The thing about these "less than a minute" problems is that it all comes down to how quickly you see how to solve the problem. I'd bet that the kids who solved this problem in under a minute had been working with geometry of shapes and were very familiar with this sort of problem.
I'm willing to bet that they might've solved it but not rigorously. You can see two sets of horizontal triangles both adding up to half the parallelogram horizontally, and then ignore the fact that only some areas are labeled yellow, and come up with the same equation for x. I "solved" it that way, but it's by no means rigorous. They could've also gotten the right answer of 9 by simply playing with the given numbers and choosing a result that "seems" correct. Since they're familiar with addition more than anything, you can add/subtract the two numbers in two different ways to get x=9 or x=11 and then choose between the two (50% of students with this method would've guessed correctly)
It's very possible that the ones who can solve it in less than a minute have already been taught how to solve this type of problems beforehand. In China it's very common for kids to get lessons for math olympiads and they study problem sets like this. They could be simply performing learned routines like "place the shapes on a checkerboard pattern, add up the ones at odd positions and subtract the ones at even positions".
As time goes by and thinking about giving up Chinese and math studies. It's time for me to admit that you are better than me and I should probably keep on just pretending to be good a school
Math questions like these are usually super easy when you know the trick. It doesnt make you smarter than someone else to know a trick that they do not. The students who understood the trick could do so without trouble in less than a minute, everyone else takes time to discover the trick for themselves.
Yeah, it's like asking an adult to cite a poem 5thgraders have just learned. Ofcourse I can't do that... That being said, we don't know how much the students had worked on the subject beforehand. I mean "a few" 5thgraders solved it? With what level of prior knowledge and of how many? There are some "gifted" children and if out of like 1 million 5thgraders a few find the trick on their own within a minute, I'd call that "gifted".
Well pattern recognition is absolutely a measure of intelligence. You are right of course, that the circumstances of the task is important, but that does not take away from the sophistication of mathematical knowledge required, nor the brilliance needed to recognize the answer right away.
Yes I agree with you. Knowing the trick anyone can do this as long as they understand basic geometry. If you gave a different question (still about geometry) to the same Chinese kid. He wouldn't get it in 1 minute. However, it's still fun to see if you can find the answer yourself. I'm not proud but it took me a little over 15 minutes to understand this.
It doesn't matter how old you are. It matters if you are taught to solve right or not. Chinese math teachers go through everything in detail so the students have no gaps in their knowledge. I remember being 9th grade and I hardly understood basic math, when I changed school my new math teacher explained math very well and she also gave plenty of examples. Now I'm one of the top mathematicians in my state currently studying mathematicians in uni.
Dear TH-cam, I am currently reviewing for my math exam tomorrow. I was so deeply focused and felt like I could finally ace the exam. How horrible of your algorithm to show me this. Me, a high schooler cannot cope with rational algebraic expressions yet you recommend a video of Chinese fifth graders solving a hard problem in just a minute. I will be reporting you, TH-cam for lowering my self esteem.
Yo, remember, the smartest 25% of China is equal to the entire USA population... seeing that they have a such a large population it is not surprising they have a large amount of geniuses they must teach
Basics? I'm sure I was never taught about the relationships between triangles in a parallelogram like this... Simple enough to understand, but wasn't taught it in school.
@@MultiChrisjb 'Simple enough to understand, but wasn't taught it in school.' Exactly this. I highly suspect the 5th graders who solved it were simply already exposed to the idea of solving in this manner. You could tell just looking at all the overlap and similar and partial triangles the 'easy' answer was going to be some partial arrangement trick like this. Realize something like this is very poor as an IQ test question, it is really a measurement of specific knowledge, not general IQ.
@ModelLights agree, iq tests don't measure your intelligence either. I have gotten scores from 100 in 1 test to over 140 in other tests, but most in 120 to 135. Maybe after a lot of iq tests you can calculate your average, but idk , it still seems silly to me. I don't believe my iq to be 100 but I don't believe it to be 140 either so I will stick with 120ish. But in the end it just is a score about your ability in solving puzzles imo. You cannot measure your iq like that.
Im asian. Took me 10 minutes to admit defeat... couldn't open up my mind enough to different possible methods... i tried doing what he did except i didnt think of substituting the white for letters and put em in an equation... sigh
I was born yesterday and I solved it in half a Planck time. Took me that long cuz I was working on Riemann zeta function and building a quantum computer at the same time.
Only a day old and already making up excuses for your laggardness. I really hope you see a change in your behaviour soon, for your family's honour's sake.
I read this comment many times before I realized the genius of this. This is the best thing I have read today. Thank you for this blessed comment. It is comedy gold.
exactly my point! it's all memorization. Ever since high school and most of my undergrad in engineering, the chinese visa students did exceedingly well on written tests, but would always fail to explain WHY things were a certain way. they admitted to just memorizing things and not knowing why it's done. this is very common, i mean it's not a bad thing, but application problems turned out to be hard for them since they wouldn't be comfortable with elaborating a memorized concept.
Hanyue Li or if that was the case then why didn't most of them solve in under 1 minutes? What about bill he solved it in 10. It's about perspective. There are simply more Chinese with a better education system then the US
Btd Pro you don’t see exact same problem, they are similar. It just builds up the experience on what direction to think in order solve this type of problems. 100 is little too much, but still quite a lot, spending several hours a day.
jim halpert application problems probably aren’t hard to them, more likely they aren’t used to solve them. Chinese students aren’t encouraged to ask questions. They learn how to do things but not so great at figuring out what happened, and they don’t care much, they just need to get to the end, the good grades.
Classic. Unfortunately, Chinese children who do that are pointed just the same at the nearest rice field with the same words and then they can't change their fate for 60 years. So they are motivated to joke less. =P
There you go. The people who solve this in less than a minute were considered gifted, people who answered like you were deemed as geniuses for thinking out of the box.
It took me a an hour to solve. I don’t care how quickly someone else can solve a problem; I have always been a slow and methodical thinker. I found it fun and challenging and solving it for myself was satisfying- what more could one want from a math puzzle? The important thing is personal growth and the joy that brings. Comparing oneself to others is unproductive and leads often to either arrogance or discouragement- neither are life enhancing. Thank you Presh Talwalkar for all these fun and challenging puzzles; I really enjoy your TH-cam channel. You have a great voice for narration; it is clear, friendly and enthusiastic.
Lovely video! I'm disappointed in myself for not being able to solve this one. Didn't know about the 2 triangles using the full base still have the same area as half of the parallelogram.
I also didn't know about the '2 triangles on the full base' before I solved the puzzle, (I didn't even 'know' (or remember, if I ever did know) the rule about one triangle on the base being half the area, but that seemed fairly obvious once you consider a triangle formed on the parallelogram's diagonal) then it occurred to me that if the triangles whose bases spanned the parallelogram's base shared a side they would combine to make one triangle - with an area of half the area of the parallelogram. Now if you split the triangles apart again, sliding their 'summits' along the opposite edge of the parallelogram, you're not changing the vertical height of either triangle, and since their bases remain the same, their areas don't change either. They will still have a combined area of half the parallelogram's area. And by that logic, you could have any number of non-overlapping triangles (not just two) whose bases span an entire edge of the parallelogram and whose vertices opposite their bases touch the opposite edge of the parallelogram, and they would have a combined area of half the parallelogram's area!
LoL - SPOILERS! After puzzling for 10 mins I noticed your comment at the top of the list and then immediately solved it. As is very often the case, it's knowing (or at least finding) the trick that's the key. In hindsight any number of triangles with bases spanning any side would have half the area.
I'm a gifted student from Spain. I'm from second grade in secondary school, and it barely took me 35 seconds to solve it. But it was great fun! Thank you MindYourDecisions for your math puzzles!
@@ravager12 Estee... te agradezco el esfuerzo, pero ni se ve nada (entre la calidad de la imagen y que ademas pasas la pagina en dos segundos...) ni ver un cuaderno lleno de garabatos demuestra un carajo acerca de que lo hayas resuelto a la velocidad cómica que aseveras Un consejo: no te tomes tan en serio los comentarios de YT, resulta vagamente grimoso que te hayas tomado todo este esfuerzo en convencer a un extraño, y ademas no consiguiendolo... Que tengas un buen dia... al menos cortes has sido, que menos que corresponder
TH-cam is a source of entertainment. if his challenges were impossible, not many people would find it appealing. which in my opinion, makes him smart, in a way that he has an audience.
or they had just had a lesson on this sort of problem, or it's an apocryphal back story, and I'm sure there are math prodigies out there that really could do the problem that fast... China is a very big country. So who know, in the end it doesn't matter. It's fun to work these problems and learn something new. The satisfaction is in the journey not being "right".
I did figure it out but I am shocked that in China 5th graders are taught such geometric principles. That's great. I did this with 4 equations and then substituted. Didn't take much time. Thanks Presh for making such videos and providing us with great math questions. It's just fun.
@@anshsharma1357 To all mind blowing solvers here, I'd like to share the most difficult Geometry (and overall Math) problem (for me) I've solved so far in my life. It's TetraSpheres from contestcen.com/geom.htm I took about 15 days with 5 attempts to finally arrive at the correct answer to TetraSpheres problem. I'd like to know how much time you take to crack this one. Also if interested, do solve the following on same page contestcen.com/geom.htm • Circles in a square • 4 Tangents • 5 Spheres Submit your answer explanation to contestcen@aol.com and check out the correctness accordingly. Thank you so much.
I don't know why people think it can't be solved by the kids under a minute. First , those are competition for gifted kids . Second , at least in my country these theorems are taught to 7th graders , so anyone preparing for Olympiad stuff could naturally get the solution fast enough . And no , it's not like there is a pattern that add the even ones and subtract the odd ones or something, after considerable practice of general , difficult geometry problems , you can come up with the solution real quick.
Once you see it it's simple. Got hopelessly entangled in trying to find similar triangles and maybe some angle. Doh! Thanks, you made me smile the first time today!
More than 20 years ago these things came easy for me. I even used triangles to solve physics math problems that I didn’t necessarily know how to solve otherwise, for as I was lazy and didn’t always attend classes. Probably would have done it in my head back then, now I definitely would have needed pen and paper and put some real effort into it. Mathematics is like a muscle when used a lot it’s strong, not using it it will become sloppy. This is my view of why young kids can solve these problems quickly, they have a talent and that tallent is put to work. Geometry was always my strongest field.
as someone who is significantly stronger than algebra, and using it to hack my way through physics wihtou paying attention, I find it fascinating that you were able to do the same with geometry.
There are 10 types of internet mathematicians, those who understand binary, and those who don't... ...And those who didn't expect this joke to be in base 3! XD
Finally, a MindYourDecisions problem that I solved "accidentally"! I am no genius, but when I got it in 2 mins, I was amazed at myself! This is one of those moments where you don't have to "overthink" and you'll reach the final answer.
Those are rookie numbers. I knew the answer was 9 before opening the video. It took me that long because I was solving Fermat's Last Theorem for the fifth time.
Exactly. "actually, the younger you are to around 4 to 6 grader, the more likely you are able to solve this types of basic questions. If you are a high schooler, you most likely had forgotten some of the rules of basic maths. Just familiarise with them again and you'll get it in record time. "
@@Marnige Moreover, these types of problems are recurrent. The reason most of these guys got it so fast was because they have already done this type of exercise 100 times over. It's like a rubix cube. Impossible to solve on the fly but if you know the method to solve it then a fifth grader will solve it and not you.
@@bourdiergustave1506 Exactly, I am graduate and remember such problems but wasn't sure how to do them. I automatically did 79+10-72-8 without even knowing why and voila
Dang. I was close before I gave up. I actually guessed that the answer was 9, although I couldn't prove it to myself. I was looking at triangles sharing a side with the paralellogram, and mentally sliding the opposite point along the opposite side of the parallogram, knowing that the triangle's area wouldn't change by doing that. But it didn't help. Very clever puzzle! Thanks for sharing.
Hey! I actually figured this one out! I'd say it took me about 10 to 15 minutes of 'fiddling' then when I realized you could make triangles with areas equal to exactly half the area of the parallelogram it was a pretty quick calculation. I wonder about the 5th graders who solved it in under a minute. It's certainly not impossible but very impressive. I wonder if they had seen similar problems before? I mean, if I ever see a similar problem, I'd be able to solve it now pretty quickly.
I suspect it is one of those "killer problems" where the solution depends on seeing a (non obvious) pattern, like the other one on the "non existing triangle" sides. If one has been taught/shown the pattern before, it becomes obvious when it is/it can be repeated. And of course on a population of 1.5 billion individuals there are bound to be a few fifth graders that are enormously talented.
The key in this problem is to be aware of the fact that one set of triangles are exactly to equal another set. It is difficult to realize that without being familiar with the nuances of the parallelogram area. Until then, you are uncertain whether they are exactly equal or not although you suspect they might be equal.
I solved it thinking differently. The parallelogram can be thought of as 6 identical separate triangles. Area of 1 of these triangles is 0.5x + 36 + 4 Area of 2 of these triangles is x + 72 + 8 Area of 3 of these triangles is 1.5x + 108 + 12 There area of 3 of these triangles equals half of the total area of the parallelogram. Therefore 1.5x + 108 + 12 = y + 79 + x + 10. y = 0.5x + 31. y is the area adjacent north to unknown (?) The area of y + ? equals the area of 1 of the separate triangles of the parallelogram. Therefore y + ? = 0.5x + 36 + 4 We already calculated y previously and we substitute it in. 0.5x + 31 + ? = 0.5x + 36 + 4 31 + ? = 40 ? = 9
I'll be honst here. I can definitely solve this, but It'll take me a minute to remember what math functions I need. Another minute to write it down and then quadruple time to work it through and then go back because I made a human math error when solving.
I remember this question in an International exam and I'm pretty sure I got it right, when the results came back I was in the top 1% worldwide and I also know with certainty that I am not Chinese (I am Australian)
This is a very recognisable pattern. Once you know how to do it with one set of numbers, it's easy to see how to apply it to any comparable alternating pattern inscribed in a parallelogram. Chances are, the children who solved this were drilled in the method. The quick, accurate ones were just average at arithmetic.
The answer is 9, I solved this almost instantly, because this problem uses a trick that I teach my students in geometry problems. I write this before watching the video, but I'm pretty sure every simple solution must use this method. The idea is that of comparing areas of shapes by comparing their heights and bases separately. If we denote the area of the quadrilateral in the middle as S, and as X the area of the triangle to the left of the 79 triangle, we have a large triangle who's area is 10+S+79+X. This triangle has the same base and height as the parallelogram so it has half of its area. Similarly, we can count the area of the two triangles resting on the parallelogram's upper edge, their combined area is 8+S+72+X+RED, where RED denotes the red area we're looking for. The sum of these two areas is also half the parallelogram, since they have the same height and their bases add up to that of the parallelogram. Equating these two expressions yields RED=9.
Wow, I've tried to solve this many times and only now figured it out. In the end I solved it pretty much the same way as in this video. It took me probably an hour altogether but I made it. :)
Took me about two minutes, basically you have to be able to observe and find the two equivalent half parallelograms and it's pretty easy from then on. I definitely remember doing these kinds of problems in fifth and sixth grade when I was in China about a decade ago. I was really good at these being the maths kid of the class lmao.
I figured it out, but slowly, by writing down the area of the parallelogram in terms of h1 and h2, substituting h2*AD with h1*DC, and solving multiple equations until I was left with x. Seeing that two triangles that use the whole base (and height) of the parallelogram have the same area as one is a genius idea.
Well, I did it in the same way(used triangles cuz the test was for kids and the way this problem was provided gave me an idea that one would need to get hold of the correct intersecting figures that would provide the area without any effort. It didn't take any more than 5 seconds to be able to see the triangles). This doesn't require any genius but I'm feeling great today, lol. You just need to have experience.
after 12 years I finally realised why 'A' level math was so easy. in primary school, we were tested on creativity but as we learned more techniques, we were tested on our mastery at using them. the former is what we are born with while the latter comes with practice.
It is based on a property known as Triangle and Parallelogram on Same Base and between Same Parallels. It is taught in quite early age so no wonder a smart kid was able to figure it out.
They don't teach first about the 2 triangle thing at all. They understand simply how the area of a triangle is derived, and applied the understanding to the parallelogram and solved it as well. How about you, do you know how to derive 0.5 * base length * height? Or do you simply remember and regurgitate it?
I was so excited once you showed the method that I had to pause the video, grab a pencil and paper and solve it myself! Absolutely delightful! I can totally see how it could be solved in 2 minutes, (dang those are some sharp 5th graders though, I'm a pre-grad school college student, and I couldn't get it until you told me the trick!)
I came back to this a few days after watching it. It still took me 5 minutes to do the math. But a lot of these problems are simple to someone just learning the concepts. If you were just practicing triangles in a parallelogram, this problem is really simple.
Took me about three minutes. One minute to recognize the way the upwards and downwards triangle sums equal each other as do the leftwards and rightwards triangles, one minute to assign letters to the blank spaces and solve it algebraically with two equations, and another minute to realize that I don't need two equations because each part is HALF the parallelogram, so they're all equal and I can just subtract it visually without algebra. Then I skimmed through the video just to see how well explained it was, to see if anyone would be able to make sense of my comment. :) Nice puzzle.
Wow it was an awesome problem I knew that area of the triangle with one of its sides same as the base or width of a parallelogram and the other vertex rests on the opposite side has half the area of the parallelogram. But I didn't knew that the sums of area of 2 triangle will be same too. Nice learning...
I figured out the concept to use, but I did not know that it worked with two triangles on the same plane. The moment you said that, I solved it the first way.
͡ ͡i legit guessed the answer right just a few seconds of looking at the triangles and I was like bruh that red triangle seems bigger then the 8 but looked almost the same as 10 so I said 9 I guess some problems just need common sense Idk
@@qdsw I measured it out with a ruler. Two sides of the triangle 8 have got the same length, so I took the third as a base. That way I could say 8 equals about 3.3 centimeters on my monitor. Then I measured out the height and the base of the red one, and got it by the formula 0.5 x base x height. I didn´t get exactly nine, but I was close to it.
I think the prerequisites are a cheeky way of saying what you ACTUALLY need to know, namely that one is half the other, but it takes _experience_ to see these triangles and instantly think "I can subdivide this figure into sets of halves and equate them" Then finding the two triangles you need is a matter of labelling the regions and seeing what combination of known areas and overlaps that cancel out to get the red area, which is not that hard since there's only four ways of halving the figure. It's very impressive to do so in under a minute though! I think the only way of solving this one under a minute is by already knowing the form of the riddle and knowing the answer is going to lie in the difference between sets of areas. A quick glance at the figure tells you the 79 can only be "triangle buddies" with the 10, ditto the 72 and the 8, so if you've been doing puzzles like these all day you might just wing it, add the "triangle buddies", work out the difference, write that and move on without proving anything.
Ans is 79+10-(72+8)=9 I first let multiple unknowns include the red region. I then create two equations consist of Half parallelogram area=Half parallelogram area Plus, The sum areas of triangles in one side is equal to the the sum of triangles area areas in another side (AB and CD respectively). Since the height is the same, (AB)h=(CD)h , for AB and CD have two line segments seperately. Eventually, I use simultaneous equation (Equ. 1-Equ. 2) to eliminate other terms except the red region unknown. Let red region be A, after calculation, you will get 2A=18. Ends out A=9. Some special relationships can be deduced.
I thought I was bad at math too,then I just did these type of stuff for fun, cuz oh well, I'm bad so I have nothing to lose, cuz tbh even though u don't get them, they are interesting and fun. And now, I'm better at this type of thinking and it's not an insecurity of mine cuz I learnt to solve these on my own time :) dw dude u can relax and not take these too serious. Forget the terror of math class!! They are just innocent puzzles. America likes to make us feel like math is impossible but it's just gentle and fun 😊😁🤗
i’m usually pretty good at these, i’d be embarrassed to say how long some of them take me, but i usually come to answer after a long pause. this one completely stumped me though, i was trying to find the side lengths from D to the top right corner of the triangle we were looking for the area of and the parallel side length from B to the bottom corner of the area 8 triangle. I knew i couldn’t find any side lengths from the given area triangles as they weren’t drawn to scale and didn’t have 90 degree angles but i thought by reflecting the area 10 triangle to the left of the line from B to the bottom corner of the area 8 triangle that i was getting somewhere. one of my longest pauses to date and none of the work i was doing was getting me anywhere, i could have had a year to work on this and i don’t think i’d have arrived at an answer. the 2 triangle within a parallelogram where the 2 of the sides make up the length/width of the parallelogram and they share a height/length being equal to half the parallelogram area part is crazy to me. after it was explained it seems so simple, but it’s far from intuitive as i’ve never seen something similar used in other problems. seems crazy that children can spot a way to set up equal equations from this to cancel like unknowns and solve.
I almost got it. I started in a sort of correct path, in the sense of "halves of the parallelogram" but changed my mind later and strayed into a path that couldn't give any real answer back. Looking at the equation that gives the solution and the key to solve the problem, but but before looking at the explanation I tried myself to get to the solution, and I think I got how the kids could solve the problem in less than a minute. What I mean is, they probably didn't use the solution method directly as shown in the video. I believe the method may have been more direct as, for example, at 5:18, all the numbers that are not part of the yellow rectangle are exactly one half of the parallelogram, as the yellow triangle itself is one half of the triangle. Then all values *only* in the blue triangles is positive (also because is where what we want it included in), everything that is part of the blue and yellow areas is reduced to zero (cancel themselves), finally everything only in the yellow triangle becomes negative. At this point one can write in a paper each of the numbers taking into account wether they are zeroes (ignored), positive or negative. I know it becomes a bit of a lucky stretch to make such assumptions going on the bare minimum of logic and proof. I would think that's how one could try and solve in so little time with some sort of confidence enough to make a final decision on it.
ah, yes. it was kinda easy, but i wasn’t sure on my answer. it took me at least 20 seconds when he made the parallelogram bigger (or whatever it’s called). may i ask, what grade are you in?
I love the problem and the solution. I'm sure you've presented a problem that uses the concept of the triangles that contains the half of the area of the parallelogram. Thanks for sharing, Presh
To be honest, this problem isn’t even hard it you were taught this at school in 5th grade. Why would you lower your self esteems just on a simple math problem that wouldnt’t get you no where in life.
79+10-72-8=9. The left and mid white regions and the 79 and 10 together form a triangle that is half the area of the parallelogram; On the other hand, red+left white+mid white+72+8 is also the half area of it. Subtract the common left white and mid white regions, you get red.
First, I transfer the parallelogram to a rectangle because you can arrange the points inside to preserve all the areas Then calculate the area of 2 big triangles, 1 containing 72 and 8, and the other containing 79 and 10 with the edge AD. Then calculate the triangle next to the red and yellow ( contain the edge AD). THen you will see the same answer (maybe it will take 1 min )
This is the first time in my life.That somebody has actually explained that math problem, the area of a triangle or a parallelogram. Nobody ever explained this to me. Ever time I came across these problems I had to skip them. No matter how many times I asked. They just wouldn't give me a clear answer. Still manged to graduate with a B+ average tho. :)
I knew (a+x) + (72+b+8) must have the same area as (c+79+e) + (f+10+d), because of the two isosceles triangles, and the smaller parallelogram, and figured (a+79+b+10) must have the same area as (x+c+72+d) + (e+8+f). But not knowing the two-triangle rule (possibly forgot), didn't put two-and-two together. With paper, I may have realised - seeing the common variables. Neat question.
Thanks!
Twelve minus 8 is ..
@@Justsaying-.12 - 8 = 8
The thing about these "less than a minute" problems is that it all comes down to how quickly you see how to solve the problem. I'd bet that the kids who solved this problem in under a minute had been working with geometry of shapes and were very familiar with this sort of problem.
Right, I doubt they were coming at it cold, as we were.
I'm willing to bet that they might've solved it but not rigorously. You can see two sets of horizontal triangles both adding up to half the parallelogram horizontally, and then ignore the fact that only some areas are labeled yellow, and come up with the same equation for x. I "solved" it that way, but it's by no means rigorous.
They could've also gotten the right answer of 9 by simply playing with the given numbers and choosing a result that "seems" correct.
Since they're familiar with addition more than anything, you can add/subtract the two numbers in two different ways to get x=9 or x=11 and then choose between the two (50% of students with this method would've guessed correctly)
It's very possible that the ones who can solve it in less than a minute have already been taught how to solve this type of problems beforehand. In China it's very common for kids to get lessons for math olympiads and they study problem sets like this. They could be simply performing learned routines like "place the shapes on a checkerboard pattern, add up the ones at odd positions and subtract the ones at even positions".
I agree with you. It's very true
i could take the same person 1 or 20 minutes to solve, you need a bit of luck to see it
I am a chinese person, and now I am thinking am I really Chinese?
dy an 不知道，你真的是中国人吗？-
你问我，我问谁？
@@texas2957 你问你自己呗 XD
＂我＂说你是中国人
As time goes by and thinking about giving up Chinese and math studies. It's time for me to admit that you are better than me and I should probably keep on just pretending to be good a school
Mayosski ok?
Math questions like these are usually super easy when you know the trick. It doesnt make you smarter than someone else to know a trick that they do not. The students who understood the trick could do so without trouble in less than a minute, everyone else takes time to discover the trick for themselves.
That's right, and that's why youtube will kill teacher's job. In this quarantine my son is advancing more here at home than at school...
Yeah, it's like asking an adult to cite a poem 5thgraders have just learned. Ofcourse I can't do that...
That being said, we don't know how much the students had worked on the subject beforehand. I mean "a few" 5thgraders solved it? With what level of prior knowledge and of how many? There are some "gifted" children and if out of like 1 million 5thgraders a few find the trick on their own within a minute, I'd call that "gifted".
Well pattern recognition is absolutely a measure of intelligence. You are right of course, that the circumstances of the task is important, but that does not take away from the sophistication of mathematical knowledge required, nor the brilliance needed to recognize the answer right away.
Yes I agree with you. Knowing the trick anyone can do this as long as they understand basic geometry. If you gave a different question (still about geometry) to the same Chinese kid. He wouldn't get it in 1 minute. However, it's still fun to see if you can find the answer yourself. I'm not proud but it took me a little over 15 minutes to understand this.
@@evastronomy8048
That's normal lmao
I always thought I was smart and really good at math and logic problems
Until I started watching this channel
I still am , even after watching these.
Hear, hear
Eriola Allmetaj
I never considered myself smart or good at math and logic problems.
Finding this channel just reinforced that fact.
Eriola Allmetaj same
Same 😂😂
It doesn't matter how old you are. It matters if you are taught to solve right or not. Chinese math teachers go through everything in detail so the students have no gaps in their knowledge. I remember being 9th grade and I hardly understood basic math, when I changed school my new math teacher explained math very well and she also gave plenty of examples. Now I'm one of the top mathematicians in my state currently studying mathematicians in uni.
Dear TH-cam,
I am currently reviewing for my math exam tomorrow. I was so deeply focused and felt like I could finally ace the exam. How horrible of your algorithm to show me this. Me, a high schooler cannot cope with rational algebraic expressions yet you recommend a video of Chinese fifth graders solving a hard problem in just a minute. I will be reporting you, TH-cam for lowering my self esteem.
thanks for ruining the joke at the end
Instead of commenting, you could have studied!
The prerequisites for the problem was to know how to solve these types of problems already
@@ajety no he was saying it to the other guy
kat pxinshu How’d the exam go?
A 5th grader in China is solving under a minute.... But in US grown adult are still arguing that the earth is FLAT.... RIP America.... lol
it's laurel
Or are they a boy or girl.
Yo, remember, the smartest 25% of China is equal to the entire USA population... seeing that they have a such a large population it is not surprising they have a large amount of geniuses they must teach
William Lewis
No you mean the dumbest 25% is equal to America
William Lewis it's to scale meaning it is not surprising because lthay doesn't matter, it should be harder to be smart with so much ppl
This is one of the most beautiful solutions on this channel. No trig, no calculus - just the very basics.
Basics? I'm sure I was never taught about the relationships between triangles in a parallelogram like this... Simple enough to understand, but wasn't taught it in school.
It is called Genius
@@MultiChrisjb 'Simple enough to understand, but wasn't taught it in school.' Exactly this. I highly suspect the 5th graders who solved it were simply already exposed to the idea of solving in this manner.
You could tell just looking at all the overlap and similar and partial triangles the 'easy' answer was going to be some partial arrangement trick like this.
Realize something like this is very poor as an IQ test question, it is really a measurement of specific knowledge, not general IQ.
@ModelLights agree, iq tests don't measure your intelligence either. I have gotten scores from 100 in 1 test to over 140 in other tests, but most in 120 to 135. Maybe after a lot of iq tests you can calculate your average, but idk , it still seems silly to me. I don't believe my iq to be 100 but I don't believe it to be 140 either so I will stick with 120ish. But in the end it just is a score about your ability in solving puzzles imo. You cannot measure your iq like that.
@@MultiChrisjb You can figure out their relationship!
The red triangle looks a little bit bigger than the triangle with an 8. I guessed 9.
Muffin Button same but i guessed based on the 10 being bigger
yeah but the question said not drawn to scale
nice logic dude
Same, thats what i said aswell
“Not to scale”
"the diagram is not to scale" oh damn I was really going to take out my ruler
I remember a question on the SAT saying it wasn't to scale, then I solved it and it was perfectly to scale. I'm getting scammed here.
Difficulty: ASIAN.
I"m not asian and I solved it in 10 sec.
Najdorf sicilian aha...
That's racist.
Mike :] im an asian myself
Najdorf sicilian i know, im kidding. You're all so smart.
Got you going though, huh? :]
Im asian. Took me 10 minutes to admit defeat... couldn't open up my mind enough to different possible methods... i tried doing what he did except i didnt think of substituting the white for letters and put em in an equation... sigh
I was born yesterday and I solved it in half a Planck time. Took me that long cuz I was working on Riemann zeta function and building a quantum computer at the same time.
Only a day old and already making up excuses for your laggardness. I really hope you see a change in your behaviour soon, for your family's honour's sake.
you must be an average asian kid
Nxn908xxx Well smart kid, Chuck Norris solved it ten days before he was conceived.
Don't you mean plank instant? Also, no human could solve it in half a plank instant because nothing is shorter than a plank instant.
Pathetic. How can someone be that slow? What a pitiful situation.
"Can You Solve A 5th Grade Math Problem From China?"
can't i just try and solve it from my living room instead?
I read this comment many times before I realized the genius of this. This is the best thing I have read today. Thank you for this blessed comment. It is comedy gold.
BlankTom Amazing comment
Lol
The English language is so buggy
loooooooool
The Chinese students can solve it in less than one minute because they literally encountered 100’s of problems like this one EVERY SINGLE DAY.
exactly my point! it's all memorization. Ever since high school and most of my undergrad in engineering, the chinese visa students did exceedingly well on written tests, but would always fail to explain WHY things were a certain way. they admitted to just memorizing things and not knowing why it's done. this is very common, i mean it's not a bad thing, but application problems turned out to be hard for them since they wouldn't be comfortable with elaborating a memorized concept.
Hanyue Li or if that was the case then why didn't most of them solve in under 1 minutes? What about bill he solved it in 10. It's about perspective. There are simply more Chinese with a better education system then the US
Btd Pro you don’t see exact same problem, they are similar. It just builds up the experience on what direction to think in order solve this type of problems. 100 is little too much, but still quite a lot, spending several hours a day.
jim halpert application problems probably aren’t hard to them, more likely they aren’t used to solve them. Chinese students aren’t encouraged to ask questions. They learn how to do things but not so great at figuring out what happened, and they don’t care much, they just need to get to the end, the good grades.
yeah you're right
Find the area of the red triangle
Point the finger to the screen and say:
”that’s the area”
Classic. Unfortunately, Chinese children who do that are pointed just the same at the nearest rice field with the same words and then they can't change their fate for 60 years. So they are motivated to joke less. =P
If it was in a math test and I didn’t understand, I’d try to be smart and do that. It’s not technically wrong
There you go. The people who solve this in less than a minute were considered gifted, people who answered like you were deemed as geniuses for thinking out of the box.
Unfortunately the area is not to scale. The image is not equal to the actual area.
It took me a an hour to solve. I don’t care how quickly someone else can solve a problem; I have always been a slow and methodical thinker. I found it fun and challenging and solving it for myself was satisfying- what more could one want from a math puzzle? The important thing is personal growth and the joy that brings. Comparing oneself to others is unproductive and leads often to either arrogance or discouragement- neither are life enhancing.
Thank you Presh Talwalkar for all these fun and challenging puzzles; I really enjoy your TH-cam channel. You have a great voice for narration; it is clear, friendly and enthusiastic.
I'm happy for you buddy, but...
Bruhhh
Lovely video! I'm disappointed in myself for not being able to solve this one. Didn't know about the 2 triangles using the full base still have the same area as half of the parallelogram.
Well now you know
Shrey Rupani That’s certainly the key insight I was missing. I think I was on the right track, but I was weak and watched the answer.
Yeahh me too. I figured out the half triangle(79+10) but thats it.
I also didn't know about the '2 triangles on the full base' before I solved the puzzle, (I didn't even 'know' (or remember, if I ever did know) the rule about one triangle on the base being half the area, but that seemed fairly obvious once you consider a triangle formed on the parallelogram's diagonal) then it occurred to me that if the triangles whose bases spanned the parallelogram's base shared a side they would combine to make one triangle - with an area of half the area of the parallelogram. Now if you split the triangles apart again, sliding their 'summits' along the opposite edge of the parallelogram, you're not changing the vertical height of either triangle, and since their bases remain the same, their areas don't change either. They will still have a combined area of half the parallelogram's area.
And by that logic, you could have any number of non-overlapping triangles (not just two) whose bases span an entire edge of the parallelogram and whose vertices opposite their bases touch the opposite edge of the parallelogram, and they would have a combined area of half the parallelogram's area!
LoL - SPOILERS! After puzzling for 10 mins I noticed your comment at the top of the list and then immediately solved it. As is very often the case, it's knowing (or at least finding) the trick that's the key. In hindsight any number of triangles with bases spanning any side would have half the area.
I'm a gifted student from Spain. I'm from second grade in secondary school, and it barely took me 35 seconds to solve it. But it was great fun! Thank you MindYourDecisions for your math puzzles!
Nah, tu perfil absolutamente vacio y nuevo dice que las posibilidades de que estés mintiendo como un bellaco son altísimas
Pics or it didn't happen
@@TheChzoronzon Sí que lo hice y si quieres que te lo demuestre lo haré. Cómo quieres que te envie fotos que lo demuestren?
Ya está
@@ravager12 Estee... te agradezco el esfuerzo, pero ni se ve nada (entre la calidad de la imagen y que ademas pasas la pagina en dos segundos...) ni ver un cuaderno lleno de garabatos demuestra un carajo acerca de que lo hayas resuelto a la velocidad cómica que aseveras
Un consejo: no te tomes tan en serio los comentarios de YT, resulta vagamente grimoso que te hayas tomado todo este esfuerzo en convencer a un extraño, y ademas no consiguiendolo...
Que tengas un buen dia... al menos cortes has sido, que menos que corresponder
Chinese gifted kids: You cant defeat me.
Me: I know, but he can.
The guy who guessed it right in 2 seconds:
The legend
Unknown
The MADLAD
legend27
Senpai
bruh das me lmfao
less than a minute? i've lost any will to compete in anything
Alex yeah sure. He got it in 1 minute, but someone who has a TH-cam channel devoted to problem solving can't solve it at all. Sounds fishy...
Max I.
MindYourDecisions is not a smart guy. There’s a reason very few of his puzzles are real challenging.
TH-cam is a source of entertainment. if his challenges were impossible, not many people would find it appealing. which in my opinion, makes him smart, in a way that he has an audience.
Alex maybe they just guessed
or they had just had a lesson on this sort of problem, or it's an apocryphal back story, and I'm sure there are math prodigies out there that really could do the problem that fast... China is a very big country. So who know, in the end it doesn't matter. It's fun to work these problems and learn something new. The satisfaction is in the journey not being "right".
I did figure it out but I am shocked that in China 5th graders are taught such geometric principles. That's great. I did this with 4 equations and then substituted. Didn't take much time. Thanks Presh for making such videos and providing us with great math questions. It's just fun.
Did you miss the 5th grade geometry problem from *Soviet Union* ?
th-cam.com/video/m5evLoL0xwg/w-d-xo.html
got this in about five minutes. you don't need to know any special theorems you just need good geometric intuition
yep, took me like a minute by just looking at the screen. was easy to spot the triangle areas that are half of the parallelogram
Nerd
@@anshsharma1357
To all mind blowing solvers here, I'd like to share the most difficult Geometry (and overall Math) problem (for me) I've solved so far in my life.
It's TetraSpheres from contestcen.com/geom.htm
I took about 15 days with 5 attempts to finally arrive at the correct answer to TetraSpheres problem. I'd like to know how much time you take to crack this one.
Also if interested, do solve the following on same page contestcen.com/geom.htm
• Circles in a square
• 4 Tangents
• 5 Spheres
Submit your answer explanation to contestcen@aol.com and check out the correctness accordingly.
Thank you so much.
Yeah, me too.
@@anandk9220 which one exactly is it?
I don't know why people think it can't be solved by the kids under a minute.
First , those are competition for gifted kids . Second , at least in my country these theorems are taught to 7th graders , so anyone preparing for Olympiad stuff could naturally get the solution fast enough . And no , it's not like there is a pattern that add the even ones and subtract the odd ones or something, after considerable practice of general , difficult geometry problems , you can come up with the solution real quick.
Once you see it it's simple. Got hopelessly entangled in trying to find similar triangles and maybe some angle. Doh!
Thanks, you made me smile the first time today!
OMG sameeee, I was trying also to find similar triangles but no triangle was similar as their areas are different lol
This really wasn’t that hard. You just need to know that specific theorem
Ye, its always easy when you know how its done
In no parallel universe would you and I be friends
@@Chief_Miles_OBrien xD..
by "specific theorem" you mean the area formula for a triangle? because that is literally everything used to derive the answer.
Android Sheldon LMAO
are you smarter than a fifth grader?
apparently not...
kkim5000 talented*
Becuz their teacher teach that to their student
Bro this is not about smartness. These were already taught to them. So they could solve it in under a minute.
I solved it in 20 secs so i guess i am
@@godbeerus2202 I know for example my school doesn t teach me to think that way so I would be able to solve this
Me: "maybe what it between 8 and 10, so i guess its 9"
Answer : 9
Me : ftw?
lol same here and I still don't know why it's 9.
It's the average
8+10÷2=18÷2=9
Now you have passed the test!
Me too lol
SARV's guessing brain: mY gOaLs ArE bEyOnD yOuR uNdErStAnDiNg.
More than 20 years ago these things came easy for me. I even used triangles to solve physics math problems that I didn’t necessarily know how to solve otherwise, for as I was lazy and didn’t always attend classes.
Probably would have done it in my head back then, now I definitely would have needed pen and paper and put some real effort into it.
Mathematics is like a muscle when used a lot it’s strong, not using it it will become sloppy.
This is my view of why young kids can solve these problems quickly, they have a talent and that tallent is put to work. Geometry was always my strongest field.
as someone who is significantly stronger than algebra, and using it to hack my way through physics wihtou paying attention, I find it fascinating that you were able to do the same with geometry.
Hmm..every 3 out of 2 students struggle with maths.
I get it! XD
There are 10 types of internet mathematicians, those who understand binary, and those who don't...
...And those who didn't expect this joke to be in base 3!
XD
Finally I'm special at math's lol
2 out of 3 you mean. oh god.
use your brain.
@@user-yp2kp1mj2q that joke went right over your head
6:26 was my OHHHH moment
A 5th grader didn't do this in their head in less than a minute.
@MKMW C lol
Finally, a MindYourDecisions problem that I solved "accidentally"! I am no genius, but when I got it in 2 mins, I was amazed at myself! This is one of those moments where you don't have to "overthink" and you'll reach the final answer.
I am a 3rd grade student, and I solved in 3 seconds. The reason I took that long is because I was in the middle of working on a Laplace Transform.
Shango Not impressive. I solved this under a second and I'm still in the second trimester.
Shango and the only reason it took that long was because i was trying to do a backflip.
LOL!
Those are rookie numbers. I knew the answer was 9 before opening the video. It took me that long because I was solving Fermat's Last Theorem for the fifth time.
Trolling like hell
I solved it under 3 minutes but I am a 12th grader in India....and I was able to solve it because I was practicing geometry problem for a while.
Exactly. "actually, the younger you are to around 4 to 6 grader, the more likely you are able to solve this types of basic questions. If you are a high schooler, you most likely had forgotten some of the rules of basic maths. Just familiarise with them again and you'll get it in record time. "
@@Marnige Moreover, these types of problems are recurrent. The reason most of these guys got it so fast was because they have already done this type of exercise 100 times over. It's like a rubix cube. Impossible to solve on the fly but if you know the method to solve it then a fifth grader will solve it and not you.
@@bourdiergustave1506 Exactly. I bet they did it in a minute because for them, it's just a matter of addition and subtraction now.
@@will123134 Did it take more than a minute though?
@@bourdiergustave1506 Exactly, I am graduate and remember such problems but wasn't sure how to do them. I automatically did 79+10-72-8 without even knowing why and voila
Dang. I was close before I gave up. I actually guessed that the answer was 9, although I couldn't prove it to myself. I was looking at triangles sharing a side with the paralellogram, and mentally sliding the opposite point along the opposite side of the parallogram, knowing that the triangle's area wouldn't change by doing that. But it didn't help.
Very clever puzzle! Thanks for sharing.
Hey! I actually figured this one out! I'd say it took me about 10 to 15 minutes of 'fiddling' then when I realized you could make triangles with areas equal to exactly half the area of the parallelogram it was a pretty quick calculation.
I wonder about the 5th graders who solved it in under a minute. It's certainly not impossible but very impressive. I wonder if they had seen similar problems before? I mean, if I ever see a similar problem, I'd be able to solve it now pretty quickly.
I suspect it is one of those "killer problems" where the solution depends on seeing a (non obvious) pattern, like the other one on the "non existing triangle" sides. If one has been taught/shown the pattern before, it becomes obvious when it is/it can be repeated. And of course on a population of 1.5 billion individuals there are bound to be a few fifth graders that are enormously talented.
ParaLOLogram
Love your videos
BRAWLLL STARS
Hello
Does it count as right if i skipped the equasion part and just thought "9"
Sup3rN0va i hope so
The people liking your comment also thought of 9
maybe you are a savant
Same here
In term of result,yes.
In term of methodology,no.
Oh my, I made it in two minutes because I remembered about the triangles within a paralleogram rule!! Very nice one, it is simple but still fun!!!!
Dad: you doctor yet?
Son: Im 12.
Dad: talk to me when you doctor.
Son: then we will never talk again, dad.
The key in this problem is to be aware of the fact that one set of triangles are exactly to equal another set. It is difficult to realize that without being familiar with the nuances of the parallelogram area. Until then, you are uncertain whether they are exactly equal or not although you suspect they might be equal.
I solved it thinking differently. The parallelogram can be thought of as 6 identical separate triangles.
Area of 1 of these triangles is 0.5x + 36 + 4
Area of 2 of these triangles is x + 72 + 8
Area of 3 of these triangles is 1.5x + 108 + 12
There area of 3 of these triangles equals half of the total area of the parallelogram.
Therefore 1.5x + 108 + 12 = y + 79 + x + 10.
y = 0.5x + 31.
y is the area adjacent north to unknown (?)
The area of y + ? equals the area of 1 of the separate triangles of the parallelogram.
Therefore y + ? = 0.5x + 36 + 4
We already calculated y previously and we substitute it in.
0.5x + 31 + ? = 0.5x + 36 + 4
31 + ? = 40
? = 9
Big Brain
I'll be honst here. I can definitely solve this, but It'll take me a minute to remember what math functions I need. Another minute to write it down and then quadruple time to work it through and then go back because I made a human math error when solving.
I remember this question in an International exam and I'm pretty sure I got it right, when the results came back I was in the top 1% worldwide and I also know with certainty that I am not Chinese (I am Australian)
I didn’t knew what was a parallelogram in 5th grade…
Damn I was like there is 8 and 10 but no nine so I’m guessing nine and it was correct
Shady Directions Studios same! Haha
i like that out of the box thinking
You will go very far in life :)👌👌👌
sAME i saw it was less than 10 but more than 8 so i thought 9
I did 10 plus 8 divided by 2 to get 9
I remember those days of dealing with these questions. I probably lost my talent after adulting. Haha.
This is a very recognisable pattern. Once you know how to do it with one set of numbers, it's easy to see how to apply it to any comparable alternating pattern inscribed in a parallelogram. Chances are, the children who solved this were drilled in the method. The quick, accurate ones were just average at arithmetic.
The answer is 9, I solved this almost instantly, because this problem uses a trick that I teach my students in geometry problems. I write this before watching the video, but I'm pretty sure every simple solution must use this method.
The idea is that of comparing areas of shapes by comparing their heights and bases separately. If we denote the area of the quadrilateral in the middle as S, and as X the area of the triangle to the left of the 79 triangle, we have a large triangle who's area is 10+S+79+X. This triangle has the same base and height as the parallelogram so it has half of its area.
Similarly, we can count the area of the two triangles resting on the parallelogram's upper edge, their combined area is 8+S+72+X+RED, where RED denotes the red area we're looking for.
The sum of these two areas is also half the parallelogram, since they have the same height and their bases add up to that of the parallelogram. Equating these two expressions yields RED=9.
I solved it in 10 seconds by guessing 9 based on observation and extreme luck lmao
Great problem! I like problems with an aha moment. As opposed to tedious calculation.
Wow, I've tried to solve this many times and only now figured it out. In the end I solved it pretty much the same way as in this video. It took me probably an hour altogether but I made it. :)
this question is so simple, still I didn't get for the first moment and I had to hear his explanation...
anyway I just need to practice for 40h/day
If you can do geometry slowly, you can do it quickly!
Mathematical intuitions are born not created!
Ling ling
another ling ling wanna be
Took me about two minutes, basically you have to be able to observe and find the two equivalent half parallelograms and it's pretty easy from then on. I definitely remember doing these kinds of problems in fifth and sixth grade when I was in China about a decade ago. I was really good at these being the maths kid of the class lmao.
I figured it out, but slowly, by writing down the area of the parallelogram in terms of h1 and h2, substituting h2*AD with h1*DC, and solving multiple equations until I was left with x. Seeing that two triangles that use the whole base (and height) of the parallelogram have the same area as one is a genius idea.
Well, I did it in the same way(used triangles cuz the test was for kids and the way this problem was provided gave me an idea that one would need to get hold of the correct intersecting figures that would provide the area without any effort. It didn't take any more than 5 seconds to be able to see the triangles). This doesn't require any genius but I'm feeling great today, lol. You just need to have experience.
Didn't know the whole 2 triangle = half of parallelogram area thing. But once i watched the trick part, i was able to solve it pretty easily
after 12 years I finally realised why 'A' level math was so easy. in primary school, we were tested on creativity but as we learned more techniques, we were tested on our mastery at using them. the former is what we are born with while the latter comes with practice.
It is based on a property known as Triangle and Parallelogram on Same Base and between Same Parallels. It is taught in quite early age so no wonder a smart kid was able to figure it out.
well,i f in china they teach first about the 2 triangle thing, then is not big deal
Mariano Gaston ,right, try that in U.S.
They don't teach first about the 2 triangle thing at all. They understand simply how the area of a triangle is derived, and applied the understanding to the parallelogram and solved it as well. How about you, do you know how to derive 0.5 * base length * height? Or do you simply remember and regurgitate it?
well, i am from hong kong, so i think i can
after watching video: yeah if only i remember what i learnt from 5th grade
Stay safe tho
@@user-ck1zi8qf4i stay safe?
@@realperson9951 Hong Kong riots?
@@user-ck1zi8qf4i oh yea, thanks
@@user-ck1zi8qf4i theres chaos near my home today
i think i need your "stay safe tho" now
I was so excited once you showed the method that I had to pause the video, grab a pencil and paper and solve it myself! Absolutely delightful! I can totally see how it could be solved in 2 minutes, (dang those are some sharp 5th graders though, I'm a pre-grad school college student, and I couldn't get it until you told me the trick!)
I feels like I almost forgot everything I learnt from school
Really you forgot everything because it will be I "feel" like I almost...... Not I feels 🤣🤣🤣🤣🤣🤣🤣
This just popped up on my suggestions but I'm amazed how that problem can be easily solved.
Easy one
After watching video😂
I came back to this a few days after watching it. It still took me 5 minutes to do the math. But a lot of these problems are simple to someone just learning the concepts. If you were just practicing triangles in a parallelogram, this problem is really simple.
Took me about three minutes. One minute to recognize the way the upwards and downwards triangle sums equal each other as do the leftwards and rightwards triangles, one minute to assign letters to the blank spaces and solve it algebraically with two equations, and another minute to realize that I don't need two equations because each part is HALF the parallelogram, so they're all equal and I can just subtract it visually without algebra.
Then I skimmed through the video just to see how well explained it was, to see if anyone would be able to make sense of my comment. :) Nice puzzle.
Wow it was an awesome problem
I knew that area of the triangle with one of its sides same as the base or width of a parallelogram and the other vertex rests on the opposite side has half the area of the parallelogram.
But I didn't knew that the sums of area of 2 triangle will be same too.
Nice learning...
I figured out the concept to use, but I did not know that it worked with two triangles on the same plane. The moment you said that, I solved it the first way.
these people really smart at making math questions...
I legit got 9 but I did it another way.... And I thought I was doing this wrong... But ya see I am in highshcool so ig.
How'd you get it, I got it correct but in my knowledge theres no other way to solve it
͡ ͡i legit guessed the answer right just a few seconds of looking at the triangles and I was like bruh that red triangle seems bigger then the 8 but looked almost the same as 10 so I said 9 I guess some problems just need common sense Idk
@@x011unknown You got lucky with this question then, it says not drawn to scale, which means there is percentage of error permitted.
problem is this is for 5th graders :(
@@qdsw I measured it out with a ruler. Two sides of the triangle 8 have got the same length, so I took the third as a base. That way I could say 8 equals about 3.3 centimeters on my monitor. Then I measured out the height and the base of the red one, and got it by the formula 0.5 x base x height. I didn´t get exactly nine, but I was close to it.
I think the prerequisites are a cheeky way of saying what you ACTUALLY need to know, namely that one is half the other, but it takes _experience_ to see these triangles and instantly think "I can subdivide this figure into sets of halves and equate them"
Then finding the two triangles you need is a matter of labelling the regions and seeing what combination of known areas and overlaps that cancel out to get the red area, which is not that hard since there's only four ways of halving the figure. It's very impressive to do so in under a minute though!
I think the only way of solving this one under a minute is by already knowing the form of the riddle and knowing the answer is going to lie in the difference between sets of areas. A quick glance at the figure tells you the 79 can only be "triangle buddies" with the 10, ditto the 72 and the 8, so if you've been doing puzzles like these all day you might just wing it, add the "triangle buddies", work out the difference, write that and move on without proving anything.
I guessed 9 in 5 seconds does that count
Ans is 79+10-(72+8)=9
I first let multiple unknowns include the red region.
I then create two equations consist of
Half parallelogram area=Half parallelogram area
Plus, The sum areas of triangles in one side is equal to the the sum of triangles area areas in another side (AB and CD respectively). Since the height is the same, (AB)h=(CD)h , for AB and CD have two line segments seperately.
Eventually, I use simultaneous equation (Equ. 1-Equ. 2) to eliminate other terms except the red region unknown.
Let red region be A, after calculation, you will get 2A=18. Ends out A=9.
Some special relationships can be deduced.
Me: *whos never been good at math or even like it*
TH-cam: *recommends this video*
Me: Oh cool, why am I still watching this?
I thought I was bad at math too,then I just did these type of stuff for fun, cuz oh well, I'm bad so I have nothing to lose, cuz tbh even though u don't get them, they are interesting and fun. And now, I'm better at this type of thinking and it's not an insecurity of mine cuz I learnt to solve these on my own time :) dw dude u can relax and not take these too serious. Forget the terror of math class!! They are just innocent puzzles. America likes to make us feel like math is impossible but it's just gentle and fun 😊😁🤗
i’m usually pretty good at these, i’d be embarrassed to say how long some of them take me, but i usually come to answer after a long pause. this one completely stumped me though, i was trying to find the side lengths from D to the top right corner of the triangle we were looking for the area of and the parallel side length from B to the bottom corner of the area 8 triangle. I knew i couldn’t find any side lengths from the given area triangles as they weren’t drawn to scale and didn’t have 90 degree angles but i thought by reflecting the area 10 triangle to the left of the line from B to the bottom corner of the area 8 triangle that i was getting somewhere. one of my longest pauses to date and none of the work i was doing was getting me anywhere, i could have had a year to work on this and i don’t think i’d have arrived at an answer. the 2 triangle within a parallelogram where the 2 of the sides make up the length/width of the parallelogram and they share a height/length being equal to half the parallelogram area part is crazy to me. after it was explained it seems so simple, but it’s far from intuitive as i’ve never seen something similar used in other problems. seems crazy that children can spot a way to set up equal equations from this to cancel like unknowns and solve.
I solved it man ,yes!!. It took about 10 minutes for Me to do it
Good job 🤗👍👏
I'm Chinese and I do remember having encountered such kind of math problems in some of my homework when I was a kid.
Asian Parents: _”great, now finish your 100 question calculus test in 30 seconds”_
I almost got it. I started in a sort of correct path, in the sense of "halves of the parallelogram" but changed my mind later and strayed into a path that couldn't give any real answer back.
Looking at the equation that gives the solution and the key to solve the problem, but but before looking at the explanation I tried myself to get to the solution, and I think I got how the kids could solve the problem in less than a minute. What I mean is, they probably didn't use the solution method directly as shown in the video.
I believe the method may have been more direct as, for example, at 5:18, all the numbers that are not part of the yellow rectangle are exactly one half of the parallelogram, as the yellow triangle itself is one half of the triangle. Then all values *only* in the blue triangles is positive (also because is where what we want it included in), everything that is part of the blue and yellow areas is reduced to zero (cancel themselves), finally everything only in the yellow triangle becomes negative.
At this point one can write in a paper each of the numbers taking into account wether they are zeroes (ignored), positive or negative.
I know it becomes a bit of a lucky stretch to make such assumptions going on the bare minimum of logic and proof. I would think that's how one could try and solve in so little time with some sort of confidence enough to make a final decision on it.
I solved this in 2 minutes but im “proud” im so proud that im chinese lel
Weird flex but ok
KelsmonsterHamburgeresser3001 Schnitzelbrötchen ikr
When you're the 7th grade kid that is the smartest in Math and almost always get a perfect score, but take half an hour to solve this: *bruh*
ah, yes. it was kinda easy, but i wasn’t sure on my answer. it took me at least 20 seconds when he made the parallelogram bigger (or whatever it’s called). may i ask, what grade are you in?
KelsmonsterHamburgeresser3001 Schnitzelbrötchen lol
I love the problem and the solution. I'm sure you've presented a problem that uses the concept of the triangles that contains the half of the area of the parallelogram. Thanks for sharing, Presh
I mean anyone can solve it if they actually learned the method on solving. Just the education system is _great_
I am so greatful to you!
I’m a gifted 8th grader in America and I was never taught how to do that. That just shows how much better China’s education system is
You are average if you don’t expand your knowledge.
i figured out the problem as soon as you said that you can have 2 triangles = to 1 half of the shape
To be honest, this problem isn’t even hard it you were taught this at school in 5th grade. Why would you lower your self esteems just on a simple math problem that wouldnt’t get you no where in life.
Me encantó! Ejercicios como este y los de tu canal amplían nuestra mente! Gracias! ❤
The dislikes are adults that think earth is flat
Roses are violets are blue
Even if ur smart
There’s another asian better than u
Roses are violets are blue, lmao never thought it that way
謝謝誇獎~~ XD
79+10-72-8=9.
The left and mid white regions and the 79 and 10 together form a triangle that is half the area of the parallelogram;
On the other hand, red+left white+mid white+72+8 is also the half area of it.
Subtract the common left white and mid white regions, you get red.
I paused the video to solve it. I did get the right answer, but I took an exaggerated long time to do so; this video was very useful.
You know I didn't know the answers BUT I learned , thank you sir.
8:47 Yes but you can be sure the 5th graders were shown this method and practiced, it is not like they discovered on their own.
First, I transfer the parallelogram to a rectangle because you can arrange the points inside to preserve all the areas
Then calculate the area of 2 big triangles, 1 containing 72 and 8, and the other containing 79 and 10 with the edge AD. Then calculate the triangle next to the red and yellow ( contain the edge AD). THen you will see the same answer (maybe it will take 1 min )
Gosh, I wish I watched it during leisure time, not at midnight!
Damn I identifed the way of solving this problem but I was to lazy to calculate further.
This is the first time in my life.That somebody has actually explained that math problem, the area of a triangle or a parallelogram. Nobody ever explained this to me. Ever time I came across these problems I had to skip them. No matter how many times I asked. They just wouldn't give me a clear answer. Still manged to graduate with a B+ average tho. :)
Presh, if you were my teacher when I was in school I might have made it passed the 9th grade!! Love the Videos! Thanks
This was so simple and mind boggling!
Thankyou 🙏🏼
I knew (a+x) + (72+b+8) must have the same area as (c+79+e) + (f+10+d), because of the two isosceles triangles, and the smaller parallelogram, and figured (a+79+b+10) must have the same area as (x+c+72+d) + (e+8+f). But not knowing the two-triangle rule (possibly forgot), didn't put two-and-two together. With paper, I may have realised - seeing the common variables. Neat question.