Let the 10 consecutive multiples of 7 be:

$$
\(7 k, 7(k+1), 7(k+2), \ldots, 7(k+9)\)
$$

where $k$ is an integer.

We can write the set X as:

$$
\(\begin{aligned}
& X=\{7 k, 7 k+7,7 k+14,7 k+21,7 k+28,7 k+35,7 k+42,7 k+49,7 k+56,7 k+ \\
& 63\}
\end{aligned}\)
$$


Quantity A: Mean of set X
The mean of an arithmetic progression is the average of its first and last term.

$$
\(\begin{aligned}
& \text { Mean }=\frac{\text { First Term }+ \text { Last Term }}{2} \\
& \text { Mean }=\frac{7 k+(7 k+63)}{2} \\
& \text { Mean }=\frac{14 k+63}{2} \\
& \text { Mean }=7 k+\frac{63}{2} \\
& \text { Mean }=7 k+31.5
\end{aligned}\)
$$


Quantity B: Median of set X
Since there are 10 numbers in the set (an even number of terms), the median is the average of the two middle terms.
The middle terms are the 5th and 6th terms when the set is ordered.
The 5 th term is $\(7(k+4)=7 k+28\)$.
The 6th term is $\(7(k+5)=7 k+35\)$.

$$
\(\begin{aligned}
& \text { Median }=\frac{5 \text { th Term }+6 \text { th Term }}{2} \\
& \text { Median }=\frac{(7 k+28)+(7 k+35)}{2} \\
& \text { Median }=\frac{14 k+63}{2} \\
& \text { Median }=7 k+\frac{63}{2} \\
& \text { Median }=7 k+31.5
\end{aligned}\)
$$

Comparison:
Quantity A: $7 k+31.5$
Quantity B: $7 k+31.5$

Both Quantity A and Quantity B are equal.
General Rule:
For any arithmetic progression (a set of numbers where the difference between consecutive terms is constant), the mean and the median are always equal. Since consecutive multiples of 7 form an arithmetic progression, their mean and median must be equal.

The final answer is The two quantities are equal.