To determine how long it will take the students at Vienna Elementary to earn more than $1,000 total, we first need to calculate the total amount earned in the first 6 weeks and then find the pattern of their earnings.
From the data provided:
- Week 1: $110
- Week 2: $100
- Week 3: $90
- Week 4: $110
- Week 5: $100
- Week 6: $90
Now let's calculate the total earnings for the first 6 weeks:
\[ \text{Total for 6 weeks} = 110 + 100 + 90 + 110 + 100 + 90 = 600 \]
Next, we observe the pattern of earnings over 6 weeks:
- The pattern is: $110, $100, $90$
- This pattern repeats every 3 weeks.
Total earnings for one complete pattern (3 weeks):
\[ \text{Total for 3 weeks} = 110 + 100 + 90 = 300 \]
Now we will find out how many complete 3-week cycles (groups of 3 weeks) are needed to exceed $1,000.
Starting with the total earnings after 6 weeks:
\[ \text{Earnings after 6 weeks} = 600 \]
And we need to know how many additional cycles of 300 it takes to exceed $1,000.
\[ 1000 - 600 = 400 \]
Now, divide the remaining amount by the total of one cycle (300):
\[ \frac{400}{300} \approx 1.33 \]
This means they need more than 1 complete cycle. Since they can't do a fraction of a cycle, they will need 2 more cycles (3 weeks each), adding a total of 6 additional weeks.
This gives us:
\[ 6 \text{ weeks} + 6 \text{ weeks} = 12 \text{ weeks} \]
After those 12 weeks, the students will have earned over $1,000.
Therefore, the answer would be: 12 weeks.
However, since the provided options are not beyond 10 weeks, it could be misinterpreted. As such, based on the provided options, the best one reachable with a bit further context is 10 weeks (which would be the total of 8 or some approximation), but not on its perfect return as we calculated above - please check on the options about any possible missing weekly cyclic aspects or approximations to $1,000.
Since the answer to the question of your options must be reconsidered if based on limits of 11 reported weeks, the best might be confusing but computed beyond would rule up at least to 10 at that limit if cycles are structured only in a special way on counts of 2.