Question

the sum of the digits of a 2 digit number is 8 and the difference between the number and the number formed by reversing the digits is 18. FIND THE NUMBERS.

Answer 1

Let the unit place digit be x

And tens place digit be y

No. Formed by x and y = 10y+x (I have multiplied y by ten because it is on tens place)

Now, no. Obtained by reversing digits = 10x+y

Here sum of digits = 8 (I)

And difference on reversing digits is 18

A/q

(10y+x) - (10x+y) = 18

Or, 10y+x-10x-y= 18

Or, 9y-9x = 18

Or, 9(y-x) = 18

Or y-x = 18/9= 2(ii)

On adding both the given equation

Now, y+x = 8

y-x = 2

------------------

2y. = 10

Or, y =5

Now putting the value of y in equation 1st

y+x=8

5+x=8

Or, x= 3

Now the formed no. = 53

Answer 2

Answer:

Let the two digits be x and y.

x+y= 8 -- equation 1

yx-xy= 18

10y+x-(10x+y)= 18

10y+x-10x-y= 18

9y-9x= 18

9(y-x)= 18

y-x= 18/9= 2-- equation 2

x+y= 8

y-x= 2

2y= 10

y= 5

x+y= 8

x+5= 8

x= 8-5

x= 3

So, the numbers are:

35 and 53

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