OA - E
so this problem is basically concerned with EXTREMES: you're trying to figure the least, or greatest, number of representatives (or both) that could be sent in certain situations, in order to determine the range of possibilities. remember, then, if you want to figure extreme values, you have to consider extreme situations. here's one way you can progress through the problem:
-- (1) alone --
clearly 41 representatives = greatest number sent by any one country.
therefore, the countries from second place on down sent a total of 75 - 41 = 34 representatives.
EXTREME CASE 1: smallest possible # for the second country
in this case, you want to spread the remaining 34 representatives out as evenly as possible, so that the 2nd, 3rd, 4th, 5th, and 6th place countries are as near each other as possible.
34/5 = 6.8, so try to cluster the numbers around this average: the distribution with the least possible amount of variation is 9, 8, 7, 6, 4 (you can't get consecutive integers - try it for yourself)
therefore, the second greatest number of representatives must be at least 9
EXTREME CASE 2: largest possible # for the second country
in this case, you want to make the 3rd, 4th, 5th, and 6th values as small as possible. this is straightforward: make them 4, 3, 2, and 1 respectively.
this means that the 2nd place country sent 34 - 4 - 3 - 2 - 1 = 24 representatives
therefore, the second greatest number of representatives must be 24 or less
9 < second highest number < 24
insufficient
-- (2) alone --
in this case, there are no further restrictions on the numbers of representatives.
the highest number of representatives that country a could send is clearly 11.
therefore, the second greatest number of representatives must be 11 or less
to make the number as small as possible, just let the 2nd, 3rd, 4th, 5th, 6th place numbers be 5, 4, 3, 2, 1 respectively, and give all the rest of the representatives to the first place country.
therefore, the second greatest number of representatives must be at least 5
5 < second highest number < 11
insufficient
-- together --
we have
9 < second highest number < 24
AND
5 < second highest number < 11
therefore
9 < second highest number < 11
still insufficient
answer = e
