Problem: a certain number of two digits is 3 times the sum of its digits and if 45 be added to it , the degits are reversed . the number is .
Rationale: let the ten ' s digit be x and unit ' s digit be y . then 10 x + y = 3 ( x + y ) = = > 7 x - 2 y = 0 = = > 7 x - 2 y = 0 - - - - > ( i ) 10 + y + 45 = 10 y + x = = > y - x = 5 = = = > - 2 x + 2 y = 10 - - - - - - ( ii ) by adding ( i ) and ( ii ) we get 5 x = 10 so x = 2 and y = 7 so the required number is 27 . so the correct option is b ) 27 .
annotated_formula: add(multiply(subtract(divide(subtract(45, const_10), subtract(subtract(const_10, 3), const_2)), subtract(subtract(const_10, 3), const_2)), const_10), divide(subtract(45, const_10), subtract(subtract(const_10, 3), const_2)))