Problem One: Aliens
Nov 16th, 2006 by Allen
100 aliens attended an intergalactic meeting on earth.
73 had two heads,
28 had three eyes,
21 had four arms,
12 had two heads and three eyes,
9 had three eyes and four arms,
8 had two heads and four arms,
3 had all three unusual features.
How many aliens had none of these unusual features?
4 ?
i think 4 is the answer.. using the venn diagram. but i think jason knows it.. difficult to explain.. ryt..
Four.
let h=heads, e=eyes, a=arms
there are:
3 hea
9 he only (12 total he - 3hea already counted)
6 ae only ( 9 total ae - 3hea already counted)
5 ha only ( 8 total ae - 3hea already counted)
56 h only (73 total h - (3hea + 9he + 5ha)counted)
10 e only (28 total e - (3hea + 9he + 6ae)counted)
7 a only (21 total a - (3hea + 6ae + 5ha)counted)
—-
96 of 100 aliens with features. Four remain.
With the help of venn diagram the alliens left with none of the above features are 13. How can we get 4 as the answer, please let me know.
Using venn diagram you get the exact same result, check you drawings :). Here are the groups:
2 heads only: 56
3 eyes only: 10
4 arms only: 7
9 (12-3) had two heads and three eyes (no 4 arms)
6 (9-3) had three eyes and four arms (no 2 heads)
5 (8-3) had two heads and four arms (no 3 eyes)
this one remains the same:
3 had all three unusual features.
Let’s verify the 3 first definition:
73 had two heads = 56 + 5 + 3 + 9
28 had three eyes = 10 + 6 + 3 + 9
21 had four arms = 7 + 5 + 6 + 3
…
Those groups forms a partition with total:
56
+ 10
+ 7
+ 9
+ 6
+ 5
+ 3
—-
96
yea - got 4 (with Venn Diagrams)