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September 8th, 2015, 10:03 AM   #1
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Find The Density of the body to find fat precentage

From one of my assignments. If someone could work this out and explain step by step, I would gratefully appreciate it.

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Another use of density is determining the percentage of fat in a human body. With a well-engineered weighing system, body density can be determined with great accuracy by completely submerging a person in water and calculating the volume of the displaced water from the weight of the displaced water. Fat cells in humans are composed almost entirely of pure triglycerides with an average density of about 0.9 kg/L. (kg/L are equivalent to g/mL, but easier to use with larger objects like the human body.) Most modern body composition laboratories today use the value of 1.1 kg/L for the density of the “fat free mass”, a theoretical tissue composed of 72% water (density = 0.993), 21% protein (density = 1.340) and 7% mineral (density = 3.000) by weight. A correction is made for the buoyancy of air in the lungs and other gases in the body spaces. To find the percentage of fat we can assume that, if we set x to the fat percentage, 100 – x is the rest of the “fat free mass” of the body. The calculation is:

Density of the body = (0.9)x + (1.1)(1-x)

Rearranging the equation to solve for x allows the calculation of the fraction of the body that is fat. Multiplying by 100 gives the percent fat. A 35 year-old female has a mass of 78 kg. When placed in a tank containing 1000 L of water, the volume in the tank increases to 1075 L.

Last edited by skipjack; September 8th, 2015 at 03:59 PM.
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September 8th, 2015, 01:12 PM   #2
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body density = $\displaystyle \frac{78 \, kg}{75 \, L} = 1.04 \, kg/L$

Quote:
Density of the body = (0.9)x + (1.1)(1-x)
1.04 = 0.9x + 1.1 - 1.1x

1.1x - 0.9x = 1.1 - 1.04

0.2x = 0.06

x = 0.3

30% body fat ... sounds reasonable to me
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September 8th, 2015, 08:11 PM   #3
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Thanks Skeeter. I was reading too much into the problem and kept getting lost. Thanks for the help.
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