Two years ago a man’s age was three times the square of his son’s age. In three yrs time, his age Will be 4 times his son’s age. Find their present ages. answer properly...
Answers
ANSWER:
Given:
- 2 years ago, a man's age was three times the square of his son's age.
- In 3 years time, his age will be 4 times his son’s age.
To Find:
- Present ages of man and his son.
Solution:
Let us assume that, man's present age is x years and son's present age be y years.
So, we are given that,
⇒ 2 years ago, a man's age was three times the square of his son's age.
So,
⇒ Man's age 2 years ago = x - 2
⇒ Son's age 2 years ago = y - 2
Hence, according to the condition,
⇒ x - 2 = 3(y - 2)²
⇒ x = 3(y - 2)² + 2 - - - -(1)
We are also given that,
⇒ In 3 years time, his age will be 4 times his son’s age.
So,
⇒ Man's age 3 years later = x + 3
⇒ Son's age 3 years later = y + 3
Hence, according to the condition,
⇒ x + 3 = 4(y + 3)
⇒ x = 4(y + 3) - 3
⇒ x = 4y + 12 - 3
⇒ x = 4y + 9 - - - -(2)
Equating (1) & (2),
⇒ 3(y - 2)² + 2 = 4y + 9
⇒ 3(y² - 4y + 4) + 2 = 4y + 9
⇒ 3y² - 12y + 12 + 2 = 4y + 9
⇒ 3y² - 12y + 14 = 4y + 9
Transposing RHS to LHS,
⇒ 3y² - 12y + 14 - 4y - 9 = 0
⇒ 3y² - 16y + 5 = 0
⇒ 3y² - 15y - y + 5 = 0
⇒ 3y(y - 5) - 1(y - 5) = 0
⇒ (y - 5)(3y - 1) = 0
So,
⇒ y = 5, 1/3
As, y = 1/3 isn't possible.
Hence,
⇒ y = 5
Substituting this value in (2),
⇒ x = 4y + 9
⇒ x = 4(5) + 9
⇒ x = 20 + 9
⇒ x = 29
Therefore, the present ages of man and his son is 29 years and 5 years respectively.
Answer is in Attachment !
