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

$65 kg \frac{3.0 kg}{100 kg} x \frac{2.20 lb}{1 kg} = 4.29 lbs$

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

$3 Liters \frac{1000 mL}{1 L} x \frac{0.94 grams}{1 mL} x \frac{1 kg}{1000 g} x \frac{2.20 lbs}{1 kg} = 6.20 lbs$

Answer 2

Answer:

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

6.2170284 pounds of fat were removed from the patient

Explanation:

Percentage of fat in the body = 3.05 of mass

Let the amount of fat be x

Mass of the athlete = 65 kg

$3.0\%=\frac{x}{65 kg}\times 100$

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

$Density=\frac{Mass}{Volume}$

$0.94 g/mL=\frac{y}{3000 mL}$

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

6.2170284 pounds of fat were removed from the patient