mridul savio Pinto (articled trainee ) (1477 Points)
09 June 2011dear , friends
1. out of 2000 employees in an office 48% preferred coffee , 54 % liked tea and 64 % used to smoke (s ) out of total 28% C& t 32 % used t & s .... and 30 % preferred c & s only .... 6 % did none of these
1 ... the number having all three ??
2. no of emoployees having t & s but not c
3.. no of preferring coffee only ??
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 09 June 2011
answers
1) 420
2) 220
3) 220
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 09 June 2011
De Morgan’s law
n(A u B u C) = n(A) + n(B) + n(C) - n(A n B ) - n(B n C) - n(A n C) + n (A n B n C)
n(C u T u S) = n(C) + n(T) + n(S) - n(C n T) - n(T n S) - n(C n S) + n (C n T n S) -------- (1)
n(C) = 960
n(T) = 1080
n(S) = 1280
n(C n T) = 560
n(T n S) = 640
n(C n S) = 600
n(C c n T c n S c) = 60
n(C c n T c n S c) = n(C u T u S) c
total number of employees = n(C u T u S) + n(C u T u S) c ------------------------------(2)
substitute values in eqn (1) & (2), we get
2000 = 960 + 1080 + 1280 – 560 - 640 - 600 + n (C n T n S) + 60
n (C n T n S) = 2000 – 1580
the number having all three, n (C n T n S) = 420
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 09 June 2011
2) no of emoployees having t & s but not c =
n(T n S) - n (C n T n S) = 640- 420
= 220
3) no of preferring coffee only =
n(C) - n(C n T) - n(C n S) + n (C n T n S) = 960 – 560 – 600 + 420
= 220
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 10 June 2011
venn diagram for more explanation
CAclubindia
