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pls tell me how to solve this.

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pls tell me how to solve this.

by PAB2706 » Sun Aug 28, 2011 12:41 am
Set A is made up of every 1/n
, where n is
a positive integer and 100 ≤ n ≤ 150.
If S is the sum of all the elements in A,
then:
(A) S < 1/3

(B)1/3 < S < 1/2


(C) 1/2 < S < 1

(D) 1 < S < 1 1/2
(E) 1 1/2 < S < 2
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by force5 » Sun Aug 28, 2011 7:36 am
should be B between 1/3 and 1/2.
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by PAB2706 » Sun Aug 28, 2011 8:30 am
OA given is C and i dnt knw how.. bcos even i got B
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by force5 » Sun Aug 28, 2011 10:14 am
it has to be B, the sum will never exceed 0.5
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by gmatboost » Sun Aug 28, 2011 9:25 pm
Set A is made up of every 1/n, where n is a positive integer and 100 ≤ n ≤ 150.
If S is the sum of all the elements in A, then:
If every number in the set was as big as the biggest element (1/100) the sum would be:
51*(1/100) = 51/100 = about 1/2

If every number in the set was as small as the smallest element (1/150) the sum would be:
51*(1/150) = 51/150 = about 1/3

The actual sum lies somewhere in between, since the terms start out at 1/100 and slowly make their way to 1/150. So, the sum is somewhere in between 1/2 and 1/3.
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by vinod192 » Mon Aug 29, 2011 11:24 am
hi gmatboost,

can u please explain why have u taken 51 and not any other number????
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by gmatboost » Mon Aug 29, 2011 11:32 am
Yes:
Set A is made up of every 1/n, where n is a positive integer and 100 ≤ n ≤ 150.
There are 51 numbers in the range 100 ≤ n ≤ 150.
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by navami » Mon Aug 29, 2011 12:17 pm
got to be 1/3< <1/2
This time no looking back!!!
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by saketk » Tue Aug 30, 2011 3:29 am
vinod192 wrote:hi gmatboost,

can u please explain why have u taken 51 and not any other number????
I think he used-- divergence of the harmonic series
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by gmatboost » Tue Aug 30, 2011 7:23 am
It definitely had nothing to do with anything related to a harmonic series. That kind of stuff is NOT on the GMAT.

It had to do with just counting the number of terms, from the biggest one (1/100) to the smallest one (1/150). There are 51 terms because there are 51 numbers from 100 to 150, inclusive.
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