Let $x$ be the cost of 1 bottle of water and $y$ be the cost of 1 bottle of cola.
- Cost of 2 bottles of water and 5 bottles of cola:

$$
\(2 x+5 y\)
$$

- Cost of 6 bottles of water and 15 bottles of cola:

$$
\(6 x+15 y\)
$$


The problem states that the ratio of these two costs is $1: 3$.
So, we can write the equation:

$$
\(\frac{2 x+5 y}{6 x+15 y}=\frac{1}{3}\)
$$


Now, let's simplify the denominator of the left side:
Notice that $6 x+15 y$ can be factored as $\(3(2 x+5 y)\)$.
Substitute this back into the equation:

$$
\(\frac{2 x+5 y}{3(2 x+5 y)}=\frac{1}{3}\)
$$


Assuming that the cost $2 x+5 y$ is not zero (which is true since costs are positive), we can cancel $\((2 x+5 y)\)$ from the numerator and denominator:

$$
\(\frac{1}{3}=\frac{1}{3}\)
$$


This equation is an identity. It means that the ratio will always be $1: 3$ regardless of the actual values of $x$ and $y$, as long as they are positive. The given ratio provides no specific information about the relationship between $x$ and $y$.


Let's test with examples:
- Example 1: $\(x=1, y=1\)$

Cost of 2 water +5 cola $\(=2(1)+5(1)=7\)$
Cost of 6 water +15 cola $\(=6(1)+15(1)=21\)$
Ratio $\(=7: 21=1: 3\)$. In this case, Quantity A (\$1) = Quantity B (\$1).
- Example 2: $\(x=10, y=1\)$

Cost of 2 water +5 cola $\(=2(10)+5(1)=25\)$
Cost of 6 water +15 cola $\(=6(10)+15(1)=75\)$
Ratio $\(=25: 75=1: 3\)$. In this case, Quantity A ( $\$ 10$ ) > Quantity B ( $\$ 1$ ).
- Example 3: $\(x=1, y=10\)$

Cost of 2 water +5 cola $\(=2(1)+5(10)=52\)$
Cost of 6 water +15 cola $\(=6(1)+15(10)=156\)$
Ratio $\(=52: 156=1: 3\)$. In this case, Quantity A (\$1) < Quantity B (\$10).

Since we can find scenarios where Quantity A is equal to, greater than, or less than Quantity B, the relationship cannot be determined from the information given.