Since \(t_9 = t_7 + t_8\), we need to use substitution and algebra to find \(t_7\).

We're given that \(t_5 = 18\). We aren't given what \(t_4\) is, but we can treat it as an unknown variable for now.

\(t_6 = t_4 + t_5 = t_4 + 18\)
\(t_7 = t_5 + t_6 = 18 + (t_4 + 18) = t_4 + 36\)
\(t_8 = t_6 + t_7 = (t_4 + 18) + (t_4 + 36) = 2t_4 + 54\)

Given that \(t_8 = 76\), solve for \(t_4\):
\(2t_4 + 54 = 76\)
\(t_4 = 11\)

Substitute \(t_4 = 11\) back in to find \(t_7\):
\(t_7 = 11 + 36 = 47\)

Now we can calculate \(t_9\):
\(t_9 = t_7 + t_8 = 47 + 76 = 123\)

Since 123 < 126, Quantity B is greater.

Just consider T5, T6, T7 and T8. T5 is 18, and T8 76. Give T6 and T7 a and b. Then a+b = 76 and 18+b=a. Through cancelling operation leads us to have 47 for b and 29 for a, the sum of which is exactly 76. More important is the sum of the 47 and 76 that is less than 126.