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I would like to confirm that what I have done with this basic problem is fine. It says:

540 GB are downloaded during a forenoon at 1900 MB/min.

In the afternoon, the same amount of data took twice the time.

What was the average data download during the whole day?

My solution:

  • Forenoon -> 1900 MB/min -> (540*1024)/1900 ≈ 291 min

  • Afternoon -> 800 MB/min -> 291 * 2 = 582 min

  • 540 GB = 552960 MB

  • Average speed: 552960/(291+582) ≈ 633.40 MB/min

Did I miss something?

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Yes, I would say so. In the afternoon the download took $582$ minutes, so that's what you should divide by at the end. Not $(291 + 582)$. – Arthur Feb 1 '14 at 21:22
@Arthur That is not right. Questions asks for the average during the whole day. – John Habert Feb 1 '14 at 21:28
Ahh, I misread. Sorry. But then you got to add the downloads from the morning as well in the numerator. So it's supposed to be $2\cdot 552,960$. – Arthur Feb 1 '14 at 21:30

Eveything looks good, up until the end.

Try finding the total amount of data downloaded (540 MB in the forenoon plus 540 MB after that = 1080 MB) and then dividing by the total amount of time taken to find the average rate for the day:

$$\dfrac {2\cdot 552960}{(291+582)}$$

So you need to simply double the rate you ended with: $2 \times 633.40$ MB/min = $1266.80$ MB/min.

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I need to be about 16 seconds faster. ;) – John Habert Feb 1 '14 at 21:30

Average speed is $\dfrac{\mathrm{total\ data\ downloaded}}{\mathrm{total\ time}}$. You've got the total time part correct since it is the 291 minutes from forenoon plus twice that afternoon. The total data downloaded though is not 540 GB. It is 1080 GB (540 GB forenoon and 540 afternoon).

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