**S**

*LOCAL VS EXPRES*FACTS:

*- 30 mph, 2/3 of the distance from DC to Richmond this train's speed drops to half*

**LOCAL***- 60mph, leaves later than Local*

**EXPRESS**Express usually catches Local right at DC (total distance). Because of difficulty, Express catches Local after 27 and 1/9 miles.

Here we go, full steam ahead:

I will used an algebraic solution, so let

**x**be the distance Local travels from the time the problem arises to the time the Express catches up with it. Also let the distance from DC to Richmond be one unit (this will be explained in a minute). We will now construct an equation.

Let's use our facts. Express normally travels twice as fast as the local but after the mechanical problem it is traveling 4 times as fast. Also since the Express normally catches the Local at the end, the Local has traveled half the distance when the Express departs. Sooooooo, when the Local has traveled 2/3 of the distance the Express has traveled 1/3. Thus the equation:

2/3 + x = 1/3 + 4x.

Solving

x = 1/9 of the way (remember the total distance was one unit).

Now the Local has only 1/3 of the distance still to go, therefore at the time the Express catches the Local there is 1/3 - 1/9 = 2/9.

Remember though the Express catches the Local after 27 and 1/9 miles

Dividing, we get

27 1/9 X 9/2 (invert and mutiply)

We get 122 miles from Richmond to Washington DC

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