Question

Approximate the percent increase in waist size that occurs when a 160 lb person gains 20.0 lb of fat. Assume that the volume of the person can be modeled by a cylinder that is 4.0 ft tall. The average density of a human is about 1.0 g/cm^3 and the density of fat is 0.918 g/cm^3.

Answer 1

Mass of person (m1) = 160 lb = 72574.78 grams

Height of person (h) = 4 feet = 122 cm

Volume of man (v1) = pi*r^2*h ( volume of cylinder )

v1 = mass/density of person = 72574.78 / 1 = 72574.78 cm^3

72574.78 = 3.14*r^2*122

r = 13.764 cm

So, waist of man (w1)= 2*pi*r = 2*3.14*13.764 = 86.44 cm

Mass of fat = 20 lb = 9071.85 grams

Volume of fat (v2) = mass / density = 9071.85 / 0.918 = 9882.2 cm^3

Total volume = volume of fat + volume of man

V = v2 + v1 = 9882.2 + 72574.78 = 82456.98 cm^3

Now let the radius of cylinder is x.

Then, volume of cylinder = pi*x^2*h (which will be equal to total volume)

82456.98 = 3.14*x^2*122

x = 14.671 cm

New waist of man (w2) = 2*pi*x = 2*3.14*14.671 = 92.134 cm

% increase in waist = (w2 - w1)*100 / w1

% increase = (92.134 - 86.44)*100 / 86.44 = 6.587% ..... Answer