We are given the $x$-th term:

$$
\(a_x=(-1)^{x+1} \cdot \frac{1}{2^x}\)
$$

for $\(x=1,2, \ldots, 10\)$.
Let's write the first few terms:

$$
\(\begin{aligned}
& x=1:(-1)^2 \cdot \frac{1}{2}=+\frac{1}{2} \\
& x=2:(-1)^3 \cdot \frac{1}{4}=-\frac{1}{4} \\
& x=3:(-1)^4 \cdot \frac{1}{8}=+\frac{1}{8} \\
& x=4:(-1)^5 \cdot \frac{1}{16}=-\frac{1}{16}
\end{aligned}\)
$$

and so on.
So the sequence is:

$$
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\cdots\)
$$

up to 10 terms.

Step 1: Recognize as geometric series
Group terms in pairs:

$$
\(\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{32}-\frac{1}{64}\right)+\cdots\)
$$


Each pair:

$$
\(\begin{gathered}
\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \\
\frac{1}{8}-\frac{1}{16}=\frac{1}{16} \\
\frac{1}{32}-\frac{1}{64}=\frac{1}{64}
\end{gathered}\)
$$


So sum of 10 terms $\(=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\frac{1}{1024}\)$.

Step 2: Sum this new series
This is geometric with first term $\(\frac{1}{4}\)$, ratio $\(\frac{1}{4}, 5\)$ terms:

$$
\(\begin{gathered}
S_{\text {pairs }}=\frac{1}{4} \cdot \frac{1-(1 / 4)^5}{1-1 / 4}=\frac{1}{4} \cdot \frac{1-1 / 1024}{3 / 4} \\
\quad=\frac{1}{4} \cdot \frac{1023 / 1024}{3 / 4}=\frac{1}{4} \cdot \frac{1023}{1024} \cdot \frac{4}{3}
\end{gathered}\)
$$


Cancel $\(\frac{1}{4} \cdot 4=1\)$ :

$$
\(S_{\text {pairs }}=\frac{1023}{1024 \cdot 3}=\frac{1023}{3072}\) .
$$


So $\(S=\frac{1023}{3072}\)$.

Step 3: Compare with $\(\frac{1}{4}\)$

$$
\(\frac{1}{4}=\frac{768}{3072}\)
$$


We have $\(\frac{1023}{3072}>\frac{768}{3072}\)$, so $\(S>\frac{1}{4}\)$.

Step 4: Conclusion
Quantity A $\(=S\)$
Quantity B $\(=\frac{1}{4}\)$
Quantity A is greater.