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27 Mar 2024, 01:32
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B
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Difficulty:
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Question Stats:
61% (02:14) correct
39%
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If S is the sum of reciprocals of a list of multiples of 3 from 300 to 396, inclusive, S is approximately equal to
A. 0.05
B. 0.1
C. 0.15
D. 0.2
E. 0.3
A. 0.05
B. 0.1
C. 0.15
D. 0.2
E. 0.3
05 Apr 2024, 07:30
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Official Solution:
If S is the sum of reciprocals of a list of multiples of 3 from 300 to 396, inclusive, S is approximately equal to
A. 0.05
B. 0.1
C. 0.15
D. 0.2
E. 0.3
\(S = \frac{1}{300} + \frac{1}{303} + \frac{1}{309} + ... + \frac{1}{396} = \frac{1}{3}(\frac{1}{100} + \frac{1}{101} + \frac{1}{103} + ... + \frac{1}{132})\)
In the brackets, we have the sum of 33 terms, which are reciprocals of consecutive integers from 100 to 132, inclusive. Notice that the largest term among the 33 is \(\frac{1}{100}\), and the smallest term is \(\frac{1}{132}\).
If all 33 terms were equal to the largest term \(\frac{1}{100}\), the sum in the brackets would be \(\frac{33}{100}\), hence the entire expression would be \(\frac{1}{3} * \frac{33}{100} = 0.11\). Since the actual sum must be less than that, then \(S < 0.11\)
If all 33 terms were equal to the smallest term \(\frac{1}{132}\), the sum in the brackets would be \(\frac{33}{132}=\frac{1}{4}=0.25\), hence the entire expression would be \(\frac{1}{3} *0.25 \approx 0.083...\). Since the actual sum must be more than that, then \(S > 0.083...\)
Therefore, \(0.083... < S < 0.11\)
Hence, S is approximately equal to 0.1
Answer: B
If S is the sum of reciprocals of a list of multiples of 3 from 300 to 396, inclusive, S is approximately equal to
A. 0.05
B. 0.1
C. 0.15
D. 0.2
E. 0.3
\(S = \frac{1}{300} + \frac{1}{303} + \frac{1}{309} + ... + \frac{1}{396} = \frac{1}{3}(\frac{1}{100} + \frac{1}{101} + \frac{1}{103} + ... + \frac{1}{132})\)
In the brackets, we have the sum of 33 terms, which are reciprocals of consecutive integers from 100 to 132, inclusive. Notice that the largest term among the 33 is \(\frac{1}{100}\), and the smallest term is \(\frac{1}{132}\).
If all 33 terms were equal to the largest term \(\frac{1}{100}\), the sum in the brackets would be \(\frac{33}{100}\), hence the entire expression would be \(\frac{1}{3} * \frac{33}{100} = 0.11\). Since the actual sum must be less than that, then \(S < 0.11\)
If all 33 terms were equal to the smallest term \(\frac{1}{132}\), the sum in the brackets would be \(\frac{33}{132}=\frac{1}{4}=0.25\), hence the entire expression would be \(\frac{1}{3} *0.25 \approx 0.083...\). Since the actual sum must be more than that, then \(S > 0.083...\)
Therefore, \(0.083... < S < 0.11\)
Hence, S is approximately equal to 0.1
Answer: B
27 Mar 2024, 01:41
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Bunuel
Assume that the series, represented by \(X_1\) = {\(300 \quad 303 \quad 306 \quad ......\quad 393 \quad 396\)}
The first term in the series = 300
The last term in the series = 396
Number of terms = \(\frac{396 - 300 }{ 3 }+ 1\) = 33
The middle term of this series is the 17th term and is equal to 348
The above series is in arithmetic progress, the sum of such a series is given by middle term * number of terms
However, the series that is relevant to this question is
\(X\) = {\(\frac{1}{300} \quad \frac{1}{303} \quad \frac{1}{306}\quad ......\quad \frac{1}{393} \quad \frac{1}{396}\)}
The middle term of the series X is \(\frac{1}{348}\)
The sum of this series = Middle Term * Number of terms = \(33 * \frac{1}{348} = \frac{11}{116} \approx \frac{11}{110} \approx 0.1\)
Option B
