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.
Let the present age of the father be 'x' and that of the son be 'y'. Then x/y = 8/3∴ 3x = 8y Further, x + 12/y + 12 = 2/1 ∴ x + 12 = 2y + 24 ∴ x - 2y = 12 ......(ii)From eqn (i) and (ii), x = 48, y = 18 ∴ sum = 66 yrs.
EXAMIANS