Problem Solving

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

OA C

We see that 1/91 > 1/92 > 1/93 ... > 1/100

There are 10 terms in p.

We see that

--1/91 + 1/91 + 1/91 ... + 1/91 > 1/91 + 1/92 + 1/93 ... + 1/100;
--1/100 + 1/100 + 1/100 ... + 1/100 < 1/91 + 1/92 + 1/93 ... + 1/100

10*(1/100) < p < 10*(1/91);

1/10 < p < 1/9.1

=> 10 > 1/p > 9.1

Thus, we see that only 10 is more than 1/p.

The correct answer: C

Hope this helps!

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Hi stevecultt,

Based on the wording of the prompt, you might think that you should add up the fractions 1/91 + 1/92 + .... 1/100, but the GMAT would NEVER require that you do that math. Instead, lets do some real basic estimation of what that sum would be LESS than and GREATER than....

There are 10 total fractions and 9 of them are GREATER than 1/100. So, at the 'lower end', let's just say that all 10 fractions are equal to 1/100....

(10)(1/100) = 10/100 = 1/10

Thus, we know that the sum of those 10 fractions will be GREATER than 1/10.

Similarly, we know that all 10 of those fractions are LESS than 1/90. So, at the 'higher end', let's just say that all 10 fractions are equal to 1/90...

(10)(1/90) = 10/90 = 1/9

Thus, we know that the sum of those 10 fractions will be LESS than 1/9.

With those two deductions, there's only one answer that 'fits'...

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

This question is similar in style to the one posted here:
https://www.beatthegmat.com/m-is-the-sum ... tml#793900

The big takeaway in both cases is that you don't actually need to do the math! The question structure indicates that you just need to approximate.

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