Question

Four year ago the father age was three times the age of his don the total of the age of the father and son after 4 year will be 64year what is the fatger

Answer 1

Solution

Four years ago,

Son's age=x

Father's age=3x

After Four year's(present ages)

Son's age=(x+4)

Father's age=(3x+4)

So, (x+4)+(3x=4)=64

= x+3x+4+4=64

4x+8=64

4x=64 - 8

4x=56

x=56/4

x=14

Son's age=(x+4)

14+4=18 years

Father's age=(3x+4)

3*14 + 4

42+4

46 years

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Answer 2

Let the age of son before 4 years  = x years

So,       father's age   ''  ''   ''      ''     =  3 x years

Present age ,

son = x + 4

Father = 3 x + 4

After 4 years .

Son = x + 4 + 4 = x +8

father = 3 x + 4 +  4 =  3 x + 8

A/Q, x + 8 + 3 x + 8 = 64

⇒ 4 x = 64 - 16 = 48

∴ x = 48/4 = 12

Present age of.

son = x + 4 = 12 + 4 = 16 years

father = 3 x + 4 = 12 x 3 + 4 = 40