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Manager  Joined: 10 Dec 2005
Posts: 88
If S is the sum of the reciprocals of the consecutive  [#permalink]

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Question Stats: 62% (01:44) correct 38% (01:46) wrong based on 1495 sessions

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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

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JAI HIND!
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Re: If S is the sum of the reciprocals of the consecutive  [#permalink]

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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

Given that $$S=\frac{1}{91}+\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}+\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}$$. Notice that 1/91 is the larges term and 1/100 is the smallest term.

If all 10 terms were equal to 1/91, then the sum would be 10/91, but since actual sum is less than that, then we have that S<1/91.

If all 10 terms were equal to 1/100, then the sum would be 10/100=1/10, but since actual sum is more than that, then we have that S>1/10.

Therefore, 1/10 < S < 10/91.

Also, notice that 10/91 < 1/9 < 1/8, thus we have that 1/10 < S < 10/91 < 1/9 < 1/8.

Therefore only 1/10 is less than S.

Similar questions to practice:
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Manager  Joined: 12 Nov 2005
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This question is a repetition here.
Anyways lemme explain the solution

1/91+1/92+..........1/100 > 1/100+1/100......10 times
or Summation S >10/100
i.e S>1/10

Similarly
1/91+1/92+..........1/100 < 1/91+1/91......10 times
Summation < 10/91 <1/9

Hence summation is only greater than 1/10.........Hence C
##### General Discussion
Director  Joined: 17 Dec 2005
Posts: 530
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I think it's C, but I'm not quite sure.

Since we summarize the reciprocals from 100 to 91, we can say also that we add ten numbers who are all (with one exception 1/100) greater than 1/100, so that the sum must be greater than 1/10.

On the other side we can say that we add the reciprocals from 91 to 100, so that the sum has to be less than the sum of ten times 1/91.

We can conclude that the sum has to be less than 1/9 but more than 1/10. That leaves us C as the only possible answer.
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Re: PS: Summation  [#permalink]

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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

S = 1/91 + 1/92 + 1/93.....+ 1/100
S = (1/91 + 1/100)(10/2) = 191/1820 = approx 0.1049.....

I. 1/8 = 0.12...
II. 1/9 = 0.11....
III. 1/10 = 0.10
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Re: PS: Summation  [#permalink]

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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

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.
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Re: If S is the sum of the reciprocals of the consecutive  [#permalink]

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2
1
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

We know that there are 10 numbers in the sum: (100-91)+1=10
Take the mean of the sum and times it by 10 to get our sum: (1/ ((100+91)/2)) x 10 = 10/95.5 = 1/9.55

From here we know that the only number which will be smaller than our sum must be divisible by >9.55

Hence only III satisfies. Answer: C
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If S is the sum of the reciprocals of the consecutive  [#permalink]

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2
Hi All,

The other explanations in this thread have properly explained the "math" behind this prompt - it's essentially about figuring out the "minimum" and "maximum" value that the sum COULD be, then realizing the sum is between those two values. Any time you find yourself reading a Quant question and you think "the math will take forever", then you're probably right AND there should be another way to get to the correct answer. The Quant section of the GMAT is NOT a "math test" (at least not in the way that you might be used to thinking about it). Yes, you will do plenty of small calculations and use formulas, but the Quant section is there to test you on LOTS of other non-math related skills: organization, accuracy, attention to detail, ability to prove that you're correct, pattern-matching, pacing, etc. To maximize your performance on Test Day, you have to be a stronger 'strategist' and 'pattern-matcher' than 'mathematician.'

GMAT assassins aren't born, they're made,
Rich
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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Manager  S Joined: 22 Jan 2014 Posts: 174 WE: Project Management (Computer Hardware) Re: If S is the sum of the reciprocals of the consecutive [#permalink] ### Show Tags 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 + ... + 1/100 or S > 1/100 + 1/100 +... + 1/100 S > 1/10 or S < 10/91 S < 0.109 => 0.1 < S < 0.109 hence C. _________________ Illegitimi non carborundum. Intern  Joined: 31 Oct 2015 Posts: 32 Re: If S is the sum of the reciprocals of the consecutive [#permalink] ### Show Tags 1 See attachment for response. Attachments SmartSelectImage_2016-01-20-20-29-20.png.png [ 56.1 KiB | Viewed 30564 times ] Retired Moderator B Status: I Declare War!!! Joined: 02 Apr 2014 Posts: 234 Location: United States Concentration: Finance, Economics GMAT Date: 03-18-2015 WE: Asset Management (Investment Banking) Re: If S is the sum of the reciprocals of the consecutive [#permalink] ### Show Tags hey! callig all experts... bunnuel / chetan4u / egmat etc to help in solving this with some less math richi do help plz EMPOWERgmat Instructor V Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13791 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If S is the sum of the reciprocals of the consecutive [#permalink] ### Show Tags 1 Hi Celestial09, 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: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Follow Special Offer: Save$75 + GMAT Club Tests Free
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If S is the sum of the reciprocals of the consecutive  [#permalink]

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1/91+1/100=.021
.021/2=.011
10*.011=.11
only 1/10<.11
C. III only
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Re: If S is the sum of the reciprocals of the consecutive  [#permalink]

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1
1/91 + 1/92 +...+ 1/100..
I took 1/95 as the median of the sequence (approximately, the real median is 1/95.5).
So 10*1/95= approximately 1/9.5.
So the only answer possible is C.
It took me like 15 seconds for this question approaching the problem this way.
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Re: If S is the sum of the reciprocals of the consecutive  [#permalink]

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Top Contributor
C. The sum of the reciprocals is 10/955, which is derived from the sum of consecutive integers formula n(n+1)/2, and that number is less than 1/8 and 1/9, both. Tested via cross multiplication.

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Re: If S is the sum of the reciprocals of the consecutive  [#permalink]

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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

we can solve this problem using the Harmonic progression rule

(highest value)(# of terms) > (sum of all terms in the HP) > (lowest value)(# of terms)
=> (1/91)(10) > sum of terms > (1/100)(10)
therefore, (1/8) > (1/9) > (1/9.1) > sum of the HP > (1/10)

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Re: If S is the sum of the reciprocals of the consecutive  [#permalink]

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1
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

This problem is a max/min one in disguise. We know that 1/91 > 1/100. So if all of the numbers were 1/91, then the upper bound for addition would be 10/91

10/91 < 1/9 = 10/90 because the denominator is larger. So 10/91 < 1/9. So we know that S, the sum, is less than 1/9. 1/9 is smaller than 1/8, so I. and II. are out.

Lastly, we know that the smallest possible sum is 10*1/100 = 10/100 = 1/10 = III. We know that the actual sum is larger than 1/10, so 1/10<S<10/91<1/9<1/8. C is the right answer.
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Re: If S is the sum of the reciprocals of the consecutive  [#permalink]

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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

Let's first analyze the question. We are trying to find a potential range for S, and S is equal to the sum of the reciprocals from 91 to 100, inclusive. Thus, S is:

1/91 + 1/92 + 1/93 + …+ 1/100

The easiest way to determine the RANGE of S 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 together 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 S and a low estimate of S, respectively. Again, we are adding 1/90 and then 1/100, ten times, 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 S is between 1/10 and 1/9, i.e., 1/10 < S < 1/9. Of the three numbers given in the Roman numerals, only 1/10 is less than S.

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Manager  G
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If S is the sum of the reciprocals of the consecutive  [#permalink]

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We need to apply Minimax concept in this question.

Sum of this Series= 1/91 + 1/92....1/100

To Maximize the sum of this series, we must consider the lowest possible denominator as common to all terms= 91
Hence, Sum Maximized= 1/91 + 1/91.....1/91= 10/91= 1/9.1

To Minimize the sum of this series, we must consider the highest possible denominator as common to all terms= 100
Hence, Sum Minimized= 1/100 + 1/100.....1/100= 10/100= 1/10

Hence, 1/9.1 >Sum of series >1/10

Therefore, Only option III matches with above range.

So, correct Ans choice: C
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Joined: 10 May 2018
Posts: 29
If S is the sum of the reciprocals of the consecutive  [#permalink]

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Hi,
Can anybody confirm my method?

Sum = Avg * Nr of terms
= (first+Last)/2 * 10
= 10/955 * 10
= 100/955

now just divide 100/955 to find the decimal If S is the sum of the reciprocals of the consecutive   [#permalink] 22 Oct 2018, 04:38
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# If S is the sum of the reciprocals of the consecutive

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