We are given four conditions involving averages of groups of three numbers from a set of six numbers a,b,c,d,e,f:
We add these equations to get = a+2b+3c+3d+2e+f = 450 Then, using the known sums a+b+c =180 and d+e+f=120, we substitute them into the equation: 180+120+(b+2c+2d+e) = 450 This simplifies to:b+2c+2d+e=150 The total sum of all six numbers is 180+120 = 300. Therefore, the average of the first three and last three numbers is: 300/6 = 50
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