Question

Five years ago, the age of a father was twice the age of his son. The sum of their present ages is 70 years. Find their present ages.

Answer 1

Answer:

Five years ago : Let the age of son be x, then age of father = 2x

Now : Son's age = x+5, Father's age = 2x+5

        Sum of their ages = 3x+10=70 ⇒ x=20

Present age of son = x+5 = 20+5 =25

Present age of father = 2x+5 = 40+5 = 45

Answer 2

Step-by-step explanation:

5 years ago = father = 2 × son

father + son = 70

son = father/2 (before 5 years)

So,

father ( present age ) = (2 × son) + 5

= (2× (father/2)) +5

= father + 5

son (present age) = son + 5

= (father/2) + 5

sum of both their present ages = son + father (present age)

70 = (father + 5) + ((father/2) + 5)

70 = x + 5 + x/2 +5 ( father = x)

70 = 10 + x + x/2

70 - 10 = x + x/2

60 = x + x/2

60 = (2x + x)/2 ( LCM OF 1 AND 2 IS 2)

60 × 2 = 3x

120 = 3x

120/3 = x

40 = x

Therefore, father's present age = father + 5 = 40 + 5 = 45

son's present age = father/2 + 5 = 40/2 +5 = 25

45 + 25 = 70