For a Chemical reaction, 
$4\mathrm{A}+3\mathrm{B}\to 6\mathrm{C}+9\mathrm{D}$
Rate of formation of C is $6\times {10}^{-2}{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$ and rate of disappearance  of A $4\times {10}^{-2}{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$

The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be. 

Rate of reaction $=\frac{1}{4}\times$ Rate of disappearance of $A=\frac{1}{4}\times 4\times {10}^{-2}=1\times {10}^{-2}{\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}$
Rate of reaction $=\frac{1}{3}$ Rate of disappearance of B
∴ Rate of disappearance of
B=3× Rate of reaction
=3×1×10^-2
Now, Rate of disappearance of
$\begin{array}{l}\mathrm{B}=\frac{\text{ change in concentration of }\mathrm{B}}{\text{ Time taken }}\\ 3\times {10}^{-2}=\frac{\text{ change in concentration of }\mathrm{B}}{10}\end{array}$
∴ change of conc. of B=3 x 10^-2 x 10 = 30 x 10^-2 mol L^-1