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If S is the sum of the reciprocals of the consecutive [#permalink]
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23 Dec 2005, 02:19
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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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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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23 Dec 2005, 03:14
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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
Answer is C
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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23 Dec 2005, 05:26
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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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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



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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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25 Nov 2013, 08:43
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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. Answer: C. Similar questions to practice: misthesumofthereciprocalsoftheconsecutiveintegers143703.htmlifkisthesumofreciprocalsoftheconsecutiveintegers145365.html
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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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10 Jan 2014, 11:40
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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 know that there are 10 numbers in the sum: (10091)+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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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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21 Feb 2015, 23:50
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 nonmath related skills: organization, accuracy, attention to detail, ability to prove that you're correct, patternmatching, pacing, etc. To maximize your performance on Test Day, you have to be a stronger 'strategist' and 'patternmatcher' than 'mathematician.' GMAT assassins aren't born, they're made, Rich
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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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22 Mar 2015, 12:10
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.
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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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20 Jan 2016, 18:30
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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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17 Jul 2016, 04:27
hey! callig all experts... bunnuel / chetan4u / egmat etc to help in solving this with some less math richi do help plz



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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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18 Jul 2016, 12:10
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
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If S is the sum of the reciprocals of the consecutive [#permalink]
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18 Jul 2016, 14:55
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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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1/91 + 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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18 Jul 2016, 18:10
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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01 Apr 2017, 01:02
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) answer is C.



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Re: If S is the sum of the reciprocals of the consecutive [#permalink]
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22 Jul 2017, 18:17
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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 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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11 Dec 2017, 07:29
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. Answer: C
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If S is the sum of the reciprocals of the consecutive [#permalink]
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19 Dec 2017, 06:57
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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