document.write( "Question 885833: If the average (arithmetic mean) of six numbers is 28 and the average of two of these numbers is 18, what is the average of the other four numbers?
\n" ); document.write( "a) 29
\n" ); document.write( "b) 30
\n" ); document.write( "c) 31
\n" ); document.write( "d) 32
\n" ); document.write( "e) 33
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Algebra.Com's Answer #535485 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
WITH CALCULATOR:
\n" ); document.write( "For the six numbers:
\n" ); document.write( "\"average=SUM%2F6=28\"--->\"SUM=28%2A6=168\"
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\n" ); document.write( "For the two numbers averaging 18:
\n" ); document.write( "\"average=sum%2F2=18\"--->\"sum=18%2A2=36\"
\n" ); document.write( "
\n" ); document.write( "The sum of the other four numbers is
\n" ); document.write( "\"168-36=132\" , and the average of those four numbers is
\n" ); document.write( "\"152%2F4=highlight%2833%29\"
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\n" ); document.write( "IN MY HEAD:
\n" ); document.write( "Those two numbers averaging 18 are (on the average)
\n" ); document.write( "\"28-18=10\" lower than average.
\n" ); document.write( "The \"2%2A10=20\" units missing must have been made up by the other \"4\" numbers
\n" ); document.write( "That means that the average of those other \"4\" numbers must be
\n" ); document.write( "\"20%2F4=5\" units higher than the average of all six numbers.
\n" ); document.write( "So the average of the other four numbers is
\n" ); document.write( "\"28%2B5=33\"
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