Question

12 years ago a father was seven times as old as his son that time At present, the
father's age is three times his son's age What is the sum of their present ages ?

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Answer 1

HeRe iS thE soLuTioN:-

let the present age of father = x

let the present age of son = y

therefore,

x= 3y( according to given statement)......(i)

Now,......12 years ago

Son's age = y -12

father's age = x - 12

Given,

x - 12 = 7( y-12)

x - 12 = 7y - 84

7y - x = 72...........(ii)

put (i) in (ii)-

7 y - 3y = 72

4y= 72

y = 18

now,x= 3y

.........= 3*18

.........=54

Therefore, the son's age is 18 years.

father's age is 54 years.

Sum of their present ages......

= 18 + 54

=72.

So, the final answer is 72.

Answer 2

Answer:

let father's age be x

let son's age be y

Step-by-step explanation:

so first equation ⇒ ( x- 12) = ( y - 12)7

                           ⇒ x - 12 = 7y - 84

                           ⇒ x = 7y - 84 + 12

                           ⇒ x - 7y = -72

second equation ⇒ x = 3y

                           

x - 7y = -72

3y - 7y = -72             (∴x -3y)

-4y = -72

y = -72/-4

y = 18

so,

x = 3 ×18

x = 54 (age of father)

age of son ⇒x = 3y

                  ⇒54 = 3y

                   ⇒ y = 54/3

                     ∴ y = 16

SUM OF THEIR AGES

=54 + 16

=70