Question

You know that a rope is no more then a hundred feet long. You need to cut the rope into three pieces. The second piece is to be three times as long as the first piece, and the third piece must be 18 ft long. What is the maximum length of the second piece? Show work.

Answer 1

Answer:

The maximum length of the second piece is 61.5 ft.

Step-by-step explanation:

We know that the rope is 100 ft long.

There are three pieces that make up the 100 ft. The first is length x, the second is three times as long as the first, so it would be 3x, and the third is 18ft.

So if we were to write this as an equation it would look something like this: x + 3x + 18 = 100. We can simplify this some more.

4x + 18 = 100

4x = 82

x = 20.5

Now that we know the length or the first piece / x. We can easily find the length of the second (3x) by plugging in 20.5 for x.

We find that the maximum length of the second piece of rope is 61.5 feet.

Good luck, hope that helps!

Answer 2

Answer:

61.5 feet

Step-by-step explanation:

Given:  a rope is no more then a hundred feet long. You need to cut the rope into three pieces. The second piece is to be three times as long as the first piece, and the third piece must be $18$ $\text{feet}$ long.

To Find: What is the maximum length of the second piece.

Solution:

maximum length of rope $(\text{m})=100$ $\text{feet}$

let first piece be $=\text{x}$

let second piece be $=\text{y}$

let third piece be $=\text{z}$

Now,

it is given that

$\text{y}=3\text{x}$

$\text{x}=\frac{\text{y}}{3}$

$\text{z} =18$ $\text{feet}$

$\text{x}+\text{y}+\text{z}\leq 100$

putting value

$\frac{\text{y}}{3}+\text{y}+18\leq 100$

$\frac{4\text{y}}{3}\leq 82$

$\text{y}\leq\frac{3\times82}{4}$

$\text{y}\leq 61.5$ $\text{feet}$

maximum length of second part of rope is $61.5$ $\text{feet}$