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------------------------------------------------ Ed and Sana run at constant speeds. Ed can run 1 mile in 9 minutes, and Sana can run 3 miles in 45 minutes. Ed and Sana have the same starting point and run in the same direction, but Ed starts running 36 minutes after Sana.

Determine when Ed will catch up to Sana.

This is a mathematical problem about ratios and time differentials, comparing the speeds of two runners. Ed runs faster but starts later than Sana. Using calculations, we find that Ed will catch up with Sana 60 minutes after he starts running.

Explanation:

The problem presented is a classic mathematics problem involving ratios and time differentials. First, we need to understand the speeds of Ed and Sana. Ed can run 1 mile in 9 minutes, thereby his speed is 1/9 miles per minute. Sana, on the other hand, runs 3 miles in 45 minutes, therefore her speed is 3/45 miles per minute, simplifying to 1/15 miles per minute.

Now, we know Sana starts running 36 minutes before Ed, so we can calculate how far she runs in that time (1/15 * 36 = 2.4 miles). When Ed starts running, they are 2.4 miles apart.

Now, we calculate when they'll meet. Ed is catching up with Sana at a rate of 1/9 - 1/15 miles per minute.

So, the time they will meet is the distance Sana is ahead divided by the rate Ed is catching up, i.e. (2.4 / ((1/9) - (1/15))) which equals 60 minutes. Hence, Ed will catch up with Sana 60 minutes after he starts running.

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The complete question is: Ed and Sana run at constant speeds. Ed can run 1 mile in 9 minutes, and Sana can run 3 miles in 45 minutes. Ed and Sana have the same starting point and run in the same direction, but Ed starts running 36 minutes after Sana. is: