The sum of the ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man’s age at the time. Find their present ages.
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Answers
answer :
Let the present age of the son be x years
Given that, sum of present ages of man and his son is 45 years.
⇒ Man’s present age = (45 – x)years
And also given that, five years ago, the product of their ages was four times the man’s age at the time.
⇒ Man’s age before 5 years = (45 – x – 5) years = (40 – x) years
And son’s age before 5 years = (x – 5) years
But, given that (40 – x) (x – 5) = 4(40 – x)
⇒ x – 5 = 4
⇒ x = 9 years
⇒ Son’s present age ⇒ x = 9 years
Now, Man’s present age
⇒ (45 – x) years = (45 – 9) years =36 years
∴ The present ages of man and son are 36 years and 9 years respectively.
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Answer:
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Explanation:
Let the present age of the son be x years
Given that, sum of present ages of man and his son is 45 years.
⇒ Man’s present age = (45 – x)years
And also given that, five years ago, the product of their ages was four times the man’s age at the time.
⇒ Man’s age before 5 years = (45 – x – 5) years = (40 – x) years
And son’s age before 5 years = (x – 5) years
But, given that (40 – x) (x – 5) = 4(40 – x)
⇒ x – 5 = 4
⇒ x = 9 years
⇒ Son’s present age ⇒ x = 9 years
Now, Man’s present age
⇒ (45 – x) years = (45 – 9) years =36 years
∴ The present ages of man and son are 36 years and 9 years respectively.