Step 1: Simplify Each Square Root
First, simplify each square root term by factoring out perfect squares.

Quantity A:

$$
\(\begin{aligned}
\sqrt{160} & =\sqrt{16 \times 10}=4 \sqrt{10} \\
\sqrt{40} & =\sqrt{4 \times 10}=2 \sqrt{10}
\end{aligned}\)
$$


So,
Quantity $\(\mathrm{A}=4 \sqrt{10}+2 \sqrt{10}=6 \sqrt{10}\)$
Quantity B:

$$
\(\sqrt{110}\)
$$


So,

$$
\(\text { Quantity } B=\sqrt{110}+3 \sqrt{10}\)
$$


Step 2: Approximate the Values
To compare $\(6 \sqrt{10}$ and $\sqrt{110}+3 \sqrt{10}$, let's approximate $\sqrt{10}$ and $\sqrt{110}\)$.

1. Calculate $\(\sqrt{10}\)$ :

$$
\(\sqrt{9}=3 \quad \text { and } \quad \sqrt{16}=4\)
$$


Since 10 is closer to $9, \sqrt{10} \approx 3.162$.

2. Calculate $\(\sqrt{110}\)$ :

$$
\(\sqrt{100}=10 \quad \text { and } \quad \sqrt{121}=11\)
$$


Since 110 is closer to $\(100, \sqrt{110} \approx 10.488\)$.

3. Compute the Quantities:

Quantity A: $\(6 \sqrt{10} \approx 6 \times 3.162=18.972\)$

Quantity B: $\(\sqrt{110}+3 \sqrt{10} \approx 10.488+3 \times 3.162=10.488+9.486=19.974\)$

Step 3: Compare the Approximations

- Quantity $\(\mathrm{A} \approx 18.972\)$
- Quantity B $\(\boldsymbol{\approx} 19.974\)$

Thus, Quantity B > Quantity A.