Question

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 .....

Answer 1

Let Father Age 2 years Ago=3x years

Son Age 2 years ago=X years

Present Age of Father=3x+2 years

Present Age of son=X+2 years

2 years Hence,

Father Age=(3x+2+2) years

Son Age=(X+2+2) years

A/Q

Age will TWICE as Five times Of son

So,

2(3x+2+2)=5(x+2+2)

=6x+4+4=5x+10+10

6x+8=5x+20

6x-5x=20-8

X=12

Present Age of son=X+2

=12+2

=14 years

Present Age of Father=3x+2

=3×12+2

=38 years

Present Ages

Father=38 years

Son=14 years

Answer 2

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$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
Age of Father two years ago = 3 × Age of his son two years ago. ----------[1]

2 × Age of Father after two years = 5 × Age of his Son after two years. ----------[2]

Present Age of Father = ?

Present Age of son = ?

$\underline{\large{\textbf{Step-1}}}$:
Find the Age of Father and his son, two years ago.

Let Age of son, 2 years ago = x

Using [1]
$\implies$ Age of his Father = 3x

$\underline{\large{\textbf{Step-2}}}$:
Find the Present Age of Father and his son

Present Age of Son =(Age of Son 2 years ago) + 2 = x+2

Present Age of His Father = (Age of Father 2 years Ago) +2 = 3x+2

$\underline{\large{\textbf{Step-3:}}}$:
Find the Age of Father and his son, after two years.

Age of Son, after two years = (Present age of son) + 2 = (x+2)+2 = x+4

Age of Father, after two years = (Present age of Father) +2 = (3x+2)+2 = 3x+4

$\underline{\large{\textbf{Step-4:}}}$:
Find the present ages of Father and Son Using the relation given as per equation [2]

Substituting Values in [2]

$\implies 2(3x+4) = 5 \times (x+4)$

$\implies 6x+8 = 5x+20$

$\implies 6x-5x = 20 - 8$

$\implies x= 12$

$\therefore$
Present Age of Son = x+2= 12+2 = 14 years.
Present Age of his Father = 3x+2 = 3(12)+2 = 36+2 = 38 years.

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