Simple math questions stumping the internet

[@Weerknuffelbeer]

This half-dog problem reminds me of another math joke I once heard... "Two mathematicians are taking a walk and see a house. They see 1 person entering the house and 2 persons leaving the house. Then one of the mathematicians says to the other: "Now 1 more person needs to enter the house and it's empty again.""

[@donaldbiden7927]

Lol but that’s wrong obviously. If one person enters, the net change will become 0 lol.

[@shoobadoo123]

I don’t understand can you explain

[@hendrikusscherphof7348]

@@shoobadoo123 They see 1 person going into the house, and 2 out. That leaves 2-1 = -1 people in the house. So if one more person walks in it will be empty (0 people) again. I guess the joke is that mathematicians like to ignore reality if it makes the math easier sometimes

[@shoobadoo123]

@@hendrikusscherphof7348 I’m assuming you meant 1-2 = -1 If another person walks in, the net change would be 0, but that doesn’t mean there would be zero people. 2 people coming out of the house: 2-0 = 0 1 person coming in: 0+1 = 1 1 more person enters: 1+1 = 2 And that would be the proper arithmetic given the joke. To me it sounds like the mathematicians are just bad at math haha

[@hendrikusscherphof7348]

@@shoobadoo123 oh yeah, that was a typo. It's a rather silly joke, I agree

[@jeffbrunton3291]

1) no mathematical answer 2) easy mathematically, not so easy for the dog in the middle

[@brucetucker4847]

Well, two dogs, discarding half of each one. Because half a large dog is not half a small dog, and 1/4 of a large dog isn't half a small dog either.

[@todabsolute]

@densch123 furry

[@joeybulford5266]

Maybe one dog just competed twice and crushed all the competition

[@lukesmith684]

Well for the first it does have an answer it's just a range as opposed to a single value

[@BlacksmithTWD]

@@brucetucker4847 Half a large dog is half a dog, and half a small dog is also half a dog, no problem there, I just wonder why the owners of those two halve dogs think they stand a chance in the competition.

[@sutfolsemaj]

Maybe there were 6 large dogs, 42 small dogs, and a Frankendog with the torso of a german sheppard and the head and legs of a chihuahua.

[@meschoui760]

Take 49 biscuits. Then you have 36 more small biscuits than big biscuits. How many small biscuits you have? Answer is 36.

[@jshariff786]

Melch1Ball: Someone is not too bright. If I have 36 small biscuits (the answer you gave), and 49 total, then the number of large biscuits I would have is 49-36 = 13. But then I would have 36-13 = _23 more_ small biscuits than large, directly contradicting the statement given in the problem that I had 36 more small than large.

[@meschoui760]

jtron84 oh yeah right

[@meschoui760]

jtron84 you are right, there is 23 more small biscuits THAN large ones. This is the important thing I’ve totally missed. My bad

[@larrystuder6378]

I THOUGHT it added up to 42 or 43 on a rough estimate. Is 42.5 correct?

[@davetye]

Reminds me of a joke: A ship carrying a cargo of blue and red paint struck a remote island. The crew was marooned.

[@arnehermann3417]

Um, that would have to be red, blue and yellow paint.... red and blue paint give you purple.

[@dadoctah7978]

@@arnehermann3417 Heck with that noise. I'm colorblind. I hear that some people think that purple and violet are two different things, but I have no experience of such matters myself.

[@Leopoldshark]

The captain was purplexed. How could this happen?

[@Dhalin]

For the 2nd problem, I think what they *MEANT* to say is: "There are 49 dogs signed up for a dog show. If there are 36 small dogs signed up for the show, how many large dogs were signed up for the show?" Because that would be a simple subtraction problem (13 is the obvious answer) which seems right for 2nd grade. I have my high doubts they expect a 2nd grader to bust out algebra equations, lol.

[@sallycarroll6329]

Can you explain to me why this is not the answer anyway. If there are 36 small dogs signed up for the show or 36 more small dogs than large dogs signed up and we have got 49 dogs why is the answer not 36? 36 small dogs are obviously competing out of 49. So there must be 13 large dogs competing.

[@bonogiamboni4830]

No, the original is 36 small dogs *more* than the large dogs, so whatever the number of large dogs is, the number of small dogs is 36 more than that. If the total was something reasonable like 48 instead of 49 it would mean there are 6 large dogs and 6+36=42 small dogs so that 6+42=48 total.

[@dmandustinreacts2362]

I think I understand why its not 36 small dogs and 13 large dogs. It's because if there were 36 more small dogs than large dogs if there's 13 large dogs than that would mean that there has to be 49 small dogs which adding to the 13 large dogs would be more than the original amount of dogs competing which is wrong. The 49 number doesn't really matter here because of what it means by "36 more" means that it has to match the original amount of the large dogs first before applying the 36 more, so in the flawed example there would be 13 small dogs to match the 13 large dogs but an additional 36 more small dogs so that would mean 49 small dogs which clearly cannot be the case if there's only 49 dogs competing. That's the problem/issue, there's enough small dogs to match the number of large dogs but its asking that there's an additional 36 more on top of that than that unknown number which obviously can't be 13, it has to be 6.5 large dogs and large dogs but 36 more would mean that there's now a total of 42.5 small dogs total plus the 6.5 large dogs and BAM that's how you get 49.

[@medexamtoolscom]

Yeah, you're probably right, it does say it was 2nd grade. I remember the trivial tripe they had me doing in 2nd grade, they definitely never would have given me anything interesting like solving for something even slightly indirect. It was just page after page of simple arithmetic problems, I don't think there was even a single word problem, but if there was, it would have just been isomorphic to a simple arithmetic question. God I hated Ms. Eisenbarth. She was enamored in the ideology of tabula rasa and couldn't acknowledge me as being better than the other kids in any way, she just forced me to do worksheets of basic arithmetic all year.

[@cascadesouthernmodeltrains7547]

I was taught problems just like this in second grade, but we solved it like 3 separate problems. It would have went something like this. We know there are 36 more small dogs so you need to subtract that from the total. Ok you get 13. Now of those 13 and half will be large and half will be small. Ok 6.5 and since we are looking for the total number of small dogs we add the 6.5 to the 36 and get 42.5. I feel sorry for the 2 people that brought half a dog. But that’s how we would have figured this problem out in 1982 when I was in second grade. And when I did get into algebra, in jr high, it came really easy as learning problems like the dog question simply was a building block for more complex math.

[@A8nton]

I know a similar problem to 1): Peter has 3 Apples, calculate the volume of the sun. This question left me stomped.

[@neeldas8955]

14 ×10^26 cubic metre you get it by plotting the average masses of Peter's apple in ratio with speed of light in vacuum raised to power of sun's radii

[@dankerlordx2884]

Hey 🅱eter

[@jasonshoraka1098]

What the f***? So you can calculate the volume of the sun because Peter has three apples? So what if I had 7 apples does that change the volume of the Sun? LOL LMAO

[@Lord_Skeptic]

A8nton purple. because aliens do not wear hats.

[@imhulki463]

A8nton lmao

[@submandave1125]

Actually, it seems the answer to the second is 36

[@NorfolkCatKickers]

the right answer

[@NihongoWakannai]

In that case wouldn't you essentially just be rewriting the question in different words without making any real deductions?

[@submandave1125]

@@NihongoWakannai No. The problem expressed the relationship between small dogs and large dogs, but did not specify that these were the exclusive categories (there may also be medium or extra-large, or tiny dogs, for example). Since the assumption that these were the only two categories produced a non-sensical answer, this clearly invalidated that assumption. Without additional information regarding the relationship between non-large-or-small dogs with either the whole population or any identified sub-population, it becomes a boundary problem.

[@mtaur4113]

The question asks how many, not for a set of all possible values consistent with the information. This is a partial answer, and the question as asked either is unanswerable in full, or if we assume that it must be solvable, it's implied that we must define two exhaustive mutually exclusive categories and allow for half-dogs. As such, it's a pretty garbage "guess what I'm thinking" style of question.

[@submandave1125]

@@mtaur4113 "There are somewhere between 36 and 42 small dogs, inclusive," is a much more useful answer than "I can't tell you the exact number." When asked a question, I have always tried to provide as complete an answer as possible. I have generally found this to be better appreciated than "I dunno..."

[@wcbfg]

As someone from China, I've seen a "puzzle" like this: "You have a ship. There are X sheep and Y goats (or any number of any items) on the ship. How old is the ship’s captain?" The answer is your own age since it assumes you have the ship. But it's really a brain teaser rather than a math problem, and it seems unlikely to appear on a school exam.

[@iconicon5642]

absurd

[@HailtotheKiin]

But what if you own the captain and hire someone else out to captain it?

[@metamorphicorder]

Petes mom has 4 children. Nickle quarter and dime. What the name of her fourth child?

[@iiromutanen5719]

metamorphicorder Pete. But it depends on who is youngest. Then the youngest is the fourth child

[@larrystuder6378]

AHA! "You" have a ship..ok

[@jongilbertson2106]

There are between 0 and 6 large dogs, between 36 and 42 small dogs, and between 1 and 13 medium dogs. The ship's captain is old enough to know better.

[@MyReligionIs2DoGood]

The question implies that there is at least one large dog, otherwise the question would contain the answer.

[@johnpotts8308]

That was my answer, too

[@bolwinklemoose1999]

@@MyReligionIs2DoGood The question implies no such thing. If there are 13 medium dogs, 36 small dogs and 0 large dogs, that is still 36 more small dogs than large dogs. If we are to assume that there cannot be any "half" dogs, then there must be at least one additional catagory other than large or small dogs with at least one dog in that additional catagory.

[@tncookies]

@@bolwinklemoose1999 Correctamundo!

[@BiharyGabor]

You got the point but we don't know anything about the rest so you cannot call them "medium". All we know is they are neither small nor large.

[@Noximien]

When the media goes crazy about something like this, there really is nothing interesting happening anywhere that day.

[@simonhumby323]

I didn't notice the media going particularly crazy about this. But it is an interesting issue.

[@haskomeyer4924]

The media actually does not go crazy. If here and there it is mentioned in some of the media, "the media goes crazy". It is just the general urge to have something "important" to say.

[@paulcrosby4752]

Surely the answer to the 2nd question is zero. Since the question doesn't explicitly say that the 36 more small dogs are even entered in the competition. So 49 large dogs entered into the competition, and then 85 small dogs just hanging around!

[@Kristianne12111992]

With that logic, you are saying math is not interesting at all. Well, not to me. I love math.

[@leonidsedov9680]

I think that in the first problem correct answer is "we have not enough data to draw any conclusion". And in the official answer they write about critical thinking. I believe it is critical enough to give the answer: "no answer".

[@Macieks300]

Well I don't think you can be an actual ship's captain if you are, let's say, 4 years old.

[@leonidsedov9680]

Macieks300 maybe in some countries you can. I think it is possible that somewhere age limit for captains is not specified by any law. I don't know. And it is not stated in the problem that the ship is in our "real world" setting. Maybe they talk about space ships? Unless we have any specific information, we can only guess.

[@FoggyReloaded]

Maybe it is a toy ship and the "captain" is a child/baby. So my answer would be: "The captain's age is > 0." Atleast there is some math in it. ^^

[@justinnanu4338]

Without thinking too much into it, if I'm giving this problem to students and trying to evaluate if they understand that they don't have enough information to solve it, I'd give "we don't have enough information" a lvl 3 (B). A lvl 4 answer would be that, plus some extra information. For example, "While we don't have enough information to solve the problem, I could narrow down the possibilities by considering the amount of time it takes to gain the necessary training and experience to be a ship captain, and considering the age at which most ship captains retire. Based on this, I think a reasonable range for the age might be between 3o and 70 years old."

[@leonidsedov9680]

Justin How would you grade the answer, if the guess is wrong? For example, your guess is wrong, I know a guy who had yacht captain license when he was

[@joseph-fernando-piano]

The second question reminds me of my 2nd year engineering thermodynamics final, one of the questions was calculating the efficiency of a given pump, but because of a typo in one of the temperatures, doing the problem correctly would give you an efficiency of over 900%... the professor realized this halfway through the exam, and made an announcement, so many of us went back and corrected the working... half an hour later, he made another announcement that they weren't technically allowed to make us do a problem twice, so just to leave it as it was and they would just mark based on the correct method... needless to say there were a lot of angry and stressed out students afterwards...

[@NS-ie2ld]

My immediate answer was 42 S, 6 L and 1 medium (non-small/ large). Then I realized you could have anywhere from 1-7 medium (non-S/L) dogs and adjust for the S and L accordingly. There don’t have to be any large dogs (0-6) The goat/ sheep question was nonsense so I skipped it.

[@markawbolton]

We had one like that in an Electronic Engineering Exam. It had an amplifier that they wanted us to analyse. The Vcc was too low to get the amplifier to work at all. So Just up it to something that will work and do the analysis. BUT it take the confidence to make that initial call. I could hear other people going through Hell. This is where you get to walk out of exan half way and get an A+ ... BTW the Lecturer did it deliberately so, knowing his style, it should have come as no surprise.

[@andrewpayette621]

1) there is no answer. 2) typo This was unfortunate

[@nakulsindhwani1841]

andrew payette lol 😂😂😂

[@NetAndyCz]

1) Older than 28 2) Anything from 36 to 42.

[@Entropy3ko]

Sometimes it happens in college textbooks: you check your answer and it's "wrong". Then later you find out the answer in the back of the book is wrong and you had it right :D

[@ImperialDrylander]

maybe for 2 there is 37 small, 1 large and 11 medium dogs. BOI that's how i solved the problem

[@hollieodell9606]

andrew payette Second one: I think it is 37 small dogs and 12 large dogs because 49 -36=13 and 13+36=49 but it says 36 more than so that would mean that there are 37 small dogs and 12 large dogs.

[@jim55price]

Hiya, Presh. Actually, the "What About Medium-Sized Dogs?" question from Kevin Knudson is more significant than you mentioned here. Most simply, the 6.5 + 42.5 answer could easily be modified to 6 + 42 + 1, the 1 being a medium-sized dog; a tiny bit less simply, 6 other solutions exist : 5 + 41 + 3, 4 + 40 + 5, 3 + 39 + 7, 2 + 38 + 9, 1 + 37 + 11, & 0 + 36 + 13. Cheers.

[@shashankkothari8066]

It makes no sense when the teacher who wrote this question has already accepted his mistake.

[@jim55price]

@@shashankkothari8066 It makes plenty of sense, given the thousands of comments on this video alone, multiplied by the dozens of videos I may watch in a day. There aren't enough hours in the day to read everything & be perfect, so I made a brief, accurate comment & left it at that. Cheers.

[@shashankkothari8066]

@@jim55price Even I thought of this but when he told that the teacher apologized we can say that it's a pretty obvious mistake. There are many wrong questions out there that students don't even realize. It's actually smart to think it this way and consider all the other possibilities but still it doesn't change the fact that it was just a silly mistake.

[@jim55price]

@@shashankkothari8066 What I don't understand here is why you're addressing this issue toward me personally. I offered the comment in the capacity of a friendly colleague who happened to stroll by at just the right time to make it, and to possibly even help. It wasn't intended to alter the universe, & there was no part of it that addressed or even implied any level of silliness on anyone's part. Please let this go. It wasn't written to you in the first place.

[@shashankkothari8066]

@@jim55price I'm sorry if I offended you my intention wasn't to be disrespectful. I just wanted to say that your solution was good but the question was obviously wrong.

[@harmenbreedeveld8026]

The best answer to the Chinese question is: "I do not know the answer, as I do not have enough information."" That would actually be a great insight: we don't know everything, we cannot know everything, and it is ok to admit that. And it is an insight we do not teach enough at school, or in life.

[@rislamov]

First problem it’s a quite old problem to check kids’ attention. But usually it starts with “You are captain of a ship...”

[@john_savage]

We used to do the "You are driving a bus and pick up x number of passengers, yadda, yadda...."

[@victorhopper6774]

i am still trying to find out who is buried in grant's tomb.

[@ammjwijesiriwardene7730]

may be they are hotdogs.

[@anshuman8949]

Now that's critical thinking.

[@webgpu]

i would surely laugh if it was funny

[@cgamejewels]

AMMJ Wijesiriwardene i was thinking the same thing. 😅😅😅

[@KatanaBart]

That leaves... 6 large dogs... 42 small dogs... and a half-eaten bun where a dog used to be.

[@dfhwze]

who made you mister buzzkill

[@rheetythepiggy6337]

“This answer said that the goats and sheep weigh 7700kg, but really it’s 7700 pounds. I have no idea where they got this number, and it’s completely wrong.”

[@XenithShadow]

yep definently a mystery guess we will never know. :p

[@CaptainDangeax]

The only error here is using weight units inherited from the time the kings marry their cousins. Metric for science

[@CaptainDangeax]

@Martin Baldwin-Edwards you got a point. Propaganda (and low mass education) is more useful to those deceiptives countries than science (at least science for everyone)

[@Kurgosh1]

I'm envisioning traumatized school kids contemplating the implications of half a dog signing up to compete.

[@protorhinocerator142]

I know right? How weird is that? Usually it's the dog owners that sign them up.

[@gregb6469]

Maybe the half-dog is a poodle!

[@TomHenksYT]

maybe there's 42 genuine small dogs, 6 genuine large dogs and then 1 that's a mix of a larger and a smaller breed?

[@grahammcfadyenhill9555]

Half of a small dog...horrible image for a second grader

[@PatentedSugar9]

@@TomHenksYT But if you allow for classifications of dogs other than "large" and "small," this leaves you with multiple potential answers. This isn't necessarily a problem or disallowed in the context of the question, but I'd argue that it's a bit ambiguous. These potential answers are: 0 large dogs, 36 small dogs, and 13 other dogs; 1 large dog, 37 small dogs, and 11 other dogs; 2 large dogs, 38 small dogs, and 9 other dogs; 3 large dogs, 39 small dogs, and 7 other dogs; 4 large dogs, 40 small dogs, and 5 other dogs; 5 large dogs, 41 small dogs, and 3 other dogs; and 6 large dogs, 42 small dogs, and 1 other dog

[@justcarcrazy]

I put a cat in the box. When I return to the box the next day, the cat is a) dead? b) alive or c) a dog?

[@avinashreddy3433]

Coherence

[@volkerallert5364]

try a different cat.

[@jegodard1]

d) has stolen away, anyhow driven by fear, e) evaporated, :-)

[@ffggddss]

f) hiding under your car, so you won't find him and put him in that box again. (Of course, if he'd found the box on his own, he would have climbed into it himself; but since you put him in there, no way is he gonna stay! ;-) Fred

[@the_phantom_cat7912]

justcarcrazy a dog

[@TyphoidBryan]

I just got as close as possible and threw in a medium. 42 small, 6 large and that lonely 1 medium.

[@sirmoonslosthismind]

it could just as easily be 41 small, 5 large, and 3 medium; 40 small, 4 large, and 5 medium; ... all the way to 37 small, 1 large, 11 medium. your choice is not supported by the text of the question. however, you are correct to note that it does not state that all dogs in the contest are either small or large, so presh is incorrect in making that assumption given that it's obviously invalid.

[@aeromodeller1]

@@sirmoonslosthismind Right, there are six possible solutions to the problem if you don't assume dogs are either large or small.

[@juttagut3695]

@@sirmoonslosthismind or 36 small, 0 large and 13 medium.

[@AndreasDelleske]

@@juttagut3695 nether-small-nor-large would be the correct answer :)

[@stevedowning3854]

Putting medium dogs in would vary answer by 11 possible answers. From 1-11 medium dogs.

[@stephenmcconnell7868]

I remember having this kind of word problems in 2nd grade. (Why do I remember this from over 60 years ago?). They moved me into 2nd grad because I was too disruptive in 1st grade. I figured out some of the “advanced” arithmetic in that class. Unfortunately, I am dyslexic (I didn’t know it till I was an adult) and my “arithmetic” skills didn’t improve much. It wasn’t until I was in University that I understood what my problem was and I found ways to overcome this by being very careful and slow in analyzing problems. I began to love math and although not a mathematician, I have kept up that love.

[@Pet_Hedgehog]

how old are you?

[@stephenmcconnell7868]

@@Pet_Hedgehog Pretty damn old… 70 but I still feel like I’m 35 (kept in good shape all my life - mentally and physically)

[@Pet_Hedgehog]

@@stephenmcconnell7868 thats damn impressive, i know some 30-40 year olds who just dont care about anything active much less math and science for the well being of their intelect

[@tessjuel]

So, the first question is intended to encourage "critical thinking" and the second one to discourage any kind of thinking.

[@Hari-vg3gy]

HAHAHAHAHAHAH.....

[@noxnocs]

I was trying to explain this problem to my 10 year old (4th grade), he has not learnt about equations and systems of equations yet. After 2 days I had a small insight and this is how it went: If there are 36 more small dogs and we remove them from total, we are left with an equal number of small and big dogs, half and half of 13. Which does not divide by 2. That makes it a 2nd grade problem!

[@AnthonyHeaton-ih6rk]

That's a good method.

[@rmsgrey]

Minor correction: to fix the second question, you need either (but not both) the total number of dogs to be even (as stated in the video), or the difference in numbers of the two types of dog to be odd. As long as the difference and the total have the same parity, there's a solution with whole dogs.

[@MichaelGGarry]

I worked in education for a few years. I would have been in DEEP trouble if errors such as the 2nd one got through to one of my exam papers - we had multiple levels of internal checking for each paper before it was signed off. This was at Polytechnic level in Singapore - so 17+ education, university style setting (but one step below university).

[@promcheg]

2) reminds me of the teacher I had in school, who has thrown away half a question and expected the answer for the whole.

[@d.kyrstede8185]

When I did the 2nd question I cam up with 36 small dogs and 13 larger dogs

[@greenisnotacreativecolour]

Do you know why that's wrong now?

[@welcomb]

The purpose of the China question is to catch students who are rote learning and simply added the two numbers together.

[@aritragupta4182]

Are pupils allowed to use WolfRam Math World in exams in China?

[@ragnkja]

I suspect Chinese school tests are no less strict on the "no communication" rule than schools elsewhere in the world are, and those ban all online tools.

[@StanKPhlaps]

The narrator said HE used it, he didn't say students were allowed. It's a critical thinking exercise and has no answer and therefore it's not possible to cheat. The question comes from France anyway.

[@passerby184]

Can the first question is ruined wordplay by translation? maybe it has more sense in original Chinese form?

[@aquawoelfly]

passerby184 it has to do with a tradition of gifting captains with animals based on thier age.....? ..... maybe?....

[@jackren295]

No. The translation is perfectly fine. People on the Internet and the media in China see it as a test for the student's ability of critical thinking and questioning of authority. The correct answer is "Not enough information to solve this problem."

[@hassanalihusseini1717]

Critical thinking as in "Let blossom 1000 flowers"?

[@orlock20]

It's not a Chinese problem. It came from France.

[@protorhinocerator142]

Original question translated more properly from Chinese: If you are goats of 10 and the 26th ship was having, how old sheeps am the goat captain? And of course the correct answer is 9 and a half. That's in dog years, obviously.

[@adwnpinoy]

If China wrote second question: “It wasn’t a mistake it was meant to encourage critical thinking after all one solution would be 6 large dogs, 42 small dogs, 1 medium dog.”

[@heliocentric1756]

Actually there is nothing wrong in question #2: 42 small dogs and 7 large dogs signed up to compete (so the total is 49). But one of the large dogs is a pregnant female that gave birth to one dog right after signing up. So now there are 43 small dogs (36 more than the number of large dogs). QED

[@TriariusMetzer]

Clearly they allow disabled dogs as social justice effects all things so that's where you get the halves... well that concludes my joke response. Liked the video pretty cool to learn math problem mistakes/oddties as well misleading/difficult math problems.

[@RuthGray8]

Robert Grubbs there’s 49 dogs in total but I still agree with the 36 small dogs. I was always hopeless at maths and that just seems too easy

[@randomguy-tg7ok]

1) Ah yes. One of these ones. *On a scale of 1-10, what is your favourite colour of the alphabet? *A tree falls over in an evergreen forest. How many bottles of milk are there? 2) Hurr, 49-36 = 13 large dogs, right...? ...Well via implications only, 6.5 large dogs and 42.5 small. ... "At least 36" is the best answer I can give.

[@rmsgrey]

I'm sure you can do better than "At least 36" - "At least 36 but not more than 42" if you assume whole dogs; "...not more than 42.5" if you don't want to risk ruling out that extra fraction.

[@dfhwze]

sounds like you applied at google?

[@randomguy-tg7ok]

Umm... what...?

[@maxblatter]

The thing with the half dogs reminds of a math problem that was published somewhere in Switzerland or Germany and leaded to results including fractions of chickens laying fractions of eggs. I sent an email to the author of the problem, calling attention to his fault. To my surprise and anger he couldn't see any fault!!! He claimed that this was pure mathematics and there was nothing wrong about the fractions. It was just the time when the animal torture of shredding male chicks in chicken farms was heatedly debated; so I wrote back to underline my point of view under this aspect. To make it short: He didn't show the least sign of discernment, let alone remorse. Poor guy ... but in such cases, we MUST freak out!!!

[@TheEpicTsube]

I really liked the animation at the end rearranging all the letters to become the ad for the books.

[@fburton8]

Why not half a dog? It could be a show of dog meat - entirely possible in China.

[@UltimateBreloom]

fburton8 It was given in the US.

[@dfhwze]

or a half dog / half wolf

[@mariosstamoulis7345]

yeah - what if one dog is half cat or something?

[@rm._7000]

Eun Jun Wang Chinese people eat cats

[@Dylann8245]

half right

[@ahmadashraf9448]

Often time, it's the language that is confusing not the math.

[@TeeVeeGames]

I think presenting only half a dog to the judges is grounds for instant disqualification.

[@Ryukai-san]

Critical Dog question thinking - half a small dog + half a large dog = 1 medium dog, therefore, there were 42 small, 1 medium, and 6 large :D

[@UngKristen]

"What is half of a large dog and half of a small dog?" Well it aint going in a dog show

[@callumross6290]

on the contrary, I’d pay to see a dog show with a freak accident of a dog like that

[@matthewcooke4011]

Reminds me of... A mother is 21 years older than her child. In 6 years from now, she will be 5 times as old as her child. Where is the father?

[@brucetucker4847]

Shtupping the mother, I would imagine, since the child is being conceived right about now.

[@whatd0605]

I saw the thumbnail and thought "This is easy how could this be stumping peop- wait a minute you can't have half a dog" lol

[@MothaLuva]

The answer is 36. The description is „more small than large“. Meaning, more or less small.

[@Cheepchipsable]

They are described as "Small dogs", so it seemed like a "What colour was George Washington's white horse?" kind of question to me.

[@jeffreygrajek583]

How the answer is not 36 is beyond me.

[@aralornwolf3140]

@@jeffreygrajek583 , That was my first thought... although, there could be a single medium sized dog and the math would work out... But.. typos and incorrect answers are the bane of math students everywhere. I had a trivia question in middle school "how many time zones are there?" The official answer was 24... however, I live in a country with 6 time zones... Pacific, Mountain, Central, Eastern, Atlantic, and... NewFoundland... which is half an hour ahead of the rest of the Atlantic Provinces... therefore, 24 couldn't be the correct answer.

[@billyt8868]

@@jeffreygrajek583 that’s the first instinct imo. pretend we have 50 dogs so it’s even. if the answer is 36 then that means 14 large dogs. which means there is only 22 MORE small dogs than large. so that’s not right. it’s like saying a shirt and pants cost 20$, the shirt cost 10$ more than the shirt, how much does the shirt cost? well obviously not 10$ because then they would both be 10$ so the shirt wouldn’t cost 10$ more.

[@vincentrobinette1507]

@@Cheepchipsable I used the analogy "Who was buried in Grant's tomb" in my reply. I also stated 36 small dogs.(13 large) The answer is right in the problem.

[@freejutube]

well, I guess that for the dog problem, the best approach in the answer - from a educational point of view - is to state first that there exist medium size dogs (definition : not small dog, not large dog), and that the question gives no immediate hint about their number. The other approach would be to begin with an hypothesis : "as the question is silent about the dogs that are not small nor large, let's first suppose that each of those dog is either small or large". Then we arrive at an impossible solution, supposing that any number of the dogs must be an integer. (or we arrive at a contradiction, supposing that our solution means that at least one dog must be partly counted as a small dog and partly counted as a large one). So now, we have to explore the other hypothesis : there are also medium size (not small, not large) dogs, and we have no data about them in the question.

[@shawnpeal1209]

Love it when these simple word problem questions get over analyzed. So, might as well join in the fun! 12 large dogs and 37 small dogs. (no, I didn't scroll thru 1900+ comments to see if someone else figured it out.) If you have 36 MORE small dogs than large dogs, then you have 1 small dog included in the dog count to equal the count of large dogs. This IS a 2nd grade question after all ...

[@ltrizzle12]

I’m a certified math teacher in 2 states here in the US. In absolutely NO way is that second question assigned in second grade classrooms here. Fat chance.

[@latable558]

critical thinking is encouraged in china?

[@orlock20]

It started out as a French problem. Just because something is found in China, doesn't make it Chinese.

[@SteveSmith-rt7wx]

Yeah, seems more like a proper communist answer for the question would be "However old the General Secretary of the CCP says he is"

[@ffggddss]

@La Table 55: Absolutely! So long as it's critical of anything China doesn't like. Fred

[@protorhinocerator142]

Ship question: The Captain is as old as his tongue, and a little older than his teeth. Dog question: How do we end up with half a dog? "Sorry, I was hungry."

[@eventhisidistaken]

I recently played Far Cry 6, and in it, one of the animal amigos you can get is a small dog with no rear legs that uses a dog wheel chair. That's half a small dog right there.

[@k9er233]

It looks to me as if the question in #2 is meant as a reading comprehension test. The answer is in the question to find out if the student is paying attention to the way the question is actually worded. The question does not reference any sizes of dogs other than small dogs and large dogs. It states that there are a total of 49 "dogs" ( not referencing size classes ) signed up for the show. 49 total. There are 36 "more" small dogs than large dogs signed up. 49 total dogs. 36 are small. Obviously 13 are large, because 49 total regardless of size, as stated in the first sentence, minus the 36 small dogs ( as stated in the second sentence ) equals wait for it... 36 small dogs. Again, the answer is in the question. It is too easy to think that this is a difficult puzzle, but if you simply read the question as stated, the second sentence already tells you that there are 36 small dogs without actually doing any math at all. Reading comprehension test, to see who is paying attention to the actual question.

[@CarolHaynesJ]

1) Over the minimum age to be allowed to be a captain (why would anyone know the minimum license age or the length it must be held to captain a bigger ship) 2) As someone who taught 7 year olds I am not surprised this caused head scratching. They don't solve simultaneous equations at that age (or do algebra) - some kids would struggle to add and subtract tens and unit

[@Ennar]

Guessing is a valid method at that age and what the teacher probably had in mind. Typo was unfortunate, though. The real problem are parents who are projecting their own inadequacies.

[@VAFFANFEDE18]

I hate when they give those kind of problems to people that are not taught even how to solve one single equation formally

[@Ennar]

I was given such problems at that age when I didn't know anything about linear systems. I'm doing my PhD in mathematics at the moment but probably wouldn't if I was never given challenging problems like these. Not every given problem is meant to be solved by everyone. That's why you have grades in school.

[@VAFFANFEDE18]

Me too and I solved them nicely but still one thing is to apply knowledge in a hard one, another is inventing knowledge

[@Ennar]

This was very likely meant to be solved by trial and error. There is no inventing knowledge necessary.

[@coxsj]

Slow news day if this stuff is making headlines. Plus, humans have a strong bias toward simple solutions.

[@AmberZak83]

For question 2 we are assuming there are only small and large dogs. There could be medium sized dogs. So the number of small dogs is between 36 (if there are no large dogs) and 42 small dogs depending on the number of medium dogs.

[@nosnibor800]

When I saw the first problem I thought it was a joke. However there is some logic once you have all the information, so its not a joke, but lacks information for a solution. But you could invent your own information: only captains over age 50 can take mixed cargo. Engineers call this a "suck and see" solution. The second problem I worked out, but there was no integer solution - hence no solution. Of course could have rounded up or down to give an "engineers" solution, sometimes called "Its as near as damn it" solution in the real world of systematic errors. This goes to show that Engineers are far cleverer than mathematicians and should earn more.

[@victorpena5217]

I doubt second graders are writing their algebraic equations to solve this and come up with decimal answers. Theyre still learning basic math so the solution they were looking for was the kids to just subtract the small number from the larger one. Easy peasy. I don't understand why people get in a huff when a kid who barely learned to tie his shoes in double knots cant master Algebra that still stumps "adults" with a good handle on life.

[@rhetoric_]

I agree. For 2nd graders they should probably have been given a question like, "A dog show has 49 dogs both large and small. 36 of the dogs are small. How many dogs are large?"

[@rmsgrey]

You don't need to know the formalism of algebra to solve a question like this - trial and error will do it: If there's 1 large dog, there are 36+1=37 small dogs and 37+1=38 dogs total ... If there are 5 large dogs, there are 36+5=41 small dogs and 41+5=46 dogs total If there are 6 large dogs, there are 36+6=42 small dogs and 42+6=48 dogs total If there are 7 large dogs, there are 36+7=43 small dogs and 43+7=50 dogs total Answer must be between 6 and 7 large dogs - 6.5 gives the right numerical answer but doesn't make sense physically.

[@timharig]

rmsgrey I'm not sure that is a particularly good lesson for second graders either. At that age they are still learning basic addition and subtraction. In the next year they will be learning basic multiplication. You want them working problems and learning to solve them using the basic methods that they are learning. Using trial and error to solve problems at that age can easily become be a bad habit. When you do actually teach estimation, trial, and error methods, you want to make sure that they employ them properly. First, they need to know when such methods are appropriate to use. Second, you want them to learn to intelligently determine which trials to evaluate rather than making outright guesses. That's a lot of distraction when some of them are still probably struggling with the basics of doing the arithmetic in the first place.

[@williamkwan6913]

Re: second item, since it’s a second grade question, and half a dog doesn’t make sense, look to the grammar. There are 13 large dogs. There are 36 more dogs that are small.

[@efosaakenbor2530]

There are 42 small dogs, 6 large dogs and a moderate sized dog signed up to compete because nothing in the question excluded medium sized dogs.

[@KrayZ90]

Well, the second question did not say that there are exclusively small and large dogs at the competition. So lets assume there is a third category of dogs that is neither small or large (lets call them x for now). That would leave us with the set of functions s+l+x=49 and still s=l+36. As we know from the given solution we know, that x has to be an odd number in order to get full integers as solutions for s, l and x. Therefore we can set x to different values and find all the possible solutions: x=1 l=6 s=42 x=3 l=5 s=41 x=5 l=4 s=40 x=7 l=3 s=39 x=9 l=2 s=38 x=11 l=1 s=37 x=13 l=0 s=36

[@greggreen5510]

This is a question for SECOND graders. They are just learning math facts. They don't know algebra. Don't make it more complicated than it is. The question gives you the answer. 36 small dogs. There are 13 large dogs. It is simple arithmetic.

[@DancerOfClouds]

@@greggreen5510 Even a medium-sized dog is LARGER than a small dog. But I agree with you 100%. No need for anything more than 2nd-grade logic here.

[@Oscar1618033]

Before watching: saw the thumbnail and said "either someone signed a cat to a dog competition, or at least one dog is medium".

[@CerealKiller2]

The answer to the second question could also be that there are 42 small dogs, 6 large dogs and 1 dog that is neither a small nor a large dog since it's not specified that there are only large and small dogs.

[@pinkmagicali]

The part that surprised me was second graders being given algebra problems. I wasn’t taught algebra until I was 12. Embarrassed I couldn’t get it but I’ll admit that half threw me.

[@spartapros2120]

It wasn’t supposed to be algebra, the problem itself was over complicated.

[@theirtemple8262]

There is also the idea that you add and remove numbers to make integer math work, by "borrowing" a dog at a particular time.

[@BenisTodbringer]

The dog problem is solvable, we can account for 1/2 a dog if: 1. The dog is a paraplegic / it's Oscar Pistorius's dog 2. The 1/2 is a conjoined twin 3. It's a medium sized dog, but the contest doesn't have a category for medium dogs, so it just counts for 1/2 in both large and small categories 4. Chainsaw accident at the dog show 3. The black dogs count for 3/5

[@stanislavbutsky8432]

The problem statement doesn't say that competing dogs consist only of small and large ones. So the possible answer is 6 large dogs 42 small dogs and 1 dog that is neither small nor large.

[@Brightstarlivesteam]

Assuming the dogs are enrolled consecutively, then up to 13 dogs pairs, they are equal, but at 14 dogs you still have 13 large and 14 small, so this is that point where small dogs exceed large dogs. Your answer 6.5 large is a fallacy as you can still add both small and large dogs until reach the maximum of large dogs, which is the difference between the maximum enrolments and the difference plus one. Algebra does not always work, as it exclude logic. If you add dog randomly at 13 large dogs, you can add small dogs until their number exceeds large dogs which occurs at 14 small dogs.

[@peterwallace8441]

If you assume that there are dogs that are neither Large nor Small (Medium dogs) you can at least get a range for Small dogs (37 to 42). I hope they gave a bonus mark to any student who did this.

[@stanhawkins1023]

On the second problem, if we assume there are only large and small dogs, there are 36 small dogs and 13 large dogs. The question supplies the answer.

[@LTVoyager]

I am glad the US response admitted their mistake. That is the right way to handle mistakes.

[@angrydalek]

The dog show question is more akin to a bar joke that focuses your attention on irrelevant rather than actually making you solve anything for a solution, which usually is the case in math.

[@pomponopete]

I think everyone is overthinking the problem it's 36 small dogs

[@Javz177]

This is my answer, and I'm sticking to it.

[@GeoRedtick]

Presh you have made one of the common blunders. The problem asked to solve for the number of small dogs not the number of large dogs. Now I am left to wonder for the rest of my life, how many small dogs where signed up.

[@quietguy-rx6kv]

One of my physics professors re-used questions, and sometimes entire exams from one year to the next. In one of the tests, there was a question that was unsolvable because of some error in how it was written. Two years later, he taught the same course and used the same test, with the same unsolvable question.

[@roberteltze4850]

Reminds me of a puzzle I heard as a child: You're a bus driver, at your first stop 5 people get on the bus. At the second stop 2 get off and 7 get on At the next stop 4 get off 16 get on (Keep this up for a while, the numbers aren't important) What is the name of the bus driver? Answer: the name of the person you are talking to, it's clearly stated at the beginning that YOU are the bus driver

[@KS-lv5ue]

If the question states there are 36 small dogs signed up to compete, then asks how many small dogs are signed up to compete, how is the answer not 36? But is instead 42.5?

[@Pet_Hedgehog]

this reminds me of our physica class jokes back in 10-12th grade: a hedgehog is walking on the left side of the road, how much does a moose weigh on the right side of the road if the sun is 30° above the horizon line?

[@SylviusTheMad]

The dog question is only unsolvable if you assume all of the dogs are either large or small. If there is one (or three, or any odd number

[@kennybridges1284]

obviously the correct answer for the second question is half of "85 minus the noneven whole number of medium dogs competing"

[@scmtuk3662]

The dogs one (in the thumbnail), this seems _really_ easy There are 49 dogs. There are 36 more small dogs than large dogs. Now, the first thing people are going to do is 36 + l = 49. But there are 36 _more_ small dogs, not 36 _total_ small dogs. What we really want, is a series of two equations: s - l = 36 s + l = 49 Since we want to find s, we can find out what l is, in terms of s, and substitute. l = s - 36, or 49 - s Substituting will give us for different equations, two of which will give us the value of s, and the other two, will just cancel all variables: s - (s - 36) = 36 36 = 36 s + (49 - s) = 49 49 = 49 s - (49 - s) = 36 s - 49 + s = 36 2s - 49 = 36 2s = 85 s = 42.5 s + (s - 36) = 49 2s - 36 = 49 2s = 49 + 36 2s = 85 s = 42.5 Therefore, there are 42.5 small dogs (and 6.5 large dogs) Alternatively, if there are x dogs, and y more dogs than z, then: y = ((y - z) + x) / 2 (or the average between the difference between the number of large dogs, and small dogs, and the total number of dogs. In this case, (36 + 49) / 2 = 42.5 Also, the number of small dogs, would be z = x - (((y - z) + x) / 2) In this case, 49 - ((36 + 49) / 2) = 6.5

[@hanzohattori9576]

42 small dogs and 6 large dogs competed. The last one was probably medium sized or other. The question doesn't say anywhere that only small and large dogs entered the competition. But then it is possible that the number of large dogs could be anything from minimum one to maximum 6. And correspondingly, the number of small dogs could be anything from minimum 37 to maximum 42.

[@NathanTenney]

In computer science, we would have just truncated the half a dog (it's no good in the skills competition anyway) because both original operands were integers. But yeah, I saw this pop up in my feed, started to do the math, realized there would be half a dog, then watched the video to see if there was something that I missed.

[@kylerivera3470]

The real answer to the second question is "36 to 42 small dogs". The reason being, there is no information about medium sized dogs (as the last headline pointed out). You can know that there must be at least 36 small dogs (assuming you can have zero large dogs participating) and you know that you can't have more than 42, but there isn't enough information to figure out more. Another potential answer is "at least 43". The reason behind this one is: The question said there were 36 more small dogs than large dogs, but it not say there were exactly 36 more. If there were 40 more small dogs than large dogs, then there would also technically be 36 more small dogs than large dogs.

[@kenlieck7756]

You could enter a half-dog in a dog show. It ain't gonna win for sh*t but that wasn't part of the puzzle.

[@stingersfury]

I worked out 42.5 but then I realised there could be 1 normal sized dog not mentioned so I just rounded down to 42

[@matthieubrilman9407]

The first problem is an adaptation of a problem written by Gustave Flaubert (a famous french writter) to his sister in 1841 : « Puisque tu fais de la géométrie et de la trigonométrie, je vais te donner un problème : Un navire est en mer, il est parti de Boston chargé de coton, il jauge 200 tonneaux, il fait voile vers Le Havre, le grand mât est cassé, il y a un mousse sur le gaillard d'avant, les passagers sont au nombre de douze, le vent souffle N.-E.-E., l'horloge marque trois heures un quart d'après-midi, on est au mois de mai…. On demande l'âge du capitaine? ». Loose translation : "As you know of geometry and trigonometry, I'm going to give you a problem : A ship is at sea, loaded with cotoon from Boston. It has a capacity of 200 barrels and is headed to Le Havre, the main mast is broken, there's a cabin boy on the forecastle desk, twelve passagers, wind is blowing from N.-E.-E., it's quarter past three in the afternoon on the clock, the month is may. How old is the captain ?". This problem was made famous in the 70s by french pedagogists, because it illustrates very well the tendency of young students to give an answer, even though there is not a single logical reasonning that can lead to an answer.

[@harrykirk7415]

I think that mistakes or awkward uses of words or ideas is probably a technique often used in advertising. I'm sure people have been taught to use such things because it captures the mind and you see it all time. It's a standard way to draw your attention. This is an example from wiki that famously used an ungrammatical phrase for the purpose of drawing attention. As we see , the same works for Math. Here's the example of a . deliberate use of a mistake "....." "Winston tastes good like a cigarette should" is an advertising slogan that appeared in newspaper, magazine, radio, and television advertisements for Winston cigarettes, manufactured by R. J. Reynolds Tobacco Company. Reynolds used the slogan from Winston's introduction in 1954 until 1972. It is one of the best-known American tobacco advertising campaigns. In 1999, Advertising Age included the "Winston tastes good like a cigarette should" jingle in its list of the 10 best radio and television jingles in the United States during the 20th century. Advertising agency William Esty Co. deliberately, and ungrammatically, used "like" rather than "as" in the slogan and jingle. Esty executives Wendell Adams and Arline Lunny were in charge of the overall campaign. Lunny produced and directed most of the campaign's content during its early years. Although Adams was a classically trained musician, Margaret Johnson (a singer, pianist, and model) ghost wrote the jingle; Johnson and her husband, Travis Johnson, recorded it with their group, the Song Spinners.[1]

[@g0dofpain100]

What has 2 legs and bleeds? Half a dog. Also the answer to the 2nd problem could be two medium dogs, one large dog, a medium dog with extra legs, two 45 years old dogs, a vegan dog, 36 small dogs, 7 hot dogs.

[@randymiller2460]

42 small dogs, 6 large dogs, and one cat that was later disqualified from competition in an act of species discrimination. How dare they!

[@johnmckown1267]

6 large + 42 small + 1 medium dog (1 medium dog = 0.5 large and 0.5 small). The problem did not state there were exactly 2 sizes, large & small, only.

[@edhartgrove7552]

Regarding the boat captain. The "problem" never said that the boat, or the captain, were in or from China. So, the Chinese boat captain licensing requirements is irrelevant.

[@bolwinklemoose1999]

Are we to assume that Large and Small dogs are the only two catagories of dogs? For example.....if there is one Medium dog, then there could be six Large dogs; if there are three Medium dogs, then there could be five Large dogs, etc. If the conditions of the question are accepted as factual, and there cannot realistically be two half-dogs entered in the competition (unless we're talking about "hot-dogs"), then it is reasonable to assume there must be at least one additional catagory of dog other than Large or Small. There must be more than two catagories for this question to be reasonable/possible, three or more total catagories. And since we do not know the quantities of the other catagories, the answer is 0 to 6 Large dogs, with 1 to 13 dogs of any catagory other than Large or Small, and because there could be 0 to 6 Large dogs, there could be 36 to 42 Small dogs. You could also say there are "at least" 36 more Small dogs than Large dogs, still making the answer.....0 to 6 Large dogs.

[@RawwkinGrimmie64]

After I tried solving the dog show problem based on the thumbnail, I thought to myself "There must be a mistake somewhere." Then, I went and ragingly clawed at my hair when I realized that you couldn't have an even difference between two sets AND an odd total for the combined sets. I just hope the second graders didn't solve this and get mental images of half a dog in their head the way I did. That poor poodle...

[@BoomSqueak]

For the second question, the answer for the number of small dogs is the range of 36 to 42 dogs. There will be 0 to 6 big dogs, and there will be 1 to 13 other-sized dogs. Ta-da!

[@r00ben]

The question doesn't specify that the only dog categories are Large & Small. There could have been 40 small dogs, 4 large dogs, and 5 medium dogs.

[@szpoti]

Not sure where the problem lies since the answer is provided by the question itself. Something either is a small or a large dog. Therefore, there are 36 small dogs.

[@orisphera]

Here's how I solved the second problem: Assuming there are no medium-sized dogs there: s+l=49 s-l=36 Adding these two equations together, I got 2s=85 s=42.5

[@PercyPanleo]

Honestly I was thinking that the second one wasn't going to be a typo and just didn't specify if there were any medium-sized dogs, leading to 7 possible answers.

[@jayluber5971]

The first one is usually presented: You are the captain of a ship, it contains.... how old is the captain....

[@tncookies]

The best answer is, "anywhere from 36 to 42 small dogs". You can't have half a dog, but you can have another category. The remainder of the dogs (medium dogs?) after counting all large and small dogs can only range from 13 (36 small dogs, 0 large dogs) down to 1 (42 small dogs, 6 large dogs).

[@colonelfredpuntridge8799]

The problem stated that 49 dogs were competing. It did not say that ONLY 49 dogs were competing.