Question

35) Father's age before a year was 4 times of his
son's age. After 6 years, father's age will be 9
years more than 2 times of his son's age. What
will be the sum of their current age?
(1) 37
(3) 48
(4142
(5) None of these
(2) 53​

Answer 1

Question:

Father's age before a year was 4 times of his son's age. After 6 years, father's age will be 9 more than 2 times of his son's age. What will be the sum of their current age?

Options:

1) 37

2) 53

3) 48

4) 42

5) None of these

Answer:

$\bigstar{\bold{Option\:4:42}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Father's age before 4 years was 4 times of his son's age
• After 6 years, father's age would be 9 more than 3 times son's age

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Sum of current ages of father and son

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Let father's age be x

→ Let son's age be y

→ Father's age a year ago = x - 1

→ Son's age a year ago = y - 1

→ By given,

  x - 1 = 4 ( y - 1 )

 x = 4 y - 4 + 1

 x = 4y - 3 ------equation 1

→ Father's age after 6 years would be x + 6

→ Son's age after 6 years would be y + 6

→ By given,

  x + 6 = 9 + 2 (y + 6)

→ Substituting value of x from equation 1

  4y - 3 + 6 = 9 + 2y + 12

 4y + 3 = 21 + 2y

  2y = 18

    y = 9

→ Hence son's age would be 9 years old

→ Substitute value of y in equation 1

  x = 4 × 9 - 3

  x = 33

→ Hence father's age is 33 years old

→ Sum of their present ages = x + y

  Sum of their present ages = 33 + 9 = 42

→ Hence option 4 is correct.