The results given in the below table were obtained during kinetic studies of the following reaction:
$$2A + B \to C + D$$

Experiment	[A]/mol L$$^{-1}$$	[B]/mol L$$^{-1}$$	Initial rate/mol L$$^{-1}$$ min$$^{-1}$$
I	0.1	0.1	$$6.00 \times 10^{-3}$$
II	0.1	0.2	$$2.40 \times 10^{-2}$$
III	0.2	0.1	$$1.20 \times 10^{-2}$$
IV	X	0.2	$$7.20 \times 10^{-2}$$
V	0.3	Y	$$2.88 \times 10^{-1}$$

X and Y in the given table are respectively:

For any chemical reaction of the type

$$2A + B \longrightarrow C + D$$

the empirical rate law is written as

$$\text{rate}=k\,[A]^m\,[B]^n$$

where $$k$$ is the rate constant and $$m,n$$ are the orders of the reaction with respect to $$A$$ and $$B$$ respectively. Our first goal is to determine $$m$$ and $$n$$ from the experimental data.

We start by comparing Experiments I and II. In these two runs the concentration of $$A$$ is kept the same while that of $$B$$ is doubled:

$$[A]_{\text I}=0.1,\;[A]_{\text{II}}=0.1$$

$$[B]_{\text I}=0.1,\;[B]_{\text{II}}=0.2=2\times0.1$$

The corresponding initial rates are

$$r_{\text I}=6.00\times10^{-3},\;r_{\text{II}}=2.40\times10^{-2}=4\times6.00\times10^{-3}$$

So the rate increases by a factor of $$4$$ when $$[B]$$ is doubled. Using the rate law:

$$\frac{r_{\text{II}}}{r_{\text I}}=\frac{k\,[A]^m_{\text{II}}\,[B]^n_{\text{II}}}{k\,[A]^m_{\text I}\,[B]^n_{\text I}} =\frac{[A]^m_{\text{II}}}{[A]^m_{\text I}}\; \frac{[B]^n_{\text{II}}}{[B]^n_{\text I}} =(1)\left(\frac{0.2}{0.1}\right)^n=2^n$$

The left-hand side equals $$4$$, so

$$2^n=4\;\Longrightarrow\;n=2$$

Next we compare Experiments I and III, where $$[B]$$ is kept constant and $$[A]$$ is doubled:

$$[A]_{\text I}=0.1,\;[A]_{\text{III}}=0.2=2\times0.1$$

$$[B]_{\text I}=0.1,\;[B]_{\text{III}}=0.1$$

The rates are

$$r_{\text I}=6.00\times10^{-3},\;r_{\text{III}}=1.20\times10^{-2}=2\times6.00\times10^{-3}$$

Thus

$$\frac{r_{\text{III}}}{r_{\text I}}=\left(\frac{0.2}{0.1}\right)^m=2^m=2$$

which immediately gives

$$m=1$$

Hence the complete rate law is

$$\boxed{\text{rate}=k\,[A]^1\,[B]^2}$$

To find the numerical value of $$k$$ we substitute data from any experiment, say Experiment I:

$$6.00\times10^{-3}=k\,(0.1)\,(0.1)^2=k\,(0.1)(0.01)=k\,(0.001)$$

$$\Rightarrow\;k=\frac{6.00\times10^{-3}}{1.00\times10^{-3}}=6.0$$

Now we use this rate constant to determine the missing concentrations.

Experiment IV

$$\text{rate}=7.20\times10^{-2},\;[B]=0.2,\;k=6.0,\;[A]=X$$

Substituting in the rate law:

$$7.20\times10^{-2}=6.0\;(X)\;(0.2)^2=6.0\;X\;(0.04)=0.24\,X$$

$$\Rightarrow\;X=\frac{7.20\times10^{-2}}{0.24}=0.30$$

Experiment V

$$\text{rate}=2.88\times10^{-1},\;[A]=0.3,\;k=6.0,\;[B]=Y$$

Substituting:

$$2.88\times10^{-1}=6.0\;(0.3)\;Y^2=1.8\,Y^2$$

$$\Rightarrow\;Y^2=\frac{2.88\times10^{-1}}{1.8}=0.16$$

$$\Rightarrow\;Y=\sqrt{0.16}=0.40$$

Thus we obtain

$$X=0.3,\qquad Y=0.4$$

Hence, the correct answer is Option C.