GMAT Question:

[ngbrian85]

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

Stuck on this one. Can any of you quant quru's help ?

[seal4913]

Answer is posted here: https://www.beatthegmat.com/good-ps-t161 ... cityevent=

[pemdas]

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

Stuck on this one. Can any of you quant quru's help ?

i must say after looking up the post copied/pasted solution is so-so
not satisfied and solved with my own

this is simple series of consecutive just reciprocal, the sum will be (first+last)/2 *no of values
(1/201 + 1/300)/2 *100 = (300+201)/300*201 *(50) = ~0.4

a

[seal4913]

Stuart Kovinsky's answer in the link above is the best way it can be explained. Your way has way too much unnecessary work. Why give an answer as a decimal when all the choices are in fractions?

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

Stuck on this one. Can any of you quant quru's help ?

i must say after looking up the post copied/pasted solution is so-so
not satisfied and solved with my own

this is simple series of consecutive just reciprocal, the sum will be (first+last)/2 *no of values
(1/201 + 1/300)/2 *100 = (300+201)/300*201 *(50) = ~0.4

a

[Brent@GMATPrepNow]

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

Stuck on this one. Can any of you quant quru's help ?

Here's another approach.

We want to find 1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300

NOTE: there are 100 fractions in this sum, since 300-201+1 = 100.

Let's examine the extreme values (1/201 and 1/300)

First consider a case where all of the values are equal to the smallest fraction (1/300)
We get: 1/300 + 1/300 + 1/300 + ... + 1/300 = 100/300 = 1/3
So, the original sum must be greater than 1/3

Now consider a case where all of the values are equal to the biggest fraction (1/201)
In fact, let's go a little bigger and use 1/200
We get: 1/200 + 1/200 + 1/200 + ... + 1/200 = 100/200 = 1/2
So, the original sum must be less than 1/2

Combine both cases to get 1/3 < M < 1/2 = A

Cheers,
Brent

[Anju@Gurome]

this is simple series of consecutive just reciprocal, the sum will be (first+last)/2 *no of values

I think his approach of getting 0.4 is much easier than the OA.

He got lucky in this case as that method of getting the sum is only valid for uniformly distributed numbers.

In this case 1/201, 1/202, ..., 1/300 are not uniformly distributed as the distance between any two consecutive terms is not a constant.

For example, consider the following problem

If M = 1/2 + 1/3 + 1/4, which of the following is true?
A. 0.9 < M < 1.0
B. 1.0 < M < 1.1
C. 1.1 < M < 1.2
D. 1.2 < M < 1.3
E. 1.3 < M < 1.4

Actual sum of the terms = (1/2 + 1/3 + 1/4) = (6 + 4 + 3)/12 = 13/12 ≈ 1.08
Hence, the correct solution is B.

But sum of the term misusing this formula = [(1/2 + 1/4)/2]*3 = (3/8)*3 = 9/8 = 1.125
And according to this the answer should be C.

Hope that helps.

[abhijitlandge]

del

+1 Thank to you for your signature line

Success doesn't come overnight

great line for those who are always in search of shortcuts and tricks

Abhijit