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Now, obviously 3*(\frac{1}{33})<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<3*(\frac{1}{31}), as {3 times the least #}<{given sum}<{3 times the largest #} --> \frac{3}{33}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31} --> \frac{1}{11}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{3}{30} --> \frac{1}{10+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{1}{10} --> n=10.

1/a > 3/33 ( i.e 1/11) ... Hence a<11 from eq 1 --- n+1 >a>11 ................ n<a<11.. hence n <11

1/a < 3/31 ( or 1/10)..... hence a>10 from eq 1 --- n+1>a>10 .... hence n+1>10 ... n> 9

Ans n=10

hi gmacforjyoab,

I guess i am lacking some mathematics in the highlighted part. Could you please throw some light. It would be great help.

Regards Atal Pandit

Since (1/n+1) < 1/a < 1/n , we can say that n+1 >a > n ( when u take the reciprocal of two numbers in an Inequality , the inequality flips ) Consider this ---- 1/4<1/3<1/2 , which would mean 4>3>2 ...

Oh and lets say - all the numbers were 1/33 , then the sum would be 3/33 , but all the numbers are not 1/33 , the other two numbers are 1/32 and 1/31 . and these two numbers are greater than 1/33 , hence the sum of 1/31 +1/32 + 1/33 would also be grater than 3/33 hence , 1/a > 3/33 i.e 1/11

Now, obviously 3*(\frac{1}{33})<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<3*(\frac{1}{31}), as {3 times the least #}<{given sum}<{3 times the largest #} --> \frac{3}{33}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31} --> \frac{1}{11}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{3}{30} --> \frac{1}{10+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{1}{10} --> n=10.

Answer: B.

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Amazing solution..... glad to learn this..... _________________

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