Question

Sum of two digit number and the number formed by interchanging its digit is 110 . If 10 is subtracted from the original number , the new no.is 4 more than 5 times the sum of the digits of the original number . Find the original no.

Answer 1

Here is your solution

Let the digit at ones place = y
Let the digit at tens place = x

original number = 10x + y 

Number formed by reversing the digits =10y + x

On adding original number and Reversed number.

10x+y+10y+x = 110
11x + 11y = 110
11 (x + y ) = 110
x + y = 10
x = 10 - y ................................................. (1)

on subtracting 10 from the original number , we get = 10x + y -10
and


4 more than 5 times the sum of digits = 5(x + y) + 4

A/q
 
10x + y -10 = 5(x + y) + 4
10x + y -10 = 5x +5y + 4
10x -5x + y -5y = 10 +4
5x - 4y = 14.................................................. (2)


putting the value of x in equation (2)

5x - 4y = 14
5×(10-y) - 4y = 14
50-5y-4y =14
-9y = 14 -50
-9y = -36
y = 4


Now

x = 10-y
x = 10 - 4
x = 6


original number = 10x + y  = 10×6+4 = 64

Hope it helps you