Question

five years a woman age was a square of her son age 10 years is hence her age be will twice that of his son find

1, find age of her son 5 years ago and

2, find the present age of women

Answer 1

$\bold{\boxed{HEY!!!}}$


$<B><I>HERE IS YOUR ANSWER :-$


Let the present age of woman be " x yrs " and that of the his son be " y yrs ".


By the first condition


( x - 5 ) = ( y - 5 )²


=> x - 5 = y² - 10y + 25


=> x - y² + 10y = 30. ... ( 1 )


By the second condition


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


=> x + 10 = 2y + 20


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


Subtracting equation ( 2 ) from equation ( 1 ) we get


- y² + 12y = 20


=> y² - 12y + 20 = 0


=> y² - 10y - 2y + 20 = 0


=> y ( y - 10 ) - 2 (y - 10 ) = 0


=> ( y - 10 )(y - 2 ) = 0


=> y - 10 = 0. or. y - 2 = 0


=> y = 10. or. y = 2 .


y = 2 is not possible because 5ago years ago his age will be negative.


So y = 10


$\bold{\boxed{\:AGE\:OF\:HER\:SON\:WAS\:5\:YRS}}$


Putting y = 10 in equation ( 2 )


x - 2( 10 ) = 10


=> x - 20 = 10


=> x = 30 .


$\bold{\boxed{\:PRESENT\:AGE\:OF\:WOMAN\:IS\:30\:YRS}}$


$<marquee>THANKS...$

Answer 2

let women present age of women =x
10 years ago her age = x+10
son age will be =1/2× x+10
=×/2+5
present age of son = x/2 +5-10
=×/2 -5

5 years ago =
then x-5 =( x/2-5- 5)^2
x-5= (x/2-10)^2




$x - 5 = \frac{ {x}^{2} }{4} + 100 - 10 \times x$
$\frac{4x - 20 = {x}^{2} + 400 - 40x}{4}$
${x}^{2} - 44x + 420 = 0$
factorise it and the value if x will be the mothers present age ..
and u can get all ages bu putting value of x in them..