In the balanced equation of the following reaction, the ratio of $\mathrm{a}/\mathrm{b}$ is 

 ${\mathrm{aCaCO}}_3+{\mathrm{bH}}_3{\mathrm{PO}}_4\to {\mathrm{pCa}}_3{\left({\mathrm{PO}}_4\right)}_2+{\mathrm{qCO}}_2+{\mathrm{rH}}_2\mathrm{O}$

${\mathrm{aCaCO}}_3+{\mathrm{bH}}_3{\mathrm{PO}}_4\to {\mathrm{pCa}}_3{\left({\mathrm{PO}}_4\right)}_2+{\mathrm{qCO}}_2+{\mathrm{rH}}_2\mathrm{O}$

First balance the `phosphate group and calcium atom,

So, $3{\mathrm{CaCO}}_3+2{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{pCa}}_3{\left({\mathrm{PO}}_4\right)}_2+{\mathrm{qCO}}_2+{\mathrm{rH}}_2\mathrm{O}$

After that balance the carbon atom,

$3{\mathrm{CaCO}}_3+2{\mathrm{H}}_3{\mathrm{PO}}_4\to {\mathrm{pCa}}_3{\left({\mathrm{PO}}_4\right)}_2+3{\mathrm{CO}}_2+{\mathrm{rH}}_2\mathrm{O}$

After that balance the oxygen and hydrogen atoms,

$3{\mathrm{CaCO}}_3+2{\mathrm{H}}_3{\mathrm{PO}}_4\to 1{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2+3{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}$

Compare this balanced equation with the given equation. So, the value of $\mathrm{a}=3 \& \mathrm{b}=2$

Hence, the value of $\raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{b}$}\right.$ is equal to the $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.