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31 Dec 2024, 01:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
49% (01:44) correct
51%
(01:43)
wrong
based on 161
sessions
History
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Time
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Not Attempted Yet
All 50 Members of Johnson's family have at least one of the following characteristics: green eyes, dark hair or pale skin. 20 members of the Johnson's family have green eyes, 30 have dark hair, and 40 have pale skin. If only 10 members of the Johnson's family have all three characteristics, How many of them at least two characteristics ?
A. 0
B. 10
C. 20
D. 30
E. 40
A. 0
B. 10
C. 20
D. 30
E. 40
31 Dec 2024, 02:29
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Bunuel
|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|
50 = 20 + 30 + 40 - |A ∩ B| - |B ∩ C| - |A ∩ C| + 10
|A ∩ B| + |B ∩ C| + |A ∩ C| = 50
|A ∩ B| + |B ∩ C| + |A ∩ C| => This contains 3 parts of |A ∩ B ∩ C| so subtracting 2 parts should give us members with atleast 2 qualities
|A ∩ B| + |B ∩ C| + |A ∩ C| - 2* |A ∩ B ∩ C| = 50 - 20 = 30
IMO: D
31 Jan 2025, 19:55
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Let x be the number of members having only 1 feature, y having 2 features, z having 3 features.
Visualize from Venn diagram, x+y+z=50
x+2y+3z=90
Subtract both the equations,
y+2z=40 Now we are given z=10
y=20.
Visualize from Venn diagram, x+y+z=50
x+2y+3z=90
Subtract both the equations,
y+2z=40 Now we are given z=10
y=20.
