Two years ago a father was five times as old as his son. Two years later his will eight than that of three times the age of his son. Find the age of the father?
Answers
Correct question :-
Two years ago a father was five times as old as his son. Two years later the father will be eight years more than that of three times the age of his son. Find the age of the father?
Solution :-
~Let,
Age of son = y
Age of father = x
~Two years ago,
Age of son = (y - 2)
Age of father = (x - 2)
According to question,
Two years ago father was 5 times as old as his son.
(x - 2) = 5(y - 2)
x - 2 = 5y - 10
x - 5y = -10 + 2
x - 5y = -8 ...................(1)
~Two years later,
Age of son = (y + 2)
Age of father = (x + 2)
According to question,
Two years later, father's age will be 8 more than 3 times his son's age.
(x + 2) = 8 + 3(y + 2)
x + 2 = 8 + 3y + 6
x - 3y = 8 + 6 - 2
x - 3y = 14 - 2
x - 3y = 12 ...................(2)
~Solving (1) and (2) by substitution method,
From (1)
x = -8 + 5y .................(3)
Substituting value of (3) in (2)
-8 + 5y - 3y = 12
2y = 12 + 8
2y = 20
y = 20/2
y = 10
Substituting value of y in (3)
x = -8 + 5y
x = -8 + 5(10)
x = -8 + 50
x = 42
~Age of father = x = 42 years
Answer:
Let the age of father is x years and son is y years.
A/q
=> x - 2 = 5 ( y-2 )
=> x - 5y = -10+2
=> x - 5y = -8
=> x = 5y-8
=> x + 2 = 3(y+2)+8
=> x - 3y = 6+8-2
=> 5y - 8 - 3y = 12
=> 2y = 12+8
=> y = 20/2
y = 10
Now,
x = 5y-8
= 50-8
= 42 yrs...
Hence,
the age of father is 42.