Question 108163
The second problem looks like you combined data from two entirely different problems.  But I can help you with the first problem.
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The difference in start times between the two runners was 1/2 hour.  That means that the first runner was 0.5 hr * 6 mph = 3 miles ahead of the second when the second runner started.  Now the difference in their speeds is 1 mph (7 mph - 6mph), which means that the second runner will close the distance between himself and the first runner at a rate of 1 mph.  That means that it will take 3 hours (3 miles / 1 mph) for the second runner to catch up.
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The relationship to remember when you are dealing with these types of problems is {{{d=rt}}}, where d is the distance travelled, r is the rate of speed, and t is the elapsed time.  Just remember to keep the units of measure consistent, i.e. if distance is in miles, and time is in hours, then rate must be expressed in miles per hour.
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Depending on what values you are given, and what you are trying to determine, you can rearrange the formula:
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{{{r=d/t}}}, or
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{{{t=d/r}}}
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The two airplane trips described in the second half of your 2nd problem could be set up as:
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Let the distance from Airport 1 to Airport 2 be d1, and the return trip distance, from Airport 2 to Airport 1 be d2, but those two distances are equal, since the airport locations are fixed.
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Let the time for the outgoing trip be t1, and the time for the return trip be t2.
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We know that {{{t1 + t2 = 5}}}.  We know that {{{d1 = 100*t1}}}, and {{{d2=150*t2}}}.
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{{{t2=5-t1}}}, so substitute that value for t2 in the return trip equation.
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{{{d2=150(5-t1)}}}
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Now, set the two right sides equal to each other because d1 = d2.
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{{{100t1=150(5-t1)}}}
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I'll let you do the solving work, but you should get t1 = 3 hours and t2 = 2 hours.