I will continue doing these on weekends. Old memories.

[@TheGamingG810]

I saw this question on the thumbnail and solved it on paper before even clicking on this video lol

[@pramoddighe3214]

You have shown three different ways in which this problem can be presented and solved. It's really fantastic. But I want to share one important point. As per the calculation, in order to reach in time, he has to cover 10 km distance within 20 minutes. If he goes at average speed of 20 km an hour, he covers the same distance within 30 minutes and thus reaches late by 10 minutes. If he covers distance of 10 km with the average speed of 25 km per hour, it takes him 24 minutes to reach college which is 4 minutes late. But for all this, one assumption is necessary and that is, in all cases, he should start at same time from his home. Difference can be in his average speed only and not in is starting time from home.

[@tomaszkisiel9915]

Actually, as far as the first method goes, there's even a quicker way: you rearrange the formula from d=r t to t = d/r. Now you do the same thing, except you have a slightly different set of equations: (1) t+1/6 = d/20; (2) t+1/15 = d/25; Now, you subtract the fractions and have (1) t=d/20-1/6; (2) t=d/25-1/15. Now, you substitute for t combining both equations and have: s/20-1/6=s/25-1/15. The rest is trivial.

[@ommadawnDK]

The equation at the beginning can't be d = rt. He normally begins at t < 0 and r < 20. The figure can't show the line d = rt. As stated, he normally starts earlier, maybe at t = -5.

[@jeske100]

1/20-1/25=0.01. 1/100 corresponds to 0.1 hours, so the entire distance is 0.1x100=10km.

[@100batman]

Good job Presh, but I think this simpler than using the 't' variable. D km = 20km/h * X hrs >> X = D/20 D km = 25km/h * Y hrs >> Y = D/25 6 min = X - Y = 1/10 hr D/20 - D/25 = 1/10 (x100) 5D - 4D = 10 D = 10 km

[@Javy_Chand]

For the sake of argument Let x(t) be the distance it took v₀ speed 1 (assume S.I mks conversion) v₁ speed 2 x₀=0, the initial distance t, the unknown time Δt, the delayed time Then, suppose he doesn't accelerate. That is, he maintains his avg velocity along the whole trayectory Finally: x(t+Δt) =v₀(t+ Δt) (1) =v₁(t + 4/10 Δt) (2) Solving for t: (1)=(2) (v₀- v₁)t +(v₀-4/10v₁)Δt=0 t=-((v₀-4/10v₁)/(v₀ -v₁))Δt As a sanity check, v₁ = 5/4 v₀, so t>0 Finally: x(t)=v₀(-(9/10v₀)/(-1/4 v₀))Δt = v₀Δt(18/5) Where v₀ is in m/s and Δt in seconds. Im... Not going to convert it back.

[@davidhopkins6946]

From the information at 0:53 you can deduce the distance between the boy's home and his college with the following equations: In general, v=d/t so let the first equation be 20=v(DayOne)=d/t(DayOne), let the second equation be 25=v(DayTwo)=d/t(DayTwo), and let the third equation be t(DayOne)=t(DayTwo)+(10-4)/60 (in which d is distance in kilometres between the boy's home and his college, t(DayOne) is his commuting distance in hours on the first day, and t(DayTwo) is his commuting distance in hours on his second day, and v is his velocities in kilometres per hour on the first day and the second day) These equations become nicer to look at if we make them linear: Equation 1: 20*t(DayOne)=d Equation 2: 25*t(DayTwo)=d Equation 3: t(DayOne)=t(DayTwo)+(10-4)/60 or (with 10 and 4 being the times in minutes he was late on these 2 days, and dividing the time in minutes by 60 to convert it to time in hours) 20*(t(DayTwo)+(10-4)/60)=d or 20*t(DayTwo)+(200-4*20)/60=d or 20*t(DayTwo)+2=d or 20*d/25+2=d or 20d+50=25d or 50=5d or d=10 So the distance between the boy's house and his college is 10 kilometres.

[@misterdubity3073]

In your head method; helps to be lucky (assuming he leaves same time both days, which was not stated, so the real answer is "not enough information given"). But ... guess & check: Suppose dist is 100 km, then 1st day trip takes 5 hours, 2nd day trip takes 4 hours, a savings of 60 minutes. But he only saved 6 minutes. 6 = 60/10; dist = 100/10. d=10 km

[@dannybodros5180]

You can also resolve the problem by converting km/h to km/min. This instantly gives you the constant value of time = 0.(3), or t = 1/3. Now that you have the constant value of time, the value of speed, and the 10min delay, you can calculate the distance. Distance = Speed x Time, so by plugging in the information you get Distance = 20 km/h x (1/3h + 1/6h) = 20km/h +1/2h = 10km.

[@muhammadsiddiqui2244]

That was one of the easiest problem. Just made these two equations: s=20t1, s=20t2. Then, t1=t+10/60 and t2=t+4/60, then substitute t1 and t2 and voila, you have two linear equations in two variables. Yes, t is the time for which if he had moved with appropriate speed would have reached just in time.

[@shlomoeshet8525]

A simpler solution is based on t as the time of the first day and t-6/60 as the the time of the second day (6 minute leass). This yields the equation 20*t = 25*(t-6/60) (both sides are d). Solving for t gives t=0.5 hours thus d=10 km.

[@fallen-il8ze]

imo, the easiest way is actually the following: 20 km/h -> 25 km/h means, he is 1/4 faster than the day before resulting in 5/4. This means, he only needs the reciproke = 4/5 of the time. Since he was 6 min faster, it means 6 min is 1/5 -> 4/5 = 24 min = 0.4 h -> 0.4 h * 25 km/h = 10 km

[@mihiguy]

Given the information without additional assumptions (in particular you don't know whether he left at the same time or at different times in both days), you only know that it takes longer than 10 minutes to reach the destination at 20 km/h and longer than 4 minutes at 25 km/h, so the distance is at least 1.6 km.

[@DarinaSouplet]

I really appreciate your work! Your videos are always of such high quality and warmth. Keep up the good work!💷🏖💞

[@thadswashesed9076]

You don't actually need to convert units and if you don't, then you aren't dealing with fractions, assuming the d r t are same units among themselves Starting with 20(x+10)= 25(x+4), expands to 20x+200=25x+100, simplifies to x=20. Then you can convert at end rather than at beginning, 30 minutes at 20km/h means 1/2 hour, so 10km

[@michaelmahoney6782]

distance/speed=time Let x = the total distance (in km), let y be the amount of time (in minutes) it would take for him to get to class exactly on time. We get: 1) x/20 = (y+10)/60 2) x/25 = (y+4)/60 Solving both equations for x (and multiplying both sides by 60) yields: 20y + 200 = 25y + 100 Which can be solved for y = 20 minutes Plugging back into 1), we get: x/20 = (20 + 10)/60 x/20 = 30/60 x/20 = 1/2 And x = 10km

[@midnighttrain-jz2my]

A simple way: He needs time t1 with v1=20km/h, and time t2=t1-6 with v2=25km/h. Time and Speed are anti-proportional. It is: t1/t2, or t1/(t1-6) = v2/v1. We solve for t1 to be 30 minutes or 0.5 hour. Then the distance is 20km/h x 0.5h = 10 km.

[@TD-er]

Way simpler way to solve this: 20 km/h is 3 minutes per km. 25 km/h is 2.4 minutes per km. Difference is 0.6 minutes per km. Difference in arrival-time is 6 minutes, so the distance is 10 km.

[@Jabbaerwocky]

I like your method! 😊

[@frank_a]

On day 1, the speed is 20km/hour. If y is the distance, and x the number of minutes of running time on day 1, then equation 1 is: 20/60 = y/x --> 1/3 x = y On day 2, the speed is 25km/hour. The distance is still y and the amount of running time is 6 minutes less than on day 1. So equation 2 is: 25/60 = 5/12 = y/(x-6) These are two equations, with two unknown variables, which can be solved --> y =10 km, x = 30 minutes (running time on day 1), running time on day 2 is 24 minutes.

[@fabianagco5902]

I think my solution is a bit more elegant. I had only one variable: The time that he had before the lecture started. -> ... So you have one equality with one variable: (10 + x) * 20/60 = (4 + x) * 25/60 or in words: The distance that he travelled is equal both to travelling the time that he had plus 10 minutes at 20km/60min and also equal to travelling the time that he had plus 4 minutes at 25km/60min. You don't even need an ugly fraction like 1/15th to solve this. x is 20min (he started travelling 20min before the lecture started) and therefore the distance is 10km.

[@davidioanhedges]

Since waking up "Late" is not a constant ... there is no answer

[@Doctor_C_Jack]

I don't think there is *no* answer as you can probably make an expression that outputs the answer as a function of difference in time when leaving late.

[@davidsorensen2116]

@@Doctor_C_Jack With departure time variable between the two days, you are left with a system of 2 equations with three variables: d=20 km/h*t_1 d=25 km/h*t_2 As an example, t_1 could be 5 hours, and t_2 could be 4 hours, making a distance of 100 km. If class starts at nine, he left at 4:10 am on the first day and 5:04 am on the second day. It could also be the case that t_1 is 75 minutes and t_2 is an hour, making d 25 km. This isn’t a very realistic scenario, but it illustrates that d doesn’t have a defined solution because there are two variables that can be changed to produce different results.

[@jansz1589]

Assuming he woke up too late by the same amount both times it's actually pretty easy. However, there might be a variation in how much he found himself late each day after waking up, which would make things vastly more complicated.

[@wmpowell8]

*Simple solution:* Assuming he left the same amount of time before class started on both days, by going 5/4 the speed and taking 4/5 the time, he cut 6 minutes off his journey. So 6 minutes is 1/5 of the journey meaning the journey is 30 minutes at 20 km/h. 30 minutes is half of an hour, so the distance is 20 km/h times half an hour which is *10 km.*

[@birds_eye_view]

I used another way, graphically/geometrically. I chose a graph with x axis is time, and y axis is speed. The distance is the area under each graph. We end up with two areas (rectangles), one lower at 20(km/h) but further, and the other higher at 25 but shorter. Both rectangles overlap, but must have the same size. Meaning the areas they don't overlap must be the same size too. The one on the right is 20*0.1 (6min diff in hours). So it is 2. The other on top must be 2 as well, so 5 (speed diff) *?=2 Thats 0.4. The rest to make the full rectangle for the actual distance using the height of 20 (the overlapping part) * 0.4 (the width of the overlapping part) = 8. Together with one of the not overlapping areas of 2 makes 10.

[@Neckhawker]

"walking to college" : at an average speed of 20km/h. Yeah "walking". 10km at 25km... yeah sure.

[@ManjeetRani-v5n]

yes, and that's the reason why he reaches 10 km after 1/3 hr or 20 mins

[@sebastien5048]

I know some people can run a marathon at around 21km/h. But is it even possible for anyone to run 10km at an average speed of 25km/h ? Sounds even more difficult. EDIT : just checked on wiki. The world record is currently at 22.91km/h. So literally no one in the world can do it at an average of 25.

[@RacingAnt]

​@ManjeetRani-v5n you can walk at 25kmh? OK, superman 😂

[@JohnDlugosz]

Good point, the running-man icon was a poor choice. It should have been a bicycle. Though the icon would apply to skates as well!

[@AzureKyle]

It's a math problem, not an example of realism. I mean, can anyone really keep the same exact pace for 20-30 minutes while walking any speed? Suspension of disbelief is expected here.

[@izzmus]

I got 10km by an entirely different method, I focused on time and not distance. The difference in speed made for a .1 hour difference in time. If he ran for one hour at 20km/h, he would have traveled 20km. If he ran that 20km at 25 km/h, he would have made the trip in .8 hours. This is a .2 hour difference in time, which is exactly twice the difference in time that we are looking for, which means that his first morning was a .5 hour long run, or 10km.

[@dhpbear2]

My algebraic solution: T1 = D/20, T2 = D/25 time difference = T1 - T2 = (10min-4min)/60 = 0.1 Hr D/20 - D/25 = 0.1 Hr Multiply both sides by (20*25): 25D - 20D = 50 5D = 50 D = 10 Km

[@hishamabuhussein6106]

i have solved this problem using the rule of three! 1/ 20km -----> 60mins distance (d)-----> t + 10mins 2/ 25km ------> 60mins d ------> t + 4mins and then using the rule of three : 1/ 60d = 20t + 200 2/ 60d = 25t + 100 therefore, 20t + 200 = 25t +100 so t = 20 mins after that you can figure out that d = 10km by using one of the 2 equations used before 60d = 20t + 200 60d = 20*20 + 200 d = 600/60 then d = 10km

[@alexvanschaik4753]

1 km @ 20 km/h takes 3 min (60/20) and 1 km @ 25 km/h takes 2,4 min (60/25). The time difference over unknown distance d is 6 min, so 3d - 2,4d = 6 or 0,6d = 6 hence d = 10 km.

[@briansimpson9318]

Solution spoiler: There is a more elegant and efficient solution than the one you show. Rather than formulating the equation where each is d=rate*(t+late amount); set the equations to d=rate*(t+the difference in late time). This reduces the number of steps in solving by one. The two equations formulate to 25*t = 20*(t+0.1h). Yields 1.25 = (t+0.1)/t, t = 0.4h. I found only using the difference in time to be a little more intuitive and it is actually pretty similar to how you solved it in excel.

[@gustavoshigueo]

I find it easier to work with minutes to find t and only then convert to hours, that results in less fractions: d / (t + 10) = 20 d = 20 (t + 10) d = 20t + 200 d / (t + 4) = 25 d = 25 (t + 4) d = 25t + 100 20t + 200 = 25t + 100 20t + 100 = 25t 100 = 5t t = 20 minutes d = 20km/h * (t + 10)minutes d = 20km/h * 30minutes ---> 30 minutes = 1/2 hour d = 20km/h * 1/2h d = 10 km

[@Ht6602-x2o]

This can be solved in 30 sec using the spreadsheet method but without the spreadsheet: He needs 1/20 = 0.05 h per km in one case, 1/25=0.04 h in the other case. So the difference in the delay is 0.01 h = 0.6 min per km distance. Since the actual delay difference is 10 times as long, the distance is 10 km.

[@57thorns]

Haven't looked too closely, but we also know the speed necessary to get there in the time allowed: 10 km in 20 minutes or 30 km/h. So going from 20 to 25 save six minutes while going from 25 to 30 save another four. And now we are moving at a paces that is hard to follow for a sprinter....

[@DB88888]

If he leaves always at the same time, the distance is 8km. Here's how I solved it. In theory there is an optimal time t0 such that if he covers the distance x between house and school exactly in t0, he's exactly on time. If he is slower, he will take an additional time dt. Given that the distance x is always given by his speed v times the ime it takes to cover x, in general one has v*(t0+dt)=x The problem gives us two instances of this general equation, both coherent with t0 and x: one with v=20km/h and dt=10 min, and the other with v=25km/h and dt=4min. So we can write the system of equations (20/60)*(t0+10)=x (25/60)*(t0+4)=x Where the division by 60 is to convert the speed from km/h to km/min. This is a simple system of two equations in two variables (t0 and x) which can be solved to find that the distance is 8km and in order to be on time he needs to cover it in exactly 20min. Notice that if he leaves the house each day at a different time, the time t0 needed to cover x to be on time would be different from day to day: say it would be t1 on the first day and t2 on the second day. In that case, you would not be able to solve the problem because you would have the equations (20/60)*(t1+10)=x (25/60)*(t2+4)=x Which is a system of 2 equations in 3 variables. So in that case it's not "seemingly impossible" but just impossible.

[@richardhole8429]

That was a fun problem. I set it up exactly the way you solved it and arrived at the same solution. Hey, everyone, we don't need to know what time he left. It must be different for the two cases. And we don't need to know how fast he would walk to get to class on time.

[@fredericvondenhoff5210]

There is a simplier way to find it : If T is the time at 20 km/h then the time at 25 km/h is T-(1/10), (1/10 hour is 6 m) --> D/20=T; D/25=t-(1/10) --> 20T=25T-2.5 --> 5T=2.5 --> T=0.5 h, As T is the time at 20KM/h ; D is 10 Km.

[@tellur808]

There is actually a simpler algebraic way. Skip the d=rt step and write it directly like this: d = 20 km/h * t_20 = 25 km/h * (t_20 - 6 min) - 6 min being the time he saved going at 25 km/h instead of 20. t_20 denotes the total time it took him going 20 km/h. (If we were really pedantic about it, we are skipping the trivial set of equations: t_20 = t_min + 10min; t_25 = t_min + 4 min, with t_min being the time between leaving the house and the bell ringing. We would express t_25 in terms of t_20 and plug it into d = 25 km/h * t_25 - but thus can also be read directly from the question) Replace 6 min by 1/10 h and you can skip a lot of bothersome fractions and shuffling of terms. Next steps would be -5 km/h * t_20 = 25 km/h * (-1/10h) t_20 = 25 * 1/5 * 1/10 h t_20 = 25/50 h = 1/2 h d = 20 km/ * 1/2 h = 10 km

[@ShawnF6FHellcat]

Interestingly enough, I'm trying to learn how to use Excel in my vocational rehab program...

[@3dplanet100]

7:32 if we draw a vertical line at the intersection of time and distance, then he traveled instantly, (which is impossible, because it would defy the physics laws). In the other hand, if we draw a horizontal line parallel to the time line) in the time line, it would never intersect the dotted horizontal line on top, because it means that time is passing but you are not moving, and therefore, you would never reach your destination.

[@richardfarrer5616]

On the second day, he travelled at 5/4 times the rate of the first day. Therefore the time he took was 4/5 of the total. If he would arrive on time after t minutes then he took t + 10 on the first day and t + 4 on the second. Thus t + 4 = 4/5(t + 10), giving t = 20 minutes.

[@rdspam]

LCM of 100 km takes 4 hours and 5 hours, gaining 1 hour. To make up 6 minutes, 1/10 of the time, takes 1/10 of the distance, or 10km.

[@VoidHybrid]

There is a simpler way you can solve it in your head. Divide 20km/h vs 25km/h (ignoring the km/h) and you get 4/5. Then take the difference between the times, (10-4=6) and multiply that by the two numbers: 6x4=24 and 6x5=30. Then simply take the 24 or 30 / 60 to turn the minutes into hours (0.4 and 0.5), then multiply that by their correct km/h. 0.4/25=10. 0.5/20=10 Figuring it out like this actually shows that in both cases, he left 20 minutes before he was supposed to be there. For example: if he was supposed to be there are 9:00 and arrived at 9:10 and 9:04, 9:10-0:30 and 9:04-0:24 both equal 8:40, or 20 minutes before the intended time of arrival. And this is just coincidence because if for example he was 10 minutes late both times, he would have left at 8:40 when he was going 20km/h and 8:46 when he was going 25km/h. So I like your answers, but you missed the explanation of a much simpler and easier solution that most anyone could follow along with without algebra.

[@CanKarakuzulu]

A faster way would be like this: 20 km/h and 25 km/h least common multiple is 100. So if distance were 100 km, at 20 km/h he would arrive in 5 hours, and at 25 km/h he would arrive in 4 hours. The delay would be 60 minutes. But the delay is 6 mins. So the distance should be 1/10th of 100 km. So it is 10 km.

[@StefanvanAalst]

I like the spreadsheet approach. Not so much the goal seek, but the initial difference was .01 given distance of 1. Since it is a linear equation and the difference had to be .1..... Of course assuming he departed at the same time and school started at the same time (yes, the latter is also assumed)

[@mannatrana9361]

I solved it.. with the only the given info.. Firstly, I converted everything to SI units Let the distance between school and home be 's' I supposed that he took 'T' time to reach school with the speed of 25km/h or (125/18)m/s So time taken to reach school with the speed of 20km/h (100/18)m/s. =( T + 360) Apply S = ut + 0.5at² ( acceleration is 0) On both speeds We get, S= 125T/18 And S=100(T+360)/18 Solving both we will get S = 10000 metres or 10km

[@DrMilk-kw4hn]

First I'm getting rid of the hours and making everything in minutes Day one: 20/60 km/m, 10 min late. Remaining distance that day, 20/6km. I got the remaining distance by multiplying how late they were with the speed. D (distance)= T (time) *20/60 (speed in minutes) + 20/6 (remaining distance) D= T*4/12 + 20/6, same equation as above, but with easier to use fractions for the other equation below Day two: 25/60 km/m, 4 minutes late. Remaining distance that day, 10/6km D = T*5/12 + 10/6 Next, substitute D from the above equation with the 'easy' one above it. T*5/12 + 10/6 = T*4/12 + 20/6 T*5/12 - T*4/12 = 20/6 - 10/6 T*1/12 = 10/6 T = (10/6)/(1/12) T = 20 Next, plug 20 in for T in any of the D='s equations D= T*4/12 + 10/3 D= 20*4/12 + 10/3 D=10km

[@johnpaterson6112]

The assumption you need to make is that on each day he left home at the same length of time before he was due at college. Those who say you need to assume he left home at the same time each day are assuming he was also due at college at the same time. Further, the simplest procedure is to equate time-differences: d/20 - d/25 = (10-4)/60

[@eletouzelallinec]

D/20 is the time it takes at 20km/h, D/25 at 25km/h; so knowing the difference is 6mns = 1/10 of an hour, solving for D/20-D/25=1/6 yields D/100=1/10 and D=10.

[@tonyennis1787]

For me, this problem is a setup problem - you have to set it up correctly, then it's easy. There are two gotchas - the hours/minutes conversion, and that the time to campus is _relative to now_ and is a constant. What's not a constant is how late you are. I think _t_ should have been stressed a little more for people trying to use the distance formula but not figuring out that the time was the start of the class plus the lateness.

[@nicholasimholte7359]

T = D/R, and since the difference in times was 6 minutes: D/20 - D/25 = 6 min = 1/10 Hr. D(25-20)/(25*20) = 1/10 D = (25*20)/(5*10) = 10.

[@randomxnp]

At T+4 (4 minutes late) on day 1 he still has 6 minutes to go at 20 km/h which is 2 km. Day 2 achieved that extra 2 km by T+4 by walking an extra 5 km/h (this is a technique I teach pilots for calculating effect of wind on distance) which would take 24 minutes. Thus the total distance is 24 minutes at 25 km/h or 10 km. To check, 24 minutes at 20 km/h is 8 km, so day 1 was still 2 km short at T+4. At 20 km/h he would achieve 10 km by T+10. This is all easy in your head (as every time is a multiple of 6 minutes, 0.1 hours) so speeds are 2.5 km/6 minutes and 2 km/6 minutes. Another trick I teach pilots, except in knots (NM/hr).

[@Jabbaerwocky]

I prefer the approach that allows me to solve it all in the head without messy equations and dealing with fractions. Recast the problem as two different persons both setting off at the same time at 20kmph and 25kmph respectively. Let's call them Mr. 20kmph and Mr. 25kmph. By the time Mr 25kmph arrives 4mins late, Mr 20kmph has 6 more minutes to go to arrive 10 minutes late. At 20kmph, 6 minutes covers a distance of 2km. For every 1km Mr 20kmph travels, Mr 25kmph travels 1.25km. When Mr 25kmph covers 10km, Mr 20kmph covers 8km with 2km more to go. These final 2km are what keeps him arriving 6 minutes later than Mr 25kmph.

[@chjialong]

Consider where he was when school started. Day 1. 20*10=200 km(min/h) Day 2. 25*4=100 km(min/h) Traveling 5km/h faster gets him 100 km(min/h) farther. That means he wake up 20 minutes before school started. So the distance is 20km/h * (20 + 10)min = 10km. The trick here is that our brain is much better handling same time than same distance.

[@Qermaq]

If we assume he left the house as long before class started on both days, then if the trip is k kilometers then (60/20)k = (60/25)k + 6. The real question is can we make this assumption? I will answer this way: if this was a real-world problem to solve, we do not have enough information to solve. But this is a test question. Unless we presume the test includes questions that can only be correctly answered "insufficient info to solve" (and that might be the correct presumption, but assuming it's not) then I think it is wise to make the assumption that the test is flawed but this assumption must be made for this to be solvable. It's the difference between logical rigor and gaming the system, and sometimes one or the other is more important.

[@viveksharma9217]

If speed is increased by 25% then the time will decrease by 20%, as the decrease in time is 6 min (20%) i.e. time spent in travelling at 20 kmph is 30 min. Hence distance is 10 km.

[@MichaelPaoli]

No need at all to solve for t: Let's presume both days he left at same time and was supposed to arrive at same time, and distance is constant. Let d be distance in km, let t be time he should (on time) be there, for t we'll use units of minutes. Distance divided by speed/rate (we use km/minutes) gives time traveled, subtract minutes late for the on-time arrival times, which are equal: d/(20/60)-10=d/(25/60)-4=t d/(20/60)-d/(25/60)=6 60d/20-60d/25=6 15d/5-12d/5=6 3d/5=6 d=6(5)/3 d=10 10km Check: 20km/h, 30 minutes, 25km/h, 24 minutes, difference of 6 minutes, as expected.

[@brucea9871]

This problem is easy if you assume he left at the same time each day (which was not stated but I presume you meant it). Assume it took him x minutes to arrive when traveling 25 km/h. Then it took him x + 6 minutes to arrive when traveling 20 km/h (he arrived 10 minutes late at 20 km/h and 4 minutes late at 25 km/h so the difference is 6 minutes). The distance he travelled in the first case is 25 X (x ÷ 60) and in the second case is 20 X ((x + 6) ÷ 60) (I divided by 60 to convert minutes to hours). These distances are equal, so we set them equal and have a simple linear equation. Solving it gives x = 24 minutes so the distance is 25 X (24 ÷ 60) = 10 km.

[@charleslivingston2256]

Way easier to just let t be the time in hours for the first day (and t-0.1 be the time in hours the second day). d = 20t = 25(t - 0.1) 20t = 25t - 2.5 5t = 2.5 t = 0.5 d = 20(0.5) = 10

[@BenDRobinson]

This was a near instant solve by thinking about ratios (once the missing conditions are assumed!)... the ratio of his two speeds is 4:5 so the times must be in same ratio (but reversed), and differing by 6 minutes they must therefore be 24 and 30 minutes. If it takes 30 minutes at 20km/hr then it's 10km.

[@HarishKini]

easiest way to solve this is to use the equation time taken = distance/speed. So solve for (D/20) - (D/25) = (1/10), which gives D=10. Here (1/10) is the time difference (10-4=6 mins in hours)

[@nitinjaganath1004]

Instead of assuming the original time as t, I did by assuming t for the speed 20km/hr as we need not find the original rate r. So the t' for 25km/ hr its t-6 mins or t-1/10. Then i got a simple equation 20t= 25(t-1/10). 20t=25t-25t/10, t=1/2. Substituting in d=20t, d=20/2, 10

[@FerdiLouw]

Need more data: The 2nd day that he overslept: did he wake up earlier or later than the 1st day?

[@corpsie666]

That's unknown. The time that class starts is also unknown. The assumption is that the student woke up with the same amount of time to get to class.

[@NotBorno]

(25×50/3)=d/(T+4) in metres per minute 20×50/3=d/(T+10) T+4=d/(25×50/3) T+10=d/(20×50/3) substracting the equations, 6=d(5)/(20×25×50/3) 6=d×(5×3)/(20×25×50) d=10000 m or 10 km

[@Aashu2631]

Here is my answer : speed. = (Distance / time) Speed is directly proportional to distance and inversely proportional to time. Ratio of speed 20:25 It means 4:5 So ratio of time will be exact opposite So ratio of time will be 5:4 Now difference in the ratio of time is equal to 1 unit (5-4=1) And that 1 unit is equal to 6 minutes. So he took 6×5=30 minutes and he was late by 10 minutes so he should reach at his place in 20 minutes. Now distance = 20 × (30/60) = 20× .5 = 10 km

[@randomguy-tg7ok]

It's a fairly easy question, surely. It's a difference of 6 minutes, so if we let X be the distance to the College in km, then: (X/25) + 0.1 = (X/20) --> X = 10. The journey takes 24 minutes @ 25km/h and 30 minutes at 20km/h.

[@christophertarry4865]

Say he's 5 km away Goes 20km/h -> gets there in (5/20)*60=15 minutes Goes 25km/h -> gets there in (5/25)*60=12 minutes 3 minute difference, so this can't be right We need the difference between these two times to be 10 - 4 = 6 minutes (x/20)*60 - 6 = (x/25)*60 -> x = 10 10 km away Goes 20km/h -> gets there in 30 minutes Goes 25km/h -> gets there in 24 minutes He actually needs to get there in 20 minutes to be perfectly on time which would require 30km/h

[@kentoutcourt]

runners actually calculate their pace in minutes/km. So the first day was at 3 minutes/km. The second day, at 2:24 (2.4 is easier to calculate so we'll use that). He shaved off 6 minutes the second day. So, 3d = 2.4d +6 you can see right away that 2.4 * 10 +6 = 30 but for the sake of it, let's do it the normal way: 3d - 2.4d = 6 0.6d = 6 d= 6/.06 d= 10 I think I have the easiest way. Took me 2 minutes.

[@RichardMellish]

The question is impossible unless we assume the same starting time. On that assumption, I made no reference to the normal travel time (your t). Distance d km at 20 km/h takes d/20 hours = 3d minutes. Distance d at 25 km/h takes d/25 hours = 2.4 d minutes. The time difference is 10 minutes - 4 minutes = 6 minutes. Therefore 3d - 2.4 d = 6, so 0.6 d = 6, so d = 10.

[@RobloxKid123]

Took me around 4 mins to solve, pretty easy problem tbh (haven't watched the vid yet) The amount of time taken on day 1 can be x+10 minutes, amount of time on day 2 can be x+4 minutes At s (speed) = 20 km/h, t = x+10 At s = 25 km/h, t = x+4 The distance is constant, and it can be found by multiplied by dividing total time by time taken for 1 km At 20 km/h, time taken is 1/20 h, or 3 minutes per kilometer At 25 km/h, time taken is 1/25 h, or 2.4 kilometer per kilometer (x+10)/3 = (x+4)/2.4 2.4x+24 = 3x+12 12 = 0.6x x = 20 Time taken at 20 km/h = 30 minutes = 0.5 hours Distance = 20 * 0.5 = 10 km

[@RobloxKid123]

I just realized I typed "2.4 kilometer per kilometer" by accident, I'm keeping it for the lols tho

[@bobross7473]

Congrats on solving it, but for others reading this, don’t get discouraged by this guy bragging his ass off. It’s a hard problem compared to most rate problems because it isn’t immediately obvious how to set up the equation. If you solved it, good job, you should feel proud. If not, you should understand Presh’s solution after a while.

[@maxhagenauer24]

It is very frustrating when he didn't say he left at the same time both days, that should absolutely be stated.

[@phr00tpie]

Anyone else just do 2 simultaneous equations and solve for time then multiply to get the distance? t= time in minutes D = total distance so it takes t minutes to go D at 20km/h and t-6 minutes at 25km/h 20t = D = 25(t-6) Distributing 25 20t = 25t - 150 collecting like terms and changing signs 5t = 150 t=30 therefore D = 30 mins at 20km/h = 10km

[@gg2023]

Assuming he wakes up at the same time difference compared to when college starts, i.e if college is at 10 he wakes up at 10-tc If college is at 9, he wakes up at 9-tc, And assuming he takes the same route so same distance = D, Then i got the answer D = 10 and he wakes up 20 mins before college through the formula: average speed= total D/ total time taken.

[@taflo1981]

That guy wakes up and just casually runs 10km more than two minutes faster than the world record. That's impressive.

[@xeuszzz]

I solved it by not considering t at all and found it easier. The latter speed (25) is 6 minutes (⅒ hours) faster, so we get the equation with only d as an unknown. d/20=d/25+1/10 d/20-d/25=1/10 (25d-20d)/500=1/10 5d/500=1/10 d/100=1/10 d=10

[@xeuszzz]

Actually multiplying both sides first by 5, gets you even easier numbers. d/20=d/25+1/10 d/4=d/5+1/2 d/4-d/5=1/2 (5d-4d)/20=1/2 d/20=1/2 d=10