# SOLUTION: At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads a

Algebra ->  Algebra  -> Rate-of-work-word-problems -> SOLUTION: At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads a      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Word Problems: Rate of work, PAINTING, Pool Filling Solvers Lessons Answers archive Quiz In Depth

 Question 95345This question is from textbook prentice hall mathematics algebra 1 : At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads after Tom. If she travels at a rate of 30 miles per hour, when will she catch up? I don't understand how to solve the problem. I dont even know where to begin. Could you please help with steps and tips of how to solve this problem?This question is from textbook prentice hall mathematics algebra 1 Answer by ankor@dixie-net.com(15745)   (Show Source): You can put this solution on YOUR website!At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads after Tom. If she travels at a rate of 30 miles per hour, when will she catch up? : Let t = T's travel time to when M catches up : Since M left 10 min later we can say: (change 10 min to hrs 10/60 = 1/6) (t-(1/6) = M's travel time : One thing we do know is that when M catches up with T they will have traveled the same distance. (This applies to all these "catch-up" problems) : That means we can make a distance equation: Dist = speed * time : M's dist = T's dist 30(t-(1/6)) = 25t 30t - (30/6) = 25t 30t - 5 = 25t 30t - 25t = +5 5t = 5 t = 1 hr + 1:30 = 2:30 Pm when M catches up with T : we can check this to see if they do, indeed, travel the same distance in the time we found. M's time = (5/6) hr and T's time = 1 hr 30*(5/6) = 25 mi 25 * 1 = 25 mi : Did this make sense to you? Any questions?