My first question, before even working with the numbers, was "are small and large the only categories". After 50+ years, I no longer trust that I get complete or even accurate information when I work on a problem.
Feels like the problem was made by someone, and then someone else went back and changed the numbers without thinking to create a "new problem", not realizing the implications of their change. Sort of thinking of math as symbolic manipulation, rather than descriptions of reality, and figuring "if it works with these numbers, it'll work with any numbers, right?"
It's like people designing algebra problems around work rate without thinking about the situation E.g. That one making the rounds on the internet: (Paraphrase) If it takes X number of people to play this symphony, how much time will it take 2X people to play it?
4:16 - Its not that it can't be solved. Its just that the answer doesn't translate to how we normally count dog (whole numbers). We can have half dogs, whether or not they are alive is a different question.
@@thenonsequitur functionally, probably not, but if your joke was over the semantics, why wouldn't you use the same verbiage as the question. Did using the word big help you at all? A medium dog is a large dog if there are only small dogs and medium dogs in the competition. See how that ruins your joke?
@@Notavailable9ag King Solomon is a king recorded in the Bible, who lived about 1015 BC. He's famous for determining the true mother of a certain baby, by threatening to cut it in two. Of the two women, one gave up because she didn't want the child to die, but the other was fine with the compromise - which was a bluff, anyway, and Solomon gave the child to the mother who preferred to surrender it rather than let it die.
Yeah, or to approach it a bit more formally, the solution to the given problem is impossible, so one of our assumptions (that there are only two categories), must be wrong.
It’s easy. There’s 42 small dogs, 6 large dogs, and one wolf that someone signed up as a dog and is hoping no one will notice. Jokes aside though, kudos to the teacher for acknowledging the mistake. I’ve met a couple of teachers that refuse to acknowledge mistakes
I had a textbook that had a mistake in it. The question was correct. The answer provided at the back of the book was wrong. Teacher never told us that this error was in the textbook... but it did show him how many of us did the homework. :D
I saw the thumbnail and thought, "That's easy, just subtract 36 from 49, then divide by half, and that's the number of large dogs." Then I realized it was a non-whole number.
Going off of what that other person said, I’m curious: is there a reason you said “divide by half?” For example, teaching in wherever you live showing the accurate grammar be “divided by half,” instead of “divide in half,” as it is where I live? Or is it just a silly mistake just because it’s easy to make?
If I had to guess, I'd say the teacher probably meant to say there are 49 *large* dogs. This would both make it a problem that would be suitable for a 7-year-old (most 2nd graders aren't learning algebra, they're learning to add two-digit numbers) and give an answer that makes sense.
I doubt it. Firstly a typo of an 8 to a 9 is an easier mistake than leaving out a whole word. Secondly there doesn't seem to be much confusion why this was given to second grade class, just that it is unsolvable. Thirdly the corrected problem is easy to solve with guess and check, which I am pretty sure was something I was doing in second grade.
@@person8064 It is a valid strategy for both solving simple problems like this one and for getting a feel for complex problems you don't yet understand. Several of my later year college classes in astrophysics involved teaching what was essentially advanced guess and check methodology.
@@MyFiddlePlayerJumping to the assumption that some necessary information is missing shouldn't be your immediate reaction when it's a question for 7 year olds.
@@jarrakulNo. When you get to the answer and it's not an integer number of dogs, the conclusion is that there must be a typo in the question, not that some necessary information like a third category of dog has been intentionally left unsaid.
@@gavindeane3670 No one said the information was intentionally left out, just that there is a lack of necessary information needed to solve the problem in a way that makes sense.
In the realm of small dogs and large dogs, the equation is simply not possible. I noticed this before clicking on the video, because it’s not as simple as 49 - 36.
I mean, mathematically there is no mistake, just logically. Mathematically, there is no issue with having half of a small dog and half of a large dog. It's only when you stop thinking of dogs as arbitrary variables that it starts becoming a scene from a horror movie. The more pressing concern is that it's an algebra problem assigned to a second grader still trying to wrap their head around basic arithmetic. This class just went directly from "adding and subtracting two digit numbers" directly to "Substitutions in systems of equations". They probably haven't even learned division yet, and not only does this require that but it has two interconnected variables.
@twobladedswordsandmauls2120 However, if it is a mathematics problem being presented as a word problem, then it must adhere to the bounds of logic, mathematically and linguistically. A word problem represents the practical application of mathematics, and fails in its purpose if it does not. If the conclusion in a raw math problem presents a fundamental error(ex. 2x = 3, x equaling 0), it means you need to check your work again. Same applies if the language of the question causes the answer to defy logic. Half of a Snickers bar is considered valid, but half of a dog is not, and that creates a fundamental error.
@@RoderickEtheria No dogs are being cut in half in this problem. Two half dogs were entered in the competition. They were already half dogs at the start of the problem.
At a minimum there are 36 small dogs. At a maximum there are 6 large dogs. If a second grader was to assume there are only small and large dogs, then they might state the for certain of 36 small dogs.
No there's only 36 small dogs. The phrase states there are 36 more small dogs than large dogs so even if there are 0 large dogs there must only be 36 small dogs as if you add more small dogs you than change aspects of the question given. If you say 42 small dogs that would not match up with the idea that there are 36 more small dogs than large dogs presented in the question as that would make it 'there are 42 more small dogs than large dogs'...there can be other types of dogs like say medium sized dogs (only other size unless you want to say extra large/extra small I guess but I've never heard that terminology before for dog size) that could make up the remaining 13 dogs (49 dogs total, 36 have to be small so there's 13 large or medium dogs). If it was to ask how many large dogs there are than without adding medium dogs into the equation there's be 13, if we throw medium sized dogs in there too there'd be 14 different possible answers though from 0 large dogs to 13 medium dogs to 13 large dogs to 0 medium dogs. The ONLY way there can be more small dogs is if it was phrased as as either 'there are at LEAST 36 more small dogs than large dogs' or 'there's 36 OR more small dogs than large dogs' however it states there IS 36 more small dogs than large which gives us a number that CANNOT be changed without changing the question as I've shown above where if you say there's 42 small dogs that would change it to 42 MORE small dogs than large dogs which obviously is NOT = to '36 MORE small dogs than large dogs.' If anything this is a question that tests reading comprehension to see if you know what the question is actually asking without changing aspects of the original question.
@benjaminmorris4962 Yes, the logic therefore must follow that to say there are 49 total dogs and there are 36 more small dogs than large dogs, there are no large dogs. The 13 other dogs are not accounted for in the original problem posed.
Trick question. One of the large "dogs" was a wolfdog hybrid and one of small "dogs" was CatDog. Thus, there was a 1/2 dog in each category. My actual answer was unsolvable due to the unknown number of medium dogs entered. (M + 2L + 36 = 49)
Close, but actually, one of the entrants came about after a liason between a great dane and the world's most ambitious chihuahua, and the judges just couldn't decide which category this abomination belonged to.
that LINGUISTICALLY is already impossible ... "x pieces more of a than of b" means u have eg. 4 of a, and 4+x of b, with a total of 2*4+x, or 2* any number ... so to get to an odd number, x must also be an odd amount, but 36 ISNT .... if there are 6 large dogs, then u will have 6+36 small dogs i.e 42, which gives a total of 48 dogs. if u have 7 large dogs, u would also have one more small dog, bringing it to 50 signed dogs. u never can have only 49 signed dogs but u can ignore the line if u just want to answer the number of small dogs and say 42 -- which is always the answer to everything either way LOL
There are _two_ half dogs; one half dog in the small set of large dogs and one half dog in the large set of small dogs. So at least the maths isn't inconsistent! If you had 98 dogs in total and 72 difference, that would work fine (13 large and 85 small).
You don’t know there are only two types of dogs competing, you know that there are two types mentioned. If the problem said something like “There are 49 LARGE AND SMALL dogs…”, then you would know. The only two things you know are 49 total dogs and if X is the number of large dogs competing, there are X+36 small dogs competing. Since the problem is poorly defined, the only thing you can deduce is that the number of small dogs competing is between 36 and 42.
“There are 49 LARGE AND SMALL dogs…” implies that every dog is both large and small. It would be more correct to say that “There are 49 dogs, each of which are large or small, but not both large and small…”
@@the_actual_lauren “something like”, it was a quick throwaway line and I hope even a 7 year old would know a dog cannot be both large and small. Now if I had said “There are 49 LARGE AND BLACK dogs” that would be a real problem… Edit: how about There are a total of 49 dogs, either large or small…
One of the large dogs is so huge that he counts as one-and-a-half large ones and one of the small ones is so tiny he‘s basically only half a small dog.
What is half of a dog? CatDog! One fine day with a woof and a purr A baby was born and it caused a little stir No blue buzzard, no three-eyed frog Just a feline canine little CatDog CatDog CatDog Alone in the world was a little CatDog!
The right question is There are 49 dogs signed up to compete in the dog show. There are 39 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?
If we can begin to just assume extra facts not in the problem then it is unsolvable, or at least unfalsifiable. 0 large dogs, 36 small dogs, 7 medium dogs, and 6 giant dogs suddenly becomes an answer that the teacher can't say is wrong.
Well, realistically some very sick individual could have cut both a large and a small sized dogs in half and signed them up for the dog show. Now, they would probably lose the dog show, unless the challenge was which one can keep still the longest, but they could technically compete, as most dog shows don't have regulations mandating that partecipants have to be whole and alive
@@spiderrabbit1556Why do you think there are 36 small dogs? The question doesn't tell us (at least not directly) how many of either size dog there is. But it does tell us that if there are 36 small dogs then there are zero large dogs.
The problem I’m immediately noticing from the thumbnail is that the even-odd parity doesn’t line up at all. If the number of large dogs is odd, then the number of small dogs is ALSO odd, because 36 is even, and odd + even = odd. But, if both the number of small dogs AND big dogs is odd, and odd + odd = even, then the total number of dogs cannot also be odd. The same contradiction arises if you assume the count of big dogs is even, as you get both counts are even, and don’t sum to an odd number. So unless the solution requires dogs missing legs or an unestablished “medium” class of dog, it can’t be solved.
It's literally a mathematics teacher's job to know that. It's a very important part of what they get paid to do, for a living. If someone not only can't solve a mathematical problem but hasn't even noticed it can't be solved, what are they even doing teaching mathematics?
@@june8599Read it again. Nowhere does the question tell you that there are 36 small dogs. What it tells you is that the number of small dogs minus number of large dogs equals 36.
It seemed to me that the original intent was simple subtraction, with a result like this, but the actual math based on how it was worded gets you a half dog. Essentially, a poorly worded math problem.
@@gavindeane3670It's depend on "There are 36 more small dogs than large dogs" Point on: the number Sum 49 ; OR ; the small dogs and large dogs itself Case 1: 36 is more than 36 of 13 for equal to 49 (49-13=36) Case 2: small dogs equal to large dogs ,plus small dog more than large dogs 36 = 6.5 and (6.5+36)42.5 = (49 of 42.5-6.5=36) Case 2 could be a real number, due to : the imagination number and function is not Base number of 49 as a integer, the 6.5 is generated from statement itself. Rational number have to generate the result from the real number of 49. (49 is a Real number ; 36 is a statement )
@@gavindeane3670 Na you're wrong, as for why: There's only 36 small dogs. It outright States there's 36 more small dogs than large dogs and there's 49 dogs in total. The equation im math would be: L (large dogs) + 36 (small dogs) = 49 or if we throw medium dogs into the equation even if they aren't mentioned nor should even be considered if not mentioned in the question especially for a 2nd grade problem: L (large dogs) + M (medium dogs) + 36 (small dogs) = 49 (total number of dogs). The phrase states there are 36 more small dogs than large dogs so even if there are 0 large dogs there must only be 36 small dogs as if you add more small dogs you than change aspects of the question given. If you say 42 small dogs that would not match up with the idea that there are 36 more small dogs than large dogs presented in the question as that would make it 'there are 42 more small dogs than large dogs'...there can be other types of dogs like say medium sized dogs (only other size unless you want to say extra large/extra small I guess but I've never heard that terminology before for dog size) that could make up the remaining 13 dogs (49 dogs total, 36 have to be small so there's 13 large or medium dogs). If it was to ask how many large dogs there are than without adding medium dogs into the equation there's be 13, if we throw medium sized dogs in there too there'd be 14 different possible answers though from 0 large dogs to 13 medium dogs to 13 large dogs to 0 medium dogs. The ONLY way there can be more small dogs is if it was phrased as as either 'there are at LEAST 36 more small dogs than large dogs' or 'there's 36 OR more small dogs than large dogs' however it states there IS 36 more small dogs than large which gives us a number that CANNOT be changed without changing the question as I've shown above where if you say there's 42 small dogs that would change it to 42 MORE small dogs than large dogs which obviously is NOT = to '36 MORE small dogs than large dogs.' If anything this is a question that tests reading comprehension to see if you know what the question is actually asking without changing aspects of the original question.
I’d like to see how creative those smart-alec kids or their parents would have gotten with their answers. “Six big dogs, 42 little dogs, and one hot dog 🌭”
A famous comedian noted that "people from New Jersey 'sawr' things."... "I sawr a dog." "What!? In half!?" "No no. I sawr him. First he was here, then he went away."
7 large and 42 small signed up, one large dog got eaten. Now you have 6 large and 42 small, that is 36 more small dogs than large ones. Initially 49 signed up
@@humanicgamerZ yeah i was thinking the same.If there were 13 big dogs and 36 more small dogs that big then that wound make 49 dogs no? Now i must admit math is not my strong suit and i may just not be getting it lol
@@humanicgamerZNo. The statement "there are 36 more small dogs than large dogs" means: number of small dogs = number of large dogs + 36 That statement doesn't tell us how many small dogs there are. It only tells us the relationship between the number of small dogs and the number of large dogs.
@@darinz136 small dogs and 13 large dogs makes a total of 49 dogs. But in that case you would only have 23 more small dogs than large dogs, so clearly it's not the correct answer.
No. In that case there would only be 23 more small dogs than large dogs, but the question requires 36 more small dogs than large dogs. The question tells us that number of small dogs = number of large dogs+ 36
What if there are even more medium sized dogs. The large dogs number could vary between 1 and 6. The difference would then be filled up with medium sized dogs.
One should expect that for this age group, the question requires simple math so the likely answer is that the word "dog" has been left out of the first sentence. There are 49 *large* dogs signed up which makes this a simple addition question of 49 + 36. A 7 year old is probably in grade 2, you only learn basic multiplication of "counting by" in grade 2 and division in grade 3 so it's not reasonable to expect that would be required to solve this problem.
The problem doesn't say the large and small dogs add up to 49. Only that there are 49 dogs in the contest and 36 more small dogs than large. Therefore, the answer is a range. From 1 to 6 large dogs is correct. All of them allow for there to be 36 more small dogs than large while not having more dogs than 49.
In a real world as half dogs don't exist, it means we have a third category of dogs that are let's say average, neither big nor small. With that new parameter, the number of small dogs must range between 36 and 42 :)
Using the link in the description to click to through to the original article reveals that the teacher themself didn't say exactly what the mistake was, but merely said: "The district worded it wrong. The answer would be 42.5, though, if done at an age appropriate grade.".
@@izeathenoellemain2733 The only way that the number of small dogs can be 36 is if there is a third category (e.g. medium dogs, or enormous dogs). And in that case, ANY number of small dogs between 36 and 42 inclusive is a solution. So 36 is possible, but you can't say it's the answer. The answer is either 42.5 (if you assume large and small are the only categories) or between 36 and 42 inclusive (if you assume we need to have an integer number of small dogs).
@@dominickeijzer5844 No, because 0 + 36 = 36, meaning that if you have 0 large dogs and the rest are medium dogs, it could work, though the question implies large dogs cometing.
Good - as an old geezer who can just about still do algebra, loving mathemtaics channels to keep my mind keen, mainly from memory of school well over 40 years ago, I did think this was wrong. Then I thought maybe I was missing something, but i got the 6.5 in about 2 seconds then spent ages tryng to work out where I had forgotten how to do mathematics. So I am relieved to see I am not losing my mind. Yet.
Mathematics is more important than dogs. The fact that half a dog makes no sense in reality has no effect on the validity of the mathematics problem. It is irksome when people try to bring the logic and reason of reality into mathemtics. The parents who complained were denying their children the experience of the aesthetic beauty of Mathematics just because half a dog doesn't exist in their drab real worlds. In the serene world of mathematics everything exists including fractions of dogs.
I completely agree with your point in general, but we do have some specific context here that is relevant. The fact that half a dog makes no sense in reality does have an effect on the validity of the question given that it is intended to be a question for 7 year olds.
You're missing the fact that the question requires there to be 36 more small dogs than large dogs, but you only have 25 more small dogs than large dogs.
@@yahoiuyaho Ah, I see I misread the question. I thought it said there are X large dogs and 36 more small dogs not 36 more small than large... in that case the answer would be 6 large dogs, 42 small and 1 medium/other sized dog.
Well done for tricking me into watching an almost 5 minute video for what could've been said in about 5 seconds - "the question is bad, as you suspected."
The answer is far simpler. It's a trick question. The answer is 36. You have 49 dogs total. There are 36 more small dogs than large dogs, so you just subtract... if you're trying to get the number of large dogs, which would be 13. But the question ultimately asks how many small dogs are there. 36. It's like the man going to St. Ives riddle as told in Die Hard 3. They throw all this data at you making you multiply 7 by itself recursively a bunch of times. But it starts with "As I was traveling to St. Ives, I met..." and the answer is one, it's only the narrator who is going there. The rest are coming from there. The trick is, the second-graders are expecting a simple subtraction problem (49-36) but it's not asking that.
x a number of big dogs and there is 36 small dogs more (x+36). So total is 2x+36, even number. Only way to make odd number is add any odd number there, so medium dogs? Obwiously question is bad. There is lack of data or given data are wrong.
There's only 36 small dogs. It outright States there's 36 more small dogs than large dogs and there's 49 dogs in total. The equation im math would be: L (large dogs) + 36 (small dogs) = 49 or if we throw medium dogs into the equation even if they aren't mentioned nor should even be considered if not mentioned in the question especially for a 2nd grade problem: L (large dogs) + M (medium dogs) + 36 (small dogs) = 49 (total number of dogs). The phrase states there are 36 more small dogs than large dogs so even if there are 0 large dogs there must only be 36 small dogs as if you add more small dogs you than change aspects of the question given. If you say 42 small dogs that would not match up with the idea that there are 36 more small dogs than large dogs presented in the question as that would make it 'there are 42 more small dogs than large dogs'...there can be other types of dogs like say medium sized dogs (only other size unless you want to say extra large/extra small I guess but I've never heard that terminology before for dog size) that could make up the remaining 13 dogs (49 dogs total, 36 have to be small so there's 13 large or medium dogs). If it was to ask how many large dogs there are than without adding medium dogs into the equation there's be 13, if we throw medium sized dogs in there too there'd be 14 different possible answers though from 0 large dogs to 13 medium dogs to 13 large dogs to 0 medium dogs. The ONLY way there can be more small dogs is if it was phrased as as either 'there are at LEAST 36 more small dogs than large dogs' or 'there's 36 OR more small dogs than large dogs' however it states there IS 36 more small dogs than large which gives us a number that CANNOT be changed without changing the question as I've shown above where if you say there's 42 small dogs that would change it to 42 MORE small dogs than large dogs which obviously is NOT = to '36 MORE small dogs than large dogs.' If anything this is a question that tests reading comprehension to see if you know what the question is actually asking without changing aspects of the original question.
I wanted to watch this video of this problem to see if there was mention of fraction of a dog entry, and I was not disappointed. I remember when this went viral and worked out the problem and based my argument on the fact that the question does not ask for an average but number of whole dog entries and therefore concluded a new size category, "medium", would have been needed to get the correct solutions.
I tried to do this, I tried algebra, other ways of algerbra I tried everything but anwser is still 42.5 😭 and then I watch the vifeo and I realised 😭😭😭
@spikey6617 In human terms I think you could be correct (or, perhaps, not questioned and forgiven) in thinking this way, but mathematicians are not human. They are rational beings with a precise language that they use explain great mysteries for the betterment of us all (so they assure me). In this case, in mathematician-speak, "more than" means the number of small dogs is _equal_ to the number of large dogs _plus_ 36. As the _difference between_ the number of small and large dogs must be 36, 36 cannot also be the answer.
So, here is why it is impossible, by solving via elimination (this 2nd grade problem is possible by algebra, where we cancel out 1 variable [making X and -X, or Y and -y] and then solving for the 2nd variable) X is large dogs and Y is small dogs. We have 2 equations. X+Y=49. (49 total dogs) X+36=y. (36 more small than large) (Minus X from both sides of 2nd equation) X+36=y = 36=-x+y, switch it up so -X+y 36 Make new equations: X+Y=49 -X+Y=36. (The Xs and cancel out bc they have opposite signs, so the Xs are ‘eliminated’) Now, solve by elimination (we don’t have to do the step where we mirror 1 of the variables, that’s done for us) Simplify it. We have this now: 2y=85. Y=42.5. Now we sub it for X. 49-42.5=6.5 So we have 42.5 small dogs and 6.5 large dogs! However, we can’t have half of a dog so no solution
36 small dogs, 0 large dogs and 13 total of not small or large dogs (medium, huge, micro, etc). 37 small dogs, 1 large dogs and 11 total of not small or large dogs (medium, huge, micro, etc). 38 small dogs, 2large dogs and 9 total of not small or large dogs (medium, huge, micro, etc). etc etc until the total reaches 49 dogs.
The explanation is that the competition happened in Springfield, Ohio and a half of a dog was eaten by Haitian immigrants. Just ask THAT candidate. 🤣🤣🤣
Ngl I feel like people who have a problem with this question really just have trouble seeing it as "just a math question". If it were me I'd just write 42.5 and, if I was feeling like making a jab that day, write that someone didn't double check the logic or something. Does the common sense of "you can't have 0.5 dog" really matter? No, it's a math question. You're lucky there's not a guy eating 52 watermelons whole.
It's not just a math question. It's a math question for 7 year olds. Obviously the mathematical answer is 42.5 small dogs. But we would not expect a question for 7 year olds to count dogs in anything other than integers, so we can see there must be a problem with the question.
I've noticed withing 10 seconds that the number of small dogs will not be an integer if the difference is even but the sum is odd (an even difference means both small and big dogs must be even or both odd but then the summ cannot be odd).
One dog is in a sealed box such that nobody can tell whether it's a large or small dog until the superposition is decohered.
A Schrödinger's dog
@@sabunkompas Mrs. Grosinger?
Schrodinger's dog?
schrodingers dog
This is the way
The solution: 6 big dogs, 42 small dogs, and 1 medium-sized dog
My first question, before even working with the numbers, was "are small and large the only categories". After 50+ years, I no longer trust that I get complete or even accurate information when I work on a problem.
or 5 big dogs, 41 small dogs and 3 medium dogs.
That's the first thing I thought of too
@@mrmimeisfunny which is again why the problem is unsolveable. The problem didn't state all of the different groups of dogs.
36 small dogs plus 13 dogs that don't fit into either the small or large categories.
The 0.5 dog might be the infamous updog.
Wassup, dawg?
what's "updog"? XD
its half a dog, split perfectly
.. or indeed Schrödinger's dog. Both dog, and not dog at the same time.
@@realmeme6 Per length or per mass? Symmetrically or asymmetrically?
Feels like the problem was made by someone, and then someone else went back and changed the numbers without thinking to create a "new problem", not realizing the implications of their change. Sort of thinking of math as symbolic manipulation, rather than descriptions of reality, and figuring "if it works with these numbers, it'll work with any numbers, right?"
It is why when setting assignments etc, that the solutions are worked through at the same time.
Could also easily end up with a solution requiring negative large dogs, if you don't sanity check your answer
It's like people designing algebra problems around work rate without thinking about the situation
E.g. That one making the rounds on the internet: (Paraphrase) If it takes X number of people to play this symphony, how much time will it take 2X people to play it?
it could just be theres 43 small dogs and 7 large dogs also and one was registered in both category's accidently double registrations do happen
fair explanation @@violetquinnlaw
4:16 - Its not that it can't be solved. Its just that the answer doesn't translate to how we normally count dog (whole numbers). We can have half dogs, whether or not they are alive is a different question.
The question is bad since you can't enter half a dog into a contest.
and half a large dog is ... small
Can amputees count as half a dog?
Schrödinger's dog
@@neondemon5137 this very much depends on the rules of the contest
36 small dogs, 13 medium dogs, no big dogs
Smaart kitty!
What? Why did you say "big" instead of large?
@@veelo Do "big" and "large" mean different things to you?
@@stephenanderle5422 Smart puppy even!
@@thenonsequitur functionally, probably not, but if your joke was over the semantics, why wouldn't you use the same verbiage as the question. Did using the word big help you at all? A medium dog is a large dog if there are only small dogs and medium dogs in the competition. See how that ruins your joke?
Who left the dog out?
🤣
From the classic Anslem Douglas original song Doggie 👍👍
who, who who?
arff arff arff arff arff
Haha. You win the funny contest.
Who let King Solomon run a dog show?
Underrated comment.
Nice
😂😂😂
I didn't get it
@@Notavailable9ag King Solomon is a king recorded in the Bible, who lived about 1015 BC. He's famous for determining the true mother of a certain baby, by threatening to cut it in two. Of the two women, one gave up because she didn't want the child to die, but the other was fine with the compromise - which was a bluff, anyway, and Solomon gave the child to the mother who preferred to surrender it rather than let it die.
Assuming there are other categories, the possible solutions are:
[36 small, 0 large, 13 other],
[37 small, 1 large, 11 other],
[38 small, 2 large, 9 other],
[39 small, 3 large, 7 other],
[40 small, 4 large, 5 other],
[41 small, 5 large, 3 other] or
[42 small, 6 large, 1 other].
Yeah, or to approach it a bit more formally, the solution to the given problem is impossible, so one of our assumptions (that there are only two categories), must be wrong.
precisely.
Cool
3:53 what is the half of a dog..... "It's dc"
claps slowly
1 confused and frightened cat!
and no platypus!
85 dogs
@@paradiselost9946 ”A PLATYPUS! PERRY The Platypus!”
Maybe there’s 7 large dogs, 43 small dogs and one negative dog.
Negative dog sez: Eh, I'm not so sure about that...
GOD?
wait til imaginary dogs starts barking
@@woozin12345√(-1) dogs
They bite more@@woozin12345
It’s easy. There’s 42 small dogs, 6 large dogs, and one wolf that someone signed up as a dog and is hoping no one will notice. Jokes aside though, kudos to the teacher for acknowledging the mistake. I’ve met a couple of teachers that refuse to acknowledge mistakes
I had a textbook that had a mistake in it. The question was correct. The answer provided at the back of the book was wrong. Teacher never told us that this error was in the textbook... but it did show him how many of us did the homework. :D
Well if they refuse to acknowledge, what was their answer?
Wolf doesn't count...he was in sheep's clothing.
...sure it wasn't Red Riding Hoods Grandma?
@@Shorty_Lickensdon't bother, this won't come in exam😂
0:27 I’m getting 42.5 small dogs. Question sets no specific number of decimal places.
I saw the thumbnail and thought, "That's easy, just subtract 36 from 49, then divide by half, and that's the number of large dogs." Then I realized it was a non-whole number.
Hi, I'm an engineer and I'm ashamed to admit that it took me breaking out pencil and paper to realize the problem
Divide IN half, not "by" half. To divide by half is to multiply by 2.
Going off of what that other person said, I’m curious: is there a reason you said “divide by half?” For example, teaching in wherever you live showing the accurate grammar be “divided by half,” instead of “divide in half,” as it is where I live? Or is it just a silly mistake just because it’s easy to make?
@@Dragoniccat Just a silly mistake. I indeed meant "divide in half", not "divide by half". Dividing by half would be the same as multiplying by 2.
I had to algebra it haha
If I had to guess, I'd say the teacher probably meant to say there are 49 *large* dogs. This would both make it a problem that would be suitable for a 7-year-old (most 2nd graders aren't learning algebra, they're learning to add two-digit numbers) and give an answer that makes sense.
49 large or small dogs. Or the 49 is a typo and it was supposed to be 48.
Yeah, that makes more sense!
I doubt it. Firstly a typo of an 8 to a 9 is an easier mistake than leaving out a whole word. Secondly there doesn't seem to be much confusion why this was given to second grade class, just that it is unsolvable. Thirdly the corrected problem is easy to solve with guess and check, which I am pretty sure was something I was doing in second grade.
@@corvididaecorax2991 if you had to guess and check in second grade, your education failed you
@@person8064
It is a valid strategy for both solving simple problems like this one and for getting a feel for complex problems you don't yet understand. Several of my later year college classes in astrophysics involved teaching what was essentially advanced guess and check methodology.
"THEY'RE CUTTING THE DOGS IN HALF!!!"
And they're eating them
Haha 😂
LMAO
And then they're going to eat me!
Damn soviet cosmonauts and their experiments 😢
i would just encourage my kid to answer "not enough information to say, but it probably involves medium dogs"
That was my immediate reaction when I read the question, because the question does not state how many categories of dogs are in the competition.
@@MyFiddlePlayerJumping to the assumption that some necessary information is missing shouldn't be your immediate reaction when it's a question for 7 year olds.
@@gavindeane3670 I mean it probably should be when the alternative suggests half a small dog and half a large dog.
@@jarrakulNo. When you get to the answer and it's not an integer number of dogs, the conclusion is that there must be a typo in the question, not that some necessary information like a third category of dog has been intentionally left unsaid.
@@gavindeane3670 No one said the information was intentionally left out, just that there is a lack of necessary information needed to solve the problem in a way that makes sense.
1. Examiner sets question; 2. Examiner fails to work question to check that it makes sense.
In the realm of small dogs and large dogs, the equation is simply not possible. I noticed this before clicking on the video, because it’s not as simple as 49 - 36.
I got to 13=2X; then thought I screwed up. glad to find out there was an error in the question.
same here
I'm glad but I'm also pissed. No satisfaction.
I mean, mathematically there is no mistake, just logically.
Mathematically, there is no issue with having half of a small dog and half of a large dog. It's only when you stop thinking of dogs as arbitrary variables that it starts becoming a scene from a horror movie.
The more pressing concern is that it's an algebra problem assigned to a second grader still trying to wrap their head around basic arithmetic.
This class just went directly from "adding and subtracting two digit numbers" directly to "Substitutions in systems of equations". They probably haven't even learned division yet, and not only does this require that but it has two interconnected variables.
Half a dog? I got the same answer...
@twobladedswordsandmauls2120 However, if it is a mathematics problem being presented as a word problem, then it must adhere to the bounds of logic, mathematically and linguistically. A word problem represents the practical application of mathematics, and fails in its purpose if it does not.
If the conclusion in a raw math problem presents a fundamental error(ex. 2x = 3, x equaling 0), it means you need to check your work again. Same applies if the language of the question causes the answer to defy logic.
Half of a Snickers bar is considered valid, but half of a dog is not, and that creates a fundamental error.
Assuming every type of dog is stated in the problem, you'd have to divide one in half.
rest in piece, also chocolate rain guy!!!
You're cutting two in half.
@@RoderickEtheria i was gonna write a comment arguing with this but then i realized no dog is exactly 50% big 50% small
Some dogs stay dry others feel the pain...
@@RoderickEtheria No dogs are being cut in half in this problem. Two half dogs were entered in the competition. They were already half dogs at the start of the problem.
The two owners who had entered their half-dogs into the competition were subsequently arrested for cruelty towards animals.
At a minimum there are 36 small dogs. At a maximum there are 6 large dogs. If a second grader was to assume there are only small and large dogs, then they might state the for certain of 36 small dogs.
No there's only 36 small dogs.
The phrase states there are 36 more small dogs than large dogs so even if there are 0 large dogs there must only be 36 small dogs as if you add more small dogs you than change aspects of the question given. If you say 42 small dogs that would not match up with the idea that there are 36 more small dogs than large dogs presented in the question as that would make it 'there are 42 more small dogs than large dogs'...there can be other types of dogs like say medium sized dogs (only other size unless you want to say extra large/extra small I guess but I've never heard that terminology before for dog size) that could make up the remaining 13 dogs (49 dogs total, 36 have to be small so there's 13 large or medium dogs).
If it was to ask how many large dogs there are than without adding medium dogs into the equation there's be 13, if we throw medium sized dogs in there too there'd be 14 different possible answers though from 0 large dogs to 13 medium dogs to 13 large dogs to 0 medium dogs.
The ONLY way there can be more small dogs is if it was phrased as as either 'there are at LEAST 36 more small dogs than large dogs' or 'there's 36 OR more small dogs than large dogs' however it states there IS 36 more small dogs than large which gives us a number that CANNOT be changed without changing the question as I've shown above where if you say there's 42 small dogs that would change it to 42 MORE small dogs than large dogs which obviously is NOT = to '36 MORE small dogs than large dogs.'
If anything this is a question that tests reading comprehension to see if you know what the question is actually asking without changing aspects of the original question.
@@ifrit1937 Are you ...? 42 small dogs and 6 big dogs would fulfil the premise of "36 more small dogs than big dogs".
@benjaminmorris4962 Yes, the logic therefore must follow that to say there are 49 total dogs and there are 36 more small dogs than large dogs, there are no large dogs. The 13 other dogs are not accounted for in the original problem posed.
@@Delibro True, with 1 other dog.
@benjaminmorris4962 why do you add them together? Small dogs and large dogs are not the same. So the answer is 36 small dogs.
How did the dogs ‘sign up’ to compete in the first place?
0.5 dog is the underdog... Everything works with the solution!
So that would make the other .5 dog the up dog then.
unterhund
@@henriknutsson8500whats updog?
@@henriknutsson8500 What's updog?
Trick question. One of the large "dogs" was a wolfdog hybrid and one of small "dogs" was CatDog. Thus, there was a 1/2 dog in each category.
My actual answer was unsolvable due to the unknown number of medium dogs entered. (M + 2L + 36 = 49)
Close, but actually, one of the entrants came about after a liason between a great dane and the world's most ambitious chihuahua, and the judges just couldn't decide which category this abomination belonged to.
Perhaps the dog show is a hotdog cookout? In which case I have point 5 dogs in my stomach.
that LINGUISTICALLY is already impossible ... "x pieces more of a than of b" means u have eg. 4 of a, and 4+x of b, with a total of 2*4+x, or 2* any number ... so to get to an odd number, x must also be an odd amount, but 36 ISNT .... if there are 6 large dogs, then u will have 6+36 small dogs i.e 42, which gives a total of 48 dogs. if u have 7 large dogs, u would also have one more small dog, bringing it to 50 signed dogs. u never can have only 49 signed dogs but u can ignore the line if u just want to answer the number of small dogs and say 42 -- which is always the answer to everything either way LOL
"In real life, we know that dogs are measured in terms of whole numbers. " Brand new sentence?
what me stumped wasn't the math, but the logical inconsistency of there being half a dog
There are _two_ half dogs; one half dog in the small set of large dogs and one half dog in the large set of small dogs. So at least the maths isn't inconsistent! If you had 98 dogs in total and 72 difference, that would work fine (13 large and 85 small).
How do dogs sign up for a dog show?
asking the real questions here
With an ink pen!
🐾
Minimum 36 small dogs, maximum 42 and we also have at least 1 medium dog.
It's simple
Zero big dogs, 36 small dogs, the other 13 are hot dogs
I was able to figure out the algebra fairly easily and was extremely confused when I got 42.5.
Wrong kind of algebra!
You don’t know there are only two types of dogs competing, you know that there are two types mentioned. If the problem said something like “There are 49 LARGE AND SMALL dogs…”, then you would know.
The only two things you know are 49 total dogs and if X is the number of large dogs competing, there are X+36 small dogs competing.
Since the problem is poorly defined, the only thing you can deduce is that the number of small dogs competing is between 36 and 42.
“There are 49 LARGE AND SMALL dogs…” implies that every dog is both large and small. It would be more correct to say that “There are 49 dogs, each of which are large or small, but not both large and small…”
@@the_actual_lauren “something like”, it was a quick throwaway line and I hope even a 7 year old would know a dog cannot be both large and small.
Now if I had said “There are 49 LARGE AND BLACK dogs” that would be a real problem…
Edit: how about There are a total of 49 dogs, either large or small…
@@the_actual_laurenThis is peak pedanticism 👌
I ended up with the same answer, just a slightly different logic, just because they signed up doesn't mean they compete.
@@sappa66 maybe except all the problem parts say “signed up to compete”, so I think you are out of the competition (you dawg!) 🤣
That "Half a dog" is Sergeant Barksy who lost all of his legs in the war. You need to be more respectful to a injured canine vet like him.
One of the large dogs is so huge that he counts as one-and-a-half large ones and one of the small ones is so tiny he‘s basically only half a small dog.
What is half of a dog? CatDog!
One fine day with a woof and a purr
A baby was born and it caused a little stir
No blue buzzard, no three-eyed frog
Just a feline canine little CatDog
CatDog
CatDog
Alone in the world was a little CatDog!
It's simple, there are 42.5 small dogs😂😂😂
That's what I got and figured the teacher made a mistake or poorly worded something.
Nowadays teacher making questions while they got drunk 😂
😅😅😅
That's what I got too😅
But can you register 0.5 of a dog in the dog show?
42 small dogs, 6 big dogs and a puppy of each size.
It's obvious! 0 small dogs, -36 large dogs, and 49 pictures of dogs!
The right question is
There are 49 dogs signed up to compete in the dog show. There are 39 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?
11?
The answer is given right in the question.
No
Where does it say that only small and large dogs are signed up? What about medium sized dogs?
What about a very large dog?
If we can begin to just assume extra facts not in the problem then it is unsolvable, or at least unfalsifiable. 0 large dogs, 36 small dogs, 7 medium dogs, and 6 giant dogs suddenly becomes an answer that the teacher can't say is wrong.
Medium size does not exist. No dog is exactly the aversge. And besides, dogs are pretty small on average.
@@seisveintiocho-x9e
Google "medium size dog breeds" and you'll find out just how wrong your statement is 😂😂👍🏼
If a bit of information like that isn't given not requested in a question it is to be assumed that it doesn't exist or matter
4:00 answer, One hot dog= 0.5 small dog
Well, realistically some very sick individual could have cut both a large and a small sized dogs in half and signed them up for the dog show.
Now, they would probably lose the dog show, unless the challenge was which one can keep still the longest, but they could technically compete, as most dog shows don't have regulations mandating that partecipants have to be whole and alive
Airbud-type loophole ftw always
Yeah, which rule says half a dog can't play basketball?
May be it is a typo? someone wanted to type 48 but typed 49?
It's been three months. I think it's about time to make this problem viral again.
Total = 49 = Small + Large
Small = Large + 36
Large = ?
49 = Large + Large + 36
13 = Large + Large
Large = 6.5
Small = 6.5 + 36 = 42.5
Ah yes, math.
Well yeah the math is mathing but the dog is probably dead
Large + large What? There are 36 small dogs signed up to compete. The question is how many small dogs are signed up to compete?
@@spiderrabbit1556read it again they literally explained it.
@@spiderrabbit1556 There are 36 MORE small dogs than large dogs competing, genius.
@@spiderrabbit1556Why do you think there are 36 small dogs?
The question doesn't tell us (at least not directly) how many of either size dog there is. But it does tell us that if there are 36 small dogs then there are zero large dogs.
"One dog was stolen and eaten." Donald Trump
My Haitian neighbor cooks it the best.
You forget the "by immigrants" part
eglol No, by illegal immigrants. Please stop conflating legal and illegal immigrants.
@daerdevvyl4314 NOBODY did steal cats and digs and eat them. That is one of Trump's many lies.
@@brotherandrew3393 bro, there are literally videos of it. I'm honestly sorry to tell you. I know you really wanted it to be a lie.
The problem I’m immediately noticing from the thumbnail is that the even-odd parity doesn’t line up at all. If the number of large dogs is odd, then the number of small dogs is ALSO odd, because 36 is even, and odd + even = odd. But, if both the number of small dogs AND big dogs is odd, and odd + odd = even, then the total number of dogs cannot also be odd.
The same contradiction arises if you assume the count of big dogs is even, as you get both counts are even, and don’t sum to an odd number.
So unless the solution requires dogs missing legs or an unestablished “medium” class of dog, it can’t be solved.
I noticed that too.
Trouble is, in 2nd grade most students just accept the bad grade as correct and never mention it.
It's literally a mathematics teacher's job to know that. It's a very important part of what they get paid to do, for a living. If someone not only can't solve a mathematical problem but hasn't even noticed it can't be solved, what are they even doing teaching mathematics?
@@bluerizlagirl I agree❣🤣
Isn't the answer in the question? 36 small, ergo 13 large.
We need 36 more small dogs than large dogs, but you've only got 23 more small dogs than large dogs.
@@june8599Read it again. Nowhere does the question tell you that there are 36 small dogs.
What it tells you is that the number of small dogs minus number of large dogs equals 36.
It seemed to me that the original intent was simple subtraction, with a result like this, but the actual math based on how it was worded gets you a half dog. Essentially, a poorly worded math problem.
@@gavindeane3670It's depend on "There are 36 more small dogs than large dogs"
Point on: the number Sum 49 ; OR ; the small dogs and large dogs itself
Case 1: 36 is more than 36 of 13 for equal to 49 (49-13=36)
Case 2: small dogs equal to large dogs ,plus small dog more than large dogs 36 = 6.5 and (6.5+36)42.5 = (49 of 42.5-6.5=36)
Case 2 could be a real number, due to : the imagination number and function is not Base number of 49 as a integer, the 6.5 is generated from statement itself.
Rational number have to generate the result from the real number of 49. (49 is a Real number ; 36 is a statement )
@@gavindeane3670 Na you're wrong, as for why:
There's only 36 small dogs. It outright States there's 36 more small dogs than large dogs and there's 49 dogs in total. The equation im math would be: L (large dogs) + 36 (small dogs) = 49 or if we throw medium dogs into the equation even if they aren't mentioned nor should even be considered if not mentioned in the question especially for a 2nd grade problem: L (large dogs) + M (medium dogs) + 36 (small dogs) = 49 (total number of dogs).
The phrase states there are 36 more small dogs than large dogs so even if there are 0 large dogs there must only be 36 small dogs as if you add more small dogs you than change aspects of the question given. If you say 42 small dogs that would not match up with the idea that there are 36 more small dogs than large dogs presented in the question as that would make it 'there are 42 more small dogs than large dogs'...there can be other types of dogs like say medium sized dogs (only other size unless you want to say extra large/extra small I guess but I've never heard that terminology before for dog size) that could make up the remaining 13 dogs (49 dogs total, 36 have to be small so there's 13 large or medium dogs).
If it was to ask how many large dogs there are than without adding medium dogs into the equation there's be 13, if we throw medium sized dogs in there too there'd be 14 different possible answers though from 0 large dogs to 13 medium dogs to 13 large dogs to 0 medium dogs.
The ONLY way there can be more small dogs is if it was phrased as as either 'there are at LEAST 36 more small dogs than large dogs' or 'there's 36 OR more small dogs than large dogs' however it states there IS 36 more small dogs than large which gives us a number that CANNOT be changed without changing the question as I've shown above where if you say there's 42 small dogs that would change it to 42 MORE small dogs than large dogs which obviously is NOT = to '36 MORE small dogs than large dogs.'
If anything this is a question that tests reading comprehension to see if you know what the question is actually asking without changing aspects of the original question.
I’d like to see how creative those smart-alec kids or their parents would have gotten with their answers. “Six big dogs, 42 little dogs, and one hot dog 🌭”
Mmm ... 5 large dogs and 44 small dogs, one of which has big dog syndrome.
As soon as I read it, I thought to myself, "2x+36 =49"...that can't work. Isolating x will leave an odd number."
How did the teachers who set this problem not spot this glaring parity error!
Ghost: What has two legs and bleeds?
Also Ghost: half a dog
I just read this comment upon reaching at 3:44 of the video 😂
A famous comedian noted that "people from New Jersey 'sawr' things."...
"I sawr a dog."
"What!? In half!?"
"No no. I sawr him. First he was here, then he went away."
7 large and 42 small signed up, one large dog got eaten. Now you have 6 large and 42 small, that is 36 more small dogs than large ones. Initially 49 signed up
i cant help but notice they just said there are *36* small dogs
Who said that? The question doesn't say that.
@@gavindeane3670 if they said there are in total 49 dogs and there are 36 more small dogs than large dogs does that not mean there are 36 small dogs
@@humanicgamerZ yeah i was thinking the same.If there were 13 big dogs and 36 more small dogs that big then that wound make 49 dogs no? Now i must admit math is not my strong suit and i may just not be getting it lol
@@humanicgamerZNo. The statement "there are 36 more small dogs than large dogs" means:
number of small dogs = number of large dogs + 36
That statement doesn't tell us how many small dogs there are. It only tells us the relationship between the number of small dogs and the number of large dogs.
@@darinz136 small dogs and 13 large dogs makes a total of 49 dogs. But in that case you would only have 23 more small dogs than large dogs, so clearly it's not the correct answer.
Is there not just 36 small dogs and 13 large dogs?
THATS WHAT IM THINKING
Or am i just stupid(i am)
No. In that case there would only be 23 more small dogs than large dogs, but the question requires 36 more small dogs than large dogs. The question tells us that
number of small dogs = number of large dogs+ 36
"0". Dogs cannot sign up.
What if there are even more medium sized dogs. The large dogs number could vary between 1 and 6. The difference would then be filled up with medium sized dogs.
Given how the question is formulated, the analytical solution would be 42.5 dogs, which is impossible. So, for me, there is no solution.
One should expect that for this age group, the question requires simple math so the likely answer is that the word "dog" has been left out of the first sentence. There are 49 *large* dogs signed up which makes this a simple addition question of 49 + 36. A 7 year old is probably in grade 2, you only learn basic multiplication of "counting by" in grade 2 and division in grade 3 so it's not reasonable to expect that would be required to solve this problem.
👍
The problem doesn't say the large and small dogs add up to 49. Only that there are 49 dogs in the contest and 36 more small dogs than large. Therefore, the answer is a range. From 1 to 6 large dogs is correct. All of them allow for there to be 36 more small dogs than large while not having more dogs than 49.
Brilliant!
The real answer is that there are 7 large dogs, 42 small dogs, 1 large wolf-dog, and 1 small wolf-dog.
43 small dogs and 7 large dogs. 1 of the small dogs is a spectator. 😂
It got disqualified
Uh, 7 year olds are not typically doing algebra. 7th graders? Maybe.
In a real world as half dogs don't exist, it means we have a third category of dogs that are let's say average, neither big nor small. With that new parameter, the number of small dogs must range between 36 and 42 :)
It’s a pity you did not say what was the mistake in the problem that was admitted by the teacher.
Using the link in the description to click to through to the original article reveals that the teacher themself didn't say exactly what the mistake was, but merely said: "The district worded it wrong. The answer would be 42.5, though, if done at an age appropriate grade.".
Huh? Why so complicated? The answer is 36 small dogs
No it isn't.
Well, it is if the number of large dogs is zero.
@@gavindeane3670so confused why can”t it be if we take the question as it state technically there would still be 36 more small dogs then large dogs
@@izeathenoellemain2733 The only way that the number of small dogs can be 36 is if there is a third category (e.g. medium dogs, or enormous dogs). And in that case, ANY number of small dogs between 36 and 42 inclusive is a solution.
So 36 is possible, but you can't say it's the answer. The answer is either
42.5 (if you assume large and small are the only categories)
or
between 36 and 42 inclusive (if you assume we need to have an integer number of small dogs).
0:24 my answer is 36 small dogs and 13 medium dogs
Assuming this isn't bait, subtract 13 from 36 to get 23. In your solution there's only 23 more small dogs than large dogs, and 23≠36.
@@dominickeijzer5844 ok
@@dominickeijzer5844 No, because 0 + 36 = 36, meaning that if you have 0 large dogs and the rest are medium dogs, it could work, though the question implies large dogs cometing.
11 medium dogs, 1 rare, 1 well done.
Good - as an old geezer who can just about still do algebra, loving mathemtaics channels to keep my mind keen, mainly from memory of school well over 40 years ago, I did think this was wrong. Then I thought maybe I was missing something, but i got the 6.5 in about 2 seconds then spent ages tryng to work out where I had forgotten how to do mathematics. So I am relieved to see I am not losing my mind. Yet.
Mathematics is more important than dogs. The fact that half a dog makes no sense in reality has no effect on the validity of the mathematics problem. It is irksome when people try to bring the logic and reason of reality into mathemtics. The parents who complained were denying their children the experience of the aesthetic beauty of Mathematics just because half a dog doesn't exist in their drab real worlds. In the serene world of mathematics everything exists including fractions of dogs.
I completely agree with your point in general, but we do have some specific context here that is relevant.
The fact that half a dog makes no sense in reality does have an effect on the validity of the question given that it is intended to be a question for 7 year olds.
Exactly! The dog fractions deserve to feel valid!
There are 36 small dogs competing. And there are 0 big dogs competing. That's it. The rest is just dog. Propably hotdog, we don't know.
Me finding the answer 134 🗿
am I missing something?
12 Large dogs + 37 small dogs = 49
36 MORE small dogs than large dogs. Basically x amount of large dogs + 36 is the amount of small dogs. :p
You're missing the fact that the question requires there to be 36 more small dogs than large dogs, but you only have 25 more small dogs than large dogs.
@@yahoiuyaho Ah, I see I misread the question.
I thought it said there are X large dogs and 36 more small dogs not 36 more small than large...
in that case the answer would be 6 large dogs, 42 small and 1 medium/other sized dog.
Well done for tricking me into watching an almost 5 minute video for what could've been said in about 5 seconds - "the question is bad, as you suspected."
I've watched two of these now and the answer both times was, "There is no rational answer." Ten minutes of my life I'll never get back.🤣
Nowadays teacher making questions while they got drunk 😂
No it was obviously a typo.
The teacher forgot to mention the microwave
The answer is far simpler. It's a trick question. The answer is 36. You have 49 dogs total. There are 36 more small dogs than large dogs, so you just subtract... if you're trying to get the number of large dogs, which would be 13. But the question ultimately asks how many small dogs are there. 36.
It's like the man going to St. Ives riddle as told in Die Hard 3. They throw all this data at you making you multiply 7 by itself recursively a bunch of times. But it starts with "As I was traveling to St. Ives, I met..." and the answer is one, it's only the narrator who is going there. The rest are coming from there. The trick is, the second-graders are expecting a simple subtraction problem (49-36) but it's not asking that.
If the number of small dogs is 36 then the number of large dogs can't be 13. The number of large dogs would have to be zero.
@@gavindeane3670 Compromise: 36 small dogs, 13 medium dogs, & 0 large dogs. Now you're both right.
@@SanchoSanchoSancho That works. Anything between 36 and 42 small dogs works.
x a number of big dogs and there is 36 small dogs more (x+36). So total is 2x+36, even number. Only way to make odd number is add any odd number there, so medium dogs? Obwiously question is bad. There is lack of data or given data are wrong.
The question is wrong. And no, theres no such thing as medium dogs. You have small and large.
@@Shorty_Lickens I called it like that, but you know what I meant. Another competition class not mentioned in question (lack of data).
There's only 36 small dogs. It outright States there's 36 more small dogs than large dogs and there's 49 dogs in total. The equation im math would be: L (large dogs) + 36 (small dogs) = 49 or if we throw medium dogs into the equation even if they aren't mentioned nor should even be considered if not mentioned in the question especially for a 2nd grade problem: L (large dogs) + M (medium dogs) + 36 (small dogs) = 49 (total number of dogs).
The phrase states there are 36 more small dogs than large dogs so even if there are 0 large dogs there must only be 36 small dogs as if you add more small dogs you than change aspects of the question given. If you say 42 small dogs that would not match up with the idea that there are 36 more small dogs than large dogs presented in the question as that would make it 'there are 42 more small dogs than large dogs'...there can be other types of dogs like say medium sized dogs (only other size unless you want to say extra large/extra small I guess but I've never heard that terminology before for dog size) that could make up the remaining 13 dogs (49 dogs total, 36 have to be small so there's 13 large or medium dogs).
If it was to ask how many large dogs there are than without adding medium dogs into the equation there's be 13, if we throw medium sized dogs in there too there'd be 14 different possible answers though from 0 large dogs to 13 medium dogs to 13 large dogs to 0 medium dogs.
The ONLY way there can be more small dogs is if it was phrased as as either 'there are at LEAST 36 more small dogs than large dogs' or 'there's 36 OR more small dogs than large dogs' however it states there IS 36 more small dogs than large which gives us a number that CANNOT be changed without changing the question as I've shown above where if you say there's 42 small dogs that would change it to 42 MORE small dogs than large dogs which obviously is NOT = to '36 MORE small dogs than large dogs.'
If anything this is a question that tests reading comprehension to see if you know what the question is actually asking without changing aspects of the original question.
I wanted to watch this video of this problem to see if there was mention of fraction of a dog entry, and I was not disappointed. I remember when this went viral and worked out the problem and based my argument on the fact that the question does not ask for an average but number of whole dog entries and therefore concluded a new size category, "medium", would have been needed to get the correct solutions.
Somehow, i ended up with 6.5 large dogs and 42.5 small dogs
Wow I was right
I tried to do this, I tried algebra, other ways of algerbra I tried everything but anwser is still 42.5 😭 and then I watch the vifeo and I realised 😭😭😭
Dude this is brainwash it's so simple
It's not 42.5??????? It's obviously 85 they don't even understand the question
85? What are you on about?
@@slaw2468 bro just added 49 and 36
@@chaosking313 i would advise reading the question again and realizing that having 85 small dogs out of the 49 dogs probably wouldn’t make sense
Sooo 42.5 (i guessed 48.5 before because i can't do math)
36 small dogs
@@Smeegheed1963 That's true as long as you ignore the concept of "more".
@@taotoo2 I knew I'd missed something important🤭
36 small dogs, 0 large dogs, and 13 medium-sized dogs.
@spikey6617 In human terms I think you could be correct (or, perhaps, not questioned and forgiven) in thinking this way, but mathematicians are not human. They are rational beings with a precise language that they use explain great mysteries for the betterment of us all (so they assure me). In this case, in mathematician-speak, "more than" means the number of small dogs is _equal_ to the number of large dogs _plus_ 36. As the _difference between_ the number of small and large dogs must be 36, 36 cannot also be the answer.
@@FlashBIOS 36 _IS_ the difference between 0 and 36.
So, here is why it is impossible, by solving via elimination (this 2nd grade problem is possible by algebra, where we cancel out 1 variable [making X and -X, or Y and -y] and then solving for the 2nd variable)
X is large dogs and Y is small dogs.
We have 2 equations.
X+Y=49. (49 total dogs)
X+36=y. (36 more small than large)
(Minus X from both sides of 2nd equation)
X+36=y =
36=-x+y, switch it up so -X+y 36
Make new equations:
X+Y=49
-X+Y=36. (The Xs and cancel out bc they have opposite signs, so the Xs are ‘eliminated’)
Now, solve by elimination (we don’t have to do the step where we mirror 1 of the variables, that’s done for us)
Simplify it.
We have this now: 2y=85.
Y=42.5.
Now we sub it for X. 49-42.5=6.5
So we have 42.5 small dogs and 6.5 large dogs!
However, we can’t have half of a dog so no solution
36 small dogs, 0 large dogs and 13 total of not small or large dogs (medium, huge, micro, etc). 37 small dogs, 1 large dogs and 11 total of not small or large dogs (medium, huge, micro, etc). 38 small dogs, 2large dogs and 9 total of not small or large dogs (medium, huge, micro, etc). etc etc until the total reaches 49 dogs.
7 large dogs, but one of them was a golden retriever who just didn't feel like ending his nap prematurely, so they never showed up to the competition.
Then question specifies that its the number of dogs signed up, not the number that actually participated.
The explanation is that the competition happened in Springfield, Ohio and a half of a dog was eaten by Haitian immigrants. Just ask THAT candidate. 🤣🤣🤣
The Haitian didn't know if he wanted to eat a small dog or a big dog, so he ate half of each.
What if theres a medium sized dog?
The dogs were hot. Hotdogs. A show with large and small sausage pies... 😅
I'm more impressed that the dogs were able to "sign up" to enter a dog show!
There had to be one medium or other size dog to make the numbers work. ☮️🇺🇸
Are you running out of problems, Presh? This was uncharacteristically uninteresting.
Ngl I feel like people who have a problem with this question really just have trouble seeing it as "just a math question". If it were me I'd just write 42.5 and, if I was feeling like making a jab that day, write that someone didn't double check the logic or something. Does the common sense of "you can't have 0.5 dog" really matter? No, it's a math question. You're lucky there's not a guy eating 52 watermelons whole.
It's not just a math question. It's a math question for 7 year olds.
Obviously the mathematical answer is 42.5 small dogs. But we would not expect a question for 7 year olds to count dogs in anything other than integers, so we can see there must be a problem with the question.
I've noticed withing 10 seconds that the number of small dogs will not be an integer if the difference is even but the sum is odd (an even difference means both small and big dogs must be even or both odd but then the summ cannot be odd).
"a dog is a dog. You can't say it's only a half"