Question

1. Some athletes have as little as 3.0% body fat. If such a person has a body mass of 65 kg, how many pounds of body fat does that person have? In liposuction, a doctor removes fat deposits from a person’s body. If body fat has a density of 0.94 g/mL and 3.0 L of fat is removed, how many pounds of fat were removed from the patient?

Answer 1

1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs


`65 kg (3.0 kg)/(100 kg) x (2.20 lb)/(1 kg) = 4.29 lbs`

2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams


`3 Liters (1000 mL)/(1 L) x (0.94 grams)/(1 mL) x (1 kg)/(1000 g) x (2.20 lbs)/(1 kg) = 6.20 lbs`

Answer 2

4.299009 pounds of fat was present in the person's body.

6.2170284 pounds of fat were removed from the patient

Step-by-step explanation:

Percentage of fat in the body = 3.05 of mass

Let the amount of fat be x

Mass of the athlete = 65 kg

`3.0\%=(x)/(65 kg)* 100`

x = 1.95 kg = 4.299009 pounds (1 kg = 2.20462 pounds)

4.299009 pounds of fat was present in the person's body.

Density of fat = 0.94 g/L

Volume of fat to be removed = V = 3.0 L = 3000 mL[/tex]

Let the mass of fat to be removed be y

`Density=(Mass)/(Volume)`

`0.94 g/mL=(y)/(3000 mL)`

y = 2,820 g = 2.820 kg=6.2170284 pounds

6.2170284 pounds of fat were removed from the patient