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First method: Let `t` represent time in hours (since the units are in miles per hour). Say the man starts running at `t=0.` After `t` hours, the man runs `12t` miles from his starting point. After `t` hours, his wife has run `10(t+1/4)` miles, because remember, she started running 15 minutes, which equals `1/4` hour, before him.
So we solve `12t=10(t+1/4).` Distribute the right side to get `12t=10t+2.5,` and bringing the variable to the left side gives `2t=2.5.` Dividing by `2` gives `t=1.25` hours, or equivalently, `75` minutes.
Second Method: These problems can often be made simpler by changing how you view them. Think about it--he runs `2` mph faster than her, so if she stops and waits after `15` minutes and he runs to her at `2` mph, he'll catch her in the same amount of time as he would if they ran `12` mph and `10` mph respectively.
So if she goes a certain distance in `15` minutes at `10` mph, he'll go that same distance in five times that long at `2` mph, since she ran five times as fast as him. Thus it will take him `75` minutes. No algebra at all if you do it that way!
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