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 s

Note by John Muradeli
4 years, 11 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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.

Satvik Golechha - 4 years, 11 months ago

Log in to reply

'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!

Hasan Kassim - 4 years, 10 months ago

Log in to reply

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!

John Muradeli - 4 years, 10 months ago

Log in to reply

@John Muradeli I think it's because :

Consider a rope such that there lies a point on the rope which divides the rope into two parts xx and yy which are the length of the two parts. The part xx burns at the rate of RxR_{x} while y at the rate of RyR_{y}. Now clearly we know the total time at which the rope burns is 60 minutes. So,yRy+xRx=60\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 tt. Leg us futher suppose that the rates at which both parts burn is nog equal , suppose Rx>RyR_{x}> R_{y}. So this would mean that after the whole xx part has completely burned , it will start burning the yy part and so the two burning streaks will meet at somewhere in yy because RyR_{y} is slower. So, making a equation out of this :tRy+x+(txRx)Ry=x+ytR_{y}+x+(t-\frac {x}{R_{x}})R_{y}=x+y Solving the equation :t=12×(yRx+xRyRxRy)t = \frac {1}{2} \times (\frac {yR_{x} +xR_{y}}{R_{x} R_{y}}) So, Substituting the first equation , t=30t=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 a1,a2.....ana_{1}, a_{2} ..... a_{n} defferent parts with rates Ra1,Ra2....RanR_{a_{1}}, R_{a_{2}}....R_{a_{n}} so we could choose a point at the end of any axa_{x} and then continue to proof it would take exactly 30 minutes.

Hence , I think , proved

QED

Utkarsh Dwivedi - 4 years, 10 months ago

Log in to reply

@Utkarsh Dwivedi 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 RxR_x to be the rate at which the xx part of the rope burns, and RyR_y to be the the rate at which the yy part of the rope burns. And yet, you then state that if Rx>RyR_x>R_y - that is, if the xx part of the rope burns faster than the yy part of the rope - then after RxR_x has burnt up the xx part of the rope it shall move onto the yy part of the rope. However, this contradicts your initial condition that RxR_x and RyR_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 xx is one-half of the rope and yy is the other half of the rope, in which this is the only case where it makes sense to say that if Rx>RyR_x>R_y then RxR_x will burn a part of yy 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.

John Muradeli - 4 years, 10 months ago

Log in to reply

@John Muradeli 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 :

.........xx.....................|............................................................................................................yy.................................................................

You may establish any relation between the parts x=y,x<yx=y, x <y or x>yx> 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=yx=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<yx <y (Case 2) then clearly the same case would hold true as discussed . When x>yx> 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=LtR=\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 , Rx=xtxR_{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 . tx+ty=60t_{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 tt . 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 Rx>RyR_{x}> R_{y}. So till time t the streak of y part would have burnt Ry×tR_{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=RtL=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 txRxt-\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+yx+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 .

Utkarsh Dwivedi - 4 years, 10 months ago

Log in to reply

@Utkarsh Dwivedi 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,Rxx, y, R_x and RyR_y. What do those things mean?

From what I can infer, you defined RxR_x to be the rate at which the xx-part of the rope burns - that is, one of the sides of the rope - and yy to be the other side, and its rate of burning hence RyR_y. But you keep confusing me when you say that the flame that starts on the xx side starts burning the yy-side. Actually, I don't even know how to make sense of ascribing lengths xx and yy 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 tt RxR_x could be >Ry>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 RxR_x, once "done with xx," can simply move on to yy - wouldn't that make it RyR_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

John Muradeli - 4 years, 10 months ago

Log in to reply

@John Muradeli I will prove the general case:

Suppose we divide our rope into nn sub-inervals, each of length a1,a2,a3,...,ana_1,a_2,a_3,...,a_n, not necesserily equal.

denote by v1,v2,v3,...,vnv_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 a1v1\frac{a_1}{v_1} to cross the first sub-interval, a2v2\frac{a_2}{v_2} to cross the second,....

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

So our first given: k=1nakvk=60\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 ithi^{th} subinterval (ii is chosen arbitrarily). In the zoomed pic, we denoted by dd the length of portion of the ithi^{th}subinterval that will burn by flame B. Consequently, the other portion, that will burn by flame A, will have length aida_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 ithi^{th} one , plus the aida_i - d portion.

Time taken by A = k=1i1akvk+aidvi\boxed{\displaystyle \sum_{k=1}^{i-1} \frac{a_k}{v_k} + \frac{a_i - d}{v_i}}

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

Flame B: It passed all subintervals infront of the ithi^{th} one, plus the dd portion.

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

As stated before: the d d portion belongs to the ithi^{th} subinterval, so the flame crossing this portion will have speed viv_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

=>k=1i1akvk+aidvi=k=i+1nakvk+dvi\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 aivi\frac{a_i}{v_i} from the fraction in (A) and merge it with the sum):

k=1iakvk=k=i+1nakvk+2dvi\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:

60=k=1nakvk=k=1iakvk+k=i+1nakvk\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:

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

=>k=i+1nakvk+2dvi+k=i+1nakvk=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

=>k=i+1nakvk+dvi=30\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 k=i+1nakvk+dvi\displaystyle \sum_{k=i+1}^n \frac{a_k}{v_k} +\frac{d}{v_i} . Hence TimetakenbyB=TimetakenbyA=30minutes\boxed{Time taken by B=Time taken by A=30 minutes}

I wish it comes clear now :)

Hasan Kassim - 4 years, 10 months ago

Log in to reply

@Hasan Kassim 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 s

John Muradeli - 4 years, 10 months ago

Log in to reply

@John Muradeli Thanks pal:)

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

Hasan Kassim - 4 years, 10 months ago

Log in to reply

@Hasan Kassim 👍👍👍👍👍👍👍👌👌👌👌👌👌👏👏👏👏👏 Hats off to you !!!!

Utkarsh Dwivedi - 4 years, 10 months ago

Log in to reply

@John Muradeli 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.

Utkarsh Dwivedi - 4 years, 10 months ago

Log in to reply

@Utkarsh Dwivedi No prob.

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

John Muradeli - 4 years, 10 months ago

Log in to reply

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

Satvik Golechha - 4 years, 10 months ago

Log in to reply

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

Utkarsh Dwivedi - 4 years, 10 months ago

Log in to reply

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

Satvik Golechha - 4 years, 10 months ago

Log in to reply

@Satvik Golechha 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.

Utkarsh Dwivedi - 4 years, 10 months ago

Log in to reply

@Satvik Golechha 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.

John Muradeli - 4 years, 10 months ago

Log in to reply

@Satvik Golechha 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.

Utkarsh Dwivedi - 4 years, 10 months ago

Log in to reply

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

Satvik Golechha - 4 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...