Let's use algebra to solve the problem.
Let's assume that there are x chairs in total.
From the first condition, we know that:
30 < x < 70
From the second condition, we know that if she put 8 chairs in one row, she would be 7 chairs short. This can be represented by the equation:
x = 8a - 7, where a is the number of rows.
Similarly, from the third condition, we know that if she put 7 chairs in one row, she would have two chairs left. This can be represented by the equation:
x = 7b + 2, where b is the number of rows.
Now we can set these two equations equal to each other, since they both represent the same number of chairs:
8a - 7 = 7b + 2
Simplifying this equation, we get:
8a - 7b = 9
We want to find a solution where a and b are both positive integers, since we are dealing with rows of chairs. We can start by trying different values of a and seeing if we get a corresponding integer value of b that satisfies the equation.
Let's try a = 2:
8(2) - 7b = 9
16 - 7b = 9
7b = 7
b = 1
This gives us a solution where there are 16 chairs (2 rows of 8 chairs) and 1 chair left over. However, we know that there are more than 30 chairs, so we need to try a larger value of a.
Let's try a = 5:
8(5) - 7b = 9
40 - 7b = 9
7b = 31
b = 31/7
This is not a whole number, so a = 5 does not work.
Let's try a = 6:
8(6) - 7b = 9
48 - 7b = 9
7b = 39
b = 39/7
This is also not a whole number, so a = 6 does not work.
Let's try a = 7:
8(7) - 7b = 9
56 - 7b = 9
7b = 47
b = 47/7
This is also not a whole number, so a = 7 does not work.
Let's try a = 8:
8(8) - 7b = 9
64 - 7b = 9
7b = 55
b = 55/7
This is also not a whole number, so a = 8 does not work.
Finally, let's try a = 9:
8(9) - 7b = 9
72 - 7b = 9
7b = 63
b = 9
This gives us a solution where there are 72 chairs (9 rows of 8 chairs) and 2 chairs left over when 7 chairs are put in each row. This satisfies all the conditions, so the answer is 72 chairs.