Two years ago a man was 6 times as old as his son, After 18 years, he will be twice as old as his son. Their present ages (in years) are?
- Let son's age 2 years ago be X
Man's age 2 years ago = 6X
2 (X + 2 + 18) = (6X + 2 + 18)
4X = 20
X = 5
Their present ages are (6X + 2) and (X + 2),
i.e., 32 years and 7 years
Let us denote the present age of the man and his son by x and y respectively.
Then, two years ago, the age of the man was x-2 and the age of his son was y-2.
According to the question,
x-2 = 6(y-2)
⇒ x-2 = 6y-12
⇒ x = 6y-10
After 18 years, the age of the man will be x+18 and the age of his son will be y+18.
According to the question,
x+18 = 2(y+18)
⇒ x+18 = 2y+36
⇒ x = 2y+18
Substituting the value of x in the equation x = 6y-10, we get
6y-10 = 2y+18
⇒ 4y = 28
⇒ y = 7
Substituting the value of y in the equation x = 6y-10, we get
x = 6y-10
⇒ x = 6×7-10
⇒ x = 32
Hence, the present ages of the man and his son are 32 years and 7 years respectively.
Therefore, Option A. 32 years and 7 years is the correct answer.
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