Here's a tough one, help!

awilhelm wrote:Two women start walking toward each other from two ends of a bridge at the same time. They both leave at dawn. When they meet, it is 12pm. When the first woman reaches the other end of the bridge, it is 4pm. When the second woman reaches the end of the bridge, it is 9pm. Both women walk at different, but constant, rates. What time is dawn?

We could plug in the answer choices and our own value for the length of the bridge.

Answer choice: dawn = 6am.
Time for first woman = 4pm - 6am = 10 hours.
Time for second woman = 9pm - 6am = 15 hours.
Plug in bridge = 30 meters.
Rate for first woman = d/t = 30/10 = 3 meters per hour.
Rate for second woman = d/t = 30/15 = 2 meters per hour.
Combined rate for the two women as they walk toward each other = 3+2 = 5 meters per hour.
Time for them to meet = d/(combined rate) = 30/5 = 6 hours.
6am + 6 hours = 12pm. Success!

Awesome, Mitch. Thank you! Could I ask you to walk me through how you would solve this problem without guessing and plugging in? I know that's kind of a pain, but I would be very grateful!

awilhelm wrote:Awesome, Mitch. Thank you! Could I ask you to walk me through how you would solve this problem without guessing and plugging in? I know that's kind of a pain, but I would be very grateful!

I'm doing this against my better judgment; I think plugging in the answers is much easier and safer.

Let t = time for the two women to meet at 12pm
Total time for first woman who finishes at 4pm = t+4
Total time for second woman who finishes at 9pm = t+9
Rate for first woman = 1/(t+4)
Rate for second woman = 1/(t+9)
Combined rate for the two women to meet after t hours = 1/t

Since (rate for first woman) + (rate for second woman) = (combined rate for the two women), we get:

1/(t+4) + 1/(t+9) = 1/t

(t+9) + (t+4) / (t+4)(t+9) = 1/t

(2t + 13) / (t² + 13t + 36) = 1/t

Cross multiplying, we get:
2t² + 13t = t² + 13t + 36
t² -36 = 0
(t+6)(t-6) = 0
t = ±6

Since t>0, t=6.

Since it takes t=6 hours for the women to meet, dawn = 12pm - 6 = 6am.

I much prefer to plug in as I did in my initial response. When you look at the two approaches, which seems easier?

Amazing, thank you! One final question: I don't understand how you are able to express each woman's rate as 1/(t + 4) and 1/(t + 9) and their combined rate as 1/t. How does dividing by 1 give us their rate when we don't know the distance?

Thanks in advance.

awilhelm wrote:Amazing, thank you! One final question: I don't understand how you are able to express each woman's rate as 1/(t + 4) and 1/(t + 9) and their combined rate as 1/t. How does dividing by 1 give us their rate when we don't know the distance?

Thanks in advance.

The numerator could be any value; the result will be the same. Using d=1 makes the algebra less complicated.

Thank you!