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27 Mar 2024, 01:29
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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
27 Mar 2024, 01:29
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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
17 Jul 2024, 17:44
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Bunuel
Hi Bunuel,
I did: There's 33 numbers so I chose the middle one (300 + 17(3)) = 351. And just aproximated (1/351) * 33, which is a little bit less than 0.1.
Is this acceptable?
