melguy wrote:Please help me with this question.

I rounded up everything to .01 so the result should be slightly less than 1/10?

Thanks

melguy, probably the best thing you could do for your score in this situation is to go beyond seeing that 1/91 is greater than 1/100 to figuring out what about your process resulted in your not seeing that that is the case.

My take is that your thought process was incomplete, in that you saw basically how to get to the answer to the question, but you did not then really logically figure out what exactly to do. Rather you just went with something that seemed ok WITHOUT FULLY CONNECTING YOUR APPROACH OR YOUR THINKING TO THE SITUATION.

So it seems that your key to scoring higher is more clearly connecting things.

melguy wrote:
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I Only
C. III Only
D. II and III only
E. I, II and III

Let's first analyze the question. We are trying to find a potential range for S, and S is equal to the sum of the reciprocals from 91 to 100, inclusive. Thus, S is:

1/91 + 1/92 + 1/93 + ...+ 1/100

The easiest way to determine the RANGE of S is to use easy numbers that can be quickly manipulated.

Note that 1/90 is greater than each of the addends and that 1/100 is less than or equal to each of the addends. Therefore, instead of trying to add together 1/91 + 1/92 + 1/93 + ...+ 1/100, we are instead going to first add 1/90 ten times and then add 1/100 ten times. These two sums will give us a high estimate of S and a low estimate of S, respectively. Again, we are adding 1/90 and then 1/100, ten times, because there are 10 numbers from 1/91 to 1/100, inclusive.

Instead of actually adding each of these values ten times, we will simply multiply each value by 10:

1/100 x 10 = 1/10

1/90 x 10 = 1/9

We see that S is between 1/10 and 1/9, i.e., 1/10 < S < 1/9. Of the three numbers given in the Roman numerals, only 1/10 is less than S.

Answer: C