`{:("Column A","The average of a set of six positive number is 30", "Column B"),("The average of the number in the set after replacing the smallest number in the set with 0",,"The average of the remaining number in the set after removing the smallest number from the set"):}`

Options

A. If column A is larger
B. If column B is larger
C. If the columns are equal
D. If there is not enough information to decide

Answer

To solve the problem, we need to compare the averages in Column A and Column B based on the given information about a set of six positive numbers.

### Step-by-Step Solution:

1. **Understand the given data**: We know that the average of a set of six positive numbers is 30. This means:
\[ \text{Average} = \frac{A_1 + A_2 + A_3 + A_4 + A_5 + A_6}{6} = 30 \]
From this, we can calculate the sum of these six numbers:
\[ A_1 + A_2 + A_3 + A_4 + A_5 + A_6 = 30 \times 6 = 180 \]

2. **Column A Calculation**: In Column A, we replace the smallest number \(A_1\) with 0. The new set of numbers becomes \(0, A_2, A_3, A_4, A_5, A_6\). The average of these numbers is:
\[ \text{Average in Column A} = \frac{0 + A_2 + A_3 + A_4 + A_5 + A_6}{6} = \frac{A_2 + A_3 + A_4 + A_5 + A_6}{6} \]
We know that:
\[ A_2 + A_3 + A_4 + A_5 + A_6 = 180 - A_1 \]
Therefore, the average in Column A can be expressed as:
\[ \text{Average in Column A} = \frac{180 - A_1}{6} \]

3. **Column B Calculation**: In Column B, we remove the smallest number \(A_1\) from the set. The remaining numbers are \(A_2, A_3, A_4, A_5, A_6\). The average of these numbers is:
\[ \text{Average in Column B} = \frac{A_2 + A_3 + A_4 + A_5 + A_6}{5} \]
Again, substituting the sum:
\[ \text{Average in Column B} = \frac{180 - A_1}{5} \]

4. **Comparison of Column A and Column B**: Now we need to compare the two averages:
- Column A: \(\frac{180 - A_1}{6}\)
- Column B: \(\frac{180 - A_1}{5}\)

Since both expressions have the same numerator \(180 - A_1\), we can compare the denominators:
- The denominator in Column A is 6.
- The denominator in Column B is 5.

Since \(5 < 6\), it follows that:
\[ \frac{180 - A_1}{5} > \frac{180 - A_1}{6} \]
Therefore, the average in Column B is greater than the average in Column A.

5. **Conclusion**: We conclude that the entry in Column B is larger than the entry in Column A.

### Final Answer:
Column B is larger than Column A.

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