2.

At least one senior= total possibilities - no seniors in the selected members

= 10C3-6C3= 100

oct07, this problem appears to me somehow interesting... I will give a try to it. Now looking at the prob,
let ,
cost of the 15 homes = x1, x2, x3, x4 and so on.

Mean = (x1 + x2 + x3 + ... + x15) / 15 = 15 x 10^4
So (x1 + x2 + x3 + ... + x15) = 225 x 10^4

Median = should be the cost of 8th homes in increasing order of cost = 13 x 10^4.

Now I -> (165 - 130) x 10^3 = 35 x 10^3 - If interval is 5 x 10^3, then the last 15th building cost will be 165 x 10^3. But we are not sure the interval will be 5 x 10^3. - To me, its not always true.
Now III -> It should be always true because at least 7 houses should cost less than 13 x 10^4.
But as I can not be there and III should be, so IMO C.

What an explanation, newmikeyork! Thank u a lot. Now I think u I have got the problem...