stevecultt wrote:If p is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is more than 1/p?
I. 8
II. 9
III. 10
(A) Only I
(B) Only II
(C) Only III
(D) Only II and III
(E) I, II and III
Let's first analyze the question. We are trying to find a potential range for 1/p, and p is equal to the sum of the reciprocals from 91 to 100 inclusive. Thus, p is:
1/91 + 1/92 + 1/93 + ...+ 1/100
The easiest way to determine the RANGE of p 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 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 p and a low estimate of p, respectively. Again, we are adding 1/90 and then 1/100, ten times each, 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 p is between 1/10 and 1/9, i.e., 1/10 < p < 1/9. We can reciprocate all three sides of this inequality by switching the inequality sign to have 10 > 1/p > 9. Thus we see that only 10 is greater than 1/p.
Answer:
C
