SOLUTION: Jack, Kay, and Lynn deliver advertising flyers in a small town. If each person works alone, it takes Jack 2 h to deliver all the flyers, and it takes Lynn 5 h longer than it takes

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Jack, Kay, and Lynn deliver advertising flyers in a small town. If each person works alone, it takes Jack 2 h to deliver all the flyers, and it takes Lynn 5 h longer than it takes       Log On


   

Question 1132886: Jack, Kay, and Lynn deliver advertising flyers in a small town. If each person works alone, it takes Jack 2 h to deliver all the flyers, and it takes Lynn 5 h longer than it takes Kay. Working together, they can deliver all the flyers in 60% of the time it takes Kay working alone. How long does it take Kay to deliver all the flyers alone?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39838) About Me  (Show Source):
Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Let k be the number of hours Kay takes to do the job alone; then k+5 is the number of hours it takes Lynn alone. We are given that it takes Jack 2 hours to do the job.

So the fractions the three of them do alone in 1 hour are 1/2, 1/k, and 1/(k+5).

Working together, it takes the three of them 60% as long as it takes Kay alone. Since it takes them 3/5 as long as Kay working alone, in 1 hour the fraction of the job they get done together is 5/3 as much as Kay alone does in 1 hour. Then the equation to solve is

1%2F2+%2B+1%2Fk+%2B+1%2F%28k%2B5%29+=+%285%2F3%29%281%2Fk%29

Multiply everything by the LCM of the denominators, 6k(k+5):

3k%28k%2B5%29%2B6%28k%2B5%29%2B6k+=+10%28k%2B5%29
3k%5E2%2B15k%2B6k%2B30%2B6k+=+10k%2B50
3k%5E2%2B17k-20+=+0
%283k%2B20%29%28k-1%29+=+0

Obviously choose the positive solution, k=1.

ANSWER: It takes Kay 1 hour to deliver all the flyers alone.