1. ## Dumb Dad Needs Help!!!

Hi all,

The older my son gets, the more stupid I feel when trying to help him with math. He is in 7th grade and here is the "Problem of the Week".

Maxine has a piece of ribbon that measures 10 feet, 1 1/2 inches.
She cuts it into 5 pieces.
The second piece is twice as long as the first.
The third piece is 5 inches less than twice the length of the second piece.
The fourth piece is half the length of the second piece
The fifth piece is equal to the sum of the lengths of the first and second piece.
How long is each piece?

This one has got me scratching my head all night and all day today. Can someone please give me a clue where I should even start with this one? There must be a common denominator, but I can't seem to find it. Thanks in advance!

2. Originally Posted by scottk2224

Hi all,

The older my son gets, the more stupid I feel when trying to help him with math. He is in 7th grade and here is the "Problem of the Week".

Maxine has a piece of ribbon that measures 10 feet, 1 1/2 inches.
She cuts it into 5 pieces.
The second piece is twice as long as the first.
The third piece is 5 inches less than twice the length of the second piece.
The fourth piece is half the length of the second piece
The fifth piece is equal to the sum of the lengths of the first and second piece.
How long is each piece?

This one has got me scratching my head all night and all day today. Can someone please give me a clue where I should even start with this one? There must be a common denominator, but I can't seem to find it. Thanks in advance!
well, let's call the first piece $a$, the second piece $b$, the third $c$, the fourth $d$ and the fifth $e$

we know that all the pieces must add up to 10feet, 1 1/2 inches, which is 121 1/2 inches. which is 243/2 inches. so we have:

$a + b + c + d + e = \frac {243}2$

since the second piece is twice as long as the first, we have:

$b = 2a$

since the third piece is 5 inches less than twice the length of the second piece, we have:

$c = 2b - 5 = 2(2a) - 5 = 4a - 5$
since the fourth piece is half the length of the second piece, we have:

$d = \frac b2 = \frac {2a}2 = a$

and finally, since the fifth piece is equal to the sum of the lengths of the first and second piece, we have:

$e = a + b = a + 2a = 3a$

so we see that, $a + b + c + d + e = a + 2a + (4a - 5) + a + 3a = \frac {243}2$

now solve for $a$. once you do that, finding the other pieces is easy. just use the formulas i gave above

3. Thanks Jhevon,

It is becoming more clear to me, but I still don't quite get it. I do understand you logic now, but in order to figure out "a", do I need to first multply 243 by 8, 16, 32, or 64 to get an equal number when I divide?

Am I correct in saying that the solution is 243 / 11 (a) - 5? I don't think so.

I am just so used to working in 10th's that I am getting confused with the inches and feet system.

Any additional hints would be most appreciated.

4. Hello, Scott!

This can be solved with one variable if we're careful.
. . That's probably what's expected in 7th grade.

Sorry . . . I'll correct it now.

Maxine has a piece of ribbon that measures 10 feet, 1½ inches.
She cuts it into 5 pieces.
Let ${\color{blue}x}$ = length of the first piece.

The second piece is twice as long as the first.
Then: . ${\color{blue}2x}$ = length of the second piece.

The third piece is 5 inches less than twice the length of the second piece.
Then: . $2(2x) - 5 \:=\:{\color{blue}4x - 5}$ = length of the third piece.

The fourth piece is half the length of the second piece.
Then: . $\frac{1}{2}(2x) \:=\:{\color{blue}x}$ = length of the fourth piece.

The fifth piece is equal to the sum of the lengths of the first and second piece.
$x + 2x \:=\:{\color{blue}3x}$ = length of the fifth piece.

How long is each piece?
The total length is: .10 ft, 1½ inches = $\left(120 +\frac{3}{2}\right)\text{ inches} \:=\:{\color{red}\frac{243}{2}}$ inches.

Our equation is: . $x + 2x + (4x-5) + x + 3x \:=\:\frac{243}{2}\quad\rightarrow\quad 11x - 5 \:=\:\frac{243}{2}$

. . $11x \:=\:\frac{253}{2}\quad\Rightarrow\quad x \:=\:\frac{23}{2}$

Therefore: . $\begin{Bmatrix}\text{first piece:} & x & = & \frac{23}{2} & = & 11\frac{1}{2} \\ \\
\text{second piece:} & 2x & = & 2\left(\frac{23}{2}\right) & = & 23 \\ \\
\text{third piece:} & 4x-5 & = & 46 - 5 & = & 41 \\ \\
\text{fourth piece:} & x & = & \frac{23}{2} & = & 11\frac{1}{2} \\ \\
\text{fifth piece:} & 3x & = & \frac{69}{2} & = & 34\frac{1}{2} \end{Bmatrix} \text{ inches}$

5. I think I get it now. I have only 2 questions with this statement;

The total length is: .10 ft, 1½ inches = inches

1- When converting inches (120) into 8ths, do I also count the "1" of the 1/8 in which I am using to divide? I was under the impression that I did not and therefore my conversion would be 960/8.

2- 10 feet equals 120 inches. But, I still have that nasty little 1 1/2 inches at the end of the ribbon. So should my initial calculation be 972/8 because I need to add an additional 12 8ths to account for the additional 1 1/2 inches?

Please let me know if I am correct or not. Thanks you very much for your help.

6. Originally Posted by Soroban
The total length is: .10 ft, 1½ inches = $\left(120 +\frac{1}{8}\right)\text{ inches} \:=\:\frac{961}{8}$ inches.
what were you doing here exactly? we have 120 inches + 1 inch + (1/2) inch. why is that 961/8 inches?

7. I got it!!!!!!!!!

Conversion into 8ths

120 inches = 960
1 inch = 8
1/2 inch = 4
Total 972

Piece 1 = x
Piece 2 = 2x
Piece 3 = 4x - 5
Piece 4 = x
Piece 5 = 3x

Equation is (x + 2x + (4x -5) + x = 3x) ---> 11x – 5 = 972

11x = (972 + 40) = 1012

1012 / 11 = 92 (x = 92)

Piece 1 = 92/8
Piece 2 = 184/8
Piece 3 = 328/8 (368 – 40)
Piece 4 = 92/8
Piece 5 = 276/8

Therefore

Piece 1 = 11 ½”
Piece 2 = 23”
Piece 3 = 41”
Piece 4 = 11 ½”
Piece 5 = 34 ½”

Thanks for your help. This old guy learned something today!

8. Originally Posted by scottk2224
I got it!!!!!!!!!

Conversion into 8ths

120 inches = 960
1 inch = 8
1/2 inch = 4
Total 972
ah, Soroban used 8ths as his units. ok

...don't really see why that was necessary though