Two years ago father was three times as old as his son and two years hence twice his age will be equal to five time that of his son .find there present ages
Answers
Let
Present age of Father = f
Present age of Son = s
Two years ago
f-2=s-2
Father was 3 times as old as his son
f-2=3(s-2)
f-2=3s-6
f=3s-6+2
f=3s-4............(1)
and 2 year hence,
f+2=s+2
then twice his age will be equal to 5 times that of his son
2(f+2)=5(s+2)
2f+4=5s+10
2f=5s+10-4
2f=5s+6.............(2)
Put the value of f in (2) from (1)
2(3s-4)=5s+6
6s-8=5s+6
6s-5s=6+8
s=14
Put the value of s in (1)
f=3s-4............(1)
f=3(14)-4
f=42-4
f=38
Present age of Father = f = 38 years old
Present age of Son = s = 14 years old
Answer:
Step-by-step explanation:
Let son's age = x
Two years ago son's age= x-2.
His Father's age at that time = 3(x-2).
Present age of Father = 3x-6+2=3x-4
Two years hence father's age=3x-4+2=3x-2.
Two years hence son's age = x+2.
Given :-5*(x+2)=2*(3x-2)
5x + 10= 6x - 4
10+4=6x-5x
14=x
SON'S PRESENT AGE : 14 years.
FATHER'S PTESENT AGE: 38 years.