9. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. Find their present ages
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9. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. Find their present ages
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Verified answer
Answer:
The present ages of Father and son are 42 Years and 10 Years respectively.
Step-by-step explanation:
Given :
To Find :
Solution :
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Two years later, father age will be 8 more than three times the age of his son.
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★ Value of (x + 2)
Son's present age = 10 Years
★ Value of 5x + 2
Father's present age = 42 years
The present ages of Father and son are 42 Years and 10 Years respectively.
Verification :-
As mentioned in the Question, two years later, father's age will be 8 more than three times the age of his son.
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The present ages of Father and son are 42 Years and 10 Years respectively.
Answer:
Two years ago :
Suppose the age of son = X
and the age of father = 5x
The Present ages :
Son present age = X + 2
Father present age = 5x + 2
After two years :
Son age = X + (2 + 2)
Father age = 5x + (2 + 2)
Two years later , his father age will be more than three times the age of her son :
5x + (2 + 2) = 8 + 3 (X + 4)
5x + 4 = 8 + 3x + 12
5x + 4 = 3x + 20
5x - 3x = 20 - 4
2x = 16
X = 8
The age of son's after 2 years = x + 2
The value of X = 8
= 8 + 2
= 10
The age of father after 2 years = 5x + 2
= 5(8) + 2
= 40 + 2
= 42
The age of son's = 10 years .
The age of father = 42 years .