Posted on 01/31/2018 10:42:17 PM PST by Morgana
FULL TITLE: Mother is left stumped by her seven-year-old daughter's maths homework question (but could YOU work it out?)
A mother was left scratching her head after trying to help her seven-year-old daughter with her maths homework.
Angie Werner was stumped when she tried answer the conundrum and turned to social media to help her out.
The problem asked how many entrants were in a dog show and while at first the answer appeared to be simple - in fact it left many parents completely miffed.
PROBLEM ON LINK
(Excerpt) Read more at dailymail.co.uk ...
Also really good at PLAY DEAD.
Exactly. It’s pretty simple algebra, although it seems a little advanced for a 7-year-old.
And I’d pay to see those half-dogs.
There are 37 small dogs, 1 large dog, and 12 medium dogs ...
Well, you’re supposed to phrase the answer as “There are 6 large dogs and a dog that isn’t small or large, it’s medium.”
I had a similar question when I was in school. It was small, medium and large (Fries or something, I don’t recall) the .5 was an exercise. The problem listed smalls and larges and that .5 carry over was “something that isn’t small or large” like extra large or medium.
Way more important!
And giving this problem to seven years olds makes even more sense
L+S=49
L+36=S
L+(L+36)=49
2L+36=49
2L=49-36
2L=13
L=6.5
S=49-6.5
S=42.5
Prefecto
Once you grasp that, the answer is simple, you treat the competing dogs as two discrete classes, small and large. One, small dogs, consists of 36 members, and the other, large dogs, consists of 13 members. You now no longer are required to compare the relative size of the small dog class to the large dog class but just determine the fact of how many more there are above the total there may be, regardless of the large dog class’ number. The reading comprehension enters in ignoring the extraneous data which is mere unnecessary "noise" to respond to the question. The problem answers itself by providing the answer, 36 small dogs.
This avoids the problem of chopped wiener dog. . .
Now, if you really want to make it a maths problem, then one has to realize the issue is significant digits. . . that one is working with non-divisible units, incapable of being divided further into fractions. I.e. absolute numbers: for example, one cannot subtract 5 dogs from 3 dogs and have a minus 2 dogs. It’s not possible in the real world. . . Just as in this case one cannot enter half-dog in a dog show.
The teacher who originally wrote this problem was stupid. He or she should have worked the math first. I do note this was a second grade problem for a seven year old, who is not yet ready to start algebra. I think the problem was a typo. The question for a word problem at this grade level would have been: “How many large dogs are competing?"
This is a good example of why math does NOT represent reality, say like climate change models, can’t really predict the future weather.
There’d be a bloody mess all over the dog arena. . . and could you imagine the owners of the half-dogs persuading their pets to heel and sit?
"Heel, Fi! Heel!"
And the other:
“Sit, Do, Sit!"
We got a ringer in there somewhere. There are now 50 mutts competing with tat answer. Rounding error?
I solved the problem correctly by using algebra. Hm?....Why are the teaching algebra, fractions, and decimals in the **second** grade?
Also...That half a dog would be competing would perplex the most gifted mathematician.
I got it right, but I couldn’t figure out if it was the front half or the back half of the one dog that was entered.
You are exactly right Swordmaker, and provide an excellent analysis of the problem.
still scratchin my head over a 7 year old doing algebra.
I was a 7 year old 2nd grader.
I didn’t learn algebra until high school.
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