document.write( "Question 1132302: In a survey of 100 out patients who were reported at a hospital one day found out that 70 complained of fever,50 complained of stomach ache and 30 were injured. All 100 patients had at least one of the complaints and 44 had exactly two of the three complaints. How many patients had all three complaints.
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Algebra.Com's Answer #749296 by Boreal(15235)\"\" \"About 
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F-70
\n" ); document.write( "S-50
\n" ); document.write( "I=30
\n" ); document.write( "two complaints=44=F+S, S+I, F+I
\n" ); document.write( "150 complaints altogether.
\n" ); document.write( "88 of them are with 44 patients
\n" ); document.write( "That leaves 56 patients.
\n" ); document.write( "x have 3 complaints
\n" ); document.write( "56-x have 1 complaint
\n" ); document.write( "The total complaints for that group is 3x+56-x, and that equals 62 complaints
\n" ); document.write( "2x+56=62
\n" ); document.write( "2x=6
\n" ); document.write( "x=3 patients with all 3 complaints ANSWER\r
\n" ); document.write( "\n" ); document.write( "3 have 3 for total of 9
\n" ); document.write( "44 have 2 for total of 88
\n" ); document.write( "53 have 1 for total of 53
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