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If S is the sum of the reciprocals of the 10 consecutive integers from

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If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   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] New post   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] New post   26 Jul 2017, 16:21
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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\(\frac{1}{6}\)


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 time-consuming 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] New post   25 Jul 2017, 23:31
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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.


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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   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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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   10 Nov 2017, 05:41
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kunalsinghNS wrote:
can we use mean = median ideology here ?

using this i got the sum and answer as A.


No. The set {1/21, 1/22, 1/23, …, 1/30} is NOT evenly spaced but the formula you apply is for an evenly spaced set.
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   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.

21-30 are 10 consecutive integers ( AP)

sum of terms S= Mean * No of terms

Therefore: \(\frac{(31+20)}{2}\) *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] New post   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 time-consuming 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] New post   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] New post   26 Mar 2018, 05:27
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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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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   19 Dec 2018, 02:17
A similar exercice

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If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   Updated on: 06 Jul 2020, 10:54
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below is my approach,

from 21 to 30 average of these numbers is 25 , and we have a total of 10 numbers

so, range will be 10/25 , which gives 1/2.5

clearly it falls under option A

Originally posted by dine5207 on 22 Dec 2018, 22:21.
Last edited by dine5207 on 06 Jul 2020, 10:54, edited 1 time in total.
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   12 Sep 2019, 14:23
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mtk10 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\(\frac{1}{6}\)



Easier way at least for me was this method.

21-30 are 10 consecutive integers ( AP)

sum of terms S= Mean * No of terms

Therefore: \(\frac{(31+20)}{2}\) *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


How is 0.003 in between 1/3 and 1/2
1/3=0.33
1/2=0.5 ??
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   18 Sep 2019, 07:17
dine5207 wrote:
i follows below approach,

from 21 to 30 average of these numbers is 25 , and we have a total of 10 numbers

so, range will be 10/25 , which gives 1/2.5

clearly it falls under option A


. I think we are not allowed to reciprocate the no. after median . as the sum is 1/21 to 1/30 . we cant assume the mean and then reciprocate it which in this case would have been 25 +26 /2 .
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If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   05 Oct 2020, 04:31
mtk10 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\(\frac{1}{6}\)



Easier way at least for me was this method.

21-30 are 10 consecutive integers ( AP)

sum of terms S= Mean * No of terms

Therefore: \(\frac{(31+20)}{2}\) *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


chetan2u is the above mentioned correct ? if yes how 0,003 is between 0.3 and 0.5 :?
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   05 Oct 2020, 05:00
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dave13 wrote:
mtk10 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\(\frac{1}{6}\)



Easier way at least for me was this method.

21-30 are 10 consecutive integers ( AP)

sum of terms S= Mean * No of terms

Therefore: \(\frac{(31+20)}{2}\) *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


chetan2u is the above mentioned correct ? if yes how 0,003 is between 0.3 and 0.5 :?



Hi,

Neither the method is correct nor the statement that 0.003 is between 1/3 and 1/2.
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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   10 Jan 2021, 02:38
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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\(\frac{1}{6}\)


The maximum sum will be \(=\frac{1}{20}*10=\frac{1}{2}\)

The least sum will be \(=\frac{1}{30}*10=\frac{1}{3}\)

So, Sum is between \(\frac{1}{3}\) and \(\frac{1}{2}\)

The answer is A
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If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   13 May 2022, 13:43
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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\(\frac{1}{6}\)


A question that is so obviously a candidate for ballparking, even the answer explanation in the OG does it that way...one of only two questions that GMAC shows ballparking in the 2022 OG Quant. By my count, of the 212 PS questions in that edition, 26 are ballparking candidates, and they cover a wide variety of math concepts. Ballparking is probably the single most useful technique on Quant (okay, maybe AD/BCE, but everybody knows that ;) ).

Since I love ballparking so much, I'll give you TWO ways to ballpark this one!

First, the differences between each item in our list and the next item in our list aren't exactly the same, but they're pretty close. If we just say the median and the mean are pretty close and roll with 10*(1/25), we should be within striking distance. That's 10/25, which is 0.4. Answer choice A.

Second, if all ten items were 1/30, we'd have 10*(1/30), which is 10/30, which is 1/3. 1/30 is the SMALLEST item in our list, so our sum should be something larger than 1/3. Answer choice A.

ThatDudeKnowsBallparking
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If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink] New post   15 Dec 2022, 11:08
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.


We don't really need to find the high end of the sum.
The low end of the sum is 1/3 (as you explained).
So the correct answer MUST be greater than 1/3.
There is only such one answer choice - option A.
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