a father is three times as old as his son, and his daughter is 4 years younger than his son. if the sum of all three ages 3 years ago was 72 years, find their present ages

Answer 1

F = present father's age

S = present son's age

D = present daughter's age

A father is three times as old as his son, and his daughter is 4 years younger than his son means:

F = 3 S , D = S - 4

3 years ago father was F - 3 yrs old

Son was S - 3 yrs old

Daughter was D - 3 yrs old

The sum of all three ages 3 years ago was 72 years means:

F - 3 + S - 3 + D - 3 = 72

Replace F with 3 S

and replace D with S - 4

in this equation.

3 S - 3 + S - 3 + S - 4 - 3 = 72

5 S - 13 = 72

Add 13 to bith sides.

5 S = 85

S = 85 / 5

S = 17 yrs old

F = 3 S = 3 • 17 = 51 yrs yrs old

D = S - 4 = 17 - 4 = 13 yrs old

Check result:

3 yrs ago father was 51 - 3 = 48 yrs old

Son was 17 - 3 = 14 yrs old

Daughter was 13 - 3 = 10 yrs old

The sum of all three ages 3 years ago was:

48 + 14 + 10 = 72 yrs