Question

eight years ago a mother's age was11 times that of son. the of their ages present ages is 40years. find the present age

Answer 1

Let mother’s age be x and the son’s age be y
So x-8=11(y-8)
Because x and y are the present the ages eight years back would be x and y minus eight
And as we know that eight years ago mother’s age was 11 time her son’s hence the equation .
So let us solve the equation
x-8=11y-88
x=11y-88+8
x=11y-80
As we know that the sum of their present ages is 40 , we can write it as x+y=40
But as mentioned above x=11y-80
So 11y-80+y=40
Now we solve the equation to get the answer
12y-80=40
12y=40+80
y=120/12
y=10
So the son’s age i.e. y = 10
And as x=11y-80
x=11.10-80
x=30

So the present ages are
Mother =30 years
Son=10 years

Answer 2

Solution:-

let , 》mothen and son age = x, y

acourding to quetion,

1st cas e 》 x-8 = 11(y-8)

》 x-11y = -80............(1)

2nd case 》x+y = 40..............(2)

substraction eq(2) in(1)
_______________________________
》 -12y = -120

》( y = 10)ans

put y's value in eq(2)

》 x + 10= 40

》 (x = 30) ans

mother's and son's present age = 30years and 10years

☆i hope it is helpfull for you☆