A certain number of two digits is three times the sum of its digits and if 45 be added to it, the digits are reversed. The number is ?

Answer: Option B

Let the ten's digit be x and unit's digit be y
Then,
$$\begin{array}{rl}& \Rightarrow 10x+y=3\left(x+y\right)\\ & \Rightarrow 7x-2y=\mathrm{0.....}(i)\\ & 10x+y+45=10y+x\\ & \Rightarrow y-x=\mathrm{5.....}(ii)\end{array}$$
Solving (i) and (ii), we get :
x = 2 and y = 7
∴ Required number = 27