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Hey Raya,
Good question - you can do this one a couple of ways. My personal preference has always been:
Use two equations for Rate = Distance/Time (one for each person/car/train) and then consolidate your D and T variables, which always have a relationship. Here:
Scott travels for 1/2 hour more than Jack, so his time could be T + 1/2 if Jack's is T
If we call Scott's distance D, then Jack's is 60 - D, since both distances have to add to 60.
You'd then have:
R(Jack) = 15 = D/T
R(Scott) = 12 = (60-D)/(T+1/2)
If we sub the first equation into the second (the first can be expressed as D = 15T, and we can then put 15T in for D in the second), we'd have:
12 = (60 - 15T)/(T + 1/2)
12T + 6 = 60 - 15T
27T = 54
T = 2 hours
If they meet after 2 hours, then they meet at 3:00.
Even though it can be slightly more time consuming, I like this method because it works for pretty much any rate/distance/work problem, so it's a one-stop shop.
To use the additive rates property on this one, you'd need to account for the fact that they're only working together for part of the 60 mile route, because Scott rides alone for 30 minutes. So you'd first need to figure out how far Scott goes in 30 minutes:
R = D/T
12 = D/(1/2)
6 = D, so Scott goes 6 miles
Then we can combine rates to see how long it takes them, together, to travel the 60-6 = 54 miles remaining.
Combined rate = 12 + 15 = 27 miles/hour
R = D/T
27 = 54/T
27T = 54
T = 54/27 = 2 hours
Again, we get 2 hours past the 1:00 start time, so the answer is B.
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