Seemingly Impossible Walking To College Riddle
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- Опубліковано 9 жов 2024
- This is a fun riddle I saw online. Can you figure it out?
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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?"
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😂😂😂😂😂😂
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.
@@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.
Presh if half knowledge is dangerous than you've made a nuclear Bomb here😂😂😂😂😂
@@mohitrawat5225 *then
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.
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
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.
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.
It doesn't matter WHEN he
started, only WHERE he started!😐😐😐
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.
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.
you don't have to make that assumption. you only have to assume he traveled the same distance.
@@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]
Exactly, and it’s only solvable if you assume that.
@@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.
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.
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
Because he's always late for training, and sports coaches don't like that.
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! 😅
@@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.
I punched these numbers into my calculator and it keeps coming back "he needs to wake up sooner".
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.
The same with the time school starts, this could also vary.
Do not complain a just enjoy and admire this beautiful math problem.
@@douglastaylor7947 there is nothing beautiful about untold assumptions
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.
@@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.
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.
Ooh that's a nice solution!
@@michaelmahoney6782it’s elegant but hard to think of on the spot. I prefer the long way
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.
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.
If you don't mind What job do you do? A teacher?
@@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!
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
this type of questions is asked in govt exams ,just apply the formula d= (S1*S2/(S1~S2)) * time difference
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
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
I just immediately went 6/15*25 in my head. Then I was amazed by the explanation in the video
The way I did it that made sense to me, the 10 minutes and 4 minutes actually helped because of the 6 minute difference
Yep, that’s how I did it too.
You call Presh's method silly, but you did the same thing except without showing the simplification steps being written out in full.
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
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.
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.
I was wondering how it would work out if he left his house three minutes late already.
@@davidsorensen2116 I agree, just pointing out exactly because of that.
That's not an *"assumption"* because it's implicitly asserted in the way that the conditions for the problem are stated.
.
@@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.
@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
Since waking up "Late" is not a constant ... there is no answer
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.
@@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.
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....
...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 .....
With sums like these there are assumptions that must made
@@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....
We even dont know college start on the samen time
"walking to college" : at an average speed of 20km/h.
Yeah "walking". 10km at 25km... yeah sure.
yes, and that's the reason why he reaches 10 km after 1/3 hr or 20 mins
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.
@ManjeetRani-v5n you can walk at 25kmh? OK, superman 😂
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!
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.
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.
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. 🙂
Not enough information. Are we supposed to assume that he left home at the exact same time each day? Or more precisely, the exact amount of time before class start (could be a different class each day). Logically this is a far stretch: even if he chronically sleeps late every day, we shouldn't expect him to wake up at the EXACT same time and run late by exactly the same amount every day.
I infer that this guy has an unpredictable daily schedule and this "riddle" is not solvable with the information provided.
Logically it's not a far stretch at all. Because, as you say, without it the puzzle is unsolvable.
We could assume that he left his house at 9 am each morning.
this type of questions is asked in govt exams ,just apply the formula d= (S1*S2/(S1~S2)) * time difference
@@updating..... Or find the time it takes him to reach the first time and the time it takes him to reach the second
time, then solve for D.
So, if D is the distance, it would take him D/20 hours to reach the first time and D/25 hrs + 6 minutes the second time.
The 6 minutes is the difference between (10 minutes, he was late the first time compared to the 4 minutes late the second time)
D/20 = D/25 + 6minutes. and solve for D
Need more data:
The 2nd day that he overslept: did he wake up earlier or later than the 1st day?
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.
Weird that it's called "an impossible riddle", it's a pretty generic speed equation
Yea. It doesn't even seems impossible at first glance, it's immediately obvious that the information is enough (if we take the assumption that college starts at the same time and the person starts going at the same time)
I mean it is impossible that he's walking 20km/h for 20 minutes, so you just have to read it as "Seemingly Impossible Walking To College" Riddle
It isn’t impossible, but it’s definitely harder than most rate problems. Most rate problems are set up exactly like the formula but this one requires some conceptual understanding.
No one will click a UA-cam video named "pretty generic speed equation riddle"
Well, it should be impossible to call a simple question like this a riddle. So calling it an impossible riddle is kind of possible.🤔
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).
you're assuming he started at the same time both days.
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.
A tiny little issue : t-6*1/60, minutes vs hours...
@russellblake9850 that's implied.
@@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.
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.
Yet another fantastic question.
Also, how did I never know about the excel trick? This can come in handy in so many situations.
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.
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.
Interestingly enough, I'm trying to learn how to use Excel in my vocational rehab program...
Wow, I really enjoyed solving this problem!
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.
I like your method! 😊
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.
Time (T) should have been defined as T = T(final) - T (initial). With this you immediately realize this problem can not be solved as you have 4 variables and only 2 equations.
I was waiting for him to expand upon the spreadsheet solution to reveal that IF assumed distance is increased from 10 to 15 km (or 20 km and so on), the difference in time between day 1 and day 2 increases accordingly.
That's wrong. We don't know at what time he got up 😄
We don't need to know it. It's ok to be 4 min late and we can use Google for this type of questions in the modern era 🙂
That guy wakes up and just casually runs 10km more than two minutes faster than the world record. That's impressive.
You dont need a paper, it is strait forward : with those speeds the difference of time for a distance of 100km is 1h, that is 60mn. To get 6mn the distance need to be 10 times less, that is 10km.
I don't get your reasoning, could you elaborate please?
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.
We not only have to know whether he left at the same time, we need to know if on-time arrival is the same time.
True. When I was in college there were Mon, Wed, Fri classes and different classes on Tue & Thurs. So very likely that on consecutive days the student had different classes that started at different times.
The shortest and simplest way is based on the fact that when the distance is fixed, the ratio of the times taken is equal to the inverse ratio of the speeds. Thus, we have-
(T+10)/(T+4) = 25/20
Subtracting 1 from both sides:
6/(T+4) = 5/20; Or T+4 = 24
Therefore, D = 24×(25/60) = 10 km.
That guy is a pretty fast runner 😂😂😂 keeping the speed 25km/h during 10km
He's using an E-Bike.
At that speed he would complete a 42.2 km marathon in 1 hour and 42 minutes! I think the word impossible in the video title refers to the 'walking' part of the puzzle. 😄😄
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.
*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.*
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
In your graphical solution, the time axis is not the time spent to travel the distance, but the time-of-day. That confused me at first.
You then show the same time-of-day start time, but that was not part of the problem description, as others has already commented.
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
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.
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.
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.
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.
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
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.
If physicists can make the assumption that everything's a sphere, then it's safe for us to assume he left the house at the same time
(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
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
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.
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.
Let distance be x.
So x/20-x/25=6/60(in hours)
So x=10 km
Actually this is the exact same way i solved it, and this method is much easier to understand
Solved it the same way in about 20 seconds. I have yet to discover why this video is more than 10 mins long 😂
Much better to eliminate the trivial variable this way and keep the required unknown to simplify the algebra. Saves time on the exam too since you are late for class. Haha.
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.
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....
A student, who is a cyclist, was told that the distance from their house to the college is 6⅔ kilometres. That is why an average speed of 20kph was chosen for the journey which then would take 20 minutes. The student then found that the distance is actually 10km which resulted in them being 10 minutes late. But, then the student knew the correct distance so they would have had to ride at 30kph to do it in the same time. Why fiddle about with 25kph when it was obvious that that was useless?
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.
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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
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.
The spreadsheet technique thing is good to know. Sometimes, you don't want a math challenge, you just want the answer.
There's not enough information since you didn't specify that he left at the same time. But assuming he does, I accidentally guessed the right answer by trying 10km first and it happened to work out perfectly.
Also, I like how he's just casually "walking" a sub-four-minute mile.
My solution:
he lives 20 m from the college, but the first day he wakes up 10 minutes late and the second day he wakes up 4 minuts late.
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.
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.
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.
What a coincidence that he woke up "late" exactly on the same time, how exciting)
I didn’t get your solution. You suppose the time of departure is the same. But you clearly said it was further late each day. Also when you say “it was 4min late” the usual meaning is that it arrived, for example, at 09:04 instead of 09:00. Instead you interpreted as taking 4min longer than usual. This is not what we normally interpret, also because in your solution the ‘normal’ speed is actually higher than when he/she is not in a hurry.
Presh didn't say the student was "further late" each day. He in both instances said that the student woke up late. In the second instance the student decides to go faster than day 1, presumably in the knowledge that the speed on day 1 was insufficient to arrive on time. He is not "further late" he is _lesser_ late the second day.
@igrim4777 sure, he didn't use those exact words, but the common interpretation if one realizes that it must run even faster than yesterday is that he is more late... IMHO
@@GabrieleCannata He realises he must run faster on the second day, because the speed he ran on the first day was not enough to get him there in time
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.
Convert km/h into the time per km.
20 km/h -> 3 minutes per km
25 km/h -> 2'24"
So you gain 36" per km.
You have to gain 360" (6 minutes), so that's 10 km.
What exactly is supposed to be "seemingly impossible" about this? It's a very simple problem with no apparent contradictions or clever solutions, just pure algebra.
Just people claiming "it isn't defined that he leaves at the same time."
Which is technically correct, but such information is always implied with every math question.
@@christopherg2347 As a Math teacher I can assure you that no, this information is not always implied in problems.
@@xicufwm Is it implied that both routes go through the same gravity well and thus neither suffers from gravitational time dilation?
Because that information seems implied too.
@@christopherg2347 MindYourDecisions clearly assumes it out of hand, though, so why's *he* claiming this is seemingly impossible?
@@christopherg2347 also no, I expect all my Grade 9 students to write a 2 page essay about gravitational wells before answering any of the questions on the test.
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.
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.
If he leaves his house 35 min before class the first day and 32 min before class the second day, the solution will be 15 km
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
Simpler: there's a bus he can catch along the route. If he walks 20kph, he catches the bus he's only four minutes late . Too slow and he doesn't make the bus stop in time, walks all the way.
I meant, obviously, he catches the bus If he walks 25 kph.
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
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.
Alternative solution: buy a new alarm clock.
It is very frustrating when he didn't say he left at the same time both days, that should absolutely be stated.
10 km in 20 minutes?! Why even bother studying for college when you've destroyed the world record by over 6 minutes? Homie needs to enter the Olympics and get that Nike sponsorship bread 😂
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
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.
I like this problem. Gets you thinking but not too difficult. Technically this is probably 8th grade level math, but I hope no 8th grade math teacher would put this question on a test lol
If the distance was 1km, then it would take him 3 minutes for the first trip and 2.4 minutes for the second - a difference of 0.6 minutes. Since the actual difference was 6 minutes, which is ten times more, then the actual distance must be 10km.
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
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.
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.
My guy, you had two simultaneous equations with d & t. Set them both to equalise the coefficients of t so that when you subtract equation 1 and 2, you get a direct equation for d = 100(1/6 - 1/5) which would give you d = 10
You get the same answer without calculating "t" in fewer steps and simpler without all these extra steps.
THIS is the simplest and most efficient way. Unsure why you needed to even calculate t at all 🤔
This problem would be more fun to solve if the second day he left 2 minutes later than the first day.
To be honest I missed the part when information has been given that on both days when he is late he had left at the same time. For all we know from the text we can only assume that on day when he was late 10 minutes he has left later than usual (otherwise he would not have to realize that he need to go really fast. On day when he has been late 4 mintes it is suggested he left even later then the day before - though it is not clearly stated.
The solution given in film works only if we assume that he left exactly on the same time on both days when he was late...
This video shows Presh is a mathematician, not a physicist. He's scratching his left ear with his right hand going under his leg and behind his head :)
I don't understand the need to have time as a solvable variable in the process if all you want to know is the distance. The way I did it was as follows. The time he takes to get to college at 20 km/h is 6 minutes (or 0.1 hours) less than the time he takes to get to college at 25 km/h. So if distance = d, then (d/20) - 0.1 = (d/25). We can easily solve this to get (d/100) = 0.1 => d = 10 km. Bish bash bosh.
Okay, running 10 km at an average of 25 km/h is pretty energy consuming
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
Using the same methods, you can also determine the student needed to run at 30 kilometers an hour to make it on time.
Call "time per distance" TPD.
TPD = 1/speed.
Speed_1 = 20 km/h.
TPD_1 = 1/20 h/km = 3 min/km.
Speed_2 = 25 km/h.
TPD_2 = 1/25 h/km = 2.4 min/km.
TPD_1 - TPD_2 = 0.6 min/km.
T_1 - T_2 = 6 min.
Distance = (T_1 - T_2)/(TPD_1 - TPD_2)
= 10 km.
A brief explanation of the logic behind "Distance = (T_1 - T_2)/(TPD_1 - TPD_2)": if travelling by one method is 0.6 min/km slower than another method, then each kilometre travelled results in a 0.6 minute difference in time taken - so 6/0.6 = 10 km results in a 6 minute difference.
WRONG: the correct answer is: not enough information to give an answer. Because no information was given about the leaving time. It was not even given that he left at the same hour on the two days, so we may not assume that.
🤖does not compute. insufficient data. 😂I felt the same way
It's perfectly fine to assume it, but it should be mentioned explicitly as part of the answer.