Please Help Fast!!!!

[Wendy Daniels]

I need to know step-by-step how to solve this problem, and I need it real fast.

"A piece of wire was cut into thirds. One-third was used. One-fifth of the remaining wire was used. The piece that remains is 16 feet long. What was the original length of the wire?"

 

[Gene]

The wire is L long.
(L/3) + 1/5*(2L/3) + 16 = L

 

[Wendy Daniels]

Gene,
I really need more help. I know the answer is 30 feet long, but I'm having a hard time seeing how to get 30 with the formula you have because I've tried something similar to yours but still did not get 30.
Thanks,
WD

 

[stapel]

What did you do, exactly?

Please reply showing your steps.

Thank you.

Eliz.

 

[Gene]

Try multiplying thru by 15 (to get rid of the fractions) and see if that helps.

 

[Wendy Daniels]

Gene and Eliz

This is what I did:

x + 1/3 + (2/3 x 1/5) = 16
But it did not result in 30.
What am I doing wrong?

 

[galactus]

You left x be the original length of the wire. You need to subtract, not add, to arrive at 30.

\(\displaystyle x-\frac{1}{3}x-(\frac{1}{5})(\frac{2}{3})=16\)

\(\displaystyle x-\frac{1}{3}x-\frac{2}{15}x=16\)

\(\displaystyle \frac{8}{15}x=16\)

\(\displaystyle 8x=240\)

\(\displaystyle x=30\)

 

[Gene]

Since galactus worked it out his way, what I gave you was
(L/3) + 1/5*(2L/3) + 16 = L
Times 15
5L+2L+15*16=15L
240=8L
L=30
Same answer, different equation.
-----------------
Gene