Jack, Kay and Lynn deliver flyers in a small town. If each person works alone, it takes Jack 4 hours to deliver all the flyers, and it takes Lynn 1 hour longer than it takes Kay. Working together, they can deliver all the flyers in 40% of the time it takes Kay working alone. How long does it take Kay to deliver all the flyers alone?
2007-09-01 01:43:58 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
For n flyers to deliver,
Jack took 4 hours, rate = n/4 flyers per hour
Kay took x hours, or rate = n/x flyers per hour
Lynn took (x+1) hours or n/(x+1) flyers per hour
Together they took 0.4x hoursn or n/(0.4x) flyers per hour
n/(0.4x) = n/4 + n/x + n/(x+1)
1/(0.4x) = 1/4 + 1/x + 1/(x+1)
1/(0.4x) = ( x(x+1) + 4(x+1) + 4x ) / (4x(x+1))
4x(x+1)/(0.4x) = x^2 + x + 8x + 4)
10(x+1) = x^2 + x + 8x + 4)
x^2 + 9x - 10x + 4 - 10 = 0
x^2 - x - 6 = 0
(x - 3) (x + 2) = 0
x = 3 or x = -2
forget x = -2 (meaningless)
Kay took x = 3 hours to deliver the flyers
2007-09-01 02:00:48 · answer #1 · answered by vlee1225 6 · 1⤊ 0⤋
I don't think you have enough information to answer the question. The fact that, working together, they can deliver all the flyers in 40% of the time one of them can do it alone doesn't tell us anything useful about their individual performances.
2007-09-01 08:56:26 · answer #2 · answered by sdc_99 5 · 0⤊ 1⤋
i think your ques is wrong because :
a + b + c = 0.4 c
4 + c-1 + c = 0.4 c
3 + 2c = 0.4 c
2c = 0.4 c - 3
so 2c never equal 0.4 only when c =0 but in this question
2c = 0.4 c - 3 so c never be a plus number
good luck
2007-09-01 09:25:25 · answer #3 · answered by $coTT flores 2 · 0⤊ 0⤋
approximately 4.7 hours
2007-09-01 09:01:27 · answer #4 · answered by Japhet 2 · 0⤊ 1⤋
2hrs
2007-09-01 08:51:54 · answer #5 · answered by Anonymous · 0⤊ 1⤋