Question

five years ago a man was 7 times as old as his son.five years hence,the father will be 3 times as old as his son.find their present ages.

Answer 1

hey

gd aftrn

here is your answer

Let the past age of son = x

Let the past age of father = 7x

So,

Present ages are x+5 and 7x+5

Future age of son = x+10

Future age of father = 7x+10

Given

7x+10 = 3*(x+10)

7x+10 = 3x+30

7x-3x = 30-10

4x = 20

x = 20/4

x = 5

7x = 35

x+5 = 10

7x+5 = 40

hope its help you

plz follow me and mark as brainlist for more beautiful answers

thanxx and be brainly.............................

Answer 2

Let 5 years ago,

Age of man's son = x years

Age of man = 7 times of sons age

                    = 7 x years

   5 years hence

Age of man's son = ( x + 5 + 5 ) years

                             = ( x + 10 ) years

Age of man = ( 7x + 5 + 5 ) years

                    = ( 7x + 10 ) years

According to the question,

Age of father 5 years hence  = 3 times the age of his son of 5 years hence

( 7x + 10 ) years = 3( x + 10 ) years

7x + 1 0  = 3x + 30

7x - 3x = 30 - 10

4x = 20

x = $\dfrac{20}{4}$

x = 5

Therefore,

Present age of son = ( x + 5 ) years

                                = ( 5 + 5 ) years

                                = 10 years

Present age of man = ( 7x + 5 ) years

                                 =  7( 5 ) + 5 years

                                 = 35 + 5 years

                                = 40 years