First discount:
Grace gets a 20\% discount on the sales price.
Price after first discount $\(=S-0.20 S=0.80 S\)$

Second discount:
Grace then gets an $x \%$ discount on the reduced price (0.80S).
Discount amount $=x \%$ of $\(0.80 S=\frac{x}{100} \times 0.80 S\)$
Price after second discount $\(=0.80 S-\frac{x}{100} \times 0.80 S=0.80 S\left(1-\frac{x}{100}\right)\)$
Net discount:
The series of two successive discounts is equivalent to a net $40 \%$ discount on the sales price.
Price after net $40 \%$ discount $=S-0.40 S=0.60 S$

Equating the final prices:
The price after the two successive discounts must be equal to the price after the net $40 \%$ discount:

$$
\(0.80 S\left(1-\frac{x}{100}\right)=0.60 S\)
$$


Divide both sides by S (assuming $\mathrm{S}>0$ ):

$$
\(0.80\left(1-\frac{x}{100}\right)=0.60\)
$$


Divide both sides by 0.80 :

$$
\(\begin{aligned}
& 1-\frac{x}{100}=\frac{0.60}{0.80} \\
& 1-\frac{x}{100}=\frac{6}{8} \\
& 1-\frac{x}{100}=\frac{3}{4}
\end{aligned}\)
$$


Now, solve for x :

$$
\(\begin{aligned}
& \frac{x}{100}=1-\frac{3}{4} \\
& \frac{x}{100}=\frac{4}{4}-\frac{3}{4} \\
& \frac{x}{100}=\frac{1}{4}
\end{aligned}\)
$$


Multiply both sides by 100:

$$
\(\begin{aligned}
& x=\frac{100}{4} \\
& x=25
\end{aligned}\)
$$


Compare Quantity A and Quantity B:
- Quantity A: $x=25$
- Quantity B: 20

Since $\(25>20\)$, Quantity A is greater.
The final answer is $\(A\)$