Question:
A group of 11 natural numbers is arranged in ascending order. The average of the first six numbers is 18 and the average of the last six numbers is 28. If the 7th number is the smallest prime number greater than 20, then what is the sum of all these numbers?
A group of 11 natural numbers is arranged in ascending order. The average of the first six numbers is 18 and the average of the last six numbers is 28. If the 7th number is the smallest prime number greater than 20, then what is the sum of all these numbers?
Updated On: Mar 9, 2025
- 253
- 276
- 299
- 241
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The Correct Option is C
Solution and Explanation
Sum of 11 Natural Numbers Calculation
- Given that the 11 natural numbers are \( x_1, x_2, ..., x_{11} \) arranged in ascending order.
- The average of the first six numbers is 18, so their sum is:
\[ \text{Sum of first 6 numbers} = 6 \times 18 = 108 \]
- The average of the last six numbers is 28, so their sum is:
\[ \text{Sum of last 6 numbers} = 6 \times 28 = 168 \]
- The 7th number \(x_7\) is the smallest prime number greater than 20, which is 23. Hence,
\[ x_7 = 23 \]
- Now, the sum of all the numbers is:
\[ \text{Sum of all 11 numbers} = 108 + 168 + 23 = 299 \]
Conclusion: The sum of all these numbers is 299.
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