# Thread: time problem. but i'm stuck when computing for the answer.

1. ## time problem. but i'm stuck when computing for the answer.

Jack, Kay and Lynn deliver advertising flyers in a small town. If each person works alone, it takes Jack 4h to deliver all the flyers, and it takes Lynn 1h 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?

2. Originally Posted by ejaykasai
Jack, Kay and Lynn deliver advertising flyers in a small town. If each person works alone, it takes Jack 4h to deliver all the flyers, and it takes Lynn 1h 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?
One way to tackle this problem is to convert the given time for each person to rates. Assuming that all 3 of then are given the same number of flyers to be delivered,

Jack takes 4 hours to deliver all the flyers so in 1 hour time, he's able to deliver 1/4 of the flyers alone. Also, Kay and Lynn deliver 1/x and 1/(x+1) of the flyers respectively in one hour. If they work together, t hours are required so in one 1 hour time they are able to deliver 1/t of the flyers.

Mathematically, solve

1/4 +1/x + 1/(x+1) = 1/(0.4x) where t=0.4x

3. (I'm assuming that all of the work is done at a constant pace.)

Jack takes $4$ hours to deliver flyers alone. In one hour, he delivers $\frac{1}{4}$ of the flyers.

Kay takes $x$ hours to deliver flyers alone. In one hour, she delivers $\frac{1}{x}$ of the flyers.

Lynn takes $x + 1$ hours to deliver flyers alone. In one hour, she delivers $\frac{1}{x + 1}$ of the flyers.

Working together, they can deliver $\dfrac{1}{4} + \dfrac{1}{x} + \dfrac{1}{x + 1}$ portion of the flyers in one hour.

We're also told that working together would take $0.4x$ hours (40% of Kay's time) to deliver the flyers, so in one hour they deliever $\frac{1}{0.4x} = \frac{5}{2x}$ of the flyers.

So the equation is:
$\dfrac{1}{4} + \dfrac{1}{x} + \dfrac{1}{x + 1} = \dfrac{5}{2x}$

Solve for x:
\begin{aligned}
4x(x + 1) \left( \frac{1}{4} + \frac{1}{x} + \frac{1}{x + 1} \right) &= 4x(x + 1) \left( \frac{5}{2x} \right) \\
x(x + 1) + 4(x + 1) + 4x &= 10(x + 1) \\
x^2 + x + 4x + 4 + 4x &= 10x + 10 \\
x^2 + 9x + 4 &= 10x + 10 \\
x^2 - x - 6 &= 0 \\
(x - 3)(x + 2) &= 0
\end{aligned}

x - 3 = 0 OR x + 2 = 0
x = 3 OR x = -2
You can reject the -2, so it takes Kay 3 hours to deliver the flyers alone.

EDIT: Much too slow! :P