mike needs 30 % to pass . if he scored 212 marks and falls short by 22 marks , what was the maximum marks he could have got ?
"if mike had scored 22 marks more , he could have scored 30 % therefore , mike required 212 + 22 = 234 marks let the maximum marks be m . then 30 % of m = 234 ( 30 / 100 ) × m = 234 m = ( 234 × 100 ) / 30 m = 23400 / 30 m = 780 answer : c"
divide(add(212, 22), divide(30, const_100))