Chimera Test Answers
This test was/is ridiculously difficult, and actually unfair due to lack of hints on some of the problems. The fact that some people attained nearperfect scores can only be attributed to extremes in one or more of these areas: intelligence, perseverance, and luck.
1. 97. "mensa" was a hint. It is Latin for "table," and the numbers are, in ascending
order, for those elements on the Periodic Table that begin with "b".
2. 1. Each number represents the number of letters in the word before it that are in order.
Example: "stammer" has r, s, and t, so the number is 3. Another acceptable answer, found by a web surfer: "The solution is 3.
The letters corresponding to the position in the words indicated by the numbers
Are the last 8 letters in William Bultas in random order."
3. 54. I proved this numerous times to myself using a diagram and a Rubik's Cube. There are probably several ways that the knight can touch all 54 squares.
4. Three acceptable answers:
A. 3. Each number is the number of homophones that exist for the word before it, plus
the word itself.
B. 0
The number after each word is equal to the number of letters in the NEXT
word in the series, minus the word itself. For ONE: The number of letters in
'TWO' (3) minus 1 is 2; for TWO: the number of letters in 'THREE' (5) minus
2 is 3; for THREE: the number of letters in 'FOUR' (4) minus 3 is 1; and for
FOUR: the number of letters in 'FIVE' (4) minus 4 is 0.
The last acceptable answer was found by a surfer:
"The solution is 1
The position of the first vowel in problem ONE is 2 (mensa)
The position of the first vowel in problem TWO is 3 (stammer)
The position of the first vowel in problem THREE is 1 (Imagine)"
5. OOXXXXXXXX. Number of X's double each time they appear in relation to the last
time that they appeared, and O's are equal to half of the X's immediately before it, plus
the total number of O's that have appeared before it in all previous sequences of O's.
6. .
.
.
The dots are reversed braille (dots appear where there should be none, and
don't appear where there should be some), and spell "GENIUS".
7. 6. Possibly the easiest problem on the test. I've never mastered juggling 4 objects.
8. S. Like my "Back to Front" problem in Chimera's Conundrums. The letters are, from
back to front, from: kingdoM (7), phyluM(6), clasS (5), ordEr (4), faMily (3), gEnus (2),
Species (1). "LIFE" is a hint as this system is used to classify just that.
9. "Prometheus stole fire from the gods". Other answers are possible, but must fit with a
'famous mythical act of heroism', and have at least 1 letter in each word underlined/used
in the phrase.
10. "silly" should be written after 'bitter'. The last 3 letters in each 'word' form the
sentence "write silly after bitter".
11. 9. Series increases 1 more each time than the time before, but starts over after
exceeding 12.
12. blank space. In each column, you have a letter or symbol at the top. In the middle of
the same column, you have that same letter or symbol with its middle section removed.
On the bottom of the same column, you have the same letter or symbol found at the top,
but with its middle AND bottom sections removed.
13. 4550. Multiply the numbers for each letter of the alphabet represented in each word.
14. Q. The first American Presidents, but counted from the opposite end of the alphabet:
Washington  'w' is 4th from the end of the alphabet, and 'd' is 4th from the beginning,
and so on.
15. Deeleh or jeetej. 'deeleh' involves
using the FOIL principle in algebra: multiply First, Outside, Inside, Last for expressions
in parentheses. 'beb' would be (2x+2), 'ged' is (7x+4). Their product is 'neeveh': 14x
(squared) + 22x + 8, and so on. Here is the answer explanation for jeetej, submitted by a web surfer:
The letters of the first words change according to a simple principle:
b e b g e d
+1 = +1 3 = 1
c e c d e c
+1 = +1 3 = 1
d e d a e b
Thus, one can expect that the letters of the second words also will change
"in a linear way":
n e e v e h
2 = = 1 = +1
l e e u e i
2 = = 1 = +1
J E E T E J
16. The sum of the letters must be 11 by the following code: odd # letters = 1, even # letters = 2, vowels = 2 even though they are odd:
1a 2b 3c 4d 5e 6f 7g 8h 9i 10j 11k 12l 13m 14n 15o 16p 17q 18r 19s
20t 21u 22v 23w 24 x 25y 26z
17. 4801047. If you subracted 3 from one of the digits, then multiplied the result by 3, you got
the next number in the series.
18. Numbers are written back to back and equal letters of the alphabet. The answer is
correct if the numbers work out to be letters representing a word that means the same as
'utterly stupid'. Examples: moronic, idiotic, inane, mindless, etc.
19. 27. First of all, you have 64 cubes.
Now, assuming the top layer always has 1 cube, there is a sequential method you can go
through to determine all of the viable objects you can make with 64 cubes or less. Top
layer has 1 cube, and all other layers have 4, 9, 16, 25, 36, or 49 cubes. First you have 1
& 4, then 1 & 9, then 1 & 16, and so on, for 2 layers. Then you try 1,4,9; 1,4,16; 1,4, 25;
and so on, then 1, 9, 16; 1,9, 25; and so on, until you have found all possible
permutations for 3 layers.
20. First acceptable answer: Meyerhof  all of them apart from Meyerhof have a moon crater named after them (or so I am told...). Second acceptable answer (actually, this is the original one): A scientist or mathematician whose last name begins with 'e' and ends with 'o'.
Eustachio, for instance. The first letter of the first word, together with the last letter of
the last word make 'do'. Last letter of the first word, together with the first letter of the
last word make 're'. Continuing on, you have mi, fa, so, la, ti. Musical...
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