Brain Shaker

Can be easily solved just by using sum of digits and options.

Let the digits be x,y. x+y=13, x<y, 10+x-y=7. So y-x=3 or y=8, x=5. xy= 58

The difference in value due to digit reversal is 9 times the difference between the digits. If the number is ab (NOT to be confused with a*b), then we have 9(a-b) =27 so a-b=3.

Moreover, a+b=13 as stated in the question so solving these 2 simultaneous equations we have a=8, b=5 and the result follows.

first number =10x+y second number=10y+x 10x+y-(10y+x)=27 9x+x+y -(9y+x+y)=27 9x-9y +(x+y)-(x+y)=27 9(x-y)=27 x-y=3 x+y=13 2x=16 x=8 y=5 xy-yx=27 yx=58

Let the number be $10x+y$ .Then $10y+x$ is the reversed number.The problem states that: $10y+x=10x+y+27\rightarrow\;10y-y+x-10x=27\rightarrow\;9y-9x=27\rightarrow\;y-x=3$ $x+y=13$ Making $x$ the subject of the formula in the second equation, we get $x=13-y$ .Substituting this into the first equation,we get: $y-(13-y)=3$ $y+y=3+13$ $2y=16\rightarrow\;y=\frac{16}{2}=\boxed{8}$ Substituting this into $x=13-y$ ,we get $x=13-8=\boxed{5}$ .So the number is $10x+y=10(5)+8=50+8=\boxed{58}$

Also, the choices only have the number 58 to fit in the description.