Example

Solving an Inverse Variation Application: Time for Ice to Melt

Apply the problem-solving strategy for inverse variation to calculate the time required for ice to melt. The problem states that the number of hours it takes for ice to melt, hh, varies inversely with the air temperature, TT. Therefore, the general formula is h=kTh = \frac{k}{T}. Given that a block of ice melts in 22 hours when the temperature is 6565 degrees Celsius, substitute these values to determine the constant of variation: 2=k652 = \frac{k}{65}, which means k=130k = 130. The specific equation relating time and temperature is h=130Th = \frac{130}{T}. To find the hours it would take for the same block of ice to melt if the temperature was 7878 degrees, substitute T=78T = 78 into the equation: h=13078h = \frac{130}{78}, which simplifies to h=53h = \frac{5}{3} or 1231 \frac{2}{3}. The ice would melt in 1231 \frac{2}{3} hours.

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Updated 2026-05-01

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