Math Olympiad | A Nice Rational Equation | 80% Failed to solve !!

#rational #matholympiad #maths #rationalequations

[@michaelcrossley4716]

2700, took me 1 minute using excel. Set all the numerators as 1 for all, then add them up. Divide 200 by the sum and you get 2700. I wish I was better at doing the proper math.

[@mathophilechirag]

kzbin.info/www/bejne/bJCpYYSBZ8msqZosi=4T2Gv2qUndo7SZoo

[@user-xj8he4gr2h]

21=3*7，77=7*11，165=11*15⋯⋯This is the most important facts. Base on this hints, 4=7-3 =11-7=15-11=…, then students could find out the way to solve by themselves. How you can find this facts is the most important thing in this problem. But you just treat it as nothing, so you did teach nothing.

[@noobbot7099]

much longer method just write nth term which is 1/(4n-1)(4n+3) and put values after creating difference

[@mathophilechirag]

kzbin.info/www/bejne/bJCpYYSBZ8msqZosi=4T2Gv2qUndo7SZoo

[@Bayerwaldler]

A very nice exercise. The trick is pretty ingenious.The execution was a little tedious, though. Under inspection it was clear that b-a was equal to 4 in every term so that the factor 1/4 could be pulled in front.

[@vijaymaths5483]

Thank you

[@SergiuCosminViorel]

solved, another way, three groups, 1&2, 3&4, 5&6, then one group, 1&3, plus one, solved

[@JPTaquari]

I know that the professor is going to use a complex method, which he knows, but I'm going to simplify it, and solve it in 5 or 10 minutes with a pencil and paper: Divide all denominators by 621, obtaining: 29.57 + 8.065 + 3.764 + 2.179 + 1.421 + 1 = 46 Then it's soft: 46 * X = 200 * 621 X = 2,700 Bingo from Brazil !

[@vijaymaths5483]

Excellent ;)

[@vishwasdeshpande3921]

Excellent

[@rangaswamyks8287]

wow

[@GWaters-xr1fv]

Not excellent. This is just a decimal approximation method ( correct to two dec. places ) NOT an exact solution. The fact that your decimal rounding errors just happened to cancel out was a happy coincidence, which will usually not be the case.

[@JPTaquari]

Man, here in Brazil, they teach that good mathematics is the one that has the shortest path. I think you must have some mental, cognitive problem. Go treat yourself!

[@beautifulmindinpuzzles7716]

At 4:59 multiply both sides by 4. RHS=800 and LHS will contain fractions like 4/(3x7) which is equivalent to [1/3 - 1/7]. Calculations will be a lot easier.

[@crandoQcat]

11:47 only use the least common denominator of 3&27 is 27。

[@mathophilechirag]

kzbin.info/www/bejne/bJCpYYSBZ8msqZosi=4T2Gv2qUndo7SZoo

[@rameshdas2450]

When at first glance it is seen that each of the denominators is a product of two numbers like n and n+4 ,then it would be simpler to multiply both sides by four at the outserand then proceeding as x/3-x/7+x/7-x/11 and so on---

[@santoshjha920]

Thanks 🙏

[@lnmukund6152]

Your atsite mental one line observation is superb, If u read solutions u can solve probs, but if u observe probs , u can create solutions thank u very much sir Mukund

[@Andrumen01]

I don't understand why the trouble of numerical problems....they are just numbers... calculators are good at getting those numbers. Real olympiad problem would be proof based!

[@demetrioalbuquerque5408]

Ótima resolução grande Mestre!! Mestre de onde é esse problema tão bonito?

[@vijaymaths5483]

Thank you ⚘️

[@user-ti2us8tj3e]

این نوع مسائل در زمانیکه سال آخر دبیرستان بودم در کتاب حساب استدلالی بود حساب استدلالی درس جالب و خوبی بود ولی حالا دیگه در دبیرستان‌ها تدریس نمیشه، حساب استدلالی در واقع درس هوش است و مسائل ابتکاری حل می‌شود

[@vijaymaths5483]

Thanks for your sweet memory 🍫

[@sumit-mn6ys]

Nice Explanation 👍

[@vijaymaths5483]

Thanks 🙂

[@luislopez4904]

Excelente explicación

[@Rocio62154]

X=2700

[@vijaymaths5483]

right!!

[@user-it6fh7hy6t]

У автора какая то мания решать простейшие задачи труднейшим способом. 14 минут решает задачу,которую можно сделать за 3 минуты.

[@vijaymaths5483]

I think method of simplification is more important

[@Ekakshpareek94]

Totally wrong way. It's rather than simplifying/solving , the teacher is making it's complicated & getting more inconvenient

[@WhiteGandalfs]

It's a trade-off: You have to decide if you want to go the way with drastically less analytic writing, thus drastically reducing the probability of errors doing the analytical reformation, or to go the the way of drastically less numeric writing (after hopefully correctly completing the analytic reformation), thus minimizing the probability of errors doing that numeric stuff. Since as long as you do not know the OUTCOME of the analytical reformation, depending on your average error probability, you can make rather many errors in that analytical reformation, and since as long as you don't know the outcome (because you invented the riddle or you happen to be familiar with that exact type of reformation by chance) you typically go at least a few unsuccessful lanes down before featuring out the successful one, if you are not truly familiar with the way the author is suggesting here, you are by far better off going the numeric way directly, since it is really trivial in this case.

[@Danieswors]

​@@WhiteGandalfs Indeed ! It's all about the whole process of solving a problem in different types of method , while the author doesn't shows the shortcut he's trying to crack the problem in alternative ways although it may be to long in competitive exams , but that shows we can also have a solution in well mannered followed steps atlast the solution is same it doesn't matter at all , it all depends on person to person by which method he/she needs to solve just stick to basic steps and bingo you will have the solution maybe you can have solution earlier , but we should know all the ways of method to end up giving a perfect answer.

[@vadim64841]

I disagree. The solution avoids massive amount of computations. It takes time to explain, but if you understand the idea you can skip all these explanations and arrive to answer pretty quickly. What exactly would be a “straightforward” solution - finding a common denominator? Lots of multiplications and additions? This would be insane, and very easy to make a mistake.

[@rainerinedinburgh5807]

@@vadim64841 Vijay's trick is brilliant, but there is a "straightforward" way that doesn't need massive amounts of computation. Instead of using the common denominator 3·7·11·15·19·23·27 of all six fractions, just add them pairwise, and as if by magic, plenty will just cancel out. 1/(3·7) + 1/(7·11) = (11+3=14)/(3·7·11) which simplifies to 2/(3·11) = A 1/(11·15) + 1/(15·19) = (19+11=30)/(11·15·19) which simplifies to 2/(11·19) = B 1/(19·23) + 1/(23·27) = (27+19=46)/(19·23·27) which simplifies to 2/(19·27) = C Now we only have to work out A+B+C = 2/(3·11) + 2/(11·19) + 2/(19·27). Again I combined just the first two to begin with instead of all three at once: A+B = 2/(3·11) + 2/(11·19) = 2(19+3=22)/(3·11·19) = 4/(3·19). And finally add C: 4/(3·19) + 2/(19·27) = (4·27+2·3)/(3·19·27) = 2(54+3=57=3·19)/(3·19·27) = 2/27. Putting that back into the original context, we get x·(2/27) = 200, and so x = 27·200/2 = 2700.

[@gerardvanwilgen9917]

"This would be insane, and very easy to make a mistake.?"@@vadim64841 That is why electronic calculators were invented, but I do not know if those are allowed in competitions.

[@ncslovers3429]

👌👌👌

[@vijaymaths5483]

thanks for watching!!

[@superiorlyrics8326]

👏👏👏

[@crazyindianvines1472]

Great explanation 😀

[@vijaymaths5483]

Glad you think so!

[@pauloenockmarquesdasilva7643]

Show. Thank you very much!!!!

[@vijaymaths5483]

You are welcome!

[@user-hr9lv4gt8i]

good!

[@tonyennis1787]

Very nice indeed

[@ToanPham-wr7xe]

😮

[@adroaldoferreira978]

Uma boa questão de fatoração.

[@vijaymaths5483]

Thank you

[@YvanMcMillan]

incredible !

[@vijaymaths5483]

Thanks you !

[@jatintrivedi5632]

Too much noise in baxkground.

[@arifustatsonmez1557]

Thanx

[@vijaymaths5483]

Thanks too for watching my Vijay Maths channel 💗

[@kalakla]

Are you serious teacher, what is this about .. it does not even get you likes or enough responses.

[@ridaoueld5939]

Intéressant pour le calcul

[@OscarGarcia-xy4xv]

Eligió el camino largo..

[@comdo777]

asnwer=1x

[@tanernays432]

нашел общий щнаменатель за 2 минуты ЫЫЫЫЫ

[@sannygs]

U r messing up d students... totally wrong way

[@yokdilyds8183]

Do you have to explain it one by one to make the video longer? I was almost afraid that he would explain the multiplication tables from the beginning. Don't treat your subscribers like such idiots. Even if there are idiots, they can't understand this problem

[@user-nd7th3hy4l]

X=2700

[@vijaymaths5483]

बिल्कुल सहि है ✅️

[@yakupbuyankara5903]

X=2700

[@vijaymaths5483]

Exactly 👍