Let's denote:
- Steve's running rate as $s$ miles per hour.
- Barney's running rate as $b$.

Barney runs 1 mph slower than twice Steve's rate:

$$
\(b=2 s-1\)
$$


Steve starts running, and Barney starts 30 minutes later ( 0.5 hour later).
Barney overtakes Steve 3 hours after Barney starts running.
Step 1: Write distance equations
- Distance Steve runs before being overtaken:

$$
\(d_s=s \times(3+0.5)=3.5 s\)
$$

- Distance Barney runs before overtaking:

$$
\(d_b=b \times 3\)
$$


Because Barney overtakes Steve, they have covered the same distance at that time:

$$
\(d_s=d_b\)
$$


Substitute:

$$
\(\begin{gathered}
3.5 s=(2 s-1) \times 3 \\
3.5 s=6 s-3 \\
6 s-3.5 s=3 \\
2.5 s=3 \\
s=\frac{3}{2.5}=1.2 \mathrm{mph}
\end{gathered}\)
$$

Step 2: Find Barney's speed

$$
\(b=2(1.2)-1=2.4-1=1.4 \mathrm{mph} .\)
$$


Step 3: Find distance Barney covered before overtaking

$$
\(d_b=b \times 3=1.4 \times 3=4.2 \text { miles }\)
$$


Step 4: Compare Quantity A and Quantity B
- Quantity A = 4.2 miles
- Quantity B = 4 miles

Quantity A > Quantity B.
Final conclusion:
Quantity A is greater than Quantity B.