Let the price of a pen be 'x' and that of pencil be 'y' Then 87x + 29y = 783Multiplying eqn (i) by 30/29 we get (87x) x 30/29 + (29y) 30/29 = 783 x 30/29∴ 90x + 30y = 810
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.
Total balls = 4 + 6 + 7 = 17∴ n(S) = 17C1 = 680Two red balls can be selected from four red balls in 4C2 = 6 ways.and the third ball can be selected from the remaining 13 balls in 13C1 = 13 ways.∴ P (E) = 13x6/680 = 39/340