Ten years before the age of father was six times his son's age at that time. After 10 years,
father's age will be twice that of his son. Find their present ages.
Answers
Answered by
8
Solution
Given :-
- Ten years before the age of father was six times his son's age at that time
- After 10 years,
- father's age will be twice that of his son.
Find :-
- Age of father & his son
Explanation
Let,
- Age of father = x years
- Age of his son = y years
According to question
Case 1.
==> (x - 10) = (y - 10) × 6
==> x - 6y = -60 + 10
==> x - 6y = -50__________(1)
Case 2.
==> (x + 10) = 2 × (y + 10)
==> x - 2y = 20 - 10
==> x - 2y = 10______________(2)
Subtract equ(1) & equ(2)
==> -6y + 2y = -50 - 10
==> -4y = -60
==> y = 60/4
==> y = 15
Keep un equ(2)
==> x - 2 × 15 = 10
==> x = 10 + 30
==> x = 40
Hence
- Age of father = 40 years
- Age of his son = 15 years.
_______________
Answered by
31
Corrected Question :-
- Ten years ago, the age of father was six times his son's age at that time. After ten years, father's age will be twice that of his son. Find their present ages.
Answer :-
Given :-
- Ten years ago, age of father was six times his son's age at that time.
- After 10 years, age of father will be twice the age of his son at that time.
To find :-
- Find the present ages of father and his son.
Solution :-
- Let the present age of father be x years.
- Let the present age of son be y years.
According to question
From equation (1)
- x - 6y = -50
- x = -50 + 6y - (3)
Put the value of x in equation (2)
- x - 2y = 10
- -50 + 6y - 2y = 10
- 4y = 10 + 50
- 4y = 60
- y =
- y = 15
Put the value of y in equation (3)
- x = -50 + 6y
- x = -50 + 6(15)
- x = -50 + 90
- x = 40
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