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If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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25 Jul 2017, 11:33
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If S is the sum of the reciprocals of the 10 consecutive integers from 21 to 30, then S is between which of the following two fractions? A. \(\frac{1}{3}\) and \(\frac{1}{2}\) B. \(\frac{1}{4}\) and \(\frac{1}{3}\) C. \(\frac{1}{5}\) and \(\frac{1}{4}\) D. \(\frac{1}{6}\) and \(\frac{1}{5}\) E. \(\frac{1}{7}\) and\(\frac{1}{6}\)
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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25 Jul 2017, 23:30
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carcass wrote: If S is the sum of the reciprocals of the 10 consecutive integers from 21 to 30, then S is between which of the following two fractions?
A. \(\frac{1}{3}\) and \(\frac{1}{2}\)
B. \(\frac{1}{4}\) and \(\frac{1}{3}\)
C. \(\frac{1}{5}\) and \(\frac{1}{4}\)
D. \(\frac{1}{6}\) and \(\frac{1}{5}\)
E. \(\frac{1}{7}\) and\(\frac{1}{6}\) \(S = \frac{1}{21} + \frac{1}{22} + \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} + \frac{1}{28} + \frac{1}{29} + \frac{1}{30}\). Notice that 1/21 is the largest term and 1/30 is the smallest term. If all 10 terms were equal to 1/30, then the sum would be 10*1/30 = 1/3, but since the actual sum is more than that, then we have that S > 1/3. If all 10 terms were equal to 1/21, then the sum would be 10*1/21 = 10/21, but since the actual sum is less than that, then we have that S < 10/21. Therefore, \(\frac{1}{3} < S < \frac{10}{21}\) (notice that 10/21 < 1/2, so 1/3 < S < 10/21 < 1/2). Answer: A.
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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25 Jul 2017, 23:31
Bunuel wrote: carcass wrote: If S is the sum of the reciprocals of the 10 consecutive integers from 21 to 30, then S is between which of the following two fractions?
A. \(\frac{1}{3}\) and \(\frac{1}{2}\)
B. \(\frac{1}{4}\) and \(\frac{1}{3}\)
C. \(\frac{1}{5}\) and \(\frac{1}{4}\)
D. \(\frac{1}{6}\) and \(\frac{1}{5}\)
E. \(\frac{1}{7}\) and\(\frac{1}{6}\) \(S = \frac{1}{21} + \frac{1}{22} + \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} + \frac{1}{28} + \frac{1}{29} + \frac{1}{30}\). Notice that 1/21 is the largest term and 1/30 is the smallest term. If all 10 terms were equal to 1/30, then the sum would be 10*1/30 = 1/3, but since the actual sum is more than that, then we have that S > 1/3. If all 10 terms were equal to 1/21, then the sum would be 10*1/21 = 10/21, but since the actual sum is less than that, then we have that S < 10/21. Therefore, \(\frac{1}{3} < S < \frac{10}{21}\) (notice that 10/21 < 1/2, so 1/3 < S < 10/21 < 1/2). Answer: A. Similar questions to practice: https://gmatclub.com/forum/misthesum ... 43703.htmlhttps://gmatclub.com/forum/ifkisthe ... 45365.htmlhttps://gmatclub.com/forum/ifsisthe ... 32690.html
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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26 Jul 2017, 16:21
carcass wrote: If S is the sum of the reciprocals of the 10 consecutive integers from 21 to 30, then S is between which of the following two fractions?
A. \(\frac{1}{3}\) and \(\frac{1}{2}\)
B. \(\frac{1}{4}\) and \(\frac{1}{3}\)
C. \(\frac{1}{5}\) and \(\frac{1}{4}\)
D. \(\frac{1}{6}\) and \(\frac{1}{5}\)
E. \(\frac{1}{7}\) and\(\frac{1}{6}\) Let's first analyze the question. We are trying to find a potential range for S in which S is the sum of the 10 reciprocals from 21 to 30 inclusive. Thus, S is: 1/21 + 1/22 + 1/23 + … + 1/30 Since we probably would not be expected to do such timeconsuming arithmetic (i.e., to add 10 fractions, each with a different denominator), that is exactly why each answer choice is given as a range of values. Thus, we do not need to know the EXACT value of S. The easiest way to determine the RANGE of values for S is to use easy numbers that can be quickly manipulated. Notice that 1/20 is greater than each of the addends and that 1/30 is less than or equal to each of the addends. Therefore, instead of trying to add 1/21 + 1/22 + 1/23 + … + 1/30, we are going to add 1/20 ten times and 1/30 ten times. These two sums will give us a high estimate of S and a low estimate of S. Again, we are adding 1/20 ten times and 1/30 ten times because there are 10 numbers from 1/21 to 1/30. Instead of actually adding each one of these values ten times, we will simply multiply each value by 10: 1/30 x 10 = ⅓. This value is the low estimate of S. 1/20 x 10 = ½. This value is the high estimate of S. We see that M is between 1/3 and 1/2. Answer: A
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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10 Nov 2017, 05:24
can we use mean = median ideology here ?
using this i got the sum and answer as A.



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10 Nov 2017, 05:41



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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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18 Dec 2017, 05:12
carcass wrote: If S is the sum of the reciprocals of the 10 consecutive integers from 21 to 30, then S is between which of the following two fractions?
A. \(\frac{1}{3}\) and \(\frac{1}{2}\)
B. \(\frac{1}{4}\) and \(\frac{1}{3}\)
C. \(\frac{1}{5}\) and \(\frac{1}{4}\)
D. \(\frac{1}{6}\) and \(\frac{1}{5}\)
E. \(\frac{1}{7}\) and\(\frac{1}{6}\) Easier way at least for me was this method. 2130 are 10 consecutive integers ( AP) sum of terms S= Mean * No of termsTherefore: \(\frac{(31+20)}{2}\) *10= 255 ( S) Now take reciprocal of this and it becomes : \(\frac{1}{255}\) by long division within few seconds, you will estimate the value to be .003 something. Only Option A went with this. Correct me if this method was wrong. Regards



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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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24 Feb 2018, 13:42
ScottTargetTestPrep wrote: carcass wrote: If S is the sum of the reciprocals of the 10 consecutive integers from 21 to 30, then S is between which of the following two fractions?
A. \(\frac{1}{3}\) and \(\frac{1}{2}\)
B. \(\frac{1}{4}\) and \(\frac{1}{3}\)
C. \(\frac{1}{5}\) and \(\frac{1}{4}\)
D. \(\frac{1}{6}\) and \(\frac{1}{5}\)
E. \(\frac{1}{7}\) and\(\frac{1}{6}\) Let's first analyze the question. We are trying to find a potential range for S in which S is the sum of the 10 reciprocals from 21 to 30 inclusive. Thus, S is: 1/21 + 1/22 + 1/23 + … + 1/30 Since we probably would not be expected to do such timeconsuming arithmetic (i.e., to add 10 fractions, each with a different denominator), that is exactly why each answer choice is given as a range of values. Thus, we do not need to know the EXACT value of S. The easiest way to determine the RANGE of values for S is to use easy numbers that can be quickly manipulated. Notice that 1/20 is greater than each of the addends and that 1/30 is less than or equal to each of the addends. Therefore, instead of trying to add 1/21 + 1/22 + 1/23 + … + 1/30, we are going to add 1/20 ten times and 1/30 ten times. These two sums will give us a high estimate of S and a low estimate of S. Again, we are adding 1/20 ten times and 1/30 ten times because there are 10 numbers from 1/21 to 1/30. Instead of actually adding each one of these values ten times, we will simply multiply each value by 10: 1/30 x 10 = ⅓. This value is the low estimate of S. 1/20 x 10 = ½. This value is the high estimate of S. We see that M is between 1/3 and 1/2. Answer: A Hello Why are you adding 1/20 ten times instead of 1/21 ? And how can 1/30 be equal to each of the addends thank you for your explanation



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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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15 Mar 2018, 01:50
My approach to this question was a little bit different. I thought it was worth to comment.
I considered a normal sequence: 1,2,3,4,5,6,7,8,9,10. So I added the first two extreme values (1,10) and did that for all the other pairs (2,9), (3,8) .... So what can we take from this?? All the pairs are equal 11. Then i thought that this should apply to reciprocals.
I did the first calculation and approximated 1/21 to 1/20. So 0.05 + 0.033=0.083. There is no need for more calculations but if you wanted you could have calculated the next pair to be sure. Then i multiplied by 5 because we have 5 pairs and got S=0.4. Which is between 1/3 and 1/2.
Just as an extra info. The real value of S, if all the calculations were done would be 0.79.
Hope it helps someone



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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]
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26 Mar 2018, 05:27
Solution \(S\)= \(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+…………….+ \frac{1}{30}\) The greatest value of \(S\):
We know, • \(\frac{1}{20}>\frac{1}{21}\). Hence, • \(\frac{1}{20}+\frac{1}{20}+……+\frac{1}{20}> \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+…….+\frac{1}{30}\) • \(\frac{10}{20}>S\) • \(S<\frac{1}{2}\) Hence, \(S\) is always smaller than \(\frac{1}{2}\). The least value of S:
We know, • \(\frac{1}{29}>\frac{1}{30}\) Hence, •\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+…………+\frac{1}{29}+\frac{1}{30}> \frac{1}{30}+\frac{1}{30}+…..+1\frac{}{30}\) •\(S> \frac{10}{30}\) •\(S\)>\(\frac{1}{3}\) Hence, \(S\) is always greater than \(\frac{1}{3}\).Thus, \(\frac{1}{3}< S<\frac{1}{2}\). Hence \(A\) is the correct answer.Answer: A
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