SOLUTION: Please help me, I have filled many scratch papers trying to solve this!
One car starts out from town C at 8pm traveling towards town D at an average rate of 40mph. A 2nd car start
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Question 719003: Please help me, I have filled many scratch papers trying to solve this!
One car starts out from town C at 8pm traveling towards town D at an average rate of 40mph. A 2nd car starts out from town D at 11am traveling towards town C at an average rate of 45mph. At what time of day will they meet if towns C and D are 460 miles apart?
Found 2 solutions by josmiceli, josgarithmetic:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The key question is:
" How much of a head start does the 1st car have?"
In hours, the 1st car's head start is 8 PM to 12 PM = 4 hrs
plus 12 PM to 11 AM = 11 hrs hrs
The 1st car's head start is
mi
This doesn't make sense. The 1st car has gone past
town D before the 2nd car leaves D, so it can't possibly
meet the 2nd car.
Unless the answer is that they don't meet, then one of the
numbers must be wrong.
This is my read, anyway.
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
BIG PROBLEM: SEE BELOW!
Deciding how to organize the data is what makes this problem more difficult than usual. My solution is not the exactly typical R T D table.
Instead of x AM or y PM, try counting time in hours starting from a 0 point. 8PM is t=0 and 11AM is t at 15. We must also use rate*time=distance.
Car_________Rate_________Distance when time 15____________Distance time 15+t
Early fast___40______________________________________
Late slow_____45_________0________________________________
TOTAL DISTANCE____________________________________________
Regardless of when either car leaves its starting point, they meet when the sum of distances is 460 miles, and the expressions to sum are now written and in the table.
This is the essential starting equation for the problem:
. Solve for and add this to the "11AM" time designation.
BIG PROBLEM: Instead of finding the equation just given as meaningful, it cannot be correct because the problem description is flawed. The the 15 hour time that the fast early car drives and before or at the time the second car begins, the fast car will have gone 600 miles. This is much farther than the distance between the two towns. Are you SURE you did not maybe give wrong start times for these two cars?
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