Question

The sum of the digits of two digit number is 8 and the difference between the number and that formed by reversing the digita is 18. Find the number

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: 53

Step-by-step explanation:

Let us assume the digits of a number as 'x' and 'y'. Since it is a two digit number, we have one's place and ten's place. ( i.e., The place value of one digit is multiple of 10 and the place value of other digit is one.)

Now going to the problem,  

Sum of two digits is 8. => x +y=8→A

Difference between number and its reverse:  

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

=>9x-9y=18

=> x-y=(18/9)=2→B

Solving A and B

x+y=8

x-y=2

=>2x=10=>x=5. Hence y = 3.  

Hence the required number is 53