Word Problem About Mowing the Lawn- Please Help

[HCL]

2. Peter takes 15 minutes longer to mow the lawn by himself than Charles does. Together they
can mow the lawn in 18 minutes. How long would it take Charles to mow the lawn by himself?
A. 9 minutes
B. 20 minutes
C. 30 minutes
D. 35 minutes



i tried to set this up where x=time it takes charles and x+15=time it takes peter but when it comes to the 18 minutes, I tried addition and multiplication but I'm not getting any of the answers...ugh, i feel like an idiot

 

[stapel]

2. Peter takes 15 minutes longer to mow the lawn by himself than Charles does. Together they can mow the lawn in 18 minutes. How long would it take Charles to mow the lawn by himself?

i tried to set this up where x=time it takes charles and x+15=time it takes peter but when it comes to the 18 minutes, I tried addition and multiplication but I'm not getting any of the answers...ugh, i feel like an idiot

Not being able to see what you've done, it is difficult to find where you might have gone wrong.

Please review at least two lessons from this list, and then re-attempt the exercise. If you get stuck, you can then reply with a clear listing of your efforts so far. Thank you!

 

[HallsofIvy]

2. Peter takes 15 minutes longer to mow the lawn by himself than Charles does. Together they
can mow the lawn in 18 minutes. How long would it take Charles to mow the lawn by himself?

When you have two people working together on a job (or two hoses filling a tank, etc.) it is their rates of work that add.
Let x be the time, in minutes, it takes Charlie to mow the lawn. Then 1/x is the rate, in "lawns per minute" at which Charllie works. It take Peter x+ 15 minutes to mow the same lawn so his rate is 1/(x+ 15) "lawns per minute. Together the work at rate 1/x+ 1/(x+ 15)= 1/18 "lawns per minute"

A. 9 minutes
B. 20 minutes
C. 30 minutes
D. 35 minutes



i tried to set this up where x=time it takes charles and x+15=time it takes peter but when it comes to the 18 minutes, I tried addition and multiplication but I'm not getting any of the answers...ugh, i feel like an idiot

 

[HCL]

When you have two people working together on a job (or two hoses filling a tank, etc.) it is their rates of work that add.
Let x be the time, in minutes, it takes Charlie to mow the lawn. Then 1/x is the rate, in "lawns per minute" at which Charllie works. It take Peter x+ 15 minutes to mow the same lawn so his rate is 1/(x+ 15) "lawns per minute. Together the work at rate 1/x+ 1/(x+ 15)= 1/18 "lawns per minute"

oh ok i get it, i should have thought of this in fractions/rates, thanks!

 

[J_Barber]

When you have two people working together on a job (or two hoses filling a tank, etc.) it is their rates of work that add.
Let x be the time, in minutes, it takes Charlie to mow the lawn. Then 1/x is the rate, in "lawns per minute" at which Charllie works. It take Peter x+ 15 minutes to mow the same lawn so his rate is 1/(x+ 15) "lawns per minute. Together the work at rate 1/x+ 1/(x+ 15)= 1/18 "lawns per minute"

Stumbled upon this and found it interesting so I thought I would try and solve for x based on the formula above. Well after looking at it I have no clue as where to even begin. Can some one give me a hint as to how to go about it?

 

[HallsofIvy]

You are asking how to solve 1/x+ 1/(x+ 15)= 1/18? Multiply both sides by x(x+ 15) to change to a quadratic equation. You can solve the quadratic equation using any of the standard methods, factoring, completing the square, or the quadratic formula.

 

[JeffM]

Stumbled upon this and found it interesting so I thought I would try and solve for x based on the formula above. Well after looking at it I have no clue as where to even begin. Can some one give me a hint as to how to go about it?

Let y = the area of the yard.

Let x = the time that Charlie takes to mow the lawn.

Let z = the time that Peter takes to mow the lawn..

Charlie's rate = \(\displaystyle \dfrac{y}{x}.\)

Peter's rate = \(\displaystyle \dfrac{y}{z}.\)

Joint rate = \(\displaystyle \dfrac{y}{x} + \dfrac{y}{z} = \dfrac{y * z + y * x}{xz} = \dfrac{y(x + z)}{xz}.\)

With me so far?

Now I want to know the time it takes for them to jointly mow the lawn.

rate = area / time so time = area / rate.

Joint time = \(\displaystyle y \div \dfrac{y(x + z)}{xz} = \dfrac{y}{1} * \dfrac{xz}{y(x + z)} = x * \dfrac{z}{x + z}.\)

Does this make sense? Consider if z = x, meaning Peter and Charlie are equally fast. Then the time it takes them jointly is one half x.

Halls example using 18 and 15 was being very specific about the actual problem, not trying to give a general method. In that problem, the joint time was given. This is why I dislike giving formulas. If you do not understand the logic behind the formula, you may use it at the wrong time. If you understand the logic, you do not need a formula. The fundamental point is, as Halls, said, that the joint rate equals the sum of the individual rates. If you remember that, you can solve any problem, whether it involves finding a rate or finding a time.

EDIT: Having seen Halls response, I realize that you may have been asking how to solve a quadratic. If so, my response is irrelevant.