Smart people! Math question

[Phaedrus]

If you want some fun math puzzles etc. Try Martin Gardner books.

www.amazon.com/s/ref=nb_ss?url=search-alias%3Daps&field-keywords=martin+gardner&x=0&y=0

[Deleted]

While this problem might not involve a lot of what we perceive as "math", it probably stimulates the brain in the same way and opens the students' minds to one more of the possible ways numbers can work.

The fact that it's being graded though seems a bit odd - it doesn't follow any of the logic that these students will probably see in math problems in the coming years...

[Deleted]

I don't know if it's being graded. It was homework, maybe even her first homework, one of six word problems. But I'm not sure what the purpose of it was. I'm waiting to hear.

I was helping her with it, but since I could not see any solution other than systematically going through the possible combinations, I gave up (and I assume she did too). I figured there was something I was missing.

Apparently not.

[mikegarrison]

Maybe it's more a test of the parents.

[Deleted]

It just gets tougher and tougher. It'd be one thing if I were taking the classes and doing the homework, but to just help with homework without knowing really what they are being taught, it's frustrating.

Hell, I was in trouble back when she brought lattice multiplication home. Where the hell did that come from? Seemed pretty simple doing it the way I always did it. This is easier?

[The Bofa on the Sofa]

Maybe it was an exercise in using Excel? Or computer programming?

Actually, I can see it as a reasonable programming assignment.

[pennstate7188]

Haha, it's not that complicated. And each term definitely doesn't represent a number as large as you're thinking lol. Simplify the equation from x/y = 1/5 to:

5x = y

using cross-multiplication (multiply the numerator of the first term by the denominator of the second term, and vice versa, to simplify).

Just plug in numbers for x and you'll find out what y equals. Let's just plug in simple numbers for now:

x = 1, so y = 5
x = 2, so y = 10
x = 3, so y = 15
x = 4, so y = 20
x = 5, so y = 25

See?

Now that we know what x and y both equal, plug those numbers back into the original equation: x/y = 1/5

Now, we can check to see if we're correct:

When x = 5, y = 25, right? So, in our original equation, replace the x with 5 and the y with 25. This is what we get:

5/25 = 1/5

Does that term negate itself or is it a true statement? As far as I can see, it's true! Let's double-check to be sure: cross-multiply the terms to check.

25(1) = 5(5)
25 = 25

It worked! Follow that for whatever other conditions the problem outlined.

[mikegarrison]

Sept 30, 2009 0:52:08 GMT -5 @pennstate7188 said:

Haha, it's not that complicated. And each term definitely doesn't represent a number as large as you're thinking lol. Simplify the equation from x/y = 1/5 to:

5x = y

using cross-multiplication (multiply the numerator of the first term by the denominator of the second term, and vice versa, to simplify).

Just plug in numbers for x and you'll find out what y equals. Let's just plug in simple numbers for now:

x = 1, so y = 5
x = 2, so y = 10
x = 3, so y = 15
x = 4, so y = 20
x = 5, so y = 25

See?

Now that we know what x and y both equal, plug those numbers back into the original equation: x/y = 1/5

Now, we can check to see if we're correct:

When x = 5, y = 25, right? So, in our original equation, replace the x with 5 and the y with 25. This is what we get:

5/25 = 1/5

Does that term negate itself or is it a true statement? As far as I can see, it's true! Let's double-check to be sure: cross-multiply the terms to check.

25(1) = 5(5)
25 = 25

It worked! Follow that for whatever other conditions the problem outlined.

Can you read? I think not.

It's the "other conditions" that are the actual problem, not dividing by 5.

[mikegarrison]

Sept 29, 2009 16:13:12 GMT -5 @ruffda said:

Hell, I was in trouble back when she brought lattice multiplication home. Where the hell did that come from? Seemed pretty simple doing it the way I always did it. This is easier?

I had no idea what this was, so I looked it up. It's functionally the same as the old normal method, just more visual. I suppose it does prevent errors due to losing track of which number is being multiplied by which other number, but it takes up a tremendous amount of space.

[pennstate7188]

I know that that isn't the EXACT answer to the problem, but those theorems and principles can be applied to whatever conditions in order to solve. I didn't see the exact problem written anywhere, so I couldn't assume. And, it's fifth grade math. It definitely isn't going to be as difficult as you think...