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If \(\mathrm{S}\) is the sum of the reciprocals of the consecutive integers from \(91\) to \(100\), inclusive, which of the following is less than \(\mathrm{S}\)?
Let's start by understanding what we're looking for in everyday terms. We have the sum S, which is adding up the reciprocals (or "one over" each number) from 91 to 100.
So S means: Take 1 divided by 91, plus 1 divided by 92, plus 1 divided by 93, and so on, all the way up to 1 divided by 100.
The question asks us to figure out which of these three fractions - \(\frac{1}{8}\), \(\frac{1}{9}\), or \(\frac{1}{10}\) - is smaller than our sum S.
In mathematical notation: \(S = \frac{1}{91} + \frac{1}{92} + \frac{1}{93} + ... + \frac{1}{100}\)
Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding
Here's the key insight: instead of calculating the exact value of S (which would be very tedious), we can figure out a range that S must fall within.
Think about it this way: We're adding up 10 fractions. The smallest fraction in our sum is \(\frac{1}{100}\) (since 100 is the largest denominator), and the largest fraction is \(\frac{1}{91}\) (since 91 is the smallest denominator).
If all 10 fractions were as small as possible, our sum would be \(10 \times \frac{1}{100}\). If all 10 fractions were as large as possible, our sum would be \(10 \times \frac{1}{91}\).
Since our actual sum falls between these extremes, we know: \(10 \times \frac{1}{100} < S < 10 \times \frac{1}{91}\)
Let's work out these boundary values to get a concrete range for S.
Lower bound: \(10 \times \frac{1}{100} = \frac{10}{100} = \frac{1}{10} = 0.1\)
Upper bound: \(10 \times \frac{1}{91} = \frac{10}{91}\)
To make this easier to work with, let's convert \(\frac{10}{91}\) to a decimal approximation: \(10 \div 91 \approx 0.11\) (since 91 is close to 90, and \(\frac{10}{90} = \frac{1}{9} \approx 0.111\))
So we have established that: \(0.1 < S < 0.11\), or equivalently: \(\frac{1}{10} < S < \frac{10}{91}\)
Now we can check each option against our established range for S.
Our range is: \(\frac{1}{10} < S < \frac{10}{91}\) (approximately \(0.1 < S < 0.11\))
Option I: \(\frac{1}{8}\)
\(\frac{1}{8} = 0.125\), which is greater than our upper bound of about 0.11
Since \(\frac{1}{8} > S\), this means \(\frac{1}{8}\) is NOT less than S.
Option II: \(\frac{1}{9}\)
\(\frac{1}{9} \approx 0.111\), which is also greater than our upper bound of about 0.11
Since \(\frac{1}{9} > S\), this means \(\frac{1}{9}\) is NOT less than S.
Option III: \(\frac{1}{10}\)
\(\frac{1}{10} = 0.1\), which is exactly our lower bound
Since \(S > \frac{1}{10}\), this means \(\frac{1}{10}\) IS less than S.
Process Skill: INFER - Drawing the non-obvious conclusion about which comparisons hold based on our bounds
Only option III (\(\frac{1}{10}\)) is less than S.
Therefore, the answer is C. III only.
This matches the given correct answer, confirming our approach and calculations are correct.
1. Attempting exact calculation instead of using bounds: Many students see the sum \(S = \frac{1}{91} + \frac{1}{92} + ... + \frac{1}{100}\) and immediately try to calculate the exact value by finding common denominators or using a calculator. This approach is extremely time-consuming and prone to arithmetic errors. The key insight is recognizing that we only need to establish rough bounds for S, not its exact value.
2. Misunderstanding the comparison direction: Students often get confused about which direction the inequality should go. The question asks "which of the following is less than S" but students may mistakenly look for options that are greater than S. This fundamental misreading of the question leads to selecting the wrong answer even with correct calculations.
3. Setting up incorrect bounds: When establishing the range for S, students might incorrectly reason that since we're adding reciprocals from \(\frac{1}{91}\) to \(\frac{1}{100}\), the sum should be between \(\frac{1}{91}\) and \(\frac{1}{100}\), forgetting that we're adding 10 terms. The correct bounds should account for the fact that we have 10 terms: \(10 \times \frac{1}{100} < S < 10 \times \frac{1}{91}\).
1. Decimal approximation errors: When converting \(\frac{10}{91}\) to a decimal, students often make rounding mistakes or use imprecise mental math. For instance, they might approximate \(\frac{10}{91}\) as 0.10 (same as \(\frac{1}{10}\)) instead of recognizing it's closer to 0.11, which significantly affects the comparison with the given options.
2. Incorrectly ordering the fractions: Students frequently struggle with comparing fractions like \(\frac{1}{8}\), \(\frac{1}{9}\), and \(\frac{1}{10}\), especially when working with decimal equivalents. They might incorrectly conclude that \(\frac{1}{10} > \frac{1}{9} > \frac{1}{8}\), when the correct order is \(\frac{1}{8} > \frac{1}{9} > \frac{1}{10}\) (larger denominators mean smaller fractions).
3. Boundary condition confusion: When S is very close to one of the boundary values (like \(\frac{1}{10}\)), students may incorrectly conclude that \(S = \frac{1}{10}\) rather than \(S > \frac{1}{10}\). This leads to wrong conclusions about whether \(\frac{1}{10}\) is less than, equal to, or greater than S.
1. Selecting based on partial analysis: Students might correctly identify that \(\frac{1}{10} < S\) but fail to verify their conclusions about \(\frac{1}{8}\) and \(\frac{1}{9}\). They could hastily select an option like "I, II, and III" or "II and III only" without systematically checking each option against their calculated bounds.
2. Misreading the Roman numeral format: The answer choices use Roman numerals (I, II, III) referring to the three given fractions, and students often mix up which Roman numeral corresponds to which fraction. For example, they might think I refers to \(\frac{1}{10}\) instead of \(\frac{1}{8}\), leading to selecting the wrong combination even with correct mathematical reasoning.