AAPL wrote:GMAT Prep
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
The OA is C
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
Cheers,
Brent
