Let $x$ be the amount invested in the fund at $5 \%$ per annum.
Let $y$ be the amount invested in the fund at $8 \%$ per annum.
From the problem statement, we know two things:
1. The total amount invested was $\$ 5000$ :

$$
\(x+y=5000 \text { (Equation 1) }\)
$$

2. The total interest earned from both funds at the end of 2001 was $\(\$ 315\)$ :

Interest from 5\% fund + Interest from 8\% fund $\(=\$ 315\)$

$$
\((x \times 0.05)+(y \times 0.08)=315(\text { Equation } 2)\)
$$


Now we have a system of two linear equations with two variables. We can solve for $x$ and $y$.
From Equation 1, we can express $y$ in terms of $x$ :

$$
\(y=5000-x\)
$$


Substitute this expression for $y$ into Equation 2:

$$
\(\begin{aligned}
& 0.05 x+0.08(5000-x)=315 \\
& 0.05 x+400-0.08 x=315
\end{aligned}\)
$$


Combine the $x$ terms:

$$
\(\begin{aligned}
& 0.05 x-0.08 x=315-400 \\
& -0.03 x=-85
\end{aligned}\)
$$


Solve for $x$ :

$$
\(\begin{aligned}
& x=\frac{-85}{-0.03} \\
& x=\frac{85}{0.03} \\
& x=\frac{8500}{3} \\
& x=2833.33 \text { (approximately) }
\end{aligned}\)
$$


Now, substitute the value of $x$ back into Equation 1 to find $y$ :

$$
\(\begin{aligned}
& y=5000-x \\
& y=5000-2833.33 \\
& y=2166.67 \text { (approximately) }
\end{aligned}\)
$$


So, Quantity A (Amount invested in fund at $5 \%$ per annum) $\(=x \approx 2833.33\)$
Quantity B (Amount invested in fund at 8\% per annum) $\(=y \approx 2166.67\)$
Comparing Quantity A and Quantity B:

$$
\(2833.33>2166.67\)
$$


Therefore, Quantity A is greater than Quantity B.
The final answer is Quantity A is greater.