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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]

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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\(\frac{1}{6}\)
[Reveal] Spoiler: OA

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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]

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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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New post 25 Jul 2017, 22:31
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.


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Re: If S is the sum of the reciprocals of the 10 consecutive integers from [#permalink]

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New post 26 Jul 2017, 15:21
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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]

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New post 10 Nov 2017, 04:24
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]

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Re: If S is the sum of the reciprocals of the 10 consecutive integers from   [#permalink] 10 Nov 2017, 04:41
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