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# Temperature conditions at the arctic pole changed from -15°F to

Temperature conditions at the arctic pole changed from -15°F to
-30°F during a 24-hour period. What was the percent of decrease?

### 1 Answer by Expert Tutors

Peter H. | Tutoring in Math, Science, and Computer EngineeringTutoring in Math, Science, and Computer ...
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Hi Missy,

"percent change" is

100 * amount of change / initial value, or

100 * (initial value - final value) / initial value

We could say, therefore, that the percent change (in °F or Fahrenheit) is

100 * (-15 - -30) / -15  =  100 * (-15 + 30) / -15  =  100 * 15 / -15  = -100

The change is -100%; that is, a decrease of 100% (because the - sign means "decrease").

This answer is fine, but not the best one. There is a big problem using Fahrenheit for calculating % change ... suppose that it was a bit warmer at the arctic that day, and instead if changing from -15 to -30°F, it changed from 0 to -15°F. We calculate %change as

100 * (0 - -15) / 0

Oh no -- division by 0 is not allowed ! We cannot give any answer. Therefore, it is better to use an "absolute" temperature scale to calculate %change. An absolute temperature scale is a scale where the coldest anything could ever be is 0. One such scale is the Rankine (R). The relationship between R and °F is

R = °F + 459.67

We need to convert -15°F and -30°F to R, and then calculate 100 * (initial value - final value) / initial value.

I'll let you do that ! But I give you a hint - the answer is very different from the -100% we got above.