Answer: the minimum number of questions that could have been on the third assignment is 18.
Explanation:
Let x be the number of questions on the first homework assignment. Then, the number of questions on the second and third assignments would be x + 1 and x + 2, respectively, since each assignment had one more question than the previous one.
The total number of questions on the three assignments would be:
x + (x + 1) + (x + 2) = 3x + 3
We know that this total is greater than 51, so we can write:
3x + 3 > 51
Subtracting 3 from both sides, we get:
3x > 48
Dividing both sides by 3, we get:
x > 16
This means that the number of questions on the first assignment must be greater than 16 for the total number of questions on the three assignments to be greater than 51.
To find the minimum number of questions that could have been on the third assignment, we want to minimize the expression x + 2, since that represents the number of questions on the third assignment. From the inequality we found earlier, we know that x must be greater than 16. Therefore, the smallest possible value for x + 2 is 18, which occurs when x = 16.
Therefore, the minimum number of questions that could have been on the third assignment is 18.
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