Question

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I need valid answer..

Q. 5 years ago, a man was 5 times the age of his son. 10 years hence the man's age will be double the age of his son. FIND THE PRESENT AGE.

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Answer 1

Hey there !

Thanks for the question !

Here's the answer !

Let the present age of the son be 'x' and that of man be 'y'.

Before 5 years:

( y - 5 ) = 5 ( x - 5 )

=> y - 5 = 5x - 25

=> 5x - y = - 20 -> Equation 1

10 years Hence,

=> ( y + 10 ) = 2 ( x + 10 )

=> y + 10 = 2x + 20

=> 2x - y = - 10 -> Equation 2

Solving Equation 1 and 2 we get,

5x - y = -20

=> y = 5x + 20

Substituting value of y in Equation 2 we get,

=> 2x - ( 5x + 20 ) = -10

=> 2x - 5x - 20 = -10

=> - 3x = - 10 - 20

=> - 3x = -30

Minus gets cancelled and therefore we get,

=> 3x = 30

=> x = 30 / 3 = 10

So the present age of son is 10 years.

Age of man can be found out as follows :

2x - y = - 10

=> y = 2x + 10

=> y = 2 ( 10 ) + 10

=> y = 20 + 10 = 30 years.

Hence the age of the man is 30 years.

Hope my answer helped !

Answer 2

Let the present age of son be x.

Let the present age of man be y.

5 years ago from now,

y - 5 = 5(x - 5)

y - 5 = 5x -25

5x - y = 20 .........equation (1)


After 10 years from now,

y + 10 = 2(x + 10)

y + 10 = 2x + 20

y - 2x = 10 ..........equation(2)


Adding equation (1) and (2),

5x - y + y -2x = 20 + 10

3x = 30

x = 10


Putting value of x in any equation

y - 2x = 10

y -2(10) = 10

y - 20 = 10

y = 30





So, present age of Son x = 10 years.

And present age of Father y = 30 years.




Thank you.