Question

Two years ago a father was five times as old as his son two years later his age will be 8 more than three times the age of son find the presnet age of father and son

Answer 1

Solution:-

Let the age of Father be "x" year and Son be "y" years.

Two years ago,

Age of Father = ( x - 2)

Age of Son = ( y - 2 )

Two year later,

Age of Father = ( x + 2)

Age of Son = ( y + 2 )

Condition I,

=) ( x - 2) = 5( y - 2)

=) x - 2 = 5y - 10

=) x = 5y - 10 + 2

=) x = 5y - 8 _____________(1)

Condition II,

=) x + 2 = 3( y + 2) + 8

=) x + 2 = 3y + 6 + 8

=) 5y - 8 + 2 = 3y + 14 [ x = 5y - 8 ].

=) 5y - 3y = 14 + 6

=) 2y = 20

=) y = 10

Substituting [ y = 10 ] in eq (1). we get,

=) x = 5(10) - 8

=) x = 50 - 8

=) x = 42

Hence,

Son's Age = y = 10 yrs.

Father's Age = x = 42 yrs.

Answer 2

2 years ago Father was 5 times the son Age.

Let age of Son 2 years ago=x years

Then Age of Father 2 years ago=5x years

Present Age will be

Son Present age=(x+2) years

Father Present age=(5x+2) years

2 years later

Son Age will be=(x+2+2)=(x+4) years

Father Age will be=(5x+2+2)=(5x+4) years

Now Question Saying

Father age will be also 8 more than 3 times son age after 2 years

Father Age after 2 years=3(son age after 2 years)+3

5x+4=3(x+4)+8

5x+4=3x+12+8

5x-3x=12+8-4

2x=20-4

2x=16

x=16÷2

x=8

Present age of Son=(x+2)=8+2=10 years

Present age of Father=(5x+4)=5×8+2=42 years

$\underline{\mathtt{\huge{Father=42\:years}}}$

$\underline{\mathtt{\huge{Son=10\:years}}}$