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?"
@@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?
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.
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.
@@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.
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.
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
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)
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?
@@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.
@0:53 “We know that there are two types of dogs: There are large dogs and there are small dogs.” How do we know that?? There’s nothing in the problem statement indicating those are the only sizes involved. Actually, we have direct evidence within the problem that those are NOT the only two sizes involved. If those were the only sizes involved, the problem would be solvable. The fact that it’s not solvable as written with only two sizes means there MUST be at least one more category of dog involved in the problem.
Well typically there is. Typically it's reasonable to assume that a maths question of this type, especially if given to school children, contains both the necessary and sufficient information to answer it. If it mentions 2 categories that it's going to be a problem in 2 variables, and further that there's usually as much information as you need to form 2 equations using those 2 variables. Typically the purpose for asking this type of question is to ascertain mathematical understanding (usually of previously taught course work - maybe even work taught the same day the homework was set - this is a big clue for students) and the question wasn't set to test the ability of school kids to be pedantic, creative or a comedian. e.g if a question asks what does someone have if they have 10 apples in one hand and 5 in the other isn't looking for "big hands" as the answer. There are, undoubtedly courses that have questions where these latter skills would come into play, but if you want your kid to pass maths teach him or her to be smart rather than a smart ass. That said rigour is an important skill, especially the higher level you study. So sometimes it does pay to adopt a more pedantic reading of a problem. In computer science too, asking the pedantic questions when you're given a set of requirements may avoid holes in the computer system that's developed from these requirements. But the kid who sits there saying "it didn't say this...it didn't say that...why can't you divide by zero?" all the time is just going to fail. For a question like this, the most likely thing is that a simple mistake has been made because there's no integer solution when considering only large and small dogs. If you questioned this and you were told the question is worded correctly, then perhaps you can start to make assumptions about extra categories of dogs and ranges of solutions for the number of small dogs. As you can see it turned out that the question setter had made a simple mistake.
No, you have to assume things in every question, in every statement, in every situation in life in general. If you assume there are medium sized dogs, you could also assume that there is a large asteroid destroying the earth at that moment, so the result would be 0 dogs. So, its pointless with assumptions beyond the stated ones.
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.
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
Maybe the purpose of the question is to help educate younger people on being able to read context and be able to recognize when a problem has no logical answer.
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…
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).
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.
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.
@@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.
The answer is 42 small dogs 6 big dogs And 1 big dog named "Small" That's why the question used "more small" instead of "smaller", it's not a typo or grammatical error, it's intentional
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.
Answer is simple: If there was a total of 49 dogs that signed up to compete and 36 more of what x is we could do 49-36 to find the large dogs. that is 13 so if we add 36 small dogs to the total dog count, then that means that there were 36 small dogs. This solution checks.
Sorry, you are not correct. Watch the video to find out why. The Word Problem does not have a solution because it demands something that is not possible.
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
Apart from the impossibility of half a dog, surely a simpler way to solve the problem is this. 49-36=13. 13÷2=6.5. 36+6.5=42.5. Ergo there are 42.5 small dogs and 6.5 large dogs.
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.
The question is not mistaken, if big/small is subjective, and therefore cannot be the criteria to judge big/small. So, one of the dogs was used as the standard to judge the other dogs. So 42 dogs smaller and 6 dogs bigger than this dog.
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.
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).
One dog might GROW during competition because that small dog either eat another dog! Or one dog was pregnant with a small dog, what do I know!?! I jusg hate this kind of ambigius math problem. Compete in what the dogshowe?!?
Technically, the answer could be any whole number from 36 to 42. There could be no large dogs, and there could be dogs that are neither considered small nor large dogs (medium dogs). So it does have an answer (multiple, in fact), but ultimately, the question is very poorly worded and has an unrealistic answer given normal assumptions.
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.
@@ifrit1937 36 + 0 + 13 = 49 (13 dogs of neither small nor large category) 37 + 1 + 11 = 49 (11 dogs of neither small nor large category) 38 + 2 + 9 = 49 (9 dogs of neither small nor large category) 39 + 3 + 7 = 49 (7 dogs of neither small nor large category) 40 + 4 + 5 = 49 (5 dogs of neither small nor large category) 41 + 5 + 3 = 49 (3 dogs of neither small nor large category) 42 + 6 + 1 = 49 (1 dog of neither small nor large category) All of these are valid answers given the question. Every last one, if you actually read the question. I'd say reading comprehension is something you need to practice yourself.
As long as there were at least two dogs of different sizes, and you lined them all up in order of size, there would definitely be a smallest dog and a largest dog. So logically, unless there are fewer than two dogs or all the dogs are exactly the same size, there must be at least one dog in the show that could be defined as "large".
Once when we were out walking our 4-pound Maltese, a neighbor saw us and wisecracked "What's the matter? Couldn't you afford a whole dog?" So, obviously, the answer is that each small dog counts as half a dog. 😀 Seriously, though, we should reflect on the societal implications when authority figures (whether teachers, textbook writers, or others) can't even provide valid material to train and assess seven-year-olds. And those folks are allowed to vote???
May we have a picture of the 49 dogs ? Something is a little hinky. According westminster kennel club dog show rules. The smallest dog counts as #0 not #1. Or first dog. It is how the questiion originated. That dog show used to be for Blue Bloods only. Thus a decision was made on a particular entrant. The entrant was number zero
0:52 Why do we know there are two types of dogs? The question only refers to small and large dogs, but there could also be mid-sized dogs. OK, I'll admit that this possibility is probably not the case in a question for 7-year-olds.
Before watching the video: There is clearly a mistake in the text of the question. If the sum of two integer numbers is odd (for example, 49) then their difference must also be odd, so it cannot be 36.
Exactly so; and it is shocking that anyone calling themself a teacher of mathematics would give such a problem to students without having a clear idea in their own head how to solve it using the techniques they have already taught their class.
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 😭😭😭
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.
Solution: 37 small dogs. There is also 1 large dog, 3 medium-sized dogs, 2 medium-small sized dogs, 2 big dogs, 2 tiny dogs, and 2 medium-large sized dogs.
Viral maths problems often enough actually have solutions, so people argue about them. As to how blatantly impossible questions gain traction instead of just being dismissed: Don't overestimate people's mathematical knowledge. This is your yearly reminder to reread the "every odd number has an E in it" Tumblr thread.
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 answer is: {36, 37, 38 .. 42} small dogs because there can be from 0-6 large dogs and 1-13 medium sized dogs... There is nowhere stated that all dogs must be either small or large....
Yup, everyone's overthinking it. Honestly never should've been in a math textbook. 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.
@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.
Before watching: x+(x+36)=49... 2x=49-36... x=13/2=6.5 (guess theres a medium dog thats half large half small.) so either 43 if the medium is grouped with the small, or 42 if they arent.
Two pipes A and B can fill a tank in 24 hours and 36 hours respectively. if both pipes are opened together find when pipe B must be turned off so that the tank just filled in 12 hours. A. 18 hours B. 15 hours C. 16 hours D. 20 hours Question from SOF IMO (International Mathematics Olympiad) question paper class 9.
Pausing at 2:00 to have a go myself: Assuming only small (S) and large (L) dogs S + L = 49 and L = S - 36 S + S -36 = 49 2*S = 85 S = 42.5 So there are 6.5 large dogs and 42.5 small dogs... Fractional dogs, makes perfect sense Assuming an unmentioned 3rd class of medium dogs S + M + L = 49 S + M + S - 36 = 49 2 S + M = 85 I guess we should assume all classes have positive integers of dogs (possibly including 0, but we already know that if M = 0 then the other two classes can't be integers. So S >= 36 Let's try with S = 36 72 + M = 85 M = 13 So the values for S + M + L is one of: 36 + 13 + 0 37 + 11 + 1 38 + 9 + 2 39 + 7 + 3 40 + 5 + 4 41 + 3 + 5 42 + 1 + 6 Any larger value for S and there has to be negative medium dogs. 43 - 1 + 7 Feels a little too advanced to have to assume an unmentioned third category; thus making it an equation with 2 unknowns and have the answer be one of 7 possible solutions. (Also if we can assume one extra category, why not 4 total categories?) Feels like theres some definite typos in the question.
Also if I had gotten this question as a young undiagnosed ADHD'er I would have just answered 36 ... or possibly 49 (depending on whether I didn't notice the "more" or the second "small" in the question). I know I would because my medicated 42 year old mind went: "Easy, it's 36 small dogs, says right there... or did they mean in total? Easy, 49, says right there! .....hang on, that's _too_ easy... doesn't math questions usually involve some kind of maths?" But I know that without Ritalin; I'd probably just go with either one of my first thoughts and carried on. Unless I had very good reason to be suspicious. (Roughly translated into English, because I don't think in language; my thoughts are more abstract and non linear than language, but it's easier to explain when I translate it into words)
The problem states that there are 36 more small dogs than large dogs. Not that there are 36 small dogs. 36 more small dogs than large dogs is represented as s = l + 36
I treated this as more of a logic problem than a math problem. As the final tally was odd, and no way to get that with two categories in question, the answer is that there was at least one other category than small dogs and large dogs (for example, medium dogs). So, we know there are at least 36 small dogs, and no indication that there are any large dogs at all. But assuming there are potentially some large dogs, you could have as many as 6 large dogs, which would make 42 small dogs, and one "other".
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.
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 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
Who left the dog out?
🤣
From the classic Anslem Douglas original song Doggie 👍👍
who, who who?
arff arff arff arff arff
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?
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.
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
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.
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
"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 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.
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😂
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.
1 confused and frightened cat!
and no platypus!
85 dogs
@@paradiselost9946 ”A PLATYPUS! PERRY The Platypus!”
3:53 what is the half of a dog..... "It's dc"
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)
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
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.
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?
@0:53 “We know that there are two types of dogs: There are large dogs and there are small dogs.” How do we know that?? There’s nothing in the problem statement indicating those are the only sizes involved. Actually, we have direct evidence within the problem that those are NOT the only two sizes involved. If those were the only sizes involved, the problem would be solvable. The fact that it’s not solvable as written with only two sizes means there MUST be at least one more category of dog involved in the problem.
assuming
assuming, we're not gonna go into reading comprehension
Well typically there is. Typically it's reasonable to assume that a maths question of this type, especially if given to school children, contains both the necessary and sufficient information to answer it. If it mentions 2 categories that it's going to be a problem in 2 variables, and further that there's usually as much information as you need to form 2 equations using those 2 variables.
Typically the purpose for asking this type of question is to ascertain mathematical understanding (usually of previously taught course work - maybe even work taught the same day the homework was set - this is a big clue for students) and the question wasn't set to test the ability of school kids to be pedantic, creative or a comedian. e.g if a question asks what does someone have if they have 10 apples in one hand and 5 in the other isn't looking for "big hands" as the answer. There are, undoubtedly courses that have questions where these latter skills would come into play, but if you want your kid to pass maths teach him or her to be smart rather than a smart ass.
That said rigour is an important skill, especially the higher level you study. So sometimes it does pay to adopt a more pedantic reading of a problem. In computer science too, asking the pedantic questions when you're given a set of requirements may avoid holes in the computer system that's developed from these requirements. But the kid who sits there saying "it didn't say this...it didn't say that...why can't you divide by zero?" all the time is just going to fail.
For a question like this, the most likely thing is that a simple mistake has been made because there's no integer solution when considering only large and small dogs. If you questioned this and you were told the question is worded correctly, then perhaps you can start to make assumptions about extra categories of dogs and ranges of solutions for the number of small dogs. As you can see it turned out that the question setter had made a simple mistake.
@@cakeplayz1476😂
No, you have to assume things in every question, in every statement, in every situation in life in general. If you assume there are medium sized dogs, you could also assume that there is a large asteroid destroying the earth at that moment, so the result would be 0 dogs. So, its pointless with assumptions beyond the stated ones.
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.
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
I was able to figure out the algebra fairly easily and was extremely confused when I got 42.5.
Wrong kind of algebra!
Maybe the purpose of the question is to help educate younger people on being able to read context and be able to recognize when a problem has no logical answer.
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.
42 small dogs, 6 big dogs and a puppy of each size.
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!) 🤣
"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).
"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.
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?
0:27 I’m getting 42.5 small dogs. Question sets no specific number of decimal places.
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
The answer is given right in the question.
No
How do dogs sign up for a dog show?
asking the real questions here
With an ink pen!
🐾
Mmm ... 5 large dogs and 44 small dogs, one of which has big dog syndrome.
Perhaps the dog show is a hotdog cookout? In which case I have point 5 dogs in my stomach.
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.
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.
There was one guy that identifies as a dog too, but nobody had the balls to disqualify him for fear of the legal repercussions.
He was a furry.
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.
4:00 answer, One hot dog= 0.5 small dog
The answer is
42 small dogs
6 big dogs
And 1 big dog named "Small"
That's why the question used "more small" instead of "smaller", it's not a typo or grammatical error, it's intentional
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!
Answer is simple: If there was a total of 49 dogs that signed up to compete and 36 more of what x is we could do 49-36 to find the large dogs. that is 13 so if we add 36 small dogs to the total dog count, then that means that there were 36 small dogs. This solution checks.
Sorry, you are not correct. Watch the video to find out why. The Word Problem does not have a solution because it demands something that is not possible.
@@CD-vb9fi I already did
Nvm, there is no solution because there is no amount of medium dogs.
@@Asianboy-Productions well... it's too ambiguous. Under such an assumption that can be 0 large dogs too. But that would be blatant cheating.
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
Apart from the impossibility of half a dog, surely a simpler way to solve the problem is this. 49-36=13. 13÷2=6.5. 36+6.5=42.5. Ergo there are 42.5 small dogs and 6.5 large dogs.
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.
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.
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!
The question is not mistaken, if big/small is subjective, and therefore cannot be the criteria to judge big/small. So, one of the dogs was used as the standard to judge the other dogs. So 42 dogs smaller and 6 dogs bigger than this dog.
It's obvious! 0 small dogs, -36 large dogs, and 49 pictures of dogs!
1. Examiner sets question; 2. Examiner fails to work question to check that it makes sense.
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.
The real answer is that there are 7 large dogs, 42 small dogs, 1 large wolf-dog, and 1 small wolf-dog.
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.".
I'm more impressed that the dogs were able to "sign up" to enter a dog show!
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).
This is not a math problem. This is wording the question correctly problem
43 small dogs and 7 large dogs. 1 of the small dogs is a spectator. 😂
It got disqualified
One dog might GROW during competition because that small dog either eat another dog!
Or one dog was pregnant with a small dog, what do I know!?!
I jusg hate this kind of ambigius math problem.
Compete in what the dogshowe?!?
Technically, the answer could be any whole number from 36 to 42. There could be no large dogs, and there could be dogs that are neither considered small nor large dogs (medium dogs). So it does have an answer (multiple, in fact), but ultimately, the question is very poorly worded and has an unrealistic answer given normal assumptions.
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.
@@ifrit1937
36 + 0 + 13 = 49 (13 dogs of neither small nor large category)
37 + 1 + 11 = 49 (11 dogs of neither small nor large category)
38 + 2 + 9 = 49 (9 dogs of neither small nor large category)
39 + 3 + 7 = 49 (7 dogs of neither small nor large category)
40 + 4 + 5 = 49 (5 dogs of neither small nor large category)
41 + 5 + 3 = 49 (3 dogs of neither small nor large category)
42 + 6 + 1 = 49 (1 dog of neither small nor large category)
All of these are valid answers given the question. Every last one, if you actually read the question. I'd say reading comprehension is something you need to practice yourself.
As long as there were at least two dogs of different sizes, and you lined them all up in order of size, there would definitely be a smallest dog and a largest dog. So logically, unless there are fewer than two dogs or all the dogs are exactly the same size, there must be at least one dog in the show that could be defined as "large".
Once when we were out walking our 4-pound Maltese, a neighbor saw us and wisecracked "What's the matter? Couldn't you afford a whole dog?" So, obviously, the answer is that each small dog counts as half a dog. 😀 Seriously, though, we should reflect on the societal implications when authority figures (whether teachers, textbook writers, or others) can't even provide valid material to train and assess seven-year-olds. And those folks are allowed to vote???
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 😂
One of them is a prairie dog
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
41 small dogs, 6 large dogs and 1 size fluid dog that can identify as anything it wants.
May we have a picture of the 49 dogs ? Something is a little hinky.
According westminster kennel club dog show rules. The smallest dog counts as #0 not #1. Or first dog.
It is how the questiion originated.
That dog show used to be for Blue Bloods only. Thus a decision was made on a particular entrant. The entrant was number zero
0:52 Why do we know there are two types of dogs? The question only refers to small and large dogs, but there could also be mid-sized dogs.
OK, I'll admit that this possibility is probably not the case in a question for 7-year-olds.
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.
Before watching the video: There is clearly a mistake in the text of the question. If the sum of two integer numbers is odd (for example, 49) then their difference must also be odd, so it cannot be 36.
Exactly so; and it is shocking that anyone calling themself a teacher of mathematics would give such a problem to students without having a clear idea in their own head how to solve it using the techniques they have already taught their class.
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
Easy. 36. 36 Small dogs, 0 large dogs, 13 Medium dogs = 49 dogs.
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.
Solution: 37 small dogs. There is also 1 large dog, 3 medium-sized dogs, 2 medium-small sized dogs, 2 big dogs, 2 tiny dogs, and 2 medium-large sized dogs.
No iguanas?
Nowadays teacher making questions while they got drunk 😂
No it was obviously a typo.
The teacher forgot to mention the microwave
Viral maths problems often enough actually have solutions, so people argue about them. As to how blatantly impossible questions gain traction instead of just being dismissed: Don't overestimate people's mathematical knowledge.
This is your yearly reminder to reread the "every odd number has an E in it" Tumblr thread.
Sooo 42.5 (i guessed 48.5 before because i can't do math)
The solution: 6 large dogs, 42 small dogs, and 1 dog that's half small, half large
0 dogs. The Springfield, Ohio residents ate them.
Total = 36+ 2*L (one L for no of large dogs) and one L needs to add to 36 because smalldogs=36+large dogs)
49 = even number (but it is not possible)..
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 answer is: {36, 37, 38 .. 42} small dogs because there can be from 0-6 large dogs and 1-13 medium sized dogs... There is nowhere stated that all dogs must be either small or large....
There had to be one medium or other size dog to make the numbers work. ☮️🇺🇸
This is not even a math question. It is just a straight up thought question.!😎
Yup, everyone's overthinking it. Honestly never should've been in a math textbook.
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.
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.
Before watching: x+(x+36)=49... 2x=49-36... x=13/2=6.5 (guess theres a medium dog thats half large half small.) so either 43 if the medium is grouped with the small, or 42 if they arent.
Me finding the answer 134 🗿
Two pipes A and B can fill a tank in 24 hours and 36 hours respectively. if both pipes are opened together find when pipe B must be turned off so that the tank just filled in 12 hours.
A. 18 hours
B. 15 hours
C. 16 hours
D. 20 hours
Question from SOF IMO (International Mathematics Olympiad) question paper class 9.
Are you running out of problems, Presh? This was uncharacteristically uninteresting.
Pausing at 2:00 to have a go myself:
Assuming only small (S) and large (L) dogs
S + L = 49 and L = S - 36
S + S -36 = 49
2*S = 85
S = 42.5
So there are 6.5 large dogs and 42.5 small dogs... Fractional dogs, makes perfect sense
Assuming an unmentioned 3rd class of medium dogs
S + M + L = 49
S + M + S - 36 = 49
2 S + M = 85
I guess we should assume all classes have positive integers of dogs (possibly including 0, but we already know that if M = 0 then the other two classes can't be integers.
So S >= 36
Let's try with S = 36
72 + M = 85
M = 13
So the values for S + M + L is one of:
36 + 13 + 0
37 + 11 + 1
38 + 9 + 2
39 + 7 + 3
40 + 5 + 4
41 + 3 + 5
42 + 1 + 6
Any larger value for S and there has to be negative medium dogs.
43 - 1 + 7
Feels a little too advanced to have to assume an unmentioned third category; thus making it an equation with 2 unknowns and have the answer be one of 7 possible solutions. (Also if we can assume one extra category, why not 4 total categories?) Feels like theres some definite typos in the question.
Also if I had gotten this question as a young undiagnosed ADHD'er I would have just answered 36 ... or possibly 49 (depending on whether I didn't notice the "more" or the second "small" in the question). I know I would because my medicated 42 year old mind went: "Easy, it's 36 small dogs, says right there... or did they mean in total? Easy, 49, says right there! .....hang on, that's _too_ easy... doesn't math questions usually involve some kind of maths?" But I know that without Ritalin; I'd probably just go with either one of my first thoughts and carried on. Unless I had very good reason to be suspicious.
(Roughly translated into English, because I don't think in language; my thoughts are more abstract and non linear than language, but it's easier to explain when I translate it into words)
What if theres a medium sized dog?
The puzzle does not exclude other classes of dogs. Let's just say medium sized dogs. Then the number of dogs could be in the range:
1 = 1
37
How do you get s equal l plus thirty six? Not so. L should total dogs. L equals thirty six plus x. X equals thirteen!
Thirty six small dogs!
S plus L equals forty nine!
The problem states that there are 36 more small dogs than large dogs. Not that there are 36 small dogs.
36 more small dogs than large dogs is represented as s = l + 36
The problem makes sense only if it is possible to divide a big dog into half a big dog and half a small dog.
With the thumbnail I would say 6 or 7 big dogs, and therefore 42 and 43 small dogs. But total dogs I'm getting is 48 and 50.
I treated this as more of a logic problem than a math problem. As the final tally was odd, and no way to get that with two categories in question, the answer is that there was at least one other category than small dogs and large dogs (for example, medium dogs).
So, we know there are at least 36 small dogs, and no indication that there are any large dogs at all. But assuming there are potentially some large dogs, you could have as many as 6 large dogs, which would make 42 small dogs, and one "other".