Question

the sum of digit of a 2 digit no. is 13.the no. formed by interchanging the digit is 45 more than original no. find original no.

Answer 1

Let the ten's digit of the number be x ,
and the unit's digit be y ;
then the number will be = 10x+y

x+y = 13
y = 13-x

(10y+x) = (10x+y)+45
10y-y = 10x-x+45
9y = 9x+45
y = x + 5

13-x = x+5
13+5 = x+x
2x = 18
x = 9

y = 13-x
y = 13-9
y = 4

Hence the original number = 10x+y
= 94
_______________________

Answer 2

let the digit at ones place be X and the digit at tens place be Y

then the number = 10Y+X

X + Y = 13 -- Eq 1

10X + Y - 45 = 10Y + X
10X - X + Y - 10Y = 45
9X - 9Y = 45 -- Eq 2

(X + Y = 13) × 9
=> 9X +9Y = 117

Using Elimination Method :-

9X + 9Y = 117
9X - 9Y = 45

9X + 9X = 117 + 45
18X = 162
X = 162 ÷ 18
X = 9

Putting this value of X in Eq 1
X + Y =13
9 + Y = 13
Y = 13 - 9
Y = 4

Number = 10Y + X
= 10 × 4 + 9
= 40 + 9
= 49 Ans

I hope it helps. If it does , then mark me as the brainliest