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Consider six distinct natural numbers such that the average of the two smallest numbers is $14$, and the average of the two largest numbers is $28$. Then, the maximum possible value of the average of these six numbers is

  1. $22 .5$
  2. $23$
  3. $23 .5$
  4. $24$
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Let the six distinct natural numbers in ascending order be a, b, c, d, e, f. The avg of two smallest number is (a+b)/2 = 14 and the avg of two largest number is (e+f)/2 = 28. For maximum possible value of (a+b+c+d+e+f)/6, we need max value of c and d. By seeing the info we can say that 14 < c < d < 28.

Possible values of (e,f) are (27,29), (26,30), (25,31), … ,(18,38). But if we take (e,f) as (27,29), then we can take d=26 and c=25 for getting maximum value.

Therefore the max possible value of (a+b+c+d+e+f)/6 is (2*14 + 25 + 26 + 2*28)/6 = 135/6 = 22.5

Ans is A.

Answer:

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