Question

Your percentage of body fat by weight is an important indicator of your general physical condition. For people in good health it should be between 10 and 15%. One way of determining this percentage is to measure your average body density. This can be done by weighing yourself underwater when you have completely exhaled. You can do this in your friendly neighbourhood pool by placing a kitchen scale on the bottom (one that is not damaged by being under water) and submerging yourself onto it when you have deeply exhaled. (You might want to use a snorkel if you are uncomfortable doing this trick.) You can then calculate your body density from the formula,  

D = W/(W-Ww)

where W is your normal weight and W_w is your weight underwater. (This formula assumes the density of water is one. If you are mathematically inclined and know a little about the physics of buoyancy from a high-school physics course you can easily derive this formula.) If as a result of such an experiment a 75 kg person is found to weigh 4.2 kg underwater, what is that person's body density?

Professionals who determine body fat by this method have developed the following mathematical relationship between percentage of body fat f and average body density D:

f = 100 x (4.20/D - 3.81)

What is the percentage of body fat in the person of the previous part?

Answer 1

a 75 kg person is found to weigh 4.2 kg underwater

Therefore, W = 75 kg, Ww = 4.2 kg

Therefore, D = 1.06 (upto two decimals)

Note that D is in the units of density of water which is taken as 1 gm/cm³, therefore units of D are gm/cm³

Therefore, the person's body density is 1.06 gm/cm³

Now,

Therefore, f = 15.48 %

Therefore, the percentage of body fat in the person of the previous part is 15.48%

Note that if you directly put D = 1.06 in the formula you will get f = 15.22% which is wrong because it has rounding errors as we have rounded D to two decimal places. Therefore you should put the expression for D in place of D in the formula for f and only put the given information to ensure correct answer.