ten years ago a father was six times as old as his daughter. after ten years he will be twice as old as his daughter. determine there present ages
Answers
Let the age 10 years ago of daughter be ( x years ) and age of father would be ( 6x years )
After 10 years, age of father = ( 6x + 10 + 10 )
= ( 6x + 20 ) years
And Age of daughter will be = ( x + 10 + 10 )
= ( x + 20 ) years
ATQ,
6x + 20 = 2( x + 20 )
=> 6x + 20 = 2x + 40
=> 6x - 2x = 40 - 20
=> 4x = 20
=> x = 5
Hence, Present age of daughter = ( x + 10 )
= ( 5 + 10 )
= 15 years
Present age of father = ( 6x + 10 )
= 6( 5 ) + 10
= 30 + 10
= 40 years
Let present age of father and daughter be X and Y .
Before 10 year's age of father = ( X - 10 ) years.
Before 10 years age of daughter = ( Y - 10 ) years.
According to the question,
X - 10 = 6 ( Y - 10 )
X - 10 = 6Y - 60
X - 6Y = -50 -------(1)
After 10 years age of father = ( X + 10 ) years.
After 10 years age of daughter = ( Y + 10 ) years.
According to the question,
X + 10 = 2 ( Y + 10 )
X + 10 = 2Y + 20
X - 2Y = 10 -------(2)
From equation (1) we get,
X - 6Y = -50
X = ( -50 + 6Y ) ------(3)
Putting the value of X In equation (2)
X - 2Y = 10
( - 50 + 6Y ) - 2Y = 10
4Y = 60
Y = 60/4
Y = 15 years.
Putting the value of Y in equation (3)
X = -50 + 6Y = -50 + 6 × 15 = 40 years.
Present age of father = X = 40 years.
And,
Present age of daughter = Y = 15 years.