Two samples of water A and B having concentrations 16.2 ppm of Ca(HCO₃)₂ and 13.9 ppm of CaCl₂ respectively. Then

A = 16.2 g of Ca(HCO₃)₂
B = 13.6 g of CaCl₂
hardness of A (molar mass of Ca(HCO₃)₂ is 162 g/mol) = $16.2 \mathrm{x} \frac{100}{162}=10 \mathrm{ppm}$
hardness of B  (molar mass of CaCl₂ is 111 g/mol) =  $13.9 \mathrm{x} \frac{100}{111}=12.5 \mathrm{ppm}$

$\frac{\mathrm{hardness} \mathrm{of} \mathrm{B}}{\mathrm{hardness} \mathrm{of} \mathrm{A}}=\frac{12.5 \mathrm{ppm}}{10 \mathrm{ppm}}=1.25$