Current Events > A Mom is going BONKERS over this MATH QUESTION..Can you solve it??

(+) Yeah! (+)

Angie Werner was left scratching her head after she was stumped on her 7 y/o daughter's math homework!!

She asked social media for help and many people gave her various responses..and while at first it seemed simple, many people miffed at the question!!

Can you solve this question? let's see if people can

http://i.dailymail.co.uk/i/pix/2018/02/01/01/48C70D0C00000578-5337705-image-a-2_1517447030132.jpg
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36?
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What am i not seeing here? it can be assumed that 49 dogs total signed up and 36 more small dogs did. So the answer is 36

It seems completely obvious, where is the discrepancy?
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36.
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Dustin1280 posted...

What am i not seeing here? it can be assumed that 49 dogs total signed up and 36 more small dogs did. So the answer is 36

36 MORE small dogs than large dogs. If the total is 49 that signed up, that means you must count the 13 difference, and cut it in half. That makes 6.5 large dogs, 6.5 small dogs, plus 36 additional dogs makes 49. It can't be solved because a half dog cannot exist.
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MabusIncarnate posted...

Dustin1280 posted...

What am i not seeing here? it can be assumed that 49 dogs total signed up and 36 more small dogs did. So the answer is 36

36 MORE small dogs than large dogs. If the total is 49 that signed up, that means you must count the 13 difference, and cut it in half. That makes 6.5 large dogs, 6.5 small dogs, plus 36 additional dogs makes 49. It can't be solved because a half dog cannot exist.

Are you sure a half dog can't exist?!? I think you meant 42.5 smalls btw.
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MabusIncarnate posted...

Dustin1280 posted...

What am i not seeing here? it can be assumed that 49 dogs total signed up and 36 more small dogs did. So the answer is 36

36 MORE small dogs than large dogs. If the total is 49 that signed up, that means you must count the 13 difference, and cut it in half. That makes 6.5 large dogs, 6.5 small dogs, plus 36 additional dogs makes 49. It can't be solved because a half dog cannot exist.

42.5
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Ahh gotcha, I see where the confusion comes from...
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It can't be 36. If there are 36 small dogs more than 36 and you say 36, then the total that signed up would be 36, not 49. If you say there's 13 large dogs, and 36 more small dogs, that means there's 13 large, 13 small plus an additional 36 which equals 62, which isn't 49.
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PiOverlord posted...

MabusIncarnate posted...

Dustin1280 posted...

What am i not seeing here? it can be assumed that 49 dogs total signed up and 36 more small dogs did. So the answer is 36

36 MORE small dogs than large dogs. If the total is 49 that signed up, that means you must count the 13 difference, and cut it in half. That makes 6.5 large dogs, 6.5 small dogs, plus 36 additional dogs makes 49. It can't be solved because a half dog cannot exist.

Are you sure a half dog can't exist?!? I think you meant 42.5 smalls btw.

Well I was saying 6.5 plus 36 additional, which equals 42.5. I just didn't total it like that.
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2 hard for a kid to solve tbh
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That's actually impossible.
If there are 36 more small dogs, than large dogs then you know there cannot be 14 large dogs, or higher, as that would be 50+. However,since it has to be 36 more than, it cannot be 13 large dogs, as that is only 23 more than. So it's a simple matter of increasing large and small each by 1 until they satisfy the requirements...however, that is impossible, as the closest match - 42 & 7, are 35 off, not 36.
41 & 8 are 33 off, and 43 and 6 are 37 off. Thus, the problem is impossible to begin with.
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SoundNetwork posted...

2 hard for a kid to solve tbh

Yes, this is what it comes down to.
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Caelthus posted...

That's actually impossible.
If there are 36 more small dogs, than large dogs then you know there cannot be 14 large dogs, or higher, as that would be 50+. However,since it has to be 36 more than, it cannot be 13 large dogs, as that is only 23 more than. So it's a simple matter of increasing large and small each by 1 until they satisfy the requirements...however, that is impossible, as the closest match - 42 & 7, are 35 off, not 36.
41 & 8 are 33 off, and 43 and 6 are 37 off. Thus, the problem is impossible to begin with.

i think its just worded horribly.
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you know what? fuck math
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This would be a great exercise in thinking if it was 48 instead of 49.

And nearly half of CE still would have failed because so many of you went with the intuitive answer of 36 with out putting any thought into it.
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MisterPengy posted...

And nearly half of CE still would have failed because so many of you went with the intuitive answer of 36 with out putting any thought into it.

....its a 7 year old's assignment,
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TheVipaGTS posted...

i think its just worded horribly.

No, that type of question is actually very common. Somebody probably either:
a)Made a mistype in the production
or
b)It's a logical test for children(looking to see if they can figure out that a question has no solution)
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MabusIncarnate posted...

Dustin1280 posted...

What am i not seeing here? it can be assumed that 49 dogs total signed up and 36 more small dogs did. So the answer is 36

36 MORE small dogs than large dogs. If the total is 49 that signed up, that means you must count the 13 difference, and cut it in half. That makes 6.5 large dogs, 6.5 small dogs, plus 36 additional dogs makes 49. It can't be solved because a half dog cannot exist.

That's not exactly correct. You can assume that for sufficiently large dogs that .5 large dogs = 1 small dog.
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DifferentialEquation posted...

MabusIncarnate posted...

Dustin1280 posted...

What am i not seeing here? it can be assumed that 49 dogs total signed up and 36 more small dogs did. So the answer is 36

36 MORE small dogs than large dogs. If the total is 49 that signed up, that means you must count the 13 difference, and cut it in half. That makes 6.5 large dogs, 6.5 small dogs, plus 36 additional dogs makes 49. It can't be solved because a half dog cannot exist.

That's not exactly correct. You can assume that for sufficiently large dogs that .5 large dogs = 1 small dog.

It states 49 dogs signed up. It doesn't state large or small. a 5lb dog or a 120 lb dog is still one dog.
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the key for this problem lies on how many medium sized dogs signed up.
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when do kids start learning to do math with variables? I can't remember, but I think it had to be at least a few years later than 7. so, these kids probably weren't meant to solve it by setting up equations. but as a guess and check addition problem? should be fine for 7.

as for half a dog signing up, I'm guessing the teacher changed the words of the question, without realizing the numbers don't work for dogs...
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HighOnSolar posted...

you know what? fuck math

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TheVipaGTS posted...

MisterPengy posted...

And nearly half of CE still would have failed because so many of you went with the intuitive answer of 36 with out putting any thought into it.

....its a 7 year old's assignment,

Yes, it's quite sad.
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DifferentialEquation posted...

That's not exactly correct. You can assume that for sufficiently large dogs that .5 large dogs = 1 small dog.

lmao
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Why are there no medium dogs?
Also that is a lot of small dogs for a contest with so few large dogs.

Based on what we are given 42.5 so two peoples dogs (one large and one small) got run over before the contest sadly and were cut in half.
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who taught you people math and english?

who taught this teacher?

the phrase 5 more than 7 would get you 12, so whats 5 more than x that equals 12?

12 = 7 + 5

whats 36 small more than large that equals 49

LavarNeedsHelp posted...

who taught you people math and english?

who taught this teacher?

the phrase 5 more than 7 would get you 12, so whats 5 more than x that equals 12?

12 = 7 + 5

whats 36 small more than large that equals 49

Wouldnt the 49 have to be 19 for your 12 problem? I mean 12 is 5 more than 7, but 5 is not 5 more than 7 nor is 7 5 more than 5. So if I said you have 5 more x's than y's that =12 total how many x's do you have? You cant say 7 or 5 as neither is 5 greater than the other. 8.5 is 5 more than 3.5 and 8.5+3.5=12
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LavarNeedsHelp posted...

whats 36 small more than large that equals 49

Who cares? It's not relevant to the question in OP's link. If you want to ask random Maths problems make your own topic.

TheVipaGTS posted...

MisterPengy posted...

And nearly half of CE still would have failed because so many of you went with the intuitive answer of 36 with out putting any thought into it.

....its a 7 year old's assignment,

That most of CE failed, yes.
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Haven't looked at the responses yet but this can't be solved unless you start getting into "half dogs" and what not, which doesn't make sense in the context of the question.
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Twin3Turbo posted...

Haven't looked at the responses yet but this can't be solved unless you start getting into "half dogs" and what not, which doesn't make sense in the context of the question.

Maybe two cats entered?
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Two cats doesn't help us. We need an odd number of some sort of third-party entity, whether that be medium dogs, cats, horses, balloons, or whatever.

The answer is either over 9000 or about tree fiddy
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Is this some Asian dog show where they show cooked dogs?
Cause otherwise it's not working.

The answer is however many god intended
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N/A Can't be solved is the correct answer. This was a comment sourced from Newsweek from a statement that the school made.

"Turns out, there's no right answer. The school district announced in a statement that they were attempting to challenge students to think critically."

http://www.newsweek.com/math-problem-homework-viral-solve-797139

This reveals the confirmed answer to this problem for those still wanting to debate. As stated by a few above, you cannot have .5 of a dog.
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There are 36 small dogs and 13 medium sized dogs
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MabusIncarnate posted...

"Turns out, there's no right answer. The school district announced in a statement that they were attempting to challenge students to think critically."

i fucking hate this kind of excuse.
like the ministry of propaganda from my country stated "a hoax is fine as long as it's constructive (read: pro government)". after being confronted by the public, he said he is "just building awareness and challenge the people to think critically".
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MabusIncarnate posted...

N/A Can't be solved is the correct answer. This was a comment sourced from Newsweek from a statement that the school made.

"Turns out, there's no right answer. The school district announced in a statement that they were attempting to challenge students to think critically."

http://www.newsweek.com/math-problem-homework-viral-solve-797139

This reveals the confirmed answer to this problem for those still wanting to debate. As stated by a few above, you cannot have .5 of a dog.

Cant have half a dog? Tell that to the poor dog I saw on the side of the road run over with its lower half missing.

But anyway the question can be answered it is just an illogical answer unless you go full Midas on the dogs, or some geneticists brings in half dog half something else specimens, or if some of Ted Turners puppies entered the show. Family guy joke
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MabusIncarnate posted...

N/A Can't be solved is the correct answer. This was a comment sourced from Newsweek from a statement that the school made.

"Turns out, there's no right answer. The school district announced in a statement that they were attempting to challenge students to think critically."

http://www.newsweek.com/math-problem-homework-viral-solve-797139

This reveals the confirmed answer to this problem for those still wanting to debate. As stated by a few above, you cannot have .5 of a dog.

are you sure that comment isn't about the sheep/goat/captain question just above it? because this is also in that link:

Despite it being impossible to have half of a dog compete in a dog show, 42.5 is actually the right answer. And the teacher who assigned the difficult question even confirmed that its 42.5.

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