Question

the tens digit of a two digit number exceeds the units digit by 5.If the digits are reserved, the new number is less by 45. If the sum of their digits is 9,find the number

Answer 1

Let the number be xy = 10x+y

when the number is reversed = yx = 10y+x

Given,

If the digits are reversed, the new number is less by 45

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

9x-9y=45

x-y=5...........(1)

Also given, the sum of their digits is 9

x+y=5..................(2)

Adding (1)&(2), we get

2x=14

x=7

y=2

So, number is 72

Answer 2

Let the units digit be y,
and the tens digit be x.
Thus the number is 10x+y
according to the condition,
(10x+y)-(10y+x) =45
9(x - y)=45
x-y=5.......(1)
also x+y=9...(given)
Adding(1) and (2)

2x=14
x=7
Putting in (1)
y=7-5
y=2

Thus the number is 72.
Hope this helps.