Posted on 12/12/2013 8:47:50 PM PST by Innovative
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:
15. Juanita wants to give bags of stickers to her friends. She wants to give the same number of stickers to each friend. She's not sure if she needs 4 bags or 6 bags of stickers. How many stickers could she buy so there are no stickers left over?
(Excerpt) Read more at dailycaller.com ...
It doesn’t matter as long as she gets a free abortion.
Pray America is Waking
Stick it, Juanita, and nothing will be left over.
OTOH, Juanita has no friends, she only has amigos so the question is moot.
Easy. the answer is 0.
Juanita wants to give bags of stickers to her friends. Bags are usually plastic. plastic is harmful to the environment. No one will be friends with someone who wants to harm the environment. Nobody will be Juanita’s friend. Juanita has no friends - she needs no stickers.
Q.E.D.
Well, she could buy 12, 24, 36, 48 etc etc
Good one!
My thought exactly. It isn’t developing an analytical thought process by being unclear.
OR
The answer may just be elementary.
Answer: None. The issue of how many friends is moot, as she has no money (so says the author of the idiotic problem).
42 was Jackie Robinsons number and it’s been retired from MLB by all teams.
bags....hmmmm....what a dopey problem
Juanita is LGBT so any answer is gender biased
If the second hand is between 12 and 3 it’s A; between 3 and 6 it’s B...
If the clock is digital and/or has no second hand you’re gonna fail.
Looking at the pdf that has the similar questions, I suppose the answer is “any multiple of twelve” because twelve can be divided evenly 4 ways, and 6 ways, and is the least number that can be so divided.
It’s kind of funny, since the implication is that she isn’t sure if she has four friends or six friends! Plus, who cares if there are stickers left over?
All these questions seem to be oriented towards testing comprehension of abstract number concepts ( LCM in particular ) in terms of “practical” problems. Nothing so very wrong with it I guess. As an aged math nerd I find them very amusing.
A+Z=X
What age/grade is that for?
I must be getting old, I've never heard of "stickers". Do you smoke, snort of shoot up that stuff?
The problem is not that hard. She is either giving out 4 bags or 6 bags (variable friendships).
She must buy at minimum of 6 stickers (not the answer), because if she chooses to give out 6 bags, then she must have 6 stickers to put one in every bag.
The next possibility is 12 stickers (first multiple of 6). She can give out 6 bags with two stickers each.
Does this also work for 4?
Yes, it does. She can give out 4 bags with three stickers in each..
needs to buy a multiple of 4 and 6.
So the first available solution is that Juanita should buy 12 stickers.
This is not the only solution. They did not say, what is the MINIMUM number of stickers she could buy. She can buy far more stickers, as long as it she buys stickers in multiples of 12. So 12*n where n is 1,2,3...
Or 12, 24, 36 ... to infinity and beyond - Buzz Lightyear
Now if we wanted to do this by multiplication.
We could have multiplied both numbers together to get 24 (6*4). That would have given us one of the many answers.
However, notice that 4 is 2 * 2 and 6 is 2*3
So we need to eliminate the common multipliers to get the lowest solution.
So drop a 2, and we are left with 2 * 2 * 3 = 12.
Therefore, 12 is the first acceptable answer.
This may be an urban legend. I read the exact same purported CC question in the comments section of a Breitbart article, except there it was “Sally” giving out “balloons”. According to Twitchy, the source is a tweeter who got it from a ‘friend’ who is a teacher. Does anyone have a link that goes to a more original source?
Basically it’s a least common multiple problem. What’s the least common multiple of 4 and 6? Answer is 12, since 3 x 4 = 12, and 2 x 6 = 12.
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