The results given in the below table were obtained during kinetic studies of the following reaction: 

$2A+B\to C+D$

Experiment	[A]/molL-1	[B]/molL-1	Initial rate/ molL-1min-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:

From rate law 

$r=\frac{1}{2}\frac{d\left[A\right]}{dt}=\frac{-d\left[B\right]}{dt}=K{\left[A\right]}^x{\left[B\right]}^y$

$6\times {10}^{-3}=K{\left(0.1\right)}^x{\left(0.1\right)}^y, 2.4\times {10}^{-2}=K{\left(0.1\right)}^x{\left(0.2\right)}^y$

$1.2\times {10}^{-2}=K{\left(0.2\right)}^x{\left(0.1\right)}^y, \left(3\right)÷\left(1\right)\Rightarrow x=1, \left(2\right)÷\left(3\right)\Rightarrow x=2$

So, order with respect to A = 1

Order with respect to B = 2

$\left(4\right)÷\left(3\right)\text{→ }\left(\frac{x}{0.2}\right)\times {\left(\frac{0.2}{0.1}\right)}^2=\frac{7.2\times {10}^{-2}}{1.2\times {10}^{-2}}, x=\frac{6\times 0.2}{4}, x = 0.3M$

$\left(5\right)÷\left(4\right)\to {\left(\frac{y}{0.2}\right)}^2=\frac{2.88\times {10}^{-1}}{7.2\times {10}^{-2}}, y^2=4\times {0.2}^2, Y = 0.4M$