Question

six years hence a man age will be three times the age of his son's and three years ago he was nine times as old as his son. Find their present ages? ​

Answer 1

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Given:

Six years man age will be three times the age of son

Three years ago man was nine times as old as his son

To find:

Present age of man and son

Method:

Let the age of man be x and age of son be y

Six years hence ,

$\\ x + 6 = 3(y + 6) \\ \\ x + 6 = 3y + 18 \\ \\ x - 3y = 18 - 6 \\ \\ x - 3y = 12 - - - - > (1)$

Three years ago,

$\\ x - 3 = 9(y - 3) \\ \\ x - 3 = 9y - 27 \\ \\ x - 9y = - 24 - - - - > (2)$

Subtracting (2) from (1)

$\\ x - 9y - (x - 3y) = - 24 - 12 \\ \\ x - 9y - x + 3y = - 24 - 12 \\ \\ - 9y + 3y = - 36 \: \: (cancel \: x ) \\ \\ - 6y = - 36 \\ \\ y = 6$

$\\ substituting \: value \: of \: y = 6 \\ in \: equation \: (1) \\ \\ x - 3(6) = 12 \\ \\ x = 12 + 18 \\ \\ x = 30$

Required answer:

$\\ Thus\: \: age \: of \: man \: (x) = 30yrs \\ age \: of \: child \: (y) = 6yrs$

Basic points:

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