Work and Time

  • Thread starter ritwik06
  • Start date
  • #1
580
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Homework Statement


Bill and Jill are hired to paint a line on the road. If Bill works by himself, he could paint the line in B hours. If Jill works by herself, she could paint the line in J hours. Bill starts painting the line from one end and Jill begins painting the line from the other end one hour later. They both work until the line is painted. Find the expression for the number of hours that Bill works.


The Attempt at a Solution



One hour work of Bill and Jill= 1/B and 1/J respectively.

Time taken by both, if both start working simultaneously= BJ/(B+J) ......(i)

As Jill starts work an hour later, work still left= 1/J

Time taken by Bill to complete the remaining work=(1/J)/(1/B)=B/J ........(ii)
Total time for which Bill had to work= (B(J^2+J+B))/(J(B+J)) ...........(i)+(ii)

But my answer doent match th given options, where is the mistake?
 

Answers and Replies

  • #2
361
21
As Jill starts work an hour later, work still left= 1/J


Wouldn't the work still left = B-1
 
  • #3
454
0
Wouldn't the work still left = B-1

No. If Jill works one hour less, an amount of work of 1/J is left undone.
What's wrong is that both Bill and Jill work to finish of this last (1/J) part of the line.
 
  • #4
361
21
The way I read it, Bill completes a total of one hour of work before Jill begins. So if you look at the remaining work after 1 hour in terms of B then it is B-1 hour. I guess that might be better if written as B-(1/B) since Bill has completed 1/B out of B. Do we agree that 1 hour of Bills work is 1/B ?
 
  • #5
580
0
Thanks Kamerling!!! and please I would ask others not to misguide me please. Thanks again kamerling
 
  • #6
361
21
and please I would ask others not to misguide me please.

Ha Ha, Ok I'll Bite.


[tex]\frac{B-(1/B)}{1/B+1/J}+1[/tex]

This is what I got. What did you end up with.
 
  • #7
580
0
Ha Ha, Ok I'll Bite.


[tex]\frac{B-(1/B)}{1/B+1/J}+1[/tex]

This is what I got. What did you end up with.

I got this:
[tex]\frac{B(J+1)}{B+J}[/tex]
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,847
964

Homework Statement


Bill and Jill are hired to paint a line on the road. If Bill works by himself, he could paint the line in B hours. If Jill works by herself, she could paint the line in J hours. Bill starts painting the line from one end and Jill begins painting the line from the other end one hour later. They both work until the line is painted. Find the expression for the number of hours that Bill works.


The Attempt at a Solution



One hour work of Bill and Jill= 1/B and 1/J respectively.

Time taken by both, if both start working simultaneously= BJ/(B+J) ......(i)

As Jill starts work an hour later, work still left= 1/J

Time taken by Bill to complete the remaining work=(1/J)/(1/B)=B/J ........(ii)
Total time for which Bill had to work= (B(J^2+J+B))/(J(B+J)) ...........(i)+(ii)

But my answer doent match th given options, where is the mistake?

I think your basic reasoning is flawed. Your first calculation is the amount of time it would take to do the entire road if they both worked the entire time. Then you add to that the time it takes Bill to do one hour's worth of Jill's work. That does not follow.

I would do it this way:
Let t be the time Bill works. Then the time Jill works is t- 1. Together they paint (1/B)t+ (1/J)(t-1) fraction of the road. Since you want the time it takes to paint the entire road, (1/B)t+ (1/J)(t-1)= 1. Solve that equation for t.
 
Last edited by a moderator:
  • #10
1,041
4
Product over sum?
 

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