# CLASS-9SIMULTANEOUS LINEAR EQUATION - PROBLEM ON AGE

Problem On Ages

Example.1) Five years ago, a man was seven times as old as his son, while 5 years hence, the man will be 3 times as old as his son. Find their personal ages.

Ans.)  Let, the present ages of the man and his son be x years and y years respectively.

Man’s age 5 years ago = (x – 5) years

Son’s age 5 years ago would be = (y – 5) years

So,  (x – 5) = 7(y – 5)

Or,  x – 7y = 5 – 35

Or,  x – 7y = - 30 …………….(i)

Man’s age 5 years hence = (x + 5) years

Son’s age 5 years hence = (y + 5) years

As per the given condition –

(x + 5) = 3 (y + 5)

=>    x – 3y = 15 – 5

=>    x – 3y = 10  ………………(ii)

On subtracting (i) from (ii), we get –

x – 7y = - 30

x – 3y = 10

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- 4y = - 40

=>       y = 10

Now we will substitute the value of y in (i), and we find –

x – 7y = - 30

or,    x – (7 X 10) = - 30

or,    x  =  70 – 30

or,    x  =  40

So, hence the man’s present age is 40 years and son’s present age is 10 years.         (Ans.)

Example.2) The present age of a woman is 3 years more than 3 times the age of her daughter, three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.

Ans.)  Let the present ages of the woman and her daughter be ‘x’ years and ‘y’ years respectively. Then,

As per the given condition –

x = 3y + 3

=>  x – 3y = 3 ………………..(i)

And, as per the given condition –

(x + 3) =  2y + 10

=>    x – 2y = 10 – 3

=>    x – 2y = 7 ……………………(ii)

Now, we will subtract (i) from (ii), and will get –

x – 2y = 7

x – 3y = 3

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y = 4

Now, we will substitute the value of y, in (i) and we will get –

x – 3y = 3

=>    x – (3 X 4) = 3

=>    x – 12 = 3

=>        x = 15

Hence, the present age of woman would be 15 years and the age of daughter would be 4 years.         (Ans.)