Ten years ago the age of father is 12 times the age of son. After ten years the
age of father is twice the age of son. Find their present age
Answers
Answer:
Son is 12 years old and Father is 34 years old.
Step-by-step explanation:
Given :
10 years ago the age of father is = 12 times the age of son
After 10 years the age of father is = twice the age of son
To find :
Thier present ages
Solution :
Let the present ages be -
- Fathers age as x
- Son's age as y
Ages 10 years later -
- Father = (x + 10)
- Son = (y + 10)
Ages 10 years ago -
- Father = (x - 10)
- Son = (y - 10)
Condition I -
10 years ago the age of father is = 12 times the age of son
⇒ x - 10 = 12(y - 10)
⇒ x = 12y - 120 + 10
⇒ x = 12y - 110 -----(Equation I)
Condition II -
After 10 years the age of father is = twice the age of son
⇒ x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
Substitute Equation I -
⇒ 12y - 110 + 10 = 2y + 20
⇒ 12y - 100 = 2y + 20
⇒ 12y - 2y = 20 + 100
⇒ 10y = 120
⇒ y = 120/10
⇒ y = 12
Son's age = 12 years old
Place the value of y in Equation I -
⇒ x = 12y - 110
⇒ x = 12(12) - 110
⇒ x = 144 - 110
⇒ x = 34
Father's age = 34 years old
Son is 12 years old and Father is 34 years old.
AnswEr:-
Present age of son = 12 years.
Present age of father = 34 years.
Step-by-step-explanation :-
Given :-
- 10 years ago,age of father = 12 times son's age.
- After 10 years,age of father = 2 times son's age.
To find :-
- Present age of son and father = ?
Solution :-
Let the present age of son be S years & age of father be F years.
*1st case:-
➩ (F - 10) = 12(S - 10)
➩ F - 10 = 12S - 120
➩ F = 12S - 120 + 10
➩ F = 12S - 110 .........(eq.i)
*2nd case:-
➩ F + 10 = 2(S + 10)
➩ 12S - 110 + 10 = 2S + 20
➩ 12S - 2S = 20 + 100
➩ 10S = 120
➩ S = 120/10
➩ S = 12 years.
- Putting value of S in (eq.i):-
⇒ F = 12(12) - 110
⇒ F = 144 - 110