Can you solve the water glass and wine bottle riddles?

[@Dragondog-Gaming]

I used a variant of the second method to calculate the volume of the liquid in the bottle. Measuring where the liquid meets in the two bottles, there is an overlap of 6 cm. Removing that you have a single full bottle. Then calculating the volume of a cylinder that is 6 cm high with a radius of 4 cm, adding that to 750, and then cutting that sum in half to get the volume of liquid in one bottle.

[@yuri8217]

Yeah, also did the same, seemed so much simpler.

[@Able89535]

Overlapping approach is much more intuitive. It also works with missing approach if the liquid were filled less as long as the liquid cover both full irregular shapes of the bottle, which we deduct the missing part from the total instead of adding

[@xraymag]

Same here. Just took three lines to calculate

[@welcomb]

Exactly. Here this is a typical grade 5 math problem. Basically (A ∪ B) - (A ∩ B) = whole bottle 750ml. So (A ∩ B) is volume of 6 cm cylinder. Granted they don't teach set theory at grade 5, but it is solved using venn diagrams.

[@nineballking06351]

Same here. It didn't need to be so complicated.

[@addicted2caffeine]

14cm... I think I worked it out before the vid started based on the thumbnail

[@nthxable]

Same here ;)

[@chinnayyakorrai8523]

Same here

[@vanillawaffle7303]

Same here

[@artemisolympian6318]

Same my man

[@bIeed]

Same, and in less than 30sec...

[@doriphor]

You don't need to drink anything! Combining the red part of each side gives you a complete full bottle that is slightly taller than the normal one and we know how tall: 33cm. So we know that double the fluid fills an entire bottle + 6cm worth of pure cylinder, therefore volume V = (750 + 6*4*4*pi)/2

[@Tiqerboy]

If there's an easier way to solve a problem than Presh's presentation, rest assured his subscriber base is going to find it. Power in numbers. Two heads are better than one or in this case a TON of smart people among Presh's subscriber base is better than Presh. It's EXTREMELY rare that I get to a Presh problem that ISN'T already solved in the comments! That's why I DO NOT read the comments if I want to get one of his problems on my own.

[@GodmanchesterGoblin]

Brilliant! Thanks for sharing.

[@Anmol_Sinha]

This is genius.

[@Bronzescorpion]

@@Tiqerboy Not only is there power in numbers, but Presh tend to over complicate problems quite often, so you don't even need more people, just one without the tendency to complicate things.

[@livehappy1415]

V is not pure cylinder so how do you know that the extra 6cm is pure cylinder?

[@nidadursunoglu6663]

I love how the first one is a primary school question and the second is pretty complex

[@Supremax67]

Well, in that case, can someone explained to me how the volume almost double when I started drinking it? 😂

[@DaTimmeh]

Mathematically, I wouldn't call the second problem more complex. Logically, absolutely. I think this is a bit of a flaw in school systems (especially America), where math is turned into 80% logic and 20% actually math. People always focus on the wording rather than the numbers. When I moved from Germany here I did the exact opposite, and had no problems, despite lacking English language skills. Between the numbers given in the problem and multiple choice answers, it was almost always obvious. Logic is of course also important, but should be in addition to math, not instead of. Like a programming class, which we really should introduce way earlier in education, and to everyone. The world depends on technology, and most people have never even looked at code in a serious manner.

[@duckner]

​@@DaTimmehnah

[@Ndreau]

Pretty complex but simple enough that I think some gifted students can do it.

[@ArwinaThePlanet]

@@Ndreau i'm not really gifted but i'm one of the best students in math in my class and all i can do for problem 2 is that 7/10 of the bottle is wine (though i'm just in year 5, and in my place students are a year late)

[@AalbertTorsius]

I love how you say "subtract from both sides" and "divide both sides by" instead of "moving over" or "cancelling out". How you say it makes clear what's actually happening, instead of some magic "cancellation". Thanks!

[@thefirminator]

its not magical its a shortcut aka common sense

[@iwantagoodnameplease]

@@thefirminator If it's common, why did it have to be taught to you in maths class?

[@templarknight7]

@@iwantagoodnameplease because children don't have common sense.

[@MarieAnne.]

@@templarknight7 Neither do a lot of adults, and some never really grasp the concept of "moving over", and I think that's what the OP was getting at.

[@joefoulger3510]

@@iwantagoodnameplease the word "common" does not mean the same thing as the word "natural." You aren't born with common sense. It is slowly taught to you over time

[@stevenz933]

For the 2nd problem, there is a simple solution that requires only a single variable. Let the displaced volume of the "dimple punt" at the bottom of the bottle equal the variable x. When the bottle is upright, the volume of liquid in the bottle equals (π)*(4^2)*14 - x = 224*π - x. When the bottle is upside down, the volume of air in the bottle equals (π)*(4^2)*8 - x = 128*π - x. The volume of liquid plus the volume of air equals the total volume of the bottle or 750. Put into an equation:(224*π - x) + (128*π - x) = 750. Solving for x = 176*π - 375. Plugging x back into the volume of liquid: L = 224*π - x = 224*π - (176*π - 375) = 48*π + 375 or about 525.8 ml of liquid.

[@OzielAlvesCavalcante]

That’s how I solve too

[@RipperJack77]

Yeah, this is also what my intuition gave me. Funnily it is late and for some reason I thought that half of 750 was 375.5 so I was off by half a ml and could not figure out why until reading your message :D

[@lumsdot]

If there was no dimple, would it just be the simple ratio of 14 divided by 14+8, multiplied by 750

[@stevenz933]

If there was no dimple, then the liquid volume would simply be the volume of a cylinder: π*r^2 *h@@lumsdot

[@HyperSnypr]

@@lumsdot If there was no dimple it would just be solved as a cylinder 14x8. We wouldn't need to know the 750ml capacity at all.

[@harryheilmann2208]

the liquid volume solutions were convoluted. simply put, the upright and inverted bottles had an overlapping cylinder section of 6cm, 96 pi volume. the air volumes being equal also equal the remaining liquid, which is half 750-96 pi, or 224.2 cc. Add that back to the overlapped section = 525.8 cc

[@dwinson]

I'm definitely not a mathematical genius but the second method was the way I solved the bottle puzzle. Seemed the more obvious approach tbh - just shows how differently people think,

[@leif1075]

Why do you think you're not a math genius ?

[@w-lilypad]

​@@leif1075because he said so?

[@ChadEnglishPhD]

I found both answers to the wine bottle as overly complicating it. There's a much simpler way. Consider the volume of air, A, in the upside down bottle. It fills 27 - 19 = 8 cm of height from the bottom of the bottle. In the right-side up bottle, the volume of wine, V, fills 14 cm from the bottom. You could imagine replacing the bottom 8 cm with the air from the upside down bottle, A, such that the total bottle volume, B, is A above the wine + A in the bottom 8 cm + a cylindrical volume, C, of wine in the middle that is 14-8 = 6 cm high. That is: (1) C = 6*pi*(4)^2 = 96pi (2) V = A + C = A + 96pi and the full bottle volume is: (3) B = 2A + C = 2A + 96pi = 750 cm^3 From (3), A = 375 - 48pi. From (2), V = 375 - 48pi + 96pi = 525.8 cm^3.

[@mike1024.]

I really liked the second solution to the problem, and I can see how the thought process results from seeing the answer the first time around. I also used a system of equations to solve it, but I partitioned the bottle into the top region, the cylindrical region, and the bottom region. I got a similar system of equations that didn't have any minus signs but there was no significant change in difficulty to solve the system.

[@bartconnolly6104]

If you yhink about it the empty space / wine in the neck and bottom bubble cancel out leaving a 6cm overlap in the middle of the regular cylinder middle when you comparecrightvway up to upsude down. Thats 3cm either side of the middle of a 4cm radius circle or 16pi area × 3cm above the middle =48pi above thevhalfway of the 750 ml bottle or 375ml. So it is 375 +(48 ×pi) . You dont need complicated equations

[@lvlarihuan0]

14cm tall, of course. Simple algebra.

[@Duke_of_Lorraine]

For the bottles : we do not have enough information to calculate the volume at the bottle base or the bottle neck. It's only reliable in the middle section with a cylindrical cross-section of 16*pi = approx 50. Adding the 2 half-filled bottles, it includes both a bottle base and a bottle neck, and we find a height of 33 cm which is 6cm more than the height of the glass. The extra volume is in the middle cylindrical part so 6*16*pi = approx 300 extra mL (rounding down) So, the total volume of both half bottles is 750 + 300 = 1050 mL. As there are 2 of them, that means each half-bottle contained 525 mL.

[@malcolmgeorge6036]

Yes I agree, no need to make it too complicated! Half the overlap volume plus half the bottle volume ( 150+375 =525 ) Simple I didn't know I was a Mathmatical Genius?

[@krzysztofmazurkiewicz5270]

My take on the first puzzle was a little different as i assigned the height to the glass x and h fror the "extending parts" as well but i just wrote bot equations of height (x+4h and x+h) and subtracted one from other to cancel X out. The rest goes basically the same.

[@markrobinson9956]

Used the same method. Solved the system by elimination.

[@Smallpriest]

For the alternative method, you can also just realise that (2 times the air in the bottle) + (6cm worth of the perfect cylinder portion) = the whole bottle (750cm^3). Then just some simple simple algebra to solve for the volume of air, then 750 minus that for the volume of liquid.

[@clivewilliams3661]

I worked out the glass problem by saying that x + h =19 and x + 4h = 34. Multiply x + h =19 by 4 to give 4x + 4h =76. Subtract the taller stack equation from this new equation to give 3x = 42 i.e. x = 14

[@PixelZoft]

The way I solved it was to find the volume "center" position in the bottle, that is where the surface of the wine in an upright bottle would be if the bottle was half full (375cm3 wine). The surface would then be exactly half way between 8cm (27cm-19cm) and 14 cm from the bottom, that is at 11cm. Since the surface in the example is at 14cm when the bottle is upright it obviously has 3cm more wine (14cm-11cm) than if it was half full. So the volume of wine in the exampel is 375cm3 + 3cm x 4cm x 4cm x Pi = 525,796cm3.

[@SiggiTh]

Glasses: 34 - 19 = 15 (3stacked glasses) 15 / 3 = 5 (1stacked glass) 19 - 5 = 14 (1single standing glass) ergo:14cm QED

[@jaideepjrx]

2nd method for bottle question was pretty good. Great questions and solutions 👏🏻👏🏻👏🏻

[@CCCompiler]

get wasted until you figure it out

[@spqr117]

​@@CCCompiler😂😂😂😂

[@BIueCat]

Easier way for the first problem with no equations. Look at the middle stack of glasses: 19 is the height of the bottom glass with one extra glass stacked in it. Subtract that from the left stack (34) and you are left with 15 for the remaining top three glasses, making the height 5 for each stacked glass. Go back to the middle stack and subtract 5 from 19 and you get 14, which is the height of a single glass.

[@Tahgtahv]

Um, that's exactly what he did, except you used words instead of symbols. The equations don't go away because you write it differently.

[@ThinkersImpossible]

Also the point of video was 2nd problem 1st one was just to lure people who would think yeah I calculated the correct answer of the thumbnail problem I m genius let's see what this measly peasent has to say of my achievements And then you would realise there is 2nd actual good question

[@iwilltubeyouall]

@@ThinkersImpossible that is exactly me.

[@mkv2718]

@@Tahgtahvhe or she is describing how to do the problem in your head. equations can be difficult for many people to visualize without pen and paper.

[@YT-Observer]

i was confused a bit since a 750 ml bottle also has some air space as well as a cork so the neck is longer than the space for the liquid

[@Noah22222]

I just imagined you have two such bottles of wine as shown in the second question--one upside down and one right side up. Freeze the liquid (don't actually do this, haha) and flip one over and put them next to each other. In these two bottles combined, you clearly have one full bottle plus an extra hockey puck of wine with a height of 6 cm (19+14-27=6). Therefore 2V=750+96*pi. You get to this point in your first solution but this just cuts right to that point without needing to define u and b. Your second solution kind of takes advantage of this trick by removing 3cm from each bottle, but it's even simpler than that. Just recognize that there is 6cm of overlap and you are done. Thanks for the great videos!

[@sandyjr5225]

I did exactly this, but freezing gives a real definition to the phenomena in action though.. nicely explained man!!

[@Dustin314]

Exactly what I did! Consider the overlap with a height of 6cm. The remaining volume in either case is the same, call it x. The overlap plus 2x is equal to the full volume of the bottle, 750 ml. The overlap is 96π ml. So 96π + 2x = 750 gives you that x = 375 - 48π. Now you just add the overlap to get the volume is 375 + 48π ml.

[@alexanderklimke6508]

I die it this way, too. Easy to calculate it mentally this way.

[@muffinconsumer4431]

A pretty entertaining and appealing approach to algebra

[@justinnitoi3227]

This is a very fast way to solve in your head: Imagine pouring the liquid from the 2nd bottle into the 3rd bottle. After filling the third bottle, that will take 8 cm from the bottom the the 2nd bottle. Since you know thr volume of a bottle is 750, you know that: 2*l = 750 + 16pi * 6 where l is the volume of liquid l = 375 + 48pi

[@kevindegryse9750]

Such a complicated way to solve the bottle problem... Ok. Lets try to remove 3 cm from each side. Wow, 11 + 16 = 27, the bottle + the bottle upside down fill the whole bottle. Wait, does it mean the bottle is half full ? Yes, definitly. But wait, we removed 3cm... No problem, we can add them to half the bottle volume with a simple cylindre volume formula. Et voilà. No need for strange miss volume strange shape. The key is to use the cylinder part.

[@sidkemp4672]

Problem 1: I figured it out from the thumbnail before I started the video. I used a different pair of equations: x + 4h = 34 x + h = 19 and got the same result. Problem 2: I got that it is solvable, but had no idea where to start. So I watched and learned. A nice pair for me, showing what I do know (and feeling good), then going on to learn what I don't know yet.

[@boomcrashbang.]

The second solution was so slick. I thought you were going to solve for each of u and b, but you didn't need to do so.

[@realitydisillusioned1544]

I'll try to describe the 2nd problem more clearly. Let V = volume of liquid, and A = volume of air. Eq.1: in the original bottle, V + A = 750. Eq.2: in the inverted bottle, A + 6*(4*4*π) = V, which is simplified to be A + 96π = V ⭐In Eq.2, "the volume of liquid with height of 19 cm" equals "the volume of air in the original bottle with height of 13 cm" plus "a cylindrical volume with height of 6 cm." 😅 Eq.3: by subtracting 96π on both sides of Eq.2, A = V - 96π. Eq.4: by substituting Eq.3 into Eq.1, 2V - 96π = 750. Eq.5: by adding 96π on both sides of Eq.4, 2V = 750 + 96π. Finally, by dividing 2 on both sides of Eq.5, V = 375 + 48π, and that's the answer!

[@moloxlavgood]

In problem 2: note that you assume that the wine's volume is more then the volume of the bottom and the volume of the neck AND is lower then the full bottle without the bottum or the neck. (and in general assume that the bottle has sub-part which is cylinder of at least 6 cm). A generalization of this problem can't be solved in a bottle which is volume isn't linear with it's height in any neighbour.

[@yurenchu]

Wow! I didn't suspect there was such a clever shortcut solution for Problem 2 ! So just do 13 + x = 19 - x ==> 2x = 19 - 13 = 6 ==> x = 3 cm ==> Volume of liquid equals {half of bottle volume} + { cross-sectional area * x } = 750/2 + (pi*4^2) * 3 = 375 + 48*pi

[@egyptian20091]

Solution of problem 1: Since the height of one whole glass and 4 semi glasses is 34cm and the height of one whole glass and one semi glass is 19cm, we can subtract the height of the two glasses from the height of the five glasses to get that the height of three semi glasses is equal to 15cm then divide that number by 3 to get that one semi glass has a height of 5cm then we subtract that 5 from the 5 glasses with a height of 19cm to get that one whole cup has a height of 14cm. Yeah, I definitely used a much simpler way.

[@obliviouz]

Second puzzle has a simpler solution: Volume of the liquid is X. 2X = top half of bottle + bottom half of bottle + 6cm overlap. We know that 1 bottle is 750cm3, and the 6cm overlap has volume 6 x 4pi2 = 6 x 16pi = 6 x 50.265 = 301.6. So 2X = 1,051.6. X = 525.8.

[@alexandros93]

When the bottle is on the right side you need 13 cm of liquid to fill it up when you flip the bottle upside down you have 19 cm so you have spare 6 cm in which the bottle is a cylinder so you have fill bottle 750 cm3 and 6 cm which is hπr2 =6*4^2*3,14= 301.6 for twice the volume.So 1051.6/2 =525.8

[@BravoNineThreeTwo]

Also took the second approach. Thought about it for a moment and realised there was a bit in the middle that would always be red and the volumes either side must be the same. No algebra required. Easier if you imagine the second bottle the correct way up.

[@awaski977]

I had an alternative solution: If we overlap the two liquids, we would fill the entire bottle and have 14 + 19 - 27 = 6 cm overlap in the middle. Meaning: 2V = 750 + 6*(4^2)*pi --> V = 375 + 48*pi I was very happy with this solution :)

[@gamingmonke1269]

I worked out the thumbnail my process was: take the taller stack of glasses (34cm) - the smaller stack of glasses (19cm) the result is 15cm so i divide 15 with 3 (because there are 3 more glasses stacked on the larger stack than the smaller stack) result is 5 cm 19-5 (smaller stack only has 1 extra glass stacked on it) = 14cm tall

[@gamingmonke1269]

took like 30 seconds lol

[@Duke_of_Lorraine]

For glasses : 14cm, didn't even need equations. Total height = height of the first glass (call it A) + the extra height of extra glasses piled up on it (B) 34 = A + 4B 19 = A + B The difference between the 2 equations gives 15 = 3B so B = 5. Substract it from the second equation and you have the height of one glass.

[@HyperSnypr]

Wasn't that the equation they used anyway?

[@Duke_of_Lorraine]

@@HyperSnypr didn't need to write it down to find the solution

[@calholli]

"Didn't even need equations.. proceeds to write equations.. :) 34 -19 = 15. 15/3= 5. So each single stack is 5. 19 -5 = 14 So you can do the first problem without even using variables.

[@Duke_of_Lorraine]

@@calholli yup, did a difference to find the extra height of piled-up glasses which is 5cm, then retracted it from the 19cm to find the height of the glass.

[@Anonymous426_]

The second one is pretty simple actually. If you subtract the volume of 6cm of height from the liquid in the cylindrical part of the second bottle, and you add the volume of liquid in the first bottle, you get the total volume of the bottle which is 750cm^3. From this you get the equation: 2x(volume of liquid) = 750 + (pie)x(radius^2)x(height) Since height and radius is known (taking height as 6cm), you’ll get 525.8 cm^3 as the volume.

[@PivDen-jv3th]

Yeah, about bottles, you need only take two bottles, take some liquid from first to second to full second bottle, and you have one full bottle and on 6 cm cylinder. Like swap bottom liquid from first and top air from second.

[@dhpbear2]

I solved the bottle problem this way: Bottle right side up: V = (224pi) - b Bottle flipped: (I subtract out the empty volume from the total volume) V = 750 - (128pi) + b Add the equations: 2V = 750 - (256pi) Solve for V: V = 375 - (96pi)

[@tvm73827]

Every once in a while you give us an easy problem, like this one, that even I can solve 😂

[@chrisroberts1440]

The puzzle states the volume of the bottle, however in a normal wine bottle there is 750ml of liquid plus the Cork and a small air gap.

[@manaspatel4679]

Your Questions are very interesting. Thanks for all of these

[@mikumikuareka]

I've just seen the thumbnail and want to try to solve it without watching: b + 5t = 34 b + 2t = 19 b + 1t = ? b + xt = y (y - 19)/(34 - 19) = (x - 2)/(5 - 2) y = 15 * (x-2) / 3 + 19 = 5x + 9 y(1) = 14

[@SRangerMtl]

Finally, one I can solve easily! The first one, I did it differently: x + 4h = 34 x + h = 19 substract the second equation from the first: 0 + 3h = 15 (=) h = 5 so from the second equation we easily get x = 14.

[@PADABOUM]

Those exercises shows how we don’t seem to all work the same way, the bottle second way was so obvious to me (and many other I believe)

[@jasonwalter-tz4qz]

So add the heights of wine in both orientations, subtract the height of the bottle, and divide by 2. Thats the height of the wine you add (or subtract) from half the volume, in the general case.

[@jeanlemire2681]

For the first one you can put X as the unknown for the glass height and A as the unknown for the extra height. Then, from the first diagram you get X + 4A = 34 and from the second diagram you get X + A = 19. Multiply the second equation by 4 to get 4X + 4A = 76. Subtract the first equation from the last one to get 3X = 42 and then X = 14.

[@_Dearex_]

my approach to the bottles seems a little less complicated: volume upright: 4^2*pi*14-x volume of Air upsidedown: 4^2*pi*8-x these two together fill the whole bottle so these added gives 750. Now solve for x ≈ 178 (volume of the notch at the bottom) Just plug that in in the volume for the liquid and you are done!

[@YaBoyUneven]

First problem is just a system of 2 equalities that the glasses give, which you can solve for either x(the length of the bottle itself) or y(the length protruding from the bottle that has been added on top

[@Quasar900]

For the piled up glass problem, It's solved in 4 mn before even clicking the video icon ! 2 piled up glasses are 19 cm high , right therefore if we subtract it from the 5 glasses pile that is 34 long , we got 34 - 19 = 15 cm . this 15 cm represent the sum of the lenght of the 3 '' top of glasses '' parts , therefore each "" top of the glass'' is 15 cm /3 = 5 cm width (they're equal ). so , if we subtract 5cm from the 2 glass pile we got 19 - 5 =14 cm which is the length of ONE GLASS ! PROBLEM SOLVED ! TATATATATAAAARAAAAAAAAANNN 🙂

[@stevecrisp5096]

Regarding the 1st puzzle. Let x represent the height of 1 glass and x+ y be the height of 2 glasses where y is the height of the 2nd glass that extends above the second glass. This means that the 5 glass stack has x +4h height and not x+3.

[@tripnick555]

Wow, how complicated can you make solving the wine puzzle? I immediately realised that if the bottle was exactly half full the wine would be at the same level when flipped, so half of the extra amount plus half of 750 is the answer, or 375+48pi. This seemed really obvious and took me less than 10 seconds in my head. I honestly struggled to follow your explanation about air gaps and incomplete cylinders. My thought process was.. Overlap is 14+19-27=6 Halve it 6/2=3 pi r ² h pi x 4 x 4 x 3 = 48pi (or 150.8 but I didn't calculate that in my head) Half the full bottle is 750/2 = 375 Answer is 375 + 48pi 375 + 150.8 = 525.8

[@Saint_Anger]

" I immediately realized that if the bottle was exactly half full the wine would be at the same level when flipped" thats only true when the bottom missing part and top missing part are equal, otherwise your assumption is wrong as you can see at 11:10

[@tripnick555]

@@Saint_Anger No. If you fill any shape object exactly half full by volume and draw a line where the liquid level is, when you turn it over the line will be in the same place. Think about it. If it is exactly half full the space with no liquid is exactly the same as the liquid filled section.

[@Saint_Anger]

@@tripnick555 ok now that is true, when you said same level I assumed you said the height of the liquid would be the same.

[@puspamadak]

I solved it like this: Let the height of 1 glass be x, and the part of one glass outside the stack be y. From first figure, there is 1 glass and 4 stacked glasses on top. So, x+4y=34. From second figure, there is 1 glass and 1 stacked glass. So x+y=19. Solving the equations, we get x to be 14cm, which is the height of 1 glass.

[@StevenDietrich-k2w]

The first puzzle was simple and took just a couple of seconds to see the solution. I failed to solve the 2nd puzzle, but I do have to say that I loved the "bonus" solution. Great video PT.

[@puliverius]

First is easy. When X is the single cup x+4y =34 and x+y=19 thus x=14. The second one is a bit tricky, but still easy. You just need to realize, that the empty parts in each rotation are of same volume, so there is 14:(27-19) = 14:8 = 7:4 ratio of the filled vs unfilled part of the bottle. Whole botle has 750 cm3 thus the filled volume is 477.27 periodic. Edit: Now I see why I got the wrong answer. The problem is that when you uniformly pour wine into the bottle, the height is not uniformly moving up because of that irregularity on the bottom of the flask. It would be usable when the bottom would be flat though. BUT, this approach is not completely useless. Height of the liquid normally is 14cm. Height of the air, when flask is upside down, is 8cm. If I combine these parts together, I get a new flask, that is same at both ends, it's volume is 750cm3. This bottle is 14+8=22cm tall. In the middle of this flask the volume is at it's half. So middle is at 11cm. This is 3cm under where the liquid now is. So all we need to do now, is add volume of a cylinder of height 3cm and r of 4 to half of the bottle's volume. Then we get the right answer.

[@a1phace]

That's the answer I got. I'm trying to check where I might have missed something but it seems to be correct.

[@jxmai7687]

Funny, just took me few seconds to find his way, but my brain told me to use formula like yours at the same time.

[@Snafuuu]

Can't believe i solved the first problem in my head. first and last time i probably solve anything in this channel lol

[@todd727300]

I think the most typical equations for the first problem would be x + h = 19 and x + 4h = 34. Subtract either equation from the other and solve for h. You kind of did that, but in a less intuitive manner when you directly substituted 19 in for x + h.

[@karcha5]

I used a variant of 2b 224π - b = volume of liquid 128π - b = volume of air after inverting Since they add to 750 we get b = 176π - 375 Substitute for b into volume of liquid before inversion we get the answer. 224π - (176π - 375) = 375 + 48π

[@barneybarney3982]

The 1st problem has 2 possible ways of solving it, the difference is only how you write it, one way is in video, other will be "h" for the part above 1st glass and "x" for glass without its upper part and you solve for "h+x" to get the result. The 2nd problem has a way easier solution than shown in video. 14+19=33 33-27=6 (6* 4^2 * pi + 750) /2 = answer we are looking for.

[@Nikioko]

2:53: I would say that x + 4h = 34 cm. Subtract x + h = 19 cm, and you get 3h = 15 cm and thus h = 5 cm. Which makes x = 19 cm - h = 14 cm.

[@quigonkenny]

Problem 1: We have two equations, where N is the height of a glass and x is rhe height of an extra glass: N + 4x = 34 ---- (1) N + x = 19 ---- (2) We're only solving for N, so multiply (2) by 4 and subtract (1). 4N + 4x - (N + 4x) = 4(19) - 34 3N = 76 - 34 = 42 N = 14 cm

[@battle00333]

The Stack of 5 and 2 glasses, can be expressed as 1 Glass + 4 parts of glass, and 1 Glass + 2 parts of glass. or 1G + 4P and 1G + 1P. If we do (1G + P4) - (1G + 1P) we get 3P. So we have that 3P = 34 - 19, which is 15. Then we find P which is 15/3 = 5. Now we can do (1G + 1P) = 19 which is 1G + 5 = 19. which results in 1G = 14. So 1 glass is 14cm tall.

[@tkrabec]

I'm liking the new format. "Regular" problems and videos with a few shorter problems.

[@thorstambaugh1520]

For the glasses. The difference between the two heights is 15 and there are three more stacks. Hence a stack adds 5cm. 4 stacks equals 20cm minus 34cm equals 14 cm for a single glass

[@MichaelPaoli]

First problem simple enough to do entirely in head easily: There's the height of glass, plus increased height of stack for additional glass(es) added to stack, subtract the height of 2nd stack from height of first stack, that's 15cm, which represents the additional height of stacking 3 more glasses, so the additional height for adding one glass to existing stack of 1 or more glasses, is 15cm/3 = 5cm, so, take 2nd stack, of 19cm, remove top glass from stack, that's 19cm-5cm=14cm so that's total height of individual glass, and thus the answer.

[@Quasar900]

For the piled up glass problem, It's solved it Mentally in 4 mn before even clicking the video icon ! 2 piled up glasses are 19 cm high , right therefore if we subtract it from the 5 glasses pile that is 34 long , we got 34 - 19 = 15 cm . this 15 cm represent the sum of the length of the 3 '' top of glasses '' parts , therefore each "" top of the glass'' is 15 cm /3 = 5 cm width (they're equal ). so , if we subtract 5cm from the 2 glass pile we got 19 - 5 =14 cm which is the length of ONE GLASS ! PROBLEM SOLVED ! TATATATATAAAARAAAAAAAAANNN 🙂

[@Quasar900]

Prove that the Area of a Triangle is equal to " Its Half Perimeter Multiplied by the Radius of the inscribed circle inside that triangle '' ! Allez Allez 🙂 (the inscribed circle of a triangle is the Circle which centre is the intersection of the 3 bisectors of each angle of the triangle ; and that circle is tangent to each of the triangle sides ) 🙂🤔🤔🤔🤔🤔🤔

[@augustgustaw7231]

Both easy, especially the glasses question. With bottle you can approach this differently. 2V = 750 + (14+19-27)xPIx4x4

[@richardhole8429]

First one: Look at the smaller amount of rise by adding each cup. 3 cups are added to get from 19cm to 34cm, 3 cups rise 15cm, so thevrise is 5cm per cup. On the right, 2 cups make 19cm, less the 5cm rise, gives 14cm for each cup.

[@andrewhawkins6754]

I did the 1st one slightly differently. x+4y = 34, x+y = 19. Subtract equation 2 from equation 1 and then finish the same way you did.

[@d7home2129]

There is a missing assumption in the problem. That the liquid stops in the perfectly cylindrical region. It seems obvious from the drawing that this is the case, however, when it isn't explicitly stated as an assumption then the problem isn't complete and can't be solved. For example if the perfectly cylindrical part is less than 6cm, then the problem isn't solvable. Even though it's well defined (i.e. possible with a bottle with less than 6 cm perfectly cylindrical to have it have the same liquid at 19cm vs 14 cm for different orientations)

[@yurenchu]

It's already guaranteed that the liquid stops in the perfectly cylindrical region. Because if that isn't the case, then either the 14 cm liquid level in the upright bottle position would have already been in the neck section, or the 19 cm liquid level in the upside-down bottle position would have already been in the dimple punt section _at __1:45__ during the problem description_ (i.e. before drinking in solution B happens). Since in the problem description at 1:45 both the 14 cm liquid level in the upright bottle position and the 19 cm liquid level in the upside-down bottle position are in the perfectly cylindrical region (no matter how close to the shoulder or the dimple tip), we can logically derive that the 50% liquid level must also be in the perfectly cylindrical region; anything else is simply physically impossible. "when it isn't explicitly stated" In addition to the drawings, the problem description explicitly states that the bottle is "in the standard shape of a wine bottle". In a standard _Bordeaux_ bottle (or similar), the dimple punt certainly doesn't reach up to more 25% of the bottle height, and the shoulder starts at about 2/3 of the bottle height. "Even though it's well defined (i.e. possible with a bottle with less than 6 cm perfectly cylindrical to have it have the same liquid at 19cm vs 14 cm for different orientations)" In that case, either the bottle height is greater than 27 cm, or we would have seen at 1:45 that either the 14 cm liquid level in the upright bottle position is somewhere in the neck, or the 19 cm liquid level in the upside-down bottle position touches the dimple (or possibly both levels are not in the perfectly cylindrical region).

[@d7home2129]

@@yurenchu That's wrong. Nothing physically impossible about a 22cm neck, 3 cm cylinder, 2 cm dimple

[@yurenchu]

​@@d7home2129 What is "wrong"? Nothing that I wrote contradicts the scenario of a 22cm neck, 3cm cylinder and 2cm dimple. In that case, the 14 cm liquid level in the upright bottle position would have already been in the neck, as I said. In other words, the 14 cm liquid level in the upright bottle position would then _not_ have been in the perfectly cylindrical region (and therefore we cannot expect the 50% liquid level to be in the perfectly cylindrical region either).

[@d7home2129]

@@yurenchu you said it is physically impossible.

[@yurenchu]

@@d7home2129 If the 14 cm liquid level in the upright bottle position and the 19 cm liquid level in the upside-down bottle position are both in the perfectly cylindrical region, then it's indeed physically impossible that the 50% liquid level is _not_ also in the perfectly cylindrical region. And your "22 cm neck, 3 cm cylinder, 2 cm dimple" scenario doesn't contradict that.

[@mrg0th1er83]

I'm too lazy to solve equations these days. So my first instinct was to equalize the bottles like you did at the end. I like these puzzle you just need to think about instead of doing equations after equations. I don't know why but it makes me fell smarter instead of doing chores.

[@totherarf]

This only works if the bottle is completely full (this is not stated, so may be not the case) You have two separate heights .... add together and minus the height of the bottle. This gives a true cylinder of 5 Cm so Pi r squared gives the volume of excess and take that off the full volume and you get your actual volume. Alternatively you can find the 5 cm figure .... half it and that is your 50% of the full bottle .... then add the additional 2.5cm x Pi R squared to get your actual volume.

[@pizza8725]

14 cm,just do x+4y=34,x+y=19 and solve for x

[@briandearing6238]

Problem 2: the upright bottle and inverted bottle have an overlapping cylinder of fluid with height of 6cm, Overlapping volume in the regular cylinder portion: pi*r^2*h=301.6ml … this means the total of the upright AND inverted bottle is the same as a completely full bottle plus the overlap: 750 + 301.6, and since we’ve counted our bottle twice, divide by 2. 375+150.8 =525.8ml

[@MichaelPaoli]

second problem also easy enough to do in head: full bottle is 27cm, partially filled bottle is more than half full (14+19)cm!=27cm, so, what to do to reduce it to half filled? remove enough liquid so it's half filled - remove from cylinder section, same amount if bottle is right side up or upside down, so, 14+19=33, which is 6 greater than 27, so remove 3 from each bottle's orientation (must match), then we have (11+16)cm=27cm, so that would be half filled bottle, which is (750cm^3)/2=375cm^3, but that's 3cm from cylinder section short of what we actually have, so have to add that back to get actual liquid in bottle. Well, volume of cylinder is cross sectional area, which is pi*r^2, multiplied by height=3cm, so liquid volumes is 375cm^3+3cm*pi*r^2, r is 1/2 diameter = 8cm/2 = 4cm, so liquid volume is 375cm^3+3cm*pi*(4cm)^2 = 375cm^3+3*16*picm^3= (375+48pi)cm^3 Anyway, slightly simpler approach variant of 2nd method given in solving that puzzle.

[@larrydickenson8922]

14 cm. Stacking 3 additional glasses adds 15 cm to the stack height. Therefore, each glass adds 5 cm. Removing 5 cm from the stack of 2 yields 14 cm for a single glass. Back in the grade school lunchroom we occasionally received juice in plastic glasses. The challenge the became seeing how high of a stack we could make. Problem was, jamming the cups together made them nearly impossible to separate. The Principal came down hard.

[@Aleblanco1987]

I hate the second problem because if you calculate the radius of the dimple in the bottom (asuming it's a sphere) it doesn't fit the bottle.

[@yurenchu]

Well-spotted! If the dimple has the shape of a _half-sphere_ , then its radius has to be about 4.396 cm , which is indeed greater than the 4 cm radius of the bottle. By the way, the neck & shoulder part is at least 8.5396 cm long.

[@Aleblanco1987]

@@yurenchu I meant half al sphere! I discovered it because I first calculated the volume as if the bottom was flat to get an aproximation. But then when I got to the answer I saw that the volume the dimple took was much greater than expected.

[@yurenchu]

@@Aleblanco1987 In the drawing in the video, the dimple doesn't appear to be half a sphere though, but rather a (half)cylindrical tunnel... :-) (How do we calculate the volume intersection of a horizontal halfcylinder and a vertical cylinder?)

[@MarieAnne.]

For the second problem, I turned the upside down bottle right side up, but with the liquid at the top. So comparing the two bottles, we have: 1st bottle : 14 cm of wine at bottom / 13 cm of air at top 2nd bottle: 8 cm of air at bottom / 19 cm of wine at top The overlapping volume of wine occurs between the 8 cm lines and 14 cm lines. So the overlapping volume is a cylinder with height = 14 cm − 8 cm = 6 cm and radius = 4 cm Overlapping volume = πr²h = π(4 cm)²(6 cm) = 96π cm³ Since the volume of wine in both bottles is the same, then the non-overlap volume of wine in 1st bottle (below 8 cm line) must be equal to the non-overlap volume of wine in 2nd bottle (above 14 cm line). Call this volume x. Since the two non-overlap regions and the overlap region take up the whole bottle (no less, no more) we get: 2x + 96π cm³ = 750 cm³ 2x = (750−96π) cm³ x = (375−48π) cm³ The volume (V) of wine in each bottle is equal to the overlapping volume + one non-overlapping volume: V = 96π cm³ + (375−48π) cm³ *V = (375+48π) cm³*

[@velloceti6898]

I solved the wine bottle problem by dividing it into 3 sections measuring 8cm and 14cm from the bottom of the bottle. The middle section is a 6cm cylinder. The top and bottom section have to be equal to each other. From there, you can solve the bottom or top section, which equals (750 - 96pi)/2. You then add the middle section back in to get (750 + 96pi)/2. More generally, Vliquid = (Vbottle + Vflipoverlap)/2

[@nathanjiang100]

before watching the video, this is my approach: x+2y=19 x+5y=34 y is the height of the part of each cup sticking out and x is the part that isn’t. there is only 1 x because only one cup has the base exposed but all of them have the tops exposed. solving for (x,y)=(9,5) and the height of the cup is the part submerged plus the part exposed so x+y=14 now time to see if I was right oh wait there’s another problem lemme work that out

[@micke_mango]

Are these 3 cms the same as ((14+19)-27)/2? I guess so... So basically, the sought volume is half the bottle_volume + πr²(height _up + height_down - bottle_height)/2. That seems to work also when liquid_volume is less than half the bottle. Would that work also for liquid volumes less than u+b, less than u, less than b? Probably not, but I'm shooting from the hip... I think that correspondingly, it wouldn't work for liquid volumes larger than 750-b-u either. The solution in this video probably also has those limits, the liquid heights must be in the straight cylindric part of the bottle...?

[@ArnabAnimeshDas]

The first problem took 2 seconds to solve. I solved the 2nd question by taking 2 times the volume of the liquid which exceeds the bottle capacity by 6 cm. height (similar logic as 2b). I calculated the total volume and then divided by 2. Took a good 30 seconds to come up with the logic and iron out the kinks.

[@charlesspringer4709]

These videos are always good. If you can train yourself to never use the word "what" unless you are asking a question, you will achieve metaphysical perfection. My crusade: free the World from "What I do now is...what you wanna do now is you wanna ... and all their illegitimate children. Imagine the man-hours that can be saved on KZbin alone! And the apparent IQ of the channels will go up 10 points.

[@kamen42]

Solved it in like 10 seconds from the thumbnail. 1 glass + 1 inserted is 19cm. 1 glass + 4 inserted are 34. That means 3 inserted glasses add 15 cm, that means 1 inserted glass adds 5 cm. Therefore a glass must be 14 cm.

[@sparshsharma5270]

I did bit similar to second method for the wine riddle. I found measure of the wine bottle by taking the case when the bottle is standing and filled with wine. I took the top part as y and next part as x and found that 6 cm was common to both the cases. From there I calculated volume from given data and it was easily done!

[@shelleyweiss9920]

My approach as well. I subtracted volume the 6 cm of common height (6*[pi]*4^2) from the 750ml, and the remainder of the volume had to be equally split between wine and air, so I simply divided that number by 2, to find the volume of air, and subtracted that from the total for the volume of wine. In short: Wine = 750 - (750 - 6*[pi]*4^2)/2 I guess this can be simplified into the equation used at the end of the video: Wine = 750/2 + 3*[pi]*4^2

[@go_gazelle]

For the second one, I added the volume right side up to the volume upside down, then divided by 2. The sum is the total volume (750 cm²) + the overlapping volume in the middle (6πr²). 2V = 750 + 6πr V = (750 + 6πr²) / 2 V = 375 + 3πr², r = 4 V = 375 + 48π For the first one, I made two equations, then subtracted the 2nd from the 1st: (1) x + 4y = 34 (2) x + y = 19 3y = 15 y = 5 x = 14

[@TheVoiTube]

What is fun about simple solution? You can count 34-19 as the amount of 3 over lap, then just remove the 1 overlap from 19. So 34-19 is 15 for 3 overlaps and 5 for 1 overlap making single glass 14cm + 20cm overlaps.

[@karlhendrikse]

I solved it kind of like the second solution, but simpler. Call the volume of liquid V. Looking at the liquid in the upright bottle, we know V reaches 14 cm up the bottle. Looking at the air in the inverted bottle, we know 750 - V reaches 8 cm up the bottle (turn the image upside down in your head). Taking the difference gives V - (750 - V) making a cylinder of height 6 cm, i.e. 2V - 750 = 6 * pi * 4², so V = 375 + 48 pi.

[@MrShysterme]

I think this is any easier way to see the second problem. "Cut" out the red liquid area of upside down and rightside up bottles and flip the upside down one over and put it on top of the the other to make an all red bottle that is larger than the 750 cc one and totally full. This new larger bottle is 2 times the volume we are trying to find. This new bottle is 14 + 19 = 33 inches tall which is 6 inches taller than the original. This extra 6 inches obviously comes from the middle, cylindrical portion being added to since the bottom and neck are unchanged. So, take the original bottle size of 750 cc and add the volume of the 6 inch added portion (pi x 4 x 4 x 6) then divide this by 2, since adding the two red portions doubled the volume you are trying to find.

[@KarlFredrik]

Solved the second by making all relations between variables. Resulted in 5 independent equations and 5 unknowns. Knew it was some smarter trick involved since only three of the equations were needed to determine the volume.

[@Englishsea24]

I kind of reasoned about it like this: If one glass piled on another glass equals 19cm then its possible the upper edge of the glass inserted equals 5cm since that would mean 19-5=14cm and the other glass with 4 glasses stacked inside would be 4x5=20 and 34-20 also equals 14cm. Just guesswork from the beginning though

[@yurenchu]

Problem 1: Removing three glasses lowers the stack with (34 - 19) = 15 cm, so removing one glass lowers the stack with (15/3) = 5 cm. Removing one glass from a stack of two glasses results in 19 - 5 = 14 cm, which is thus the height of one glass. Problem 2: A is cross-sectional area of cylindrical section = π*4² = 16π cm² V is maximum containable volume = 750 cm³ D is volume removed by bottom dimple S is volume removed by neck and shoulder. x is required liquid volume V+D+S = 27*A (eq. 1) x + D = 14*A (eq. 2) x + S = 19*A (eq. 3) ... add eq. 2 and eq. 3 ... 2x + D + S = 33*A ... subtract eq. 1 ... 2x - V = 6*A x = 3*A + V/2 = 3*16π + 750/2 = 48π + 375 ≈ 525.80 cm³ By the way, D = 14A - x = 176π - 375 ≈ 177.92 cm³ S = 19A - x = 256π - 375 ≈ 429.25 cm³

[@ปรเมศวร์เหล่าสินชัย]

@yurenchu Well calculated. What follows is not directed at you but at the way people solve math problems in general. D is equal to more than 3 cm of the cylindrical part of bottle and S is more than half the bottle. Both are impossible which means that the assumption of a very thin bottle makes no sense at all. I don't know why the one who set up the problem did not check this. It also means that all the calculation makes no sense. So, if a correct solution to a problem does not make sense, there is something wrong with the problem. A more sensible solution would be about 375 + 3*37.5 ml = 487.5 ml, well under 525 ml.

[@yurenchu]

@@ปรเมศวร์เหล่าสินชัย Both are _not_ impossible. It simply means that this bottle (assuming it has negligible wall-thickness) fits in a cylinder with a height of 27 cm (=bottle height) and a volume of approximately (177.92 + 750 + 429.25) = 1357.17 cm³ (or 432π cm³ , to be precise). If the bottle is placed upright on a table, the tip of the dimple is at _at most_ 8 cm above the table top surface (which is not impossible, because 177.92 is less than 8/27 of 1357.17), and the start of the shoulder is at _at least_ 14 cm above the table top surface (which is not impossible either, because 429.25 is less than 13/27 of 1357.17). So the tip of the dimple is _well below_ the start of the shoulder, and there is at least (14 - 8) = 6 cm of cylindrical section along the height of the bottle..

[@AVeryCooIName]

My solution to problem 1: X + 4Y = 34 cm X + Y = 19 cm Subtract equations X + 4Y - (X + Y) = 3Y 34 - 19 = 15 3Y = 15 cm Divide by 3 Y = 5 cm Put 5 back in equation X + 5 = 19 cm Subtract 5 on both sides X = 14 cm

[@Robert_H.]

Bottle: Volume Vb, height hb Liquid 1: Volume V1, height h1 Liquid 2: Volume V2, height h2 Reduce same Volume of Liquid in both bottles to get 1/2 ratio: h1 - L = hb - (h2 - L) L = (h1 + h2 - hb)/2 Vl = V1 = V2 = Vb/2 + L * A = 1/2 * [Vb + (h1 + h2 - hb) * A] With cylindrical bottle: A = Pi * R²

[@MonsterERB]

First problem is super easy, height of first glass = X, added height from a stacked glass = Y. So you have X + 4Y = 34 and X + Y = 19. Subtract the two equations, 3Y = 15, Y = 5, therefore X = 14 cm. Second was trickier but I got there using basically the first method.

[@randomdude6594]

I find the solution 2b awesome and even practical for real life. If you want to know if you emptied half of the bottle, you just measure from botton to end of the liquid once and the second time you measure that again with the bottle flipped. If both of these add up to the bottle height, you know you emptied half of the bottle. Should be applicable to every shape of bottle. Of course assuming that bottles are filled to the rim.

[@GetMeThere1]

So, the bottle problem assumes that the bottle is "full" when it's filled to the very top of the neck, and that volume is 750 ml. Is it true then (as it seems to me) that if the bottle were filled somewhat above the base of the neck, then no solution would be possible?

[@mizinoinovermyhead.7523]

There is an easier way to arrive at the second method: Over lap. If you add both bottles together you get the full bottle plus 6 cm of liquid in the center. Subtract 3 CM from both so that adding the two bottles together gets you a full bottle. Now realize that that full bottle is the same fluid, so half that comes from each of them, and then the same is for the 6cm of fluid. Then add what remains.

[@jamesalewis]

My solution in my head was closest to solution 2.b. I realized that the volume of liquid in the righted bottle and the air at the top of the inverted bottle have the same bias due to the dimple, so seeing that they are 14cm and 8cm respectively, I intuited that their difference of 6cm will bracket the 50% point, so I halved the 6cm, multiplied by 4²π, and added to the 375ml volume of 50% to get the 525.796ml.

[@joanmackie1735]

The first one is very easy, and you don’t need algebra. The difference between stack one and stack 2 is three glasses / 15 cm. So one added glass is 3cm. Subtract this from stack 2 to find the height of the glass alone. This takes seconds, but I found the second puzzle less obvious.

[@HomerErectus]

Incorrect. One added glass is 5cm.