A 100-foot rope is cut into three pieces. The first piece is three times as long as the second, and the third piece is 30 feet shorter than the second. How many feet long is the shortest piece?

Guest May 11, 2021

#1**0 **

A 100-foot rope is cut into three pieces. The first piece is three times as long as the second, and the third piece is 30 feet shorter than the second. How many feet long is the shortest piece?

Let First Piece = $p_1$

Let Second Piece = $p_2$

Let Third Piece = $p_3$

$p_1 + p_2 + p_3 = 100$

$p_1 = 3p_2$

$p_3 = p_2 - 30$

$3p_2 + p_2 + p_2 - 30 = 100$

$5p_2 = 130$

$p_2 = 26$

$p_1 = 78$

$p_3 = -4$

Where did I go wrong???

MathProblemSolver101 May 11, 2021

#2**+1 **

Say the second piece is length x. The first piece is 3 times that, so it is 3x.

The 3rd piece is 20 feet shorter than it, so it is x-20.

Adding them we get 5x-20, and that equals 100.

Solving for x we get 24.

So the lengths are 72, 24 and 4.

KOBE2482 May 11, 2021

#4**0 **

Yes....these numbers work.....BUT the question says the 3rd piece is 30 feet shorter then the second NOT 20 feet as you solved

so the question is incorrect....not possible !

But like you ...If we modify the question somewhat IF the question says third piece is 30 ft less than the FIRST piece then it would work

55.714 ft 18.5714 ft and 25.714 ft WOULD be the three pieces

ElectricPavlov
May 11, 2021

#3**+1 **

f = 3s and t = s-30 and the three pieces sum to 100 ft

3s + s + s-30 = 100

5s = 130

s = 26 then f = 3s = 78 and t = 12 Which tells me this question is incorrect !

ElectricPavlov May 11, 2021