Let the number be 10x+y

According to the question, the two digit number is three times the sum of its digits

10x+y=3(x+y)

⇒ 10x-3x+y-3y=0

⇒ 7x-2y=0---Eq1

and when 45 is added to original number, the digits are reversed

10x+y+45=10y+x

⇒ 9x-9y=-45

⇒ 9y-9x=45

⇒ y-x=5---Eq2

If we multiply Eq2 by '2' we get,

 -2x+2y=5×2---Eq3

By adding Eq1 and Eq3 we get

⇒ 5x=10

⇒ x=2

Substituting the value of x we get y=7

∴ Number is (10×2)+7=27

Hence option D is correct.