Q10 LOGICAL PUZZLES

Let T be the set of integers (3,11,19,27, …….451,459,467) and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in Sis

(a). 32

(b). 28

(c) 29

(d). 30

No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.

Now S will have atleast have of 59 terms i.e 29 .

Also the sum of 29th term and 30th term is less than 470.

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