If S is the sum of recriprocal

[melguy]

Please help me with this question.

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

Thanks

[Brent@GMATPrepNow]

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

We want the sum 1/91 + 1/92 + 1/93 + . . . + 1/100
Of these 10 fractions, 1/91 has the GREATEST value, and 100 has the SMALLEST value

So, let's examine some EXTREMES

If all of the 10 fractions were 1/91, then the sum would equal 1/91 + 1/91 + 1/91 + .... + 1/91
= 10/91
Of course most of the fractions are less than 1/91, so we can conclude that S < 10/91

If all of the 10 fractions were 1/100, then the sum would equal 1/100 + 1/100 + 1/100 + . . . + 1/100
= 10/100 = 1/10
Of course most of the fractions are greater than 1/100, so we can conclude that S > 1/10

So, we know that 1/10 < S < 10/91

Since 1/10 < S, we know that statement III works

What about S < 10/91 . What does this tell us?
First of all, 1/9 = 10/90
Second, 10/91 < 10/90, so we can conclude that 10/91 < 1/9
So, we know that S < 10/91 < 1/9
Since 1/9 is greater than S, we know that statement II does NOT work

Since 1/8 is greater than 1/9, we know that statement I does NOT work

Answer: C

[Brent@GMATPrepNow]

Please help me with this question.

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

Thanks

You are actually rounding DOWN.
1/100 < 1/99 < 1/98 < 1/97 < ....
So, for example, when you round 1/99 to 1/100, you are rounding down.

Cheers,
Brent

[MartyMurray]

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.

[Jeff@TargetTestPrep]

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