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A passenger train leave the station three hours after a freight train left the same station. The freight train travels thirty mph slower than the passenger train. Find the rate of each, if the passenger train catches up with the freight train in four hours.

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What have you done? –  Pragabhava Nov 13 '12 at 1:15
By which Pragabhava means, what are your thoughts? What have you tried so far? And not, "My god, what have you done?" :O –  Rahul Nov 13 '12 at 3:45

3 Answers 3

For fun, we first solve the problem without "algebra." The passenger train gains $30$ miles every hour, so it gains $120$ miles before catching up. Thus the freight must have had a lead of $120$ miles when the passenger train got underway. It obtained that lead in $3$ hours, so travelled at $\dfrac{120}{3}$ miles per hour.

Or else we can let $s$ be the speed of the freight train. Then the speed of the passenger train is $s+30$.

The passenger train has travelled $4$ hours, so the freight train has travelled for $7$. Thus $$4(s+30)=7s.$$ Solve for $s$.

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Hint: let $p$ be the speed of the passenger train and $f$ be the speed of the freight. Let $t$ be the time in hours from when the freight leaves. Write equations for the position of each train as a function of time. Equate those positions and solve for time.

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Assume that spped of freight train is v As given passanger train should caught that train in 4 hrs so distance traveeled should be same 4(v+30)=7v 3v=120 v=40m/h

so passanger spped is v+30=70m/h

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