Help with these Math questions

snipez90
Sat, 24 Nov 2007 20:41:47 +0000

No I am done taking it. Got a 2210 the first time and a 2230 the second time and perfect in math both times (only my CR went up the second time :-, nothing else changed).

Well the way I figured number one was by drawing 7 blanks. First we have
2 _ _ 6 _ _ 20 since 6 is right in the middle (median) and we’re given 2 is min and 20 is max. We’re also told that 3 is the mode, so it must appear twice (or else there is no mode) but since 3 can only go into the two blanks between 2 and 6, it does so. So now we have 2 3 3 6 _ _ 20, now we know that the remaining blanks can’t a 6 or 20 (since then that would also be a mode) nor can it have two of the same number (such as 7, 7) for the same reason.

So the minimum integers have to be 7 and 8 and the max 18 and 19. Average is sum divided by number of terms and indeed it can take on any value between (2+3+3+6+7+8+20)/7 = 7 and (2+3+3+6+18+19+20)/7 ~ 10.142 inclusive. So I II and III should work.