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
Let the no. be 10x + y.No. formed by the interchange of digits = 10y + xWe have y - x = 2 .........(i)y + x = 14 .........(ii)Solving (i) and (ii), we get x = 6, and y = 8∴ the no. is 68.
Let 'N' is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.Required number = (LCM of 13 and 16) - (common difference of divisors and remainders) = (208) - (11) = 197.
EXAMIANS