If the harmonic mean of two numbers is 4 and 2A + G2 = 27, where A and G are their arithmetic and geometric means, respectively, find the numbers.

The harmonic mean (H) of $x$ and $y$ is:

$$H = \frac{2xy}{x+y}$$

The arithmetic mean (A) of $x$ and $y$ is:

$$A = \frac{x+y}{2}$$

The geometric mean (G) of $x$ and $y$ is:

$$G = \sqrt{xy}$$

We are given that the harmonic mean is 4:

$$\frac{2xy}{x+y} = 4 \quad (*)$$

We are also given the relation $2A + G^2 = 27$. Substitute the formulas for A and G:

$$2\left(\frac{x+y}{2}\right) + (\sqrt{xy})^2 = 27$$ $$x+y + xy = 27 \quad (1)$$

From equation $(*)$, we can simplify:

$$2xy = 4(x+y)$$ $$xy = 2(x+y) \quad (2)$$

Substitute equation (2) into equation (1):

$$(x+y) + 2(x+y) = 27$$ $$3(x+y) = 27$$ $$x+y = \frac{27}{3}$$ $$x+y = 9$$

Now substitute $x+y=9$ back into equation (2) to find $xy$:

$$xy = 2(9)$$ $$xy = 18$$

We have the sum ($x+y=9$) and the product ($xy=18$) of the two numbers. These numbers are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$:

$$t^2 - 9t + 18 = 0$$

Factor the quadratic equation:

$$(t-3)(t-6) = 0$$

The solutions are $t=3$ or $t=6$.
Thus, the two numbers are 3 and 6.

The numbers are 3 and 6, which corresponds to option d).

The final answer is $\text{d) 3 and 6}$.