document.write( "Question 1064536: A set of 7 positive integers has a median 3 and
\n" ); document.write( "unique mode 4. If you add two 2's to the set the
\n" ); document.write( "median and unique mode are then both equal 2.
\n" ); document.write( "What is the mean of the new set? \n" ); document.write( "
Algebra.Com's Answer #679634 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
A set of 7 positive integers has a median 3 and
\n" ); document.write( "unique mode 4. If you add two 2's to the set the
\n" ); document.write( "median and unique mode are then both equal 2.
\n" ); document.write( "What is the mean of the new set?\r
\n" ); document.write( "\n" ); document.write( "A set of 7 positive integers
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\r\n" );
document.write( "Let these be\r\n" );
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document.write( "A ≤ B ≤ C ≤ D ≤ E ≤ F ≤ G\r\n" );
document.write( "

\n" ); document.write( "has a median 3
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\r\n" );
document.write( "So the middle number must be 3 \r\n" );
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document.write( "A ≤ B ≤ C ≤ 3 ≤ E ≤ F ≤ G\r\n" );
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\n" ); document.write( "and unique mode 4.
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\r\n" );
document.write( "So there must be at least 2 4's among E,F and G.\r\n" );
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document.write( "There are only 3 integers for A, B and C, which are\r\n" );
document.write( "1, 2, and 3, so there must be at least 2 1's, 2 2's or 2 3's.\r\n" );
document.write( "Since there must be more 4's than anything else, there must\r\n" );
document.write( "be more than two 4's, so there must be three 4's. So E, F \r\n" );
document.write( "and G must all be 4's.\r\n" );
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document.write( "A ≤ B ≤ C ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
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document.write( "So there are only 5 possibilities:\r\n" );
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document.write( "1 ≤ 1 ≤ 2 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
document.write( "1 ≤ 1 ≤ 3 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
document.write( "1 ≤ 2 ≤ 2 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
document.write( "1 ≤ 2 ≤ 3 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
document.write( "2 ≤ 2 ≤ 3 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
document.write( "

\n" ); document.write( "If you add two 2's to the set the
\n" ); document.write( "median and unique mode are then both equal 2.
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\r\n" );
document.write( "Since we must add two 2's, there must\r\n" );
document.write( "already be two 2's for if there were\r\n" );
document.write( "only one 2, there wouldn't be enough\r\n" );
document.write( "2's to be more than the three 4's. \r\n" );
document.write( "Only two of those have two 2's, so\r\n" );
document.write( "we've narrowed it down to these two:\r\n" );
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document.write( "1 ≤ 2 ≤ 2 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
document.write( "2 ≤ 2 ≤ 3 ≤ 3 ≤ 4 ≤ 4 ≤ 4 \r\n" );
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document.write( "Let's add two 2's to each of those:\r\n" );
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document.write( "1 ≤ 2 ≤ 2 ≤ 2 ≤ 2 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
document.write( "2 ≤ 2 ≤ 2 ≤ 2 ≤ 3 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
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document.write( "Now we can eliminate the second one because \r\n" );
document.write( "the median is not 2.  So the only possibility is\r\n" );
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document.write( "1 ≤ 2 ≤ 2 ≤ 2 ≤ 2 ≤ 3 ≤ 4 ≤ 4 ≤ 4\r\n" );
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\n" ); document.write( "What is the mean of the new set?
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\r\n" );
document.write( "So the mean is:\r\n" );
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document.write( "\"%281%2B2%2B2%2B2%2B2%2B3%2B4%2B4%2B4%29%2F9\"\"%22%22=%22%22\"\"24%2F9\"\r\n" );
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document.write( "Answer: \"24%2F9\"\r\n" );
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document.write( "Edwin
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