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OpenStudy (anonymous):

An express train travels 426 miles between two cities. During the first 120 miles of a trip, the train traveled through mountainous terrain. The train traveled 11 miles per hour slower through mountainous terrain than through level terrain. If the total time to travel between the cities was 9 hours, find the speed of the train on level terrain.

OpenStudy (anonymous):

Let S be the speed during level terrian. Let t be the time taken to go on level terrian. Then you have: S*t = 306 (S-11)*(9-t) = 120 (S-11)(9-306/S) = 120 (S-11)(9S - 306) = 120S Solve for S?

OpenStudy (anonymous):

Wait where the 306 come from?

OpenStudy (anonymous):

S = 51 mph seems like the root (or answer). t = 306/51 = 6 hours. 11 mph less through mountains means 40 mph. It will take 3 hours for 120 miles. Total travel time = 6 + 3 = 9. Tada!!

OpenStudy (anonymous):

426 minus 120.

OpenStudy (anonymous):

OK?

OpenStudy (anonymous):

I dont really understand it? :/

OpenStudy (anonymous):

Let me explain. We are given a bunch of things. But, we are missing two pieces of information. #1: How fast is the train going on "level" terrain #2: How long did it take to travel through the level terrain. To find two unknowns, you need two equations. Thankfully, there is enough information to formulate two equations.

OpenStudy (anonymous):

First equation is based on the fact that out of the 426 miles, the first 120 miles were awful (mountains). That tells us that the level terrain was 306 miles long. If the train travelled through this part of the terrian at speed S and took time T, then S*T must equal 306. So, equation #1: S*T = 306. Second equation is based on the mountain terrain. We know the train travelled 11 mph slower there. So, its speed was S-11. We also know the total journey was 9 hours. So, the train took 9-T hours through this messy terrain. We also know that the distance of 120 miles. Then, speed * time = distance gives us the second equation. So, equation #2: (S-11)(9-T) = 120

OpenStudy (anonymous):

Now, substitute for T in equation #2 as 306/S (based on equation #1). You get: (S-11)(9-306/S) = 120 That can be re-written as: (S-11)(9S - 306) = 120S That gives you a quadratic in S. Solve for S to find the root to be 51. That means S = 51 miles per hour.

OpenStudy (anonymous):

Hmm makes some more sense thank you

OpenStudy (anonymous):

51 is the answer

OpenStudy (anonymous):

\[r t = d, t=\frac{d}{r} \]Let r be the trains speed on level terrain. Solve the following for r:\[\frac{120}{r-11}+\frac{426-120}{r}=9 \]\[r=51 \text{ mph} \]The other solution for r, 22/3, is too small.

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