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 ?
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pemdas
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i must say after looking up the post copied/pasted solution is so-songbrian85 wrote: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 ?
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
Success doesn't come overnight!
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?
pemdas wrote:i must say after looking up the post copied/pasted solution is so-songbrian85 wrote: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 ?
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
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Brent@GMATPrepNow
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Here's another approach.ngbrian85 wrote: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 ?
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
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Anju@Gurome
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pemdas wrote:this is simple series of consecutive just reciprocal, the sum will be (first+last)/2 *no of values
He got lucky in this case as that method of getting the sum is only valid for uniformly distributed numbers.katerina.ajtova wrote:I think his approach of getting 0.4 is much easier than the OA.
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
Actual sum of the terms = (1/2 + 1/3 + 1/4) = (6 + 4 + 3)/12 = 13/12 ≈ 1.08If 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
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.
Anju Agarwal
Quant Expert, Gurome
Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.
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Quant Expert, Gurome
Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.
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abhijitlandge
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+1 Thank to you for your signature linepemdas wrote:del
Success doesn't come overnight
great line for those who are always in search of shortcuts and tricks
Abhijit
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Problems common to all. Attitude makes the difference
https://www.beatthegmat.com/co-ordinate- ... tml#609851
https://www.beatthegmat.com/time-speed-a ... 13760.html
https://www.beatthegmat.com/permutations ... 25328.html
