We want the sum 1/91 + 1/92 + 1/93 + . . . + 1/100If 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
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
