Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

...............

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