the age of father 8 years ago was 8 times the age of his son. after 10 years the age of the father will be twice the age of the son find their present ages?
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Answers
Answer:
Let the present age of the son be x.
Present age of the father will be 4x.
According to the question
⇒ (x – 4) / (4x – 4) = 1/8
⇒ x = 7
∴ Age of the son is 7 years.
Answer:
The father’s age four years ago was 8 times the age of his son. At present the father’s age is 4 times that of his son. What is the son’s present age?
The son is 7 years old.
The problem can be represented by a system of equations. Where 1 equation represents the relationship between their current ages and how old they were 4 years ago and one represents the current relationship of their ages.
F = Fathers Current Age
S = Sons Current Age
So 4 years ago
F-4 = 8 (S-4)
F = 4S
The first equation is the more tricky one and people will usually leave out the -4 for the son.
The second equation is already solved for the F variable so take that expression that is equivalent to F (4S) and plug it into the first equation. Now you will have a one variable equat
Son's present age is 7 years and father's age is 28 hence the present age of father equals 4 times son's age. Four years ago father's age would be 24 and son's age would be 3 years hence father's age equals son's past age X 8.
Say father’s present age is F years and son’s present age is S years.
So, 4 years back the relation between the ages of father and son was as below:
(F-4)=8(S-4)
At present the relation of father’s and son’s age is F=4S
So, substituting F by 4S in earlier relation brings following:
(4S-4)=8(S-4)
Or, 4S-4=8S-32
Or, 4S=28
Hence S=7
So, son’s present age is 7 years.
Step-by-step explanation:
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