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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?
These are Aliens not having unusual features!
Read given data first!
Let's simplify the process with Venn diagram like below.
1. We know, 3 had all three unusual features.
2. 12 had two heads and three eyes, 9 had three eyes and four arms, 8 had two heads and four arms. Here, 3 having all unusual features too need to be counted in each of Aliens' counts above.
So, Aliens having 2 unusual features of 2 heads & 3 eyes = 12 - 3 = 9.
Aliens having 2 unusual features of 3 eyes & 4 arms = 9 - 3 = 6.
Aliens having 2 unusual features of 2 heads & 4 arms = 8 - 3 = 5.
3. We know, 73 had two heads, 28 had three eyes, 21 had four arms.
Therefore, the number of Aliens having only 2 heads as an only unusual feature is 73 - (9 + 3 + 5) = 56.
The number of Aliens having 3 eyes as an only unusual feature is
28 - (9 + 3 + 6) = 10.
The number of Aliens having 4 arms as an only unusual feature is
21 - (5 + 3 + 6) = 7.
4. So, we have 56 + 9 + 10 + 3 + 5 + 6 + 7 = 96 Aliens having at least one unusual feature.
Therefore, the number of Aliens not having any of unusual features is
100 - 96 = 4.
