Suppose you decide to define your own temperature scale using the freezing point (13 ∘c) and boiling point (360 ∘c) of oleic acid, the main component of olive oil. if you set the freezing point of oleic acid as 0 ∘o and the boiling point as 120 ∘o, what is the freezing point of water on this new scale?

First, calculate for the scaling of the new setting through the given measurements for oleic acid. The difference between the actual freezing point and boiling point of oleic acid is calculated below.
   difference  = boiling point - freezing point
   difference = (360°C - 13°C) = 347°C

As per the new setting, this is equivalent only to 120°C. It can be deduced that for the actual 347°C, only 120°C is reflected. Hence the ratio is,
     ratio = 347°C/120°C = 347/120

For water with freezing point of 0°C. 
  difference from 13°C = -13°C

This would be equivalent to:
    -13°C x (120/347) = -4.49°C

ANSWER: -4.5°C

The answer is probably c. xylem