Noobs Weekly: Release I

@John M. – 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 :)

@John M. – 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.