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Ted is a long-distance driver. It took him 30 min longer to drive 275 km on the Trans-Canada Highway west of Swift current, than it took him to drive 300km due east of Swift Current. He averaged 10km/h less while travelling west of Swift Current due to more severe snow conditions. What was Ted’s average speed for each part of the trip?

User AdrienF
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Let x be Ted's east leg speed. West leg speed is x-10. Time difference for 25km is 1hr, so West = 275/(x-10) = 300/x + 0.5. Solve this quadratic for x (east speed) and subtract 10 for west speed.

Here's how to solve for Ted's average speed in each part:

1. Find the time difference: Since 30 minutes (0.5 hours) correspond to a 25-km difference (300-275 km), Ted spent 1 hour per 100 km driving west vs. the east leg.

2. Set up the equations: Let "v" be Ted's speed on the east leg. His speed on the west leg would be "v-10" due to slower conditions. We can then set up equations based on distance and time:

East leg: 300 km / v = time taken (t1)

West leg: 275 km / (v-10) = time taken (t2)

3. Solve for time: We know t2 = t1 + 0.5 (30 minutes extra). Substitute t1 from the first equation into the second:

275 km / (v-10) = 300 km / v + 0.5

4. Solve for v: This equation leads to a quadratic equation. Solving it provides two possible solutions, but only one makes sense (positive speed). This solution gives us "v" (east leg speed).

5. Calculate west leg speed: Once you have "v", subtract 10 to find Ted's average speed on the west leg (v-10).

Therefore, Ted's average speed was v km/h on the east leg and v-10 km/h on the west leg. The specific values for v can be obtained by solving the quadratic equation mentioned in step 4.

By following these steps, you can accurately determine Ted's average speed for each part of his trip, accounting for the difference in driving conditions.

User Jdmonty
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