Two years ago, father was three times as old as his son and two years hence, twice
his age will be equal to five times that of his son. Find their present ages.
Answers
Answer:
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.
Answer:
Let us consider
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
The present age of Father = 3X – 4
Two years hence father’s age = 3X – 4 + 2
Two years hence father’s age = 3X – 2
Two years hence son’s age = X + 2
Given:
5 × (X + 2) = 2 × (3X – 2)
5X + 10 = 6X – 4
X = 14
Son’s present age X = 14 years.
Father’s present age = (3 × 14) – 4
Father’s present age = 38 years.
hope this is help u..