#### poppaqball

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In a run for charity Mary runs at a speed of 4mph. John leaves 12 minutes after Mary and runs 6mph. How long will it take for John to catch up with Mary?

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In a run for charity Mary runs at a speed of 4mph. John leaves 12 minutes after Mary and runs 6mph. How long will it take for John to catch up with Mary?

Hi poppaqball,poppaqball said:

In a run for charity Mary runs at a speed of 4mph. John leaves 12 minutes after Mary and runs 6mph. How long will it take for John to catch up with Mary?

Let's let t = the time Mary runs at the rate of 4 mph.

Then, John leaves 12 minutes later. Convert 12 minutes to hours and you get .2 hours.

This means John will travel (t - .2) at the rate of 6 mph.

When he catches up with Mary, they will have both run the same distance.

Using D = rt, Mary's distance is stated as D = 4t. John's distance is stated as D = 6(t - .2).

Setting the two distances equal to each other, we have:

\(\displaystyle 4t=6(t-.2)\)

Solve for t, which is in hours, and then convert to minutes.

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This is the second time shes done things backwards...unless I'm missing something.

Nevermind, Thanks for the help! I see what the problem was, I was solving for John not Mary.

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You need to find the time John runs.

Masters is also correct - but after finding 't' in his solution (Mary's running time), you have to find John's running time.