7th grade algebra problem

[kstar]

Here's how I checked my work with the 382 solution:


Start with 382.

Give away 1/2 (191) + 1 = 192

Left with 190.

Give away 1/2 (95) + 1 = 96

Left with 94.

Give away 1/2 (47) + 1 = 48

Left with 46.

Give away 1/2 (23) + 1 = 24

Left with 22.

Give away 1/2 (11) + 1 = 12

Left with 10.

Give away 1/2 (5) + 1 = 6

Left with 4.

Give away 1/2 (2) + 1 = 3

Left with 1.

[911Rob]

Originally Posted by jyl

........ All the "A" students get "C"s on those, unless their parents help out. I'm not quite sure what the point of it is.

To prove her friggin' existence?
I had several instructors like this in college.
One guy almost got fired for trying to fail a few of us on a mid-term. 3 of us passed the exam and close to 100 failed. (statics)

[aigel]

I do not think there is an algebra solution that you can solve with a single equation. I also do not think the algebra teacher intended kids to brute force the problem bottom up without using algebra. Here is my suggested solution - probably along the lines what a good 7th grader can do:

C = number of chocolates

1st Kid gets C/2 + 1
Left are C/2 - 1

2nd kid gets (C/2 - 1)/2 + 1 = C/4 - 1/2 + 1 = C/4 + 1/2
Left are (C/2 - 1)/2 - 1 = C/4 - 1/2 - 1 = C/4 - 3/2

3rd kid
Left are (C/4 - 3/2)/2 - 1 = C/8 - 3/4 -1 = C/8 - 7/4

4th kid
Left are (C/8 - 7/4)/2 -1 = C/16 - 15/8

You can keep calculating it or you can see the pattern now ...

5
left are C/32 - 31/16

6
left are C/64 - 63/32

7
left are one choclate or by our algebra C/128 - 127/64 = 1

Solve the equation:

C/128 = 1 + 127/64 = 191/64

C = 128*191/64 = 382

Quirky teacher indeed ...

George

[jyl]

It sort of looks like art and George used the same method? I like it, you see the pattern hidden in the progression.

I don't know if this teacher will like the "work backward" answer, she may prefer your method. She and I have had some email exchanges before on what is the "appropriate" solution to a homework problem.

Confession - I first tried the "write a single equation" approach. Took about 1 minute to decide that was going tp be too tedious. Then I drew a picture of a ruler being repeatedly bisected but the cuts being offset by 1, and started to vaguely grasp the pattern George showed. But fractions confuse me, and I also realized that any intuitive pattern would have to be laboriously proved (this teacher is big on "show your work"). George, in your solution, starting at "1st kid", I only had the patience for the next three lines - I'm lazy. So the idea of working backwards seemed easier, because I could simply type a formula into excel, copy the row seven times, and voilà.

Pretty sad, to pass up the elegant solution that reveals the hidden pattern, for the expedient one that lends itself to a spreadsheet.

Of course, if she gets the hint, my poor daughter will have to work it with a calculator.

This hotel is noisy. Hard to sleep.

[jyl]

For extra credit:

Same problem, except you have "n" friends. How many chocolates did you start with? In other words, what is the general formula for the answer?

(this wasn't part of her homework, its just extra credit for PPOT)

[masraum]

Is she currently studying "series" in algebra?. This looks like a series problem more than normal algebra.

[masraum]

I suspect the teacher may be looking for something like, assuming this is what they are studying.



Not that it give the correct answer.

[sammyg2]

Originally Posted by trekkor

Now for a seventh grade question:

"How will I ever use this"?


KT

LOL like I tell my 15 year old daughter when I'm helping her with stuff like this:
it isn't important to KNOW how to do this. It's important to LEARN how to do this.

Just like lifting weights builds muscle, working out problems like this builds brain power and teaches us how to learn.
That way when we finally get out of school and go into the real world, we'll be albe to learn more quickly how things really work. That is what a good education is all about.

[Willem Fick]

Working backwards gives you x (or 0+1x); 2+2x; 6+4x; 14+8x etc.

From this follows that you have a series that can be expressed as follows:

lim (n=7) of [2^(n+1)]+x(2^n)-2 where x=1

Solves for 382

[masraum]

Originally Posted by Willem Fick

lim (n=7) of [2^(n+1)]+x(2^n)-2 where x=1

Solves for 382

That's it, that's what I was looking for.

[dhoward]

It's important to know if you've got a bunch(!) of chocolates and like each one of your friends 1/2 +1 as much as the other...(?)

[Willem Fick]

Hated math, but love a challenge in logic!

[myamoto1]

My calc teacher in college liked to do similar "tests". This of course provoked the "what will I use this for" question from another student. My prof replied, "it'll help you determine how much dirt is in a hole." To which I replied, "I don't need calculus for that. There's no dirt in a hole, that's what makes it a hole." Needless to say, that set the tone for the rest of the semester. I was also able to prove to him how he had lost nearly $250k on a real estate deal he thought he was "making a killing on." I walked out of that class with a solid "D", but I also wasn't $250k light in the pockets either....

[masraum]

Originally Posted by myamoto1

My calc teacher in college liked to do similar "tests". This of course provoked the "what will I use this for" question from another student. My prof replied, "it'll help you determine how much dirt is in a hole." To which I replied, "I don't need calculus for that. There's no dirt in a hole, that's what makes it a hole." Needless to say, that set the tone for the rest of the semester. I was also able to prove to him how he had lost nearly $250k on a real estate deal he thought he was "making a killing on." I walked out of that class with a solid "D", but I also wasn't $250k light in the pockets either....

Nice, yeah, most prof/teachers don't like students to argue or prove them wrong.

[Clay Perrine]

The real world results of this problem would be that you would get the crap beat out of you by half your friends because you were not fair in how you split the chocolates.

[kang]

Originally Posted by jyl

In case you want to test your math skills against the kids in my daughter's algebra class.

You have seven friends and an unknown quantity of chocolates. You give the first friend half of your chocolates, plus one. You give the second friend half of your remaining chocolates, plus one. You give the third friend half of your remaining chocolates, plus one. And so on. After you have given chocolates to all seven friends, you have one chocolate left. How many chocolates did you start with?

And how did you solve the problem?

If you have an unknown quantity of chocolates, how do you know what "half of your chocolates, plus one" is? How do you give the first friend one half plus one, unless you know the total quantity?

[911Rob]

I'll bet that teacher doesn't mess with JYLs kid.

I teach business to high school kids; I love to show them how what I learned in school applies to the real world. The teachers love me for it too

Just like right now.... I'm taking my pilots license course; damn happy I paid attention in school, makes it a whole lot easier to understand all of it.

[afterburn 549]

Anal BS
They call it higher learnen LOL
I will use this example of how to solve......................tragictory....nope
pies are sq...nope
logs and woodpiles and rhythms...nope
check book....nope
structural loading.......nope
I give up
that must be it !!

[jyl]

Maybe you do it by weight? You're being a troublemaker. "D" for you!

If you have an unknown quantity of chocolates, how do you know what "half of your chocolates, plus one" is? How do you give the first friend one half plus one, unless you know the total quantity?