Question

the tenth place of two digits exceeds the units place by 5.If the digits is reverse it is less by 45.if the sum of the digits is 9.find the numbers.

Answer 1

hey friend!

ur solution here!☺


Let the units digit be y
➡
and the tens digit be x.

➡
Thus the number is 10x+y
when number is reversed then

➡
10y +x

Given that reverse digit is less than by 45
so

➡
(10x+y)-(10y+x) =45

➡
9(x - y)=45

➡
x-y=5.......(1)

➡
the sum of digit is 9
so

➡
x+y=9...(given)............(2)

➡
Adding(1) and (2) we get

➡

2x=14

➡
x=7
Putting value of X in (1)

➡
y=7-2

➡
y=2

☺☺☺☺