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Question
Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.
Solution
Let the present age of the man be x years and the present age of his son be y years.
After 6 years, the man’s age will be (x+6) years and son’s age will be (y+6) years. Thus using the given information, we have
` x+ 6 = 3(y+6)`
`⇒ x+ 6 = 3y +18`
`⇒ x - 3y -12 =0`
Before 3 years, the age of the man was (x-3) years and the age of son’s was (y -3) years. Thus using the given information, we have
` x-3 =9 (y -3)`
`⇒ x -3 = 9y -27`
`⇒ x - 9y +24 =0`
So, we have two equations
` x- 3y-12 =0`
` x- 9y +24=0`
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
`x/((-3)xx24 - (-9)xx(-12))=(-y)/(1xx24-1xx(-12))=(1)/(1xx(-9)-1xx(-3))`
`⇒ x/(-72-108)=(-y)/(24+12)=1/(-9+3)`
`⇒ x/-180=y/36=1/6`
`⇒ x = 180/6,y = 36/6`
`⇒ x =30, y = 6`
Hence, the present age of the man is 30 years and the present age of son is 6 years.
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