avarage and median

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Source: — Data Sufficiency |

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by scoobydooby » Mon Feb 16, 2009 11:26 am
let the numbers be a1, M, a2 (in increasing order)

stmnt 1: M<(a1+a3)/2.......................1
A=(a1+M+a3)/3 or a1+a3=3A-M
replacing the value of a1+a3 in eqn 1 we get M<(3A-M)/2
or 2M<3A-M
or 3M<3A
or M<A sufficient


stmnt 2: a1=A-150, a3=A+250
A=(A-150+M+A+250)/3 (putting the values in the average formula)
A=(2A+M+100)/3
or 3A=2A+M+100
or A=M+100 =>A<M sufficient

hence D

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by Bidisha_800 » Mon Feb 16, 2009 11:26 am
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