We are told the average of 3 unequal quotations is Rs.110. So:
\[
\frac{x + y + z}{3} = 110 \Rightarrow x + y + z = 330
\]
We are to find whether the highest of the three exceeds 129.
From Statement I: The lowest quotation is Rs.100
Let’s assume $x = 100$ (lowest).
Then $y + z = 230$
But without knowing either $y$ or $z$, we can’t determine if the maximum exceeds 129.
Try values like $y = 100$, $z = 130$ (✓) or $y = 115$, $z = 115$ — but these could be equal. So I alone is insufficient.
From Statement II: One quotation is Rs.115
That helps us plug in one variable, but without knowing the lowest, the other two could still be anywhere.
Combining I and II:
Let $x = 100$, $y = 115$
So $z = 330 - 215 = 115$
But now $y = z = 115$, which violates the condition that all quotations are unequal. So $z \neq 115$
Try $z = 115 + 1 = 116$
Then $x + y + z = 100 + 115 + 116 = 331$ too high
So we reduce $z$: try $z = 114$ → total = 329 ⇒ average = 109.67
Eventually, we can find that the only valid setup with $x = 100$, $y = 115$ will force $z$ to be $115$, making two equal. To maintain inequality, the third must be higher than 115 and raise the total above 330.
Hence, for the total to remain 330, and values to be unequal integers, one of the values must go above 129 to balance lower entries like 100 and 115.
Therefore, both statements together are needed.