Question

If you reverse the digits of my mother's age, you get my father's age. And the sum of their ages is 11 times the difference between their ages. If my mother is younger than my father, how old is she?

Answer 1

Let mother's age be (xy)
Hence, father's age is yx

Also, xy+yx= 11*(yx-xy)
12xy=10yx
6xy=5yx

6*(10x+y)=5*(10y+x)
60x+6y=50y +5x
55x=44y
5x=4y

Hence, tens place digit in mother's age is (4/5)times the digit at ones place

It is possible only in one case,
Therefore, mother's age is 45

Answer 2

Hello friend

According to the given question
Let us suppose that the age of mother will be = xy

Now, then father's age which is reversed of mother's age will be = yx

As it is said that , mother is younger than the father

And the sum of their ages = 11 times of the difference of their ages

Hence
xy + yx = 11 ( yx - xy) ...............(I)

As
25 is written as 2×10 +5×1

Similarly, xy cab be written as 10×x+ y×1

And yx can be written as 10×y + x×1

So putting these values in the equation (I) , we get

{ 10x + y } +{ 10 y + x }= 11 [ ( 10y + xx ) - ( 10x + y)
Then ,

11x +11y = 11 (10 y + x - 10 x -y )

11(x+y) =11( 9 y -9x)

x+y = 9(y-x)

9y - y = 9x + x

8y -10 x =0

8y -10x = 0
8y = 10x
y = 10x/8
y = 5x/4

So , the age of mother will be 10x + y = 10x + 5x/4 = 45x/4
then father's age will be 10y + x = 50x/4 + x = 54x/4

I hope this may help u