2 years ago, a father was five time as old as his son.Two years later,his age will be 8more than three times the age of the son.Find the present age of father and son
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present age of father is 42 and of son is 10
sahilbhatia920:
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let the present age of father be X and that of his son be Y ......before 2 years , fathers age will be X-2 and sons age will be Y-2......it is given that fathers age is 5 times as that of his son....that means 5(Y-2)=x-2...............SIMILARLY YOU CAN DO FOR THE NEXT TWO YEARS AND SOLVE IT
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Hi ,
At present
________
Age of the son = x year's
Age of father = y years
2 years ago
_________
Son's age = ( x - 2 ) years
Father's age = ( y - 2 ) years
According to the problem given,
Father's age = 5 × son's age
y - 2 = 5 ( x - 2 )
y = 5x - 10 + 2
y = 5x - 8 -----( 1 )
After 2 years
__________
Son's age = ( x + 2 ) years
Father's age will be = ( y + 2 ) years
Father's age = 3( son's age ) +8
y + 2 = 3(x + 2 ) + 8
y = 3x + 6 + 8 -2
y = 3x + 12 ----( 2 )
( 1 ) = ( 2 )
5x - 8 = 3x + 12
5x - 3x = 12 + 8
2x = 20
x = 20 / 2
x = 10
Put x = 10 in equation ( 2 ),
y = 3x + 12
y = 3 × 10 + 12
y = 30 + 12
y = 42,
Therefore ,
At present ,
Son's age = x = 10 years
Father's age = y = 42
I hope this helps you.
:)
At present
________
Age of the son = x year's
Age of father = y years
2 years ago
_________
Son's age = ( x - 2 ) years
Father's age = ( y - 2 ) years
According to the problem given,
Father's age = 5 × son's age
y - 2 = 5 ( x - 2 )
y = 5x - 10 + 2
y = 5x - 8 -----( 1 )
After 2 years
__________
Son's age = ( x + 2 ) years
Father's age will be = ( y + 2 ) years
Father's age = 3( son's age ) +8
y + 2 = 3(x + 2 ) + 8
y = 3x + 6 + 8 -2
y = 3x + 12 ----( 2 )
( 1 ) = ( 2 )
5x - 8 = 3x + 12
5x - 3x = 12 + 8
2x = 20
x = 20 / 2
x = 10
Put x = 10 in equation ( 2 ),
y = 3x + 12
y = 3 × 10 + 12
y = 30 + 12
y = 42,
Therefore ,
At present ,
Son's age = x = 10 years
Father's age = y = 42
I hope this helps you.
:)
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