Can you solve the 4 foods puzzle?
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- เผยแพร่เมื่อ 27 ก.ย. 2024
- Thanks to James W. for the suggestion! At an amusement park, 65% of visitors ate a donut, 80% ate a soft pretzel, 80% ate pizza, and 90% ate ice cream. What is the minimum percentage of visitors that ate all 4 foods?
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I solved it by inverting the percents (100-p)%, then adding it together, to figure out what percentage at most didn’t eat at least one food
Best solution
You can make that mathematically sound by showing that the minimum percentage goes up if there are overlaps between the people who don't eat each food. It's still the best way though.
same!
Or add them together and do (mod 100).
I did the same thing and arrived at 15%, but since I did it "intuitively", I didn't have confidence in the result.
For the practical application, you actually do NOT want to target the people who bought all 4. Bundles are typically cheaper than the sum of the individual items - that's the point of the bundle. If we only sell the bundle to people who bought all 4 anyway, we just lose money.
The people we ACTUALLY want to target are the ones who bought less, but might be convinced to buy all 4. Most likely, the people who usually buy 3 out of 4 could be tempted by a bundle offer. The donut merchant would be the most likely to profit from this, as most people who only skipped out on 1 probably skipped the donut.
The information request is still useful though. We know, that with such bundle we loose with at least 15% of the customers, and up to 65%. So, looking for info about people who bought all 4 makes sense, but we should focus on the maximum number of people who did that. Then we can set up the pricing in such way, that even in worst case scenario extra spending of the part of 35% of customers will make up for the loss on 65%. This is of course a "safe bet" scenario, most companies probably could tolerate some risk and look for "most likely % of people who bought all 4" to use in estimations instead.
@@edsimnett The four numbers most certainly aren't independent variables. People have preferences and don't truly randomly buy food in the real world. Upper and lower boundaries however still apply.
You also have to measure the time people bought items. If people bought all four at once or 3 items at once (usually a family) that's different than an individual buying all four items but over a longer period of time.
Keep in mind that the *_"geniuses in marketing"_* was meant sarcastically.
@@willguggn2 true, and for a small sampling ..not very informative
..but on a larger sampling .. the errors smooth out...
as in the exaple lets assume 20 ppl...the variance from day to day will be massive
if we say saple 20000/day for a week ... i would say the error becomes so small .. we can with a quite high degree of certainty predict the sales... in the future... and what prferences ppl have .. we can then add other factors as weather, pricing etc...
...we can also just by using common sense do estimations (if its 10 deg and windy, the probability our outdoor ice cream stand sell alot of icecream is quite low)
The easiest way is calculating the inverse. The max amount of people that didn't eat all 4. You just add the percentages of people who didn't eat each food. 10 for ice-cream, 20 for pretzels and pizza, and 35 for donut. 10+20+20+35=85, which means at most 85% didn't eat all 4 foods, so at least 15% had to eat all 4.
Nice, but don't forget to also limit the answer to between 0-100% .
If it exceeds 100% subtracted, you can state that there is no portion of the populace that guaranteed ate all types of food.
Exactly! @@RoderickEtheria
@@ShaiFishmanwow i would have missed good content if i haven't come through comment section 😄
this is how I solved it
I can never solve the puzzles in the channel because when I open the video the solution is right in front of me in the top comment.
Minimum percentage can always be determined by picking the lowest percentage (in this case, 🍩 at 65%), and then subtracting away all the "did not eat X" percentages. In this case, those are 20%, 20%, and 10%, for a total of 50%. 65% - 50% = 15%. The actual amount that ate all four items is probably higher, up to maximum of the lowest percentage, 65%.
Yeah I remember being in school and struggling with a variation of this problem. Is there any flaw with that reasoning?
That's how I solved it.
You don't have to start with the lowest percentage, any of them work as a starting point.
@@knoxiegb1782 Formally speaking, what I did was just a specific application of the proof given near the end of the video. The union of WXYZ is no smaller than the whole space (100 percentage points) minus the complement of each individual set. Thing is, (whole space) - (complement of set W) = the set W, by definition.
You can, as other replies have said, pick any starting point. I just find it most natural to start from the smallest one first.
max(0,100-sum(100-x for x in percentages))
simpler.
I did it by thinking of a group of 100 people, and taking it at groups of 2 at a time. If you take the first two groups, 80 people had the pretzel, which means minus the 35 that didnt have the donut you still need 45 that did. Then for pizza you take the 55 away that didnt have both donut and pretezel but you still need 25 that had both, so then you need 25 that had all 3. Then for icecream you take 75 away that didnt have all previous 3, but that still leaves you with 15 that did, so atleast 15 had to have all 4.
I did the second method but then I realized it's easier:
Assuming that there are 100 people and no person ate the same item twice.
If all of them ate all 4 items there would be 400 sales.
But if we add up the numbers there has been 315 sales.
So at most 85 people did not eat all 4 items.
That means at minimum 15 people ate all 4 items.
> no person ate the same item twice.
The thing is, in the problem description this is left implicit. Somewhere while explaining the answers, this assumption was suddenly put into place.
Before that, nobody excluded a person eating, say, 4 slices of pizza.
@@metamud8686 Right. But maybe it's not important how much one person bought of each item. So for example if a person buys 1 donut or 10 doesn't matter. The question then would be, did they buy it or not.
I loved seeing it worked in many different ways. This is the real skill, being able to approach things in multiple ways and gauge which one will be simplest.
35% didn't eat a donut, 20% didn't eat a pretzel, 20% didn't eat a pizza, 10% didn't eat an ice cream. Worst case scenario : they are not the same people, therefore 35 + 20 + 20 +10 = 85% of people didn't eat one item, so 15% ate all 4.
I either got lucky or found a much easier way to do this:
-Adding up the percentages of those that didn't eat each food equals 85%.
-Subtract that answer from 100%, which equals 15%.
I can't explain why that worked, but it apparently did.
From a marketing perspective: if on average everyone is already buying at least three food items, there’s really no need to offer a combo discount.
Maybe, unless part of the goal is to encourage those who were getting just three items to "just pay a little bit extra" to get all 4. Particularly if the difference between 3 items and the discounted 4 items is more than the unit cost.
After my headache subsides, I'll watch this video again and see if I can follow the logic this time.
Intuitive answer is 15%. There is an algorithm to execute which makes this exceptionally easy.
Basic description:
You take the two lowest percentages to start. You have to calculate their intersection so that is going to be
65-(100-80)=45
then you take that answer and intersect the 3rd.
45 - (100-80) = 25
Take that number and intersect the 4th.
25 - (100-90) = 15.
Done; it should take you approximately 3 seconds to do in your head.
Algorithm:
var accumulator = 0
foreach(var p in list){
if accumulator ==0 accumulator = p
else accumulator -= (100-p)
}
accumulator will contain the intersection as long as there is one. A defense is necessary to prevent a disjoint set from resetting the accumulator to a positive number, but I'm not presenting this as a general solution to the problem of all intersecting sets.
You have to sort them lowest to highest. It would have been a better problem if he had presented them out of sort order
I got the first 65% 80% trick to arrive at 45% due to seeing this kind of puzzle but didn't think to extend it out further to all four, cool trick!
It's also interesting to note that, so long as each set contains less than 100%, you can keep adding more and more orthogonal sets and eventually arrive at 0% minimum overlap
Eg, for sets that contain 50% of a population, you need 2 such sets to have a minimum overlap of 0%, if they cover 75% then you need 4 sets, 90% coverage requires 10 such sets, etc etc but however many ways you have of grouping people, so long as you have enough of them you'll be able to show that the minimum possible overlap is 0%
Another shortcut...
*Sum, then subtract 300%.*
Seems like a great problem to illustrate key ideas like DeMorgan's law and the Union Bound, but I don't think the connection was made as clearly as it could have been. There's a lot of intuition that was brought out be each solution, I think the connection to formalized concepts could have been a lot stronger.
All these puzzles are so fun
I just saw a video with a math puzzle on it.
At the top of a paper write "I have $50"
Now divide the paper into two columns, Spent Balance
20. 30
15. 15
09. 06
06. 00
For each of these you spend some amount of money, then you write your new balance in the balance column.
Now total up each column.
The spent adds up to 50, which we expected. Yet the balances all add up to 51...where did that extra one dollar come from?
I used a method similar to the second, take each percentage and minus it from 100 add all the remainders together then take the result away from 100. So it becomes 100-65=35, 100-80=20, 100-80=20, 100-90=10. 35+20+20+10=85. 100-85=15%
I tried at first to follow the method of option A with weird bar graphs but gave up and found myself doing naturally going with option B.
I wondered if I had the right idea pairing donuts with pretzels and pizza and ice cream but it seemed to make sense to me. And I'm glad the video showed that I was right!
First: 90% vs 80%, the 10% that didn’t eat from the 90% could of ate part of 80%
Thus, there is a minimum of 70% overlap
Second: 80% vs 80%, since one of the 80% has a 70% overlap from before, 30% of the people who didn’t overlap could of ate from the other 80%
Thus, there is a minimum of 50% overlap
Third: 65% vs 80%, since the 80% has a 50% overlap from before, 50% of people who didn’t overlap could of ate from the 65%
Thus, there is a minimum of 15% overlap.
Didn’t see many people point out the logic, hope this helps.
the minimum percentage is 15%, because 35% didn't ate their donuts, 20% pretzels, 20% pizza, and 10% didn't ate their ice cream. While the maximum percentage is simply 65% ate all of their foods, and 10% didn't ate anything
Another way to solve.
The percentage who didn't take the food is 35+20+20+10=85
So all food are taken by
100-85=15%
Here's how I did it and I think it's simpler than the solution given. Suppose we have 100 visitors. 65 had donuts and 80 had pretzels. That's a total of 145, so we know at least 45 people had both. 80 people had pizza and at least 45 had both pretzels and donuts for a total of 125. So at least 25 people had all three. 90 people had ice cream and at least 25 had the other three for a total of 115, so at least 15 people had all four. Yes, this is pretty close to the Venn diagram solution that Presh gave, but I didn't need a diagram, and quickly arrived at the same answer.
I solved it by finding the lowest common denominator if the percentiles, which is 20, which will be our population of park goers. 13 donuts were eaten (A), 16 pretzels were eaten (B), 16 pizza slices were eaten (C), and 18 ice creams were eaten (D). Then I made a table of 20 people, and distributed A, B, C, and D as evenly as possible -- only 3 people ate all four, or 15%. Not the simplest solution, but it got there.
This is the literraly the first method shown in the video
@@flacsomtodosclas2165 Yes. Good job. You figured that out. Congrats. These videos frequently encourage people to solve the problem in the comments before watching the solutions, which is what I, and many other people commenting, did.
A more visual, but also odd way imo is the following:
Imagine a line of 100 people standing from left to right.
Now take any of the number of percentages from above, say 80% (pretzel) and color the people from left with yellow. Now, take, say 90% (ice cream) and color the people from right with blue. The overlapping amount of people will be the ones who at minimum ate both pretzels and icecream, which is 70 people.
Do that again with the other numbers and you will end up with 15 people = 15%
This is by far more complicated than it should be but its the first idea that came to my mind because i wanted to visualize it.
Edit: I wrote this comment before watching the video.
Another did it all in my head and got 15%, as follows - perhaps it's a variation on Venn diagram, but I find this much easier to visualize,
We want minimum % that ate all four - keep that in mind, so, we'll work with the four food percentages, bottom up, 90%, 80%, 80%, 65%, respectively, as follows:
think of a square
now shade it starting at left, until it's shaded 90% of the way to the right. We want minimum that ate all, so ...
now overlap that, minimizing overlap, starting from the right and shading to the left until shaded to 80% of the way back to the left.
We minimize the overlap doing it this way, so to get the overlap take the first one and how short it comes of far right, so 100% - 90% = 10%
now subtract that from the 2nd that shades right to left by 80%, 80% - 10% = 70%. That's minimum for first two foods.
Keep that result, and discard our individual tracking of the first two foods, and similarly repeat, shade from left - 70% must've eaten first two,
now shade 3rd food 80% from right to left. How much overlaps? Again, take the first (now combined from first two) of 70% from 100%, that leaves 30% (that may not have eaten the first two),
subtract that from the right to left 80% shading of the 3rd food, and that leaves 50%, so 50% must've eaten the first 3 foods.
And again, next iteration, shade 50% from left for all that must've eaten first three foods.
Now shade 4th food from right to left to 65% - there's an overlap of 15% out of the 100%,
so 15% must've eaten all four foods. Could also do similarly thinking along a line segment instead of square - unit length, or length 100, whatever's easiest to think of in one's head. But I found the mental imagery of square shading and thinking of that as % of whole easier to conceptualize, and would work for for any number of people (well, integer people if they're positive integer multiple of our 20 which is smallest that would fit positive integer number of people to all the given percentages). So ... 20 people, ... or 20 million, or 20 billion ... shade the square ... and in all cases, results in 15%.
I usually don't do this, but solved it in less than 10 seconds. More of a logic puzzle than math.
Simple solution: negate everything (35 did not eat donut, 20 pretzel, 20 pizza, 10 ice cream), add them up and you get that 85% skipped a food, so 100-85=15% who didn't skip any, thus ate all.
The way I solved it was sort of a blend of the first and second method (basically the second approach but without thinking about it in terms of venn diagrams, and more like in the first approach instead). I started from the end, figuring out what percentage ate pizza and ice cream:
::::::::.. - pizza (80%)
:::::::::. - ice cream (90%)
Now the minimum possible overlap looks like this:
::::::::.. - pizza (80%)
.::::::::: - ice cream (90%)
.:::::::.. - overlap (70%)
The I considered "pizza&ice cream" as one food and replaced them in the list with that, with 70% as amount, and repeated that until I was left with all 4 foods as one.
So I realized that I only needed to figure out how to solve it for just 2 kinds of foods, and from that I could solve it for any amount. And for just 2 kinds of foods I could easily visualize it (as shown above)
I just calculated extra after mini 65% overlapping in other three food found easy 15℅ to be minimum.
Using the pigeonhole principle is trivial. Use a 100 cells pigeonhole, now distribute 80 pretzels, 80 slizes of pizza, 65 donuts and 90 icecreams and look for how many cells have necessary the 4 items
The real question is, why is everyone eating so much food? The median guest ate at 3 distinct food stands during their trip, that's nuts
Given that they could just be getting a drink or a snack, it’s not crazy that they could be visiting 3 food stands in a day at an amusement park.
my spin on this problem before watching:
make a 10 by 10 grid
90 are for ice cream
the 10 unused are used for pizza,then 70 overlap
30 spaces have only one food,fill them up with pretzels,50 rest overlap with the rest
50 spaces with only two foods are used for dough nuts,15 left for overlap with the rest
-->15% guaranteed overlap
yay,I was right!
I think it would be 65% at the most and 15% at the least
The opening statement is that "the food vendors tell you..." the percentage of foods eaten. But from their perspective, each of the people must buy at least *one* food item (otherwise the vendors would not be able to count the person). So the maximum percentage who eat all four foods should be
There is no 'lesser than' the maximum. The least eaten food item is eaten by exactly 65% of the people (who buy at least 1 food item). Therefor, the maximum amount of possible people that bought all 4 food items is 65%. It's not at most 65%. It is exactly 65%.
@@thenonexistinghero Show me a solution using the: 20 people example...
@@DanielFSmith I mean, it's just common sense. Here's a solution with 100 people.
65% of them eat at least 1 food type.
So 65 people eat at least 1 food type.
It also means 65 people can eat all 4 food types at most.
The rest of the percentages don't matter. 65% is the lowest. Whether it's 80% or 90% of people that eat other food types, it's still 65 people at most who can eat all 4 since there's a good chunk of at least 35% that haven't eaten the least popular food and as such, cannot possibly have eaten all types.
@@thenonexistinghero Let's take an example with 60% and, say, 90% using 10 people.
Pie. aaaaaa0000
Cake. bbbbbbbb0
Doesn't work... The last person didn't eat at least one. You can only get 50% max.
@@DanielFSmith They aren't eating from a single slice of pie. The amount of pies is limitless.
OK. Before watching, I thought about it and I was trying to come up with ways to figure out overlap.
So if 90% of people ate one item, then only 10% ate something else. Those 10% could have been the people that ate food B, alone, and 20% of those that didn't eat B would have meant only 70% overlap.
So 90%=10
80%=20
80%=20
65%=35
10+20+20+35=85.
There's at LEAST 15% that ate all four foods
Exactly this problem was (in 1885!) subject of Knot X of 'Tangled Tales' by Lewis Carroll. In 'The Chelsea Pensioners', Carroll writes:
_Problem_ - If 70 per cent have lost an eye, 75 per cent an ear, 80 per cent an arm, 85 per cent a leg: what percentage, _at least_ , must have lost all four?
_Solution_ - (I adopt that of Polar Star, as being better then my own.) Adding the wounds together, we get 70+75+80+85=310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10."
Actually you ask two different questions, how many ate all the foods and what is the min percentage that ate them all. I agree with you about the minimum, it is 15%. But I think a better answer to the question, "how many?" is 37.44% based on converting the percentages to probabilities and multiplying them to get the probability that a given customer ate all items.
I don’t think he ever asked how many, just the minimum
@@pivotfever You are right that he does treat it as a question about minimum numbers, but he does start by saying that what they are really interested in is how many would buy all three, and the minimum number that ate all three in this sample is not the best statistic to use for that.
@@MichaelGowland well he says the marketing people wanna know how many people are interested, not interested in how many people. Thus the actual question, what’s the minimum percentage, not most likely
I was mentally thinking of Method 2, but was physically trying to write it out using Method 1, and using 100 people.
I don't understand why this was so hard, I just added the missing percentages to 100% from each and then substracted it from 100%. 100%-10%-20%-20%-35%=100-85%=15%, didn't even make it to the video, solved from thumbnail. Made perfect sense in my head.
Found it a lot simpler to inverse the calculations, and said No!Donut (35%) + No!Preztel (20%) + No!Pizza (20%) + No!Ice (10%), which adds up to 85% that can have not eaten all the food. and as such, since it's only possible for 85% of the guest to not have eaten all 4 things, that would mean that at least 15% doesn't fall in this group and as such have actually eaten all 4 things
i took two items and took the inverse of the bigger and subtracted from the smaller. substituting the result and repeating yielded 15 percent. after i saw that any order yields the same solution, i am pretty confident
Nice to have an easy problem every so often.
I took 65% and subtracted the percentages of people who didn't eat one of the other foods: 65 -/- 20 -/-20 -/- 10 = 15
1000th video!! Congrats man!
I can imagine an ice cream donut and a pretzel pizza but not the ones you imagined.
...actually I had two donuts, three pretzels, one slice of pizza, and two ice cream cones.
The real world conclusion of this story is that the amusement park sells the bundle and finds out that much lower than 15% buy it (probably even less than 5%). The reason being human behavior; no one wants to eat all four at the same time. Even if the bundle were like coupons to be used throughout the day, most people don't plan on eating all four of those items and/or are hesitant to part with their money up front to commit to eating all four. They'd rather just spend their money at POS when they have a craving for something.
Yeah. This is two mains and two desserts. A person might like a deal for one and one. But not all four.
The bundles might be interesting for groups that share the food.
I don't think it's a very popular idea anyway, though. A third of the people in this scenario apparently just don't like donuts, so they will be skeptical with a bundle that forces them to buy one.
Worse case scenario.... Put up a graph. The highest 90% goes one way (ltr). and the lowest 65%(rtl) The two 80% ones (pizza and pretzels) one goes from ltr and the other goes rtl, respectively.
The two 80% ones turn into only a 40% overlap. This overlap is covered by the 90% by default, thus no change. However, the 65% can only cover 15% less space from the other side, thus the lowest possible amount that all four amounts can overlap is 25%
That said, second best case is if both 80% are on the same side (ltr), while the 65% and 90% cover the other side (rtl) then it's only the 65% vs the 80% that we'd have to worry about, so that'd be 45% in the overlap.
And of course the absolute best case is if all 65% of the donut eaters had all four, so that's be a max of 65%
Darn, I got it wrong. Had the right idea, but wrong execution. Thanks!
Solved in 3 mins.. very satisfied, thank you!
I solved by considering how many % of all people ever tried one of the food, then 2, then 3 and finally all 4.
I feel kind of silly now because I just multiplied the percentages of each of the chances assuming it'd show the probability of someone being selected amongst each group (meaning they ate all of the above food items) getting 37.44%.
I am not exactly sure why our answers disagree.
I think the difference comes down to me assuming that the amount of food purchased would have no correlation to how many more items would be purchased.
But because someone would be less likely to buy food after eating it's not sufficient to merely find the chance, you have to find the minimum because you're looking for value already in the data.
An economist enters the room and knows that 90% of those who eat three pieces would buy the whole package. Calculate the number of visitors who eat at least three pieces.
My method was similar to the first one. I thought "What would people have to do in order to get these numbers with the least people getting all 4?" And the answer is everyone who didn't buy 4 is buying 3 items exactly. So I'm assuming that everyone who didn't buy ice cream bought a pizza, a pretzel, and a donut. And everyone who didn't buy a pizza bought icecream, a pretzel and a donut and so on. So I took the inverse of every percentage and added them up since assuming everyone is behaving this way, they are mutually exclusive combinations. They added up to 85%. So 85% of people could buy only 3 items and still get these percentages, so the remaining 15% have to buy all 4.
You are forgetting about the fact that some visitors could have eaten NONE of the 4 foods ... that could be as high as 10% (ice cream max 90%). Minimum who ate all 4 foods then is actually 0% if you do a 4-circle Venn diagram with all possible intersections. Put a 0 in the very middle quadruple-intersection. Put 15 in the donut/pretzel/pizza triple-intersection. Put 25 in the donut/pizza/icecream triple-intersection and another 25 in the donut/pretzel/icecream triple-intersection. And finally put 40 in the pizza/pretzel/icecream triple-intersection. Put zero everywhere else in the diagram, and 10 outside all 4 circles. I solved it with a system of equations, with 10 outside the circles, so needing the locations within the circles to total 90. Then forced the 0 in the middle. Then some algebra from there. The symmetry of the pizza and pretzels helped some. There may be other solutions.
that was easy. i found the answer in my head after 30sec thought and i'm average at these puzzles.
Here's how I solved it:
*I took 100 people as a sample size to make the calculation easier. I took the 90% first, then that means 10% of the people didn't eat ice cream.
*I took the pizza slice(80%) next, then that means 20% of them didn't eat pizza slice, and I carried over the 10% from earlier to here, so that means 30% neither ate the pizza slice nor ice cream(or both).
*Same process with the donut(20%). That means 20% of the people didn't eat donuts and I carried the 30% from earlier again here. That means 50% of people didn't eat at least one of the following: Donuts, Pizza Slice and Ice Cream.
*Finally, I took the pretzel(65%). That means 35% of the people didn't eat it then I carried over the 50% of the people from the previous proposition. That means 85% of people didn't eat at least one of these 4 items. That leaves me 15% of people who ate all of these 4 items.
I don't know how this works, but hey, I got the correct answer and that's what matters. 😅😅😅
On a side note, I realized that 100 people isn't necessary here. 😂😂😂
Sorry. I inverted the donut and the pretzel, but you get the point. 😅😅😅
the minimum possible is 15% but the probable amount is 37.44%
I did the second approach but by combining one extra food item each time.
Look at that: 35 = no (a); 20 = no (b); 20 = no (c) and 10 = no (d). So, 35 + 20 + 20 + 10 = 85 persons who do not eat at least one of the four types. Thus, (a), and (b), and (c), and (d) do not belong to this set of 85 persons. Therefore, 15 people eat all four types. The union of the four sets may be less than 85. So, the complement may be greater than 15. Therefore 15 people are the minimum.
"Let's imagine there are 20 people..."
Why on Earth did you not go for 100 people?
I solved the first one by adding the percentages all up and putting it over 100 and then reducing it
I immediately inverted. I took 20/20/10 as half of people and then the 65% gave it away.
You look at this from a maximum point of view. The best case scenario is every single person eats 3 of the foods. This gives you 300%. However, we have 315% food, so we have 15% higher than the best we can do if everyone eats 3.. 315% - 300 = 15%
For not "ez" starting percentages, the circles diagram is the best solution
I just summed up the remaining of each group (35 + 20 + 20 + 10 = 85) and deducted from the total (100). I find it hard to point why this sounds logical to me and I'm unsure if this would work on other cases or if it's just a coincidence.
65% of people ate Donuts. 35% did not.
20% didn't eat pretzels
20% didn't eat pizza
10% didn't eat ice cream
thats 50% that didn't eat something, and we can only account for 35% of people who didn't eat donuts.
that leave 15% of people (50%-35% = 15%) who had to eat all of the items
There is an easier way to think. Take the smallest number (65) and assume that 10 people who didn't eat the ice cream also didn't eat donut, that means 65 - 10 = 55, and repeat the pattern for pizza and pretzel and you will get %15. The key point to solve the puzzle is to assume that n peoplo who didn't eat a type of food also didn't eat the food you choose to operate on. The solution of the problem is that assumption. Rest is basic math.
Honestly I just added the chances of someone not having any particular option together to get 85% then subtracted that feom 100 to get the answer
I just added the percentages of people who didn't buy each food, then subtracted it from 100%
🍩 35%
🥨 20%
🍕 20%
🍦 10%
+ ----
85%
100%-85%=15%
Take the lowest percentage .. 65% .. and now work out the 'did not eat' percentage for the other three .. 20% twice and 10% once .. add them together .. 20 + 20 + 10 = 50 .. and now subtract 50 from 65 .. giving you 15 .. and telling you that the minimum number of people who ate all four is 15% .. going up to a maximum of 65%.
I managed to find a solution that allowed for 0% to eat all four. The majority ate two foods. 65% donut and ice cream, 25% ice cream and pretzel, 55% pretzel and pizza, 25% just pizza. All total percentages are satisfied and nobody ate all four foods.
But that doesn't work. If you add up all four of your numbers you get 170. You can't have 170% of people as the maximum number of people who ate something (anything) is 100%
This seemed hard for a few seconds until I realized I could invert it. 🙂
"We can imagine Donut-pretzel and pizza slice-ice creams as food."
State Fair Vendors: "Hmmmm.."
Having eaten all for foods I can confirm the ice cream was my favourite.
I thought about it for 20 - 30 sec, no calculating (except adding four numbers). Same result as video got. Now I feel intelligent. Nice😉
The last equation in simple: answer = 100-(100-w + 100-x + 100-y + 100-z)
On the other hand what's the likeliest number for those who ate all four? Start by assuming that the probabilities are independent: that someone who eats one of the items isn't more or less likely to eat any one of the three others than someone who doesn't eat the first one. Then the probability of eating all four is 65%*80%*80%*90%=37.44%.
Mind you, independent probability is a big if!
My intuitive guess was to imagine the 65% and the 80% filling a bar from opposite ends. Because 65% definitely ate doughnuts, and 80% definitely ate pretzels, then the overlap is the minimum people that ate both. Doing this for pizza and ice cream will net you the same number or higher so it had to be 15. Also side note, I just realised typing this out, this is my exact mental method for solving those nonogram puzzles
Yeah did the same as you
Haven't watched the video solution methodology yet, just checked if my result was right. The way I did it was:
1. Assuming 100 people (to make it easier), A=65, B=80, C=80, D=90.
2. How many people have NOT eaten a food? !A=35, !B=20, !C=20, !D=10.
3. Now, I compared A & !D. If A has 65 people eating it, but only 10 people did not eat D, that means that 65 - 10 people ate both A & D. So, A & D = 55.
4. I do the same for B & !C. If B has 80 people eating it, but only 20 people did not eat C, that means that 80 - 20 people ate both B & C. So, B & C = 60.
5. Lastly, I compare A & D with !(B & C). 55 people ate A & D, while 40 people did NOT eat B & C. That's 55 - 40 = 15, so 15 people ate all four foods. 15 / 100 = 15%.
@@thomasmaughan4798 Yeah I think 10% could've eaten nothing, but since the question only asks for who ate everything, I didn't look at the ones that ate nothing :)
I solved it in a bit of a different way, it's kinda the same as your first step if the 29 people, but I imagined it out of 100, then I said that if 80 people are a pizza, let's just put them over the 100 of the 80 who ate pretzels, imagine 100 circles with 80 yellow, and we theb put 80 more that are red over them, but we make sure to cover the 20 empty ones there because we want the minimum, only 60 circles are both red and yellow, then take that 60 and put it over the 90 of the icecream, but make sure to cover the 10, you'll have 50 left, now put the 50 on the donut one, but make sure to cover the 35, we'll have 15 circles that touched all 4 foods, and 15 out of 10p people is 15%, which is the answer, I hope presh sees this so pls like if u see
i came up with the visual approach but i did not do a good enough job spaceing my answers
The first method was good and intuitive.
I'm guessing you start by collecting the 4 100%s to give a relative figure of 400. Then you add up the individual percentages, giving a total of 315. Subtract that from 400 and divide by 4? So the minimum percentage would be about 20.125%. That's a rough guess.
EDIT: I see I was wrong, but in my defence, I would argue that your answer is less accurate to the truth. It answers the question of course, What is the lowest possible? But it's probably wise to dismiss that as quickly as you would dismiss the 65% maximum.
amazing explication! i will definitely write all these methods down as they will help me a lot in exams
I did it by making a rough pie chart of each food's did and did not buy percentages and seeing how you could lay them on top of each other such that the overlap of all 4 takes up the least area, whiich gave way to the delightfully simple solution that the answer is just 100% - the individual percentages of people who did not buy each food.
I just took the minimum percentage of 65 and subtracted remainings of others like 20 from 80 20 from other 80 and 10 from 90. Got 15
Solved it through Method 1.
Still trying to wrap my head around the Method 2 venn diagrams, although the Minimal Proof helped it.
The Possible trick? is an intriguing and plausible method.
I think this could be done simpler by skipping numbers and using the typical representation of percentages: a pie chart. Put the donut on the chart, add the pretzel so that the empty areas don't overlap and the answer quickly becomes obvious: We should be selling pies!
I basically used method 1, using a 1-100 number line.
That's the first time I've seen Venn diagrammes made usefull
I hate how you didnt include "minimum" on the thumbnail and i had to be confused by it being literally unsolvable
I solved this problem correctly using a different method, lost confidence in myself assuming I was wrong, and then was pleasantly surprised we got the same answer
@@thomasmaughan4798 Yes of course getting the same answer isn't proof of correctness, but in my statement when I said "solved this problem correctly" it was implied that I did a correct method. I can show you the method I used if you would like and you can evaluate if it is correct?
@@davidcarter8269 It appears I answered a question not precisely asked and i'm still thinking about it. Is it possible to insist that at least 15 percent of all visitors *must have* eaten all four foods? It is easy to show that up to 10 percent may have eaten nothing! But that's the other end of the stick. One end of the stick says from zero to 10 percent may have eaten nothing. But the other end says 15 to 65 percent ate all four.
I guess technically the answer would be somewhere between 15 and 65 percent because you can’t really know for certain with such limited info- but for the sake of the logic problem, yes at least 15 percent ate all 4 foods. Example of how to solve the problem “it’s possible that the 20% that did NOT eat pizza DID eat donuts, so we can remove their 20% from those who ate ALL 4 foods”. 65-20 = 45% then repeat this step in the problem for the for the other two foods and you get 65-20-20-10 = 15 (start with the donut value and subtract the DIFFERENCE between the other percentages of foods eaten and 100%. This tells you the lower limit of AT LEAST what percentage of people MUST have consumed all 4 foods - because that lowest percentile MUST apply to all 4 groups. This will tell you the POSSIBLE percentage of people that MIGHT have eaten the other 3 foods, but not the 4th. It is 50% that might not have eaten 3 or less foods, but not all 4. And so the most accurate answer would be a range of somewhere between 15% and the full 65% because albeit unlikely, you can not rule out that everyone who took a donut also took the other 3 foods as well. Highly unlikely maybe but still likely.
The answer is 15%, because it is the MINIMUM amount of people that have eaten all four foods. Not "How many did".
The third method is essentialy the pigeon hole theory. Which basically is if you have n "pigeons" and m "holes" and n > m, n - m holes is going to have more than 1 pigeon. A very obvious line of thought but nevertheless very useful.
The pigeon hole theory is one of the most useful and unsung parts of logic. Most games and puzzles use it in some way
@@MushookieMan Some applications of PHP just blow my mind when I see them.
i had similar question for a big test that we had in turkiye which determined where i could study in highschool and i couldn't solve it i still can't and i don't think i will ever. my dad told me to do the "imagine if everyone ate 3 of those" technique too
I solved it by eating a crap ton of unhealthy food, then weighing my next bowel movement.
15.
This kind of problem will be very familiar to anyone that has taken a genetics course.
I threw myself off so hard. The maximum percent that ate all four foods was 65%, and I was stuck on that forever, before I realized that I had the max, not the min.
I instinctively did sum mod (100) = 15 so 15% and I thought of putting the 2 foods together in a new category that's the intersection of both!