Author Topic: Is this Common Core math question the worst math question in human history?  (Read 904 times)

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Offline Rapunzel

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http://dailycaller.com/2013/12/07/is-this-common-core-math-question-the-worst-math-question-in-human-history/

Is this Common Core math question the worst math question in human history?

Posted By Eric Owens On 10:00 AM 12/07/2013

U.S. Education Secretary Arne Duncan has promised to improve education quality vastly by pushing for the implementation of the Common Core State Standards Initiative.

This year, 45 states and the District of Columbia have implemented the Common Core standards and curricula based on those standards.

Duncan doesn’t much care for the people who criticize Common Core, either. He has insisted that it’s all a bunch of “white suburban moms who — all of a sudden — their child isn’t as brilliant as they thought they were and their school isn’t quite as good as they thought they were, and that’s pretty scary.”

What, exactly, is the content of this Common Core that’s going to make American kids so much smarter? So far it appears to be a slew of worksheets and tests involving various, incomprehensible arrays of squares and circles. (RELATED: EPIC FAIL: Parents reveal insane Common Core worksheets)

There are also traditional word problems. Twitchy has found a word problem that may be the most egregiously awful math problem the Common Core has produced yet. Take a look:

    @michellemalkin a great common core question a friend,who is a teacher, posted. Cc @TwitchyTeam pic.twitter.com/LUxlyMEG3k

    — Kevin (@kevinpost) December 6, 2013

 

According to the Twitter user who posted it, the vexing problem came from a friend who is a teacher.

The problem comes from a Houghton Mifflin Assessment Guide. It appears among a larger set of basically similar math problems here. The problem involving Juanita appears on page AG102, nestled among some other problems that are similarly weak and crappy — though not nearly as harrrowing as the problem above.

Houghton Mifflin is Houghton Mifflin Harcourt, a huge textbook publisher. The company’s website promises to be “a partner who will share the responsibilities” of the Common Core: “We have created a wide range of content, curricula, and services to support school leaders, teachers and educators, parents, and especially students with this transition.”

Twitchy readers tried to tease out the answer to the Juanita problem — how can you not? – and determined that the answer is either 12, 24, 0 or 7.
“The time is now near at hand which must probably determine, whether Americans are to be, Freemen, or Slaves.” G Washington July 2, 1776

Offline Rapunzel

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http://twitchy.com/2013/12/06/unreal-check-out-this-ridiculous-common-core-math-problem-pics/

Unreal: Check out this ridiculous Common Core math problem [pics]
Posted at 2:51 pm on December 6, 2013 by Twitchy Staff | View Comment at link]

    @kevinpost @michellemalkin @TwitchyTeam So we're teaching our kids to make up answers when there's not enough data. Sounds familiar.
    —
    Todd Watts (@van_decaf) December 06, 2013

    @van_decaf that way no one is wrong.—
    Kevin (@kevinpost) December 06, 2013

Everyone’s a winner! But we’re not handing Common Core cheerleaders any participation trophies for this one.

A teacher shared this breathtakingly stupid math problem with Twitchy reader @KevinPost. It’s reportedly from a Common Core-aligned book.

    @michellemalkin a great common core question a friend,who is a teacher, posted. Cc @TwitchyTeam http://t.co/LUxlyMEG3k
    Kevin (@kevinpost) December 06, 2013


Say wha?

After seeing this kind of gobbledygook many times, we’re all out of shocked faces. Twitchy founder Michelle Malkin called her daughter’s Houghton Mifflin algebra textbook a “nightmare” when she tweeted about the error-ridden text. Juanita’s sticker predicament appears to be from a Houghton Mifflin Assessment Guide.

    @TwitchyTeam from assessment guild. This pic shows publisher http://t.co/r41Qx6PPeN
    Kevin (@kevinpost) December 06, 2013

For context, check out this chapter a fourth-grade teacher uploaded to her website (PDF). It includes the same sticker scenario and the preceding questions offer no additional information to help students solve the poorly-worded problem. So the answer is up to you!

    @TwitchyTeam no problem. Best answer so far " she buys all of them so no one is left out"—
    Kevin (@kevinpost) December 06, 2013

    @TwitchyTeam she buys zero because some kids might be allergic to sticker glue—
    Kevin (@kevinpost) December 06, 2013

Special thanks to @KevinPost for bringing this example to our attention.

[Update]

Twitchy readers responding to our tweet of this post have replied that the answer is definitely “12.” Or “24.” Or “0.” Or “7.”

It’s all clear now, huh? Of those answers, we think 12 would be the smallest number of stickers she should buy (if we’re reading the problem correctly). But if the goal was to confuse people with a strangely-worded question, then score!

[Update]

Answers continue to come in (see comments below and tweets sent to @TwitchyTeam) and there’s one thing that is clear: the phrasing is misleading enough that adults are interpreting the question in several different ways. How does that help kids learn math?
« Last Edit: December 07, 2013, 10:13:49 PM by Oceander »
“The time is now near at hand which must probably determine, whether Americans are to be, Freemen, or Slaves.” G Washington July 2, 1776

Offline truth_seeker

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She knock sumbody white upside dey haid, take all dem damn stickers.

Yo, knockout, sucka.

Onliest damn math we needs, is da free rent, da free food, da free phone, da freedom from responsibility fo da chirrens we make, etc.

Den we be eligible fo college, loans, scholarships fo sports, etc. Den Obama, he forgives da loans.


Offline flowers

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I dont kneed no dam math, i giv ma stickers to ma dop dealer!


Offline AbaraXas

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I hate to say it, but there actually is a correct answer and the question does make sense (although I do admit it is poorly worded).

The variable is the number of friends (X). She has the choice of either 4 or 6 bags (number of stickers in each unknown but we'll assume each bag has the same amount) according to the instructions so the answer is either of those. The challenge is to make sure that no matter what X is, there are no stickers left (no remainder).

If X is an even number, 4 or 6 will work as you can divide an even number by them with no remainder.
If X is an odd number, only 6 will work, you can divide an odd or even number by 6 with no remainder.

The correct answer is 6. No matter what X is, 6 will have an amount divisible by that without any remainder.

Example. If she has 2 friends, if she has 4 bags each get 2. If she has 6 bags each get 3.
However, if she has 3 friends, if she has 4 bags, each would get 1 with a remainder of 1. If she has 6 bags, each get 2.
« Last Edit: December 07, 2013, 07:22:06 PM by AbaraXas »
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Offline AbaraXas

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BTW, this is a standard pre algebra, polynomial fair division type problem. More info on this type of problem and how it is used:

http://www.colorado.edu/education/DMP/fair_division.html
Никогда не обманывайте себя мыслью, что вы «влияете» или вносите изменения в Интернет. Это эфемерное удовольствие.

Offline Oceander

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I hate to say it, but there actually is a correct answer and the question does make sense (although I do admit it is poorly worded).

The variable is the number of friends (X). She has the choice of either 4 or 6 bags (number of stickers in each unknown but we'll assume each bag has the same amount) according to the instructions so the answer is either of those. The challenge is to make sure that no matter what X is, there are no stickers left (no remainder).

If X is an even number, 4 or 6 will work as you can divide an even number by them with no remainder.
If X is an odd number, only 6 will work, you can divide an odd or even number by 6 with no remainder.

The correct answer is 6. No matter what X is, 6 will have an amount divisible by that without any remainder.

Example. If she has 2 friends, if she has 4 bags each get 2. If she has 6 bags each get 3.
However, if she has 3 friends, if she has 4 bags, each would get 1 with a remainder of 1. If she has 6 bags, each get 2.

Pretty good; however, the question itself is missing sufficient context to clue in the typical reader.

Offline Chieftain

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Offline AbaraXas

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Pretty good; however, the question itself is missing sufficient context to clue in the typical reader.

It is pretty poorly written if taken at face value, however, we don't see the context of the page to add more to it. However, on the Twitchy page, people are going the other way and reading way too much into the question.

The question can be broken down like this.

We have 1 unknown variable, the number of friends (X)

We have 3 unique factors. 1. She purchases either 4 or 6 bags (the part people are goofing on trying to factor in more bags than the question says she is choosing from). 2. Each friend receives an equal number of stickers. 3. There is no remainder.

Some of the factors not described in the question as written but we should take to the logical conclusion for argument's sake. 1. Do each bag have an equal number of stickers (assume yes, bags are equal groupings of stickers)? 2. Can you divide a sticker itself into smaller parts (assume no, rational, whole number answers are all that is allowed). 
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Offline Oceander

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It is pretty poorly written if taken at face value, however, we don't see the context of the page to add more to it. However, on the Twitchy page, people are going the other way and reading way too much into the question.

The question can be broken down like this.

We have 1 unknown variable, the number of friends (X)

We have 3 unique factors. 1. She purchases either 4 or 6 bags (the part people are goofing on trying to factor in more bags than the question says she is choosing from). 2. Each friend receives an equal number of stickers. 3. There is no remainder.

Some of the factors not described in the question as written but we should take to the logical conclusion for argument's sake. 1. Do each bag have an equal number of stickers (assume yes, bags are equal groupings of stickers)? 2. Can you divide a sticker itself into smaller parts (assume no, rational, whole number answers are all that is allowed). 

I agree with you on the analysis; it's a good question, just very poorly presented.


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