JAI HIND 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
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