Question

Ten year ago father was 12 times as old as his son and 10 year hence, he will be twice as old as his son , find their present age by elimination method

Answer 1

Solution :

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Given :

Condition 1 :

10 years ago , father was 12 times as old as his son,.

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⇒ Let the age of father be x years

&

⇒ Let the age of son be y years

Then,

we can tell that,

10 years ago,

Age of the father was : (x - 10) years

&

Age of the son was : (y - 10) years

⇒ x - 10 = 12(y - 10)

⇒ x - 10 = 12y - 120

⇒ x - 12y = -120 + 10

⇒ x - 12y = -110 ...(i)

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Condition 2 :

10 years hence, He will be twice as old as his son,.

10 years hence,

Age of the father will be : (x + 10) years

Age of the son will be : (y + 10) years

⇒ x + 10 = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x - 2y = 20 - 10

⇒ x - 2y = 10 ...(ii)

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To find :

The present of the son and his father,.

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By subtracting (ii) from (i)

We get,

⇒ (x - 12y) - (x - 2y) = -110 - (10)

⇒ x - 12y - x + 2y = - (110 + 10)

⇒ -10y = -120

⇒ y = $\frac{-120}{-10}$

⇒ y = 12 ,.

∴ Present age of the son : 12 years,.

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Substituting the value of y in equation (ii),

We get ,

⇒ x - 2y = 10

⇒ x - 2(12) = 10

⇒ x - 24 = 10

⇒ x = 10 + 24

⇒ x = 34

∴ Present age of the father = 34years,.

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                                         Hope it Helps