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
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