Math Whizzes Needed for HW Help, Please!

[majimikate]

I am embarrassed to admit this...I am usually very good at Math, majored in Accounting in college....but a 4th grade math homework problem has me stumped . I know there is some easy answer to this...?I just can't think of it . This is not supposed to be solved as an algebraic equation, they have not started algebra yet...

Here it is:

3000 books are arranged into 3 piles.
The first pile has 10 more books than the second pile.
The number of books in the second pile is twice the number of books in the third pile.
How many books are there in the third pile?

HELP!!! and Thanks

 

[Mrs.Reese]

May I never teach fourth grade....

Step 1- Divide 3000 by 3 to get 1000 in each stack.
Step 2- Take 10 from stack 2 and put it in stack one - stack one has 1010 and stack 2 now have 990
Step 3 - Divide 990 by two- stack 3 has 495 books.

I solved by drawing a picture of 3 books and then going line by line. I hope I'm right.

I am obviously not a math whiz....

 

[DisneyMomOK]

Pile 1 is 1206 books, Pile 2 is 1196 and Pile 3 is 598 books.

Third pile would be one pile (one set of 598 books), pile 2 would be 2 sets of books (pile 3 times 2/598 *2 =1196) and pile one would be pile 2 plus 10 (1196 + 10). 3000 books less the 10 books (2990) would be the number that is divided by 5.

Sharon

 

[BCDisneyFanatic]

I am embarrassed to admit this...I am usually very good at Math, majored in Accounting in college....but a 4th grade math homework problem has me stumped . I know there is some easy answer to this...?I just can't think of it . This is not supposed to be solved as an algebraic equation, they have not started algebra yet...

Here it is:

3000 books are arranged into 3 piles.
The first pile has 10 more books than the second pile.
The number of books in the second pile is twice the number of books in the third pile.
How many books are there in the third pile?

HELP!!! and Thanks

Let x represent the number of books in the third pile

Therefore, the number of books in the second pile is 2x

The number of books in the third pile is 2x + 10

therefore, x + 2x + 2x+10 = 3000
5x + 10 = 3000
5x = 2990
x = 598

There are 598 books in the third pile.

 

[Mnementh]

All of those explanations are still algebraic. I found a website that basically has the exact same problem on page three, if you can view PDF files (http://math.coe.uga.edu/tme/Issues/v14n1/v14n1.Beckmann.pdf) with an explanation of how to handle that sort of a problem non-algebraically. I couldn't really make heads or tails of what they were saying, but maybe you can. I'll continue to look to see if I can find anything more clear. Good luck!

 

[majimikate]

Excellent work on the algebra guys!! Thank you! But does anyone know how I solve it with out the algebra???? I need to explain it to a 9 year old who hasn't started algebra yet, so it seems there must be some other way...
It's in a math chapter about multiplying and dividing 2 digit numbers....thank god I already passed 4th grade !!!

 

[DisneyMomOK]

Actually, I think my solution was much like the referenced website is using. I edited my original response, I used piles too often, rather than sets, but this is a problem where logic is necessary. Using variables is algebraic, but this question uses a definite number of books (3000 less the extra 10) which is then sorted into three piles, two of which are twice as many as pile 3, a total of 5 sets of books.

Sharon

 

[Serena]

I was trying to figure it out making 4 piles and going from there. I was still figuring it out when I decided to see if anyone else had an answer.
But I see I didn't even have the question right. lol

 

[KariC]

From Kari's husband:

If the second pile is X, then the first pile is X+10, and the third pile is .5X, all summing to 3000. That means

(X+10) + X +.25X = 3000, which means
2.5X + 10 = 3000, which means
2.5X = 2990
X = 1196
so the third pile has 1196 * .5 = 598.

As a proof of this

(1196 + 10) + 1196 + 598 = 3000

From a 4th grader perspective, the trick is to say the piles are not evenly divided, and that because the 3rd pile is half the size you really only have 2.5 piles, so, 3000 / 2.5 is your base number of 1200 for a regular pile, and 600 for the half-pile. Then, you would take from 5 from the second pile and add it to the first pile to create a difference of 10. Then you would realize that you can't divide the second pile 2 and get a whole number, so you would have to add 1 back (or take one away) and do the same for the first pile. Then you would have to add or take away increments of 2 from the last pile. If you were to go up the first time, you would discover the total is greater than 3000, so you would reverse it and go down to 598. That is the way to do it without algebra, which if you are doing it would only take 2 iterations to arrive at. Yes, this way is kind of backwards, but that is all I can think of in how you would approach it without doing any formulas.

Hope this helps

 

[BCDisneyFanatic]

The number of books in pile three is equal to one unit.

The number of books in pile two is equal to two units.

The number of books in pile three is equal to two units plus 10.

Picture a number line...mark out 3000. Subtract 10 to show the last 10 added to Pile 1. The remaining 2990 books can be divided equally into one unit for Pile 1, two units for Pile 2, and two units for Pile 3 (in other words, the number line is divided into five equal parts between 0 and 2990).

Each unit then ends up being 598 books.

This is my interpretation of the explanation from the article on Singapore math listed above. Don't know if this helps, or that it makes explaining the problem any easier!

Heather

 

[majimikate]

Thank you all for your excellent help...DisneyMom OK...you've got it all going on!!!! I totally get how you all solved it...but can you imagine explaining that to a 9 year old ?!?!?!?!?!?!?!? This is a new math series this year, the Singapore series (great work finding the site Mnementh!! that was the exact problem, but what kind of help was that they were giving us ????)

Thank you all so much...the DIS family ROCKS!!!

 

[KariC]

Ok, so to show how this would work from piles (iterations)
3000 in 2.5 piles gets you started with:

1200 + 1200 + 600 = 3000, then get the difference of 10 working
1205 + 1195 + 600 = 3000, then get the 3rd pile at 1/2 of the second pile
OOPS, can't get a whole number from 1195 / 2, so adjust the piles, round down or up...let's go the wrong way for giggles
1204 + 1194 + 1194/2 (which is 597) = 2995, go the other direction, you must add 2 to the second pile to divide by 2, so you get as a trial and error
1206 + 1196 + 1196/2 (which is 598) = 3000

That is how you get there without "algebra"....i guess