R mows lawn in 1 hr less than E; finish in 5 hrs together.

[kingozzie]

Robert can mow the lawn in one hour less time than Ed. Together they can finish the job in 5 hours. How long would it take for Robert to finish the lawn alone?

I have seen two different anwsers to this problem. What do you see?

 

[Loren]

Re: Mow the Lawn

Show us your work that you think is correct. If there is something wrong, we can probably help you find it.

 

[Denis]

I have seen two different anwsers to this problem. What do you see?

I "see" that you're fishing for an answer :shock:

 

[kingozzie]

Why would the answer not be 9.5 hours? If 9.5249, wouldn't the average between the two be more than 5?

I want to porve the CWRU student question is not exact.

{-(-9)+-sqrt[(-9)^2 -4(1)(-5)]}/2(1)
[9+-sqrt(81+20)]/2
[9+-sqrt(101)]/2

 

[galactus]

You're correct. Robert does it in 9.52 hours by himself. Then, Ed does it in 10.52 hours.

\(\displaystyle \frac{1}{9.52}+\frac{1}{10.52}\approx.2=\frac{1}{5}\)

Check.

It does not work off of averages.

 

[wjm11]

Robert can mow the lawn in one hour less time than Ed. Together they can finish the job in 5 hours. How long would it take for Robert to finish the lawn alone?

Why would the answer not be 9.5 hours? If 9.5249, wouldn't the average between the two be more than 5?

{-(-9)+-sqrt[(-9)^2 -4(1)(-5)]}/2(1)
[9+-sqrt(81+20)]/2
[9+-sqrt(101)]/2

(To elaborate on Galactus' explanation

[9+sqrt(101)]/2 is correct. Robert’s rate of work is (1/9.5249) lawns per hour. Ed’s rate is (1/10.5249) lawns per hour.

(1/9.5249 lawns per hour)(5 hrs) = .5249 lawns
(1/10.5249 lawns per hour)(5 hrs) = .4751 lawns

Together they complete one lawn in 5 hrs.

Avg rate of work:

(1/9.5249 + 1/10.5249)/2 = .2/2 = 1/10 lawns per hr.

If Robert cut ½ of the lawn, then stopped working, and Ed cut the other half, then their combined times would be over 10 hours (and the avg would be over 5). BUT Robert does not stop working; he cuts more than half the lawn, while Ed cuts less than half the lawn. They both work for 5 hours for a combined time of exactly 10 hours.

Hope that helps.

 

[kingozzie]

[/quote]If Robert cut ½ of the lawn, then stopped working, and Ed cut the other half, then their combined times would be over 10 hours (and the avg would be over 5). BUT Robert does not stop working; he cuts more than half the lawn, while Ed cuts less than half the lawn. They both work for 5 hours for a combined time of exactly 10 hours.

Hope that helps.[/quote]

Then would this statement be false: " Together they can finish the job in 5 hours"? True statement or false?

 

[Deleted member 4993]

If Robert cut ½ of the lawn, then stopped working, and Ed cut the other half, then their combined times would be over 10 hours (and the avg would be over 5). BUT Robert does not stop working; he cuts more than half the lawn, while Ed cuts less than half the lawn. They both work for 5 hours for a combined time of exactly 10 hours.

Hope that helps.[/quote]

Then would this statement be false: " Together they can finish the job in 5 hours"? True statement or false?[/quote]

In 5 hours Robert cuts slightly more than half the lawn..............\(\displaystyle _(\frac{1}{2} + x)\)

In 5 hours Ed cuts slightly more than half the lawn..............\(\displaystyle _(\frac{1}{2} - x)\)

Two together would finish it in 5 hour ..............\(\displaystyle (\frac{1}{2} - x) \, + \, (\frac{1}{2} + x) \, = \, 1\)