# Noobs Weekly: Release I

Week 1: Burning Rope

You are given two ropes and a lighter. Each of the two ropes has the following property: if you light one end of the rope, it will take exactly one hour to burn to the other end. It doesn't necessarily burn at a uniform rate.

How can you measure a period of 45 minutes?

s

Note by John M.
5 years, 10 months ago

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A super-cool teaser.

Light both ends of one of the rope and one end of the other rope. By the time the first rope gets burnt, it'll be 30 minutes, and the second one would be half burnt, then light it's other end, that'll be 45 mins.

Though one of my friends had asked me this riddle earlier, so I knew the answer.

- 5 years, 10 months ago

'It doesn't necessarily burn at a uniform rate' which means that when the two burning ends of the first rope meet, it is not necesserly that 30 minutes have passed!

- 5 years, 9 months ago

Uhm I had the same complaint, but, given the original condition of the problem, that the rope burns in 60 minutes NO MATTER WHAT if it's lit from one side, then even if one side burns at the speed of light for some time interval, the net burn must come out to 30 mins when they meet.

I don't know how to mathematically prove this, but I guess it'd be a great idea to make a math problem out of this! I'll post a note later, maybe, on this.

Cheers, Hasan!

- 5 years, 9 months ago

I think it's because :

Consider a rope such that there lies a point on the rope which divides the rope into two parts $x$ and $y$ which are the length of the two parts. The part $x$ burns at the rate of $R_{x}$ while y at the rate of $R_{y}$. Now clearly we know the total time at which the rope burns is 60 minutes. So,$\frac {y}{R_{y}} + \frac{x}{R_{x}} = 60$ Let us see what happens when we burn the rope from both the ends. Suppose they meet till time $t$. Leg us futher suppose that the rates at which both parts burn is nog equal , suppose $R_{x}> R_{y}$. So this would mean that after the whole $x$ part has completely burned , it will start burning the $y$ part and so the two burning streaks will meet at somewhere in $y$ because $R_{y}$ is slower. So, making a equation out of this :$tR_{y}+x+(t-\frac {x}{R_{x}})R_{y}=x+y$ Solving the equation :$t = \frac {1}{2} \times (\frac {yR_{x} +xR_{y}}{R_{x} R_{y}})$ So, Substituting the first equation , $t=30$ In this case we have taken only two variations in the non uniform rates possible for the rope but in each case the total time required to burn the rope when lighted from both the sides will be everytime the same 30 minutes . Further thinking is requested for the viewers to clearly understand the proof.

In general suppose there exists $a_{1}, a_{2} ..... a_{n}$ defferent parts with rates $R_{a_{1}}, R_{a_{2}}....R_{a_{n}}$ so we could choose a point at the end of any $a_{x}$ and then continue to proof it would take exactly 30 minutes.

Hence , I think , proved

QED

- 5 years, 9 months ago

Hello Utkarsh,

Thanks for your proof! Though I'd like to point out that I don't clearly see your flow of reason throughout your steps. For example, you've defined $R_x$ to be the rate at which the $x$ part of the rope burns, and $R_y$ to be the the rate at which the $y$ part of the rope burns. And yet, you then state that if $R_x>R_y$ - that is, if the $x$ part of the rope burns faster than the $y$ part of the rope - then after $R_x$ has burnt up the $x$ part of the rope it shall move onto the $y$ part of the rope. However, this contradicts your initial condition that $R_x$ and $R_y$ burn their respective parts of the rope only. Hence, any further analysis from this step onward is illogical.

From what I see, you assume that $x$ is one-half of the rope and $y$ is the other half of the rope, in which this is the only case where it makes sense to say that if $R_x>R_y$ then $R_x$ will burn a part of $y$ of the rope. But I don't think that's what you meant, now, is it?

P.S. - assuming all you've said is true, I cannot follow your equations - more specifically, your first equation. Where'd you get it from? A little explanation of how you've obtained the statement would be nice.

Anyway, I think we can improve this. And we should.

Thanks for your time.

- 5 years, 9 months ago

Hi John, The thing is that I didn't and never intended to say that I divided the rope into two equal parts x and y , instead we are just assuming that as it is it given that the rope burns at non uniform rate so let us first assume the case when one part of the rope burnt at a defferent rate than the other ( later , when we have proved this we would proof for all such cases )

Visualize the rope :

.........$x$.....................|............................................................................................................$y$.................................................................

You may establish any relation between the parts $x=y, x or $x> y$ if you want but that doesn't matter in the proof . We take the rate at which per unit x burns is greater than that of y because it is given that the rope burns at a non uniform rate . If $x=y$ (Case 1)then surely x would burn up at less time than y because ther rate are defferent ( rate of x is greater) And when we shall light the rope from both the ends the x part would burn up first leading to the x streak coming to y part and burning the y part from the left side , from our veiw , while the y streak would still be at somewhere in the y part and hence the two streaks would meet at somewhere in the y part in this case. When $x (Case 2) then clearly the same case would hold true as discussed . When $x> y$ ,(Case 3) then x may or may not have completed its part till y reaches the boundary , it completely depends on the rates involved. Before coming to the equation , let me define what I mean by rate , rate is the amount or length of the rope burnt per unit time :$R=\frac {L}{t}$ Where L could be x or y burning at their respective times t. (When I say x or y I mean the length of x part or y part ) Now , $R_{x} = \frac {x}{t_{x}}$ And you may follow the same for y . Also , as the total time is (when burnt only from one end ) the sum of the time taken to burn x and y . $t_{x} +t_{y} = 60$ And when we substitute defferent values in this equation we get our first equation as I wrote in my comment . Now it would get a little trickier here .Sorry and thanks at the same time to you John as I should have written this in my comment that the second equation I wrote is for only Case 1 and Case 2. Case 3 shall be discussed later in the comment. In the equation I denoted the time when the two streaks meet as $t$ . Let us now burn the rope I said you to visualize .We have discussed that in two cases the two streaks would meet at somewhere in y part when $R_{x}> R_{y}$. So till time t the streak of y part would have burnt $R_{y} \times t$ . And x streak would have burnt the whole x plus what it burnt at y part from left which clearly is as written because the distance burnt $L=Rt$ and as the time period on which x streak was burning y part is the total time minus the time it spent in the x part $t-\frac {x}{R_{x}}$ so it is as followed written. And as adding the length of the rope burnt by x part streak and that of the y is the total length of the rope $x+y$ hence the equation. Solving the equation we obtain the value of t. This is for case 1 and 2 and this can be extended to the case 3 but Substituting y instead of x where ever it is present and also change their rates vice versa .

In the case that I explained have only two variations of the non uniform rates but it could be extended to many other cases when the variations in the rates in the rope are more than 2 .

- 5 years, 9 months ago

Utkarsh,

I see your line of reasoning much better now. But the main problem I have here is simply with your definitions of $x, y, R_x$ and $R_y$. What do those things mean?

From what I can infer, you defined $R_x$ to be the rate at which the $x$-part of the rope burns - that is, one of the sides of the rope - and $y$ to be the other side, and its rate of burning hence $R_y$. But you keep confusing me when you say that the flame that starts on the $x$ side starts burning the $y$-side. Actually, I don't even know how to make sense of ascribing lengths $x$ and $y$ to the rope if the flame can interject into the other's territory. How does that even work?

And since the rate is not uniform, I'm not quite sure, but doesn't that imply that for some $t$ $R_x$ could be $>R_y$ and vice versa (and equal)? Doesn't "not uniform" imply unequal - or - changing? Even if not, and even if ropes from left and right burn at the same rate for their own rates throughout, I still cannot see how the flame that burns the rope at the rate of $R_x$, once "done with $x$," can simply move on to $y$ - wouldn't that make it $R_y$? It's not like the flame is any stronger on the left or on the right - it's the rope that resists combustion (burn).

And I'm sorry if this is not what you meant - I just tried my best to interpret what you wrote. A little clarification on what you mean by those four parameters would do just fine. I'm sure I will interpret your explanation much better then.

Thanks

- 5 years, 9 months ago

I will prove the general case:

Suppose we divide our rope into $n$ sub-inervals, each of length $a_1,a_2,a_3,...,a_n$, not necesserily equal.

denote by $v_1,v_2,v_3,...,v_n$ the rates of fire in each sub-interval.

Now, begin with our first statement: "If the rope is lit from one end, it will take an hour to burn to the another"

Let's see: Our flame will pass through all our subintervals:It will take a time of $\frac{a_1}{v_1}$ to cross the first sub-interval, $\frac{a_2}{v_2}$ to cross the second,....

Hence, till it reaches the end of the track, this will be: $\sum_{k=1}^n \frac{a_k}{v_k}$, equals an hour.

So our first given: $\sum_{k=1}^n \frac{a_k}{v_k} = 60$ (working in minutes).

Now, see the following figure:

Let's simplify things:We lit both ends, producing two flames, A and B, propagating towards each other. They will meet at the green spot (arbitrarily chosen). We took a screenshot where the flames are about to enter the $i^{th}$ subinterval ($i$ is chosen arbitrarily). In the zoomed pic, we denoted by $d$ the length of portion of the $i^{th}$subinterval that will burn by flame B. Consequently, the other portion, that will burn by flame A, will have length $a_i - d$.

We are going now to express time, taken by each flame to reach the green spot, in terms of lengths and speeds:

Flame A: It passed all subintervals behind the $i^{th}$ one , plus the $a_i - d$ portion.

Time taken by A = $\boxed{\displaystyle \sum_{k=1}^{i-1} \frac{a_k}{v_k} + \frac{a_i - d}{v_i}}$

Just because the $a_i - d$ portion belongs to the $i^{th}$ subinterval, the flame crossing this portion will have speed $v_i$.The sum, I think you would know how it comes from, as explained above.

Flame B: It passed all subintervals infront of the $i^{th}$ one, plus the $d$ portion.

Time taken by B = $\boxed{\displaystyle \sum_{k=i+1}^{n} \frac{a_k}{v_k} + \frac{d}{v_i}}$

As stated before: the $d$ portion belongs to the $i^{th}$ subinterval, so the flame crossing this portion will have speed $v_i$.

Now, we should state the obvious statement: Since A and B meet at the green spot, then they will take equal time to reach that point!!

Time taken by A = Time taken by B

$\displaystyle => \sum_{k=1}^{i-1} \frac{a_k}{v_k} + \frac{a_i - d}{v_i} = \sum_{k=i+1}^{n} \frac{a_k}{v_k} + \frac{d}{v_i}$

Rearranging:(I will take $\frac{a_i}{v_i}$ from the fraction in (A) and merge it with the sum):

$\displaystyle \sum_{k=1}^{i} \frac{a_k}{v_k} = \sum_{k=i+1}^{n} \frac{a_k}{v_k} + 2\frac{d}{v_i}$

This is our second given!!

Now one thing left: let us rewrite our first given like this:

$\displaystyle 60=\sum_{k=1}^n \frac{a_k}{v_k} = \sum_{k=1}^i \frac{a_k}{v_k} +\sum_{k=i+1}^n \frac{a_k}{v_k}$

Substitute our second given in here:

$\displaystyle \sum_{k=1}^i \frac{a_k}{v_k} +\sum_{k=i+1}^n \frac{a_k}{v_k} =60$

$\displaystyle => \sum_{k=i+1}^{n} \frac{a_k}{v_k} + 2\frac{d}{v_i} + \sum_{k=i+1}^n \frac{a_k}{v_k} =60$

$\displaystyle => \sum_{k=i+1}^n \frac{a_k}{v_k} +\frac{d}{v_i} =30$

And proved that Time taken by B to reach green spot is $\displaystyle \sum_{k=i+1}^n \frac{a_k}{v_k} +\frac{d}{v_i}$. Hence $\boxed{Time taken by B=Time taken by A=30 minutes}$

I wish it comes clear now :)

- 5 years, 9 months ago

EXCELLENT WORK HASAN!

I've imagined an ideal solution to this to be exactly in this form! I just didn't have the time to put it all together, but you did! Magnificent.

King Integral strikes again.

s

- 5 years, 9 months ago

Thanks pal:)

Yeah it needs time a lot ( I took about 2 hours typing and designing O_o)

- 5 years, 9 months ago

👍👍👍👍👍👍👍👌👌👌👌👌👌👏👏👏👏👏 Hats off to you !!!!

- 5 years, 9 months ago

Hey John,

You got the solution by Hasan Kassim . Actually my solution is just exactly similar but as I am in 9 standard and they have not taught us the Sigma or Zeta , whatever it is I was not not able to explain it you , instead my solution used the same concept in defferent form. Today I learned about all that the solution has , Sigma and all from my elder brother and was able to figure out what the solution was . Never mind.

- 5 years, 9 months ago

No prob.

And I heard you say you're 9? Well, you're taking on some impressive problems for yourself. Good for you!

- 5 years, 9 months ago

Is your age wrong? I mean 9th class at 16?

- 5 years, 9 months ago

I'm just of your age and I think probably of your class too . It s wrong like many in Brilliant.

- 5 years, 9 months ago

Oh! I'm in 10th class, and BTW Why don' yer get y'er age chang'd?

- 5 years, 9 months ago

Actually it's okay for me . It benefits me to access many defferent sites on Internet reserved for some higher ages than mine . It's intential.

- 5 years, 9 months ago

Yeah like for example one of the stupidest things to do on YouTube, if you're under 18, is to tell your real age - and then some stupid "violent content" videos are gonna be blocked. Lying about the age gives you less restrictions which is always good.

- 5 years, 9 months ago

Well , BTW it seems to me that you have skipped some class OR you have joined the school earlier OR I have joined the school later.

- 5 years, 9 months ago

Yeah, the former. I went to Nursery for 12 days, and then to LKG. Ne'er went ter play s'cool!

- 5 years, 9 months ago