math

Oops...I made a "mistype" when I added the rates...("106" is correct, not "102"} Sorry.

(But, the answer is still correct!!)

Quote:

At noon sarah headed from elk city to Idabel at 60 mph. Two hours later, Joan headed from Idabel to elk city at 46 mph. If it is 332 miles from Elk city to Idabel, what time did they meet?

Let Idabel be at 0. Then Elk City is at 332.

Sarah's position is s = 60T

Joan's position is j = 332 - 45(T-2)

solving these we get

60T = 332 - 45(T-2)

105T = 422

T = 422/105 = 4.02 = 4hrs 1.14 minutes

At noon sarah headed from elk city to Idabel at 60 mph. Two hours later, Joan headed from Idabel to elk city at 46 mph. If it is 332 miles from Elk city to Idabel, what time did they meet?

Mmmm....let's look at this again.....

The cities are 332 miles apart - if I'm reading the question correctly. When Joan starts traveling, Sarah has already traveled 120 miles (2 hrs. @ 60 mph). So the distance between them when Joan starts is now just 332 - 120 = 212 miles. Since they're heading towards each other, their combined rates are (60 + 46) = 102 mph.. So, dividing the distance left by the combined rate we get......212 / 106 = 2 hrs.

Thus, Sarah has traveled two hours when Joan starts and it takes her two more hours to meet Joan, so they meet at 4 PM.

I believe that's correct.