Question

(1) Two years ago, the age of a father was three and a half times the age of his
daughter then. Six years hence, the age of father will be ten years more than twice
the age of his daughter then. Find their present ages.
(1) Let the present age of the father be x years and that of his daughter be y years.
(2) Form two equations from the given conditions.
(3) Solve the equations and find the answer.​

Answer 1

Answer:

Father age, x = 44 years

Daughter age, y = 14 years

Step-by-step explanation:

We are given that let the Present age of father be x years.

                               and Present age of daughter be y years.

• First condition given is;

Two years ago, the age of a father was three and a half times the age of his

daughter then i.e.

         ⇒ x - 2 = $3\frac{1}{2}$*( y - 2 )       ⇒  x - 2 = $\frac{7}{2}$*( y - 2 )

                                                ⇒  x - 2 = $\frac{7y}{2} - 7$  

                                                ⇒ x =  $\frac{7y}{2} - 5$   ----------- [Equation 1]

• Second condition given is;

Six years hence, the age of father will be ten years more than twice

the age of his daughter then i.e.,

         ⇒ x + 6 = 10 + 2*(y+6)      ⇒ x + 6 = 10 + 2*y + 12

                                                   ⇒ x = 10 + 2*y + 12 - 6

                                                   ⇒ x = 2*y + 16 ------------ [Equation 2]

equating equation 1 and 2 we get,

             ⇒  $\frac{7y}{2} - 5$  = 2*y + 16

             ⇒ $\frac{7y}{2} - 2*y$ = 16 + 5

             ⇒  $\frac{7y - 4y}{2}$ = 21   ⇒ $\frac{3y}{2}$ = 21

Hence,  y = 14 and then putting value of y in equation 2 we get, x = 2*14 + 16

                                                                                                          x = 44

Therefore, Father age, x = 44 years.

                  Daughter age, y = 14 years.