Poll of the Day > 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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x = big dogs
x+36 = small dogs
x + x + 36 = 49
x = 6.5

42.5 small dogs

Someone brought half a dog.
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Pretty sure Bonkers was a cat, not a dog.

From a mere math standpoint, 42 is the only answer that it can be. If you conveniently ignore the context of the question, in which case, you realize 42 makes no sense at all, unless this is some freakish dog show/mad scientist show that allows half dogs to participate.

As stated, the problem has no answer.
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x is the number of big dogs
x + 36 is the number of small dogs

2x + 36 = 49
2x = 49 - 36
x = (49 - 36) / 2
x = 6.5
x + 36 = 42.5

But it's a pretty morbid show if they're counting halves of dogs.
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2nd graders shouldn't be having to do algebra to solve their math. That's dumb.
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2 people brought their half dogs to the competition.
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The correct answer is 36. There were no large dogs, only small and medium-sized dogs.
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I think you might be the one who is actually going bonkers

lets see what people think
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42.5 is a correct answer but didn't make sense. But this is math and so make sense.

Am I bonkers now?
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VeeVees posted...

2 people brought their half dogs to the competition.

great minds think alike.

by which axis were they halved, do you suppose?
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The answer is 42.5. You need to use algebra to solve it, which 7-year-olds are not expected to know. The answer is also illogical. How can you have 1/2 a dog? There must have been some mistakes in the question.
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HeyImAlex posted...

2nd graders shouldn't be having to do algebra to solve their math. That's dumb.

2nd graders shouldn't have to do mathematics to solve their calculations.
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I mean it says they are 36 small dogs signed up in a dog show that has 49 dogs. Then asks how many small dogs signed up. The answer would be 36 to me as you know it already told you the answer.
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I'm guessing it was supposed to start out "There are 49 large dogs..."

The easy way to solve it is (49+36)/2.
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x+y=49
x=49-y
x=49-(x+36)
x=49-x+36
x-36=49-x
2x=85
x=85/2
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hmm..
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49 - 36 = 13
13 / 2 = 6.5
49 - 6.5 = 42.5

42.5 - 6.5 = 36
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quigonzel posted...

What am I missing?

There are 36 more. That means whatever number of large dogs there are is subtracted from the number of small dogs with 36 remaining.
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For once, the question is dumber than the angry parent.
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quigonzel posted...

Huh?

What am I missing?

49 total dogs.

36 are small.

Which leaves 13 large dogs?

Answer is 13?

EDIT: I see it's asking how many are small.

So 36?

It's not just 36 small dogs, it's 36 more small dogs than large dogs. So the answer is 42.5 small dogs (somehow there's half a dog).
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SilentSeph posted...

So the answer is 42.5 small dogs (somehow there's half a dog).

Not just one. There is half a small dog and half a large dog.
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at least it's reasonably solvable and not inventing ridiculous new rules for this specific question
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quigonzel posted...

SKARDAVNELNATE posted...

quigonzel posted...

What am I missing?

There are 36 more. That means whatever number of large dogs there are is subtracted from the number of small dogs with 36 remaining.

Ahhh, I've never been good at math. I've always been a reading and grammar person.

But...you read it wrong...
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you could say that there is 1 medium dog. then there's 6 large dogs, 1 medium dog, and 42 small dogs. but doing it that way, with some unknown number of neither-small-nor-large dogs, gives you many solutions.

probably it's supposed to be 48 instead of 49, which would seem a more likely number to use with 36.
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quigonzel posted...

It says there are 36 more small dogs signed up to compete than large.

49 total.

36 small, leaving 13 large.

If there are 13 large dogs, and 36 small dogs signed up, doesn't that make it 36 small dogs? If there's only large and small dogs competing?

Yeah, I'm stupid, don't mock me too badly >_>

=(

The 'more than' part means there has to be a difference of 36 between the small dogs and the large dogs. So whatever number of large dogs there is, you add 36 to it to find the number of small dogs. The problem doesn't state the exact number of either small dogs or large dogs, just the total between them. So the only two numbers that have a difference of 36 between them and that also add to 49 are 6.5 and 42.5.
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The correct answer is 42.5. (School standard answer)
But this is about dogs, so 42 is the correct answer. (My honest answer)

I can admit my math is really bad.
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Obviously the .5 dog was Catdog folks.
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Setsunahenry posted...

But this is about dogs, so 42 is the correct answer.

49 - 42 = 7
42 - 7 = 35 not 36

49 - 43 = 6
43 - 6 = 37 not 36

It has to be 42.5 to work.
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fishy071 posted...

The answer is 42.5. You need to use algebra to solve it, which 7-year-olds are not expected to know. The answer is also illogical. How can you have 1/2 a dog? There must have been some mistakes in the question.

Standardized work books are printed and reprinted so often by competing publishing companies that stuff like this happens way more than it should.
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SilentSeph posted...

The 'more than' part means there has to be a difference of 36 between the small dogs and the large dogs. So whatever number of large dogs there is, you add 36 to it to find the number of small dogs. The problem doesn't state the exact number of either small dogs or large dogs, just the total between them. So the only two numbers that have a difference of 36 between them and that also add to 49 are 6.5 and 42.5.

.... I still don't get it. 49 dogs signed up. They say 36 more small dogs than large dogs. That's fine as the remaining 13 dogs can be any size. It doesn't say the size of the remaining dogs so I assume they are large given the question. It doesn't matter anyway as they say "36 more small dogs" so have already said 36 small dogs entered.
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wwinterj25 posted...

They say 36 more small dogs than large dogs.

there aren't 36 small dogs unless there are 0 large dogs and 13 something-else dogs
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wwinterj25 posted...

It could very well be 13 large dogs.

no, because then there'd only be 36-13=23 more small dogs than large dogs.

hmmm
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Sahuagin posted...

there aren't 36 small dogs unless there are 0 large dogs and 13 something-else dogs

Right....

Sahuagin posted...

no, because then there'd only be 36-13=23 more small dogs than large dogs.

hmmm

I see. The wording seems to be the thing that catches me off guard in math questions. I think I get it a little bit more now. Thanks.
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If you assume the possibility that there are other dogs which make up the total but are neither small or large then the answer becomes a range.

With 1 other dog there are 6 large and 42 small.
With 13 other dogs there are 0 large and 36 small.
Thus there are 1-13 other dogs, 0-6 large dogs, and 36-42 small dogs.
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