+0

# algebra

0
283
4

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?

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???

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.

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
edited by 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 !

May 11, 2021