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number of hand shakes before the meeting = 12c2 = 66 Similarly at the end of meeting = 66 Thus total = 132

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Given that (x - 1) (x - 3) (x - 5) (x - 7) = 9

Let x - 4 = p Then the given eqn becomes (p + 3) (p + 1) (p - 1) (p - 3) = 9 (p2 - 1) (p2 - 9) = 9 p4 - 10p2 + 9 = 9 p2 (p2 - 10) = 0 p2 =0 or p2-10 =0 p = 0 or p = sqrt(10) or p = - sqrt(10) then x - 4 = 0, x - 4 = sqrt(10) or x - 4 = - sqrt(10) Now the roots of the given eqn are 4,4 + sqrt(10) and 4 - sqrt(10) The irrational roots are 4+sqrt(10) and 4 - sqrt(10) The sum of the irrational roots = 4 + sqrt(10) + 4 - sqrt(10) = 8.

Hence the answer is 8.

Let initially X grass was present there,and it is increasing by Y grass per day, then for the first condition We get, X+24*y = 24*70 ----(1) For the 2nd condition, we have, X+60*Y = 60*30----(2) Now, On solving equation (1) and (2), we get X = 1600 and Y = 10 /3 Third Condition, X+96*Y = 96 *N -----(3) [N = Number of Cows required] Putting the values of X and Y in equation (3), We get N = 20.

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x^0.8=16 x=(2^4)^(10/8) x=2^5 x=32

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