Let the number of children be 'x' Therefore, each child will receiveX x 60/100 = 3x/5 chocolates∴ X x 3x/5 = 735∴ x² x 735x5/3 = 1225∴ x = 1225 = 35∴ Number of chocolates = 3x/5 = 21
Since the difference between the divisors and the respective remainders is not constant, back substitution is the convenient method. None of the given numbers is satisfying the condition.