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If S is the sum of the reciprocals of the consecutive integers from 91

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sayan640
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Re: If S is the sum of the reciprocals of the consecutive integers from 91 [#permalink] New post   15 Jul 2019, 01:14
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JAI HIND wrote:
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I only
C. III only
D. II and III only
E. I, II, and III


The more is the value of the denominator , the lesser is the value of the fraction.
For ex :- 1/2 > 1/3 ...1/2 means 50%.... 1/3 means 33.33%...

Now here , 1/90 > 1/91
1/90 > 1/92..
.................

10 / 90 > 1/91 + 1/92 + ........1/100
Since each of the term on the left hand side i.e 1/90 is greater than the each of the term on the right hand side ....

So 1/9 > S..

Now , 1/8 > 1/9 ...So 1/8 is also greater than S.
Only one option is left . Option 3 is the answer.

Now lets check why option 3 is the answer.

1/100 < 1/91
1/100 < 1/92
..................

10/100 < 1/91 + 1/92 + ........1/100
1/10 < S

Please give me kudo s if you liked my explanation.
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Re: If S is the sum of the reciprocals of the consecutive integers from 91 [#permalink] New post   26 Jun 2020, 22:57
krisrini wrote:
JAI HIND wrote:
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I only
C. III only
D. II and III only
E. I, II, and III


From the answer choice we get 1) 1/8 = .125
2) 1/9 = .11 (Approx)
3) 1/10 = .1

1/100 = .01 The other numbers from 1/91 to 1/99 will also be close to .01 but little greater than that . so approx .01 * 10 = .1 So its C.



Thanks, I think this was the best and fastest solution

do the maths and solve
=> 1/91 & 1/92
=> 0.0101 (since all of them work out to approximately the same 0.0101 number you know that this series is going to be around the same which mean you can multiple it by 10)
=> 10*0.0101
=> 0.101

And now do the same for each of the different answers and you will see only 3 is lesser.
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Re: If S is the sum of the reciprocals of the consecutive integers from 91 [#permalink] New post   03 Aug 2020, 19:42
JAI HIND wrote:
If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S?

I. 1/8
II. 1/9
III. 1/10

A. None
B. I only
C. III only
D. II and III only
E. I, II, and III

****************************************************
S=(1/91)+(1/92)+(1/93)......(1/100)......Harmonic Progression where n=10
S= Median * n=[{(1/95)+(1/96)}/2]*10

The approach to this question will be based on approximation
S=10/(95*2) + 10/(96*2)
S= (1/19)+(1/19) (approximation)=(1/9.5)
Please note that the sum cannot be less than this and hence

The answer is (C)
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Re: If S is the sum of the reciprocals of the consecutive integers from 91 [#permalink] New post   22 Jan 2022, 05:57
I used the sum of consecutive numbers formula= first +last*n/2, there are 10 terms in this range.

1/100= 0.01 , 1/91 =slightly >0.01. 91 isnt too far from 100 so it will be slightly greater.

Hence 1/100+1/91 =>0.02 * 5 = slightly >0.1

C- Definitely less than 0.1
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Re: If S is the sum of the reciprocals of the consecutive integers from 91 [#permalink] New post   17 Feb 2024, 08:46
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Re: If S is the sum of the reciprocals of the consecutive integers from 91 [#permalink]
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