manhattan math 2

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manhattan math 2

by resilient » Sun Feb 17, 2008 11:17 pm
Peter, Paul, and Mary each received a passing score on his/her history midterm. The average (arithmetic mean) of the three scores was 78. What was the median of the three scores?

(1) Peter scored a 73 on his exam.

(2) Mary scored a 78 on her exam.






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by smushkas » Mon Feb 18, 2008 6:04 am
Hey there,
Well, let's assume tht Peter is A, Paul B and Mary is C. Then we have;
[A + B + C ] / 3 = 78 ===> [ A + B + C ] = 234
From statement (1) A = 73 ==> [ B+C ] = 234 - 73 = 161. Passing grade is assumed > 50, or it depends on the rules of the class.
In this case B could be 61, and C could be 100. Then the Median is ==> [ 61 - 73 - 100 ]
but if B = 75, C = 86, then Median is ==> [ 75 - 78 - 86 ]
(1) is not sufficient

Statement (2) ==> C=78
We now have [A + B + C ] / 3 = 78 ==> [ A + B +78 ] = 234 ==> A+B = 156.
In case if A > 78, then B must be < 78, or if A < 78, then B must be greater than 78, but the median always will be in between ( i.e. 78 ), if everyone scores the same, median won't change. Thus, you can get a unique number. Statement (2) is sufficient.

Any other ways?