Question

if a thermometer reads freezing point of water 30 degree and boiling point of water as 160 degrees , how much this thermometer reads when actual temperature is 70 degree celcius

Answer 1

Solution

$\frac{t - (30)}{160 - (30)} = \frac{c - 0}{100 - 0} \\ = > \frac{t - 30}{130} = \frac{c}{100} \\ = > 100t - 3000 = 130c \\ = > 100t = 3000 + 130c \\ = > t = 30 + 1.3c \\ now \: \: \: c = 70 \\ = > t = 30 + 1.3(70) = 30 + 91 = 121$

the actual temperature will be =121°c

hope this helps you.....xd