Question

5 years ago A man was 8 times as old as his son five year ,hence the father will be thrice the age of his son find their present age

Answer 1

Let present age of man and his son be X year's and Y year's.

Age of man before 5 year's = Present age of man - 5 = ( X - 5 ) year's.

Age of his son before 5 year's = Present age of son - 5 = ( Y - 5 ) year's.

According to question,

Age of man before 5 year's = 8 ( Age of son before 5 year's ).

X - 5 = 8 ( Y - 5 ).

X - 5 = 8Y - 40

X - 8Y = -40 + 5

X - 8Y = -35 --------------Equation (1).

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Age of man after 5 year's = Present age of man + 5 = ( X + 5 ) year's.

Age of son after 5 year's = Present age of son + 5 = ( Y + 5 ) year's.

According to question,

After 5 year's age of man = 3 ( Age of son after 5 year's ).

X + 5 = 3 ( Y + 5 ).

X + 5 = 3Y + 15

X - 3Y = 15 - 5

X - 3Y = 10 --------------Equation (2).

From equation (1) , we get

X - 8Y = -35

X = (-35+8Y ) ---------------Equation (3)

Putting the value of X in equation (2),

X - 3Y = 10

-35Y + 8Y - 3Y = 10

5Y = 10 + 35

5Y = 45

Y = 9 year's.

Putting the value of Y in equation (3),

X = (-35 + 8Y ) = (-35 + 8 × 9 )

X = 37 year's.

____________________________

Hence,

Present age of man = X = 37 year's.

and,

Present age of son = Y = 9 year's.

Answer 2

Answer:

Let Five years ago the age of son be x years and age of father be 7x years

Present age of son =x+5

Present age of father =7x+5

5 years later their age will (x+10) and (7x+10)

∴7x+10=3(x+10)

7x−3x=20

4x=20

x=5

So, the present age of son=x+5=5+5=10years

and the present age of father=7x+5=35+5=40years