Question

Lot the digit at ones place ( =x ) Thenteret digit at ten's place ( =9-x )
[
text { So, the number }=10 times(9-x)+frac{x}{1}
]
1
( 4 t )
ten's eace
[
begin{array}{l}
=90-10 x+x
=90-9 x
end{array}
]
When aligits are interchanged ten's alciat ( =x ) alcigit
[
=x
]
Lone's digit ( =9-x ). So, the mumber becomes
[
begin{array}{l}
=10 times 20+9-x
=10 x+9-x
=9 x+9
end{array}
]
According fo given condition ( (9 x+9)^{prime}-(90-9 x)=27 )
( 9 x+9-90+9 x=27 )
[
begin{array}{r}
|8 x-8|=27
|8|=|27+8|
18 x=108
end{array}
]
( Rightarrow quad x=frac{108}{18} )
[
Rightarrow x=6
]
( begin{aligned} text { The refore the two digit menuloel } i=90-9 times 6 &=90-54=36 end{aligned} )

# 12. The sum of digits of a two digit number is 9. If the digits are interchanged, the resulting nu greater than the original number by 27. What is the two-digit number?

Solution