Can you solve the water glass and wine bottle riddles?
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- เผยแพร่เมื่อ 2 มิ.ย. 2024
- These are a couple of interesting puzzles. Can you solve them?
0:00 problems
2:10 solution 1
3:38 solution 2a
8:04 solution 2b
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14cm... I think I worked it out before the vid started based on the thumbnail
Same here ;)
Same here
Same here
Same my man
Same, and in less than 30sec...
I love how the first one is a primary school question and the second is pretty complex
Well, in that case, can someone explained to me how the volume almost double when I started drinking it? 😂
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.
@@DaTimmehnah
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.
Yeah, also did the same, seemed so much simpler.
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
Same here. Just took three lines to calculate
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.
Same here. It didn't need to be so complicated.
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
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.
Brilliant! Thanks for sharing.
This is genius.
@@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.
V is not pure cylinder so how do you know that the extra 6cm is pure cylinder?
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!
its not magical its a shortcut aka common sense
@@thefirminator If it's common, why did it have to be taught to you in maths class?
@@iwantagoodnameplease because children don't have common sense.
@@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.
@@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
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.
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.
@@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.
@@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?)
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
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.
That’s how I solve too
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
If there was no dimple, would it just be the simple ratio of 14 divided by 14+8, multiplied by 750
If there was no dimple, then the liquid volume would simply be the volume of a cylinder: π*r^2 *h@@lumsdot
@@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.
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.
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
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.
A pretty entertaining and appealing approach to algebra
Your Questions are very interesting. Thanks for all of these
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?
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.
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.
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,
Why do you think you're not a math genius ?
@@leif1075because he said so?
Every once in a while you give us an easy problem, like this one, that even I can solve 😂
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.
2nd method for bottle question was pretty good. Great questions and solutions 👏🏻👏🏻👏🏻
get wasted until you figure it out
@@CCCompiler😂😂😂😂
Heisenberg strikes again. You've changed the volume by measuring it.
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.
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.
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.
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
@@ThinkersImpossible that is exactly me.
@@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.
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
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.
Used the same method. Solved the system by elimination.
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!
I did exactly this, but freezing gives a real definition to the phenomena in action though.. nicely explained man!!
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.
I die it this way, too. Easy to calculate it mentally this way.
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.
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
Can't believe i solved the first problem in my head. first and last time i probably solve anything in this channel lol
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 :)
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
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.
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!
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.
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.
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.
I'm liking the new format. "Regular" problems and videos with a few shorter problems.
Great Job.
Elegant
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.
Wasn't that the equation they used anyway?
@@HyperSnypr didn't need to write it down to find the solution
"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.
@@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.
Maths is just awesome. Thank you!
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.
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.
14cm tall, of course. Simple algebra.
Finally a riddle that I can actually solve and understand!
ive learn something new today, thanks
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.
Beautiful !!
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.
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.
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!
Did this in my head before the video started. This one was easy.
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.
the 2b solution is beautiful one!
thanks for the 1st problem, finally an easy one I could figure in my head, lol
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.
That's the answer I got.
I'm trying to check where I might have missed something but it seems to be correct.
Funny, just took me few seconds to find his way, but my brain told me to use formula like yours at the same time.
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)
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).
@@yurenchu
That's wrong. Nothing physically impossible about a 22cm neck, 3 cm cylinder, 2 cm dimple
@@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).
@@yurenchu you said it is physically impossible.
@@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.
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
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.
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
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.
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.
First time in my life, I was able to solve all the questions without the solution
"from the bottom to the top" is superfluous in "the height --- is"
That first one is just the easiest simultaneous equations question, anyone who went to school until year 9 should be able to do that easily, and even if you didn’t, it’s still basic logic and maths with possibly a little bit of algebra
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
This is a good question to learn a sequence for my students. Thank you
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!
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
Solved the glasses problem straight from the thumbnail 😄
I literally used the 2nd method, with slight difference. Was a fun question.
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.
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.
Both easy, especially the glasses question. With bottle you can approach this differently. 2V = 750 + (14+19-27)xPIx4x4
When I saw the second question, I just solved using a slight variation of the second solution and didn't even think about the first method, probably because I always try solving these questions without a pen, just in my mind
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
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)
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.
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
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.
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
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 🙂
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.
Centimeters? That takes out all U.S. citizens from the question😂
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
I got 14. Here was my work before the answer is explained...
5 glasses is 34 cm
2 glasses is 19 cm
The difference between the two stacks is 15 cm, which is also a 3 glass difference.
If every glass added goes up by 5 (15 cm / 3 glasses being added) then we can subtract 5 from the two glass stack. 19-5 is 14.
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.
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.
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.
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.
Read the question before clicking the video.. Just saw the thumbnail... Internet was slow... Video took a few seconds to load.... Solved it even before listening : "HEY THIS IS PRESH TALWALKER" 😂😎
I did the first problem more directly: The difference between the 2 stacks is 15 cm, while stack 1 has 3 extra glasses. Therefore any extra glass adds 5 cm to the height. Therefore the height of one glass must be 14 cm, since one glass plus one extra (5 cm) is 19 cm. No algebra, just logic.
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.
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.
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.
The 1st puzzle I got it right, the second too, but I've used a different method, I calculated the area of the liquid as a perfect cylinder with 14*4²π then I calculated the air considering the reversed bottle as a perfect cylinder with (27-19)4²π then I added the results and subtracted 750 and I removed half of this result from the volume of the liquid calculated as a perfect cylinder, since Liquid + air = 750cm³
The 1st problem is something similar of that on the SBAC practice performance task
there is just one thing a standard wine bottle has 750ml of liquid. but a wine bottle is never filled to the top . a small amount is left for expansion so that the bottle will not explode if the wine has to much pressure. therefore the bottle capacity will be slightly more than 750ml.
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.
I solved the first one pretty easily, though I set up the equations a little differently. I didn't get the second one though.
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.
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
For the 2nd qn, can't we just replace the empty space in the second bottle with the empty space in the third bottle as they are equivalent in volume. Then we take linear proportion to find the volume of liquid. ie. 27-19=8cm in third bottle's empty space. Vol. of liquid=14/(14+8) x 750cm3.