Question

plz solve the problem
Sum of digits of a 2 digit no. is 9 .No.formed by reversing the digits is 45 more than original no. Find the no.

Answer 1

let the digit at tens place be x
the digit at unit place be y
the original number= 10x + y
the no. after reversing the digit = 10y + x
x +y =9 .....eq.1

A T Q

10x + y +45 = 10y + x

10x + y _10y + x = _45

9x_ 9y = 45

x _ y= 5 ....eq 2

add eq. 1 and 2
x + y = 9

x _ y = 5

+y and _y cancel

2x = 14

x = 7 put in eq.1 or 2

x + y = 5

7 _ y= 5
_y =5_ 7
_y = -2
y = 2

the no. 10x+ y
10×7 + 2
72