Question:

A set \(S\) contains 12 consecutive integers. The sum of the 6 smallest integers of \(S\) is 321. The mean of the largest six integers of \(S\) is

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The mean of consecutive integers equals the average of the first and last terms.
Updated On: Jun 15, 2026
  • 59.5
  • 62.5
  • 64.2
  • 65.8
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The Correct Option is A

Solution and Explanation

Let the smallest integer be \(n\). Then the six smallest integers are \[ n,n+1,n+2,n+3,n+4,n+5. \] Their sum is \[ 6n+15=321 \] \[ 6n=306 \] \[ n=51. \] Thus the 12 integers are \[ 51,52,53,\ldots,62. \] The largest six integers are \[ 57,58,59,60,61,62. \] Their mean is \[ \frac{57+58+59+60+61+62}{6} \] \[ =\frac{357}{6} \] \[ =59.5. \]
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