NCERT/CBSE MATHEMATICS CLASS 11 textbook
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MISCELLANEOUS EXERCISES
Answers to NCERT/CBSE MATH (Class XI)textbook
SETS
Solution:
Note : The Venn diagram is used for a clearer understanding of the solution.
Let A be the set of people reading the newspaper H , so n (A) = 25(Red circle)
Let B be the set of people reading the newspaper T , so n (B) = 26(Blue circle)
Let C be the set of people reading the newspaper I , so n (C) = 26 ( Green circle )
Also n ( A ∩ B ) = 11(Shaded blue and red region),
n ( A ∩ C ) = 9 (Shaded purple and red region)
n ( B ∩ C ) = 8(Shaded green and red region)
and n ( A∩ B∩ C) = 3(Shaded red region )
(i) The number of people who read atleast one of the three newspapers
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NCERT/CBSE MATHEMATICS CLASS 11 textbook
http://www.TutorBreeze.com
= n ( A∪ B∪ C)
= n(A) + n(B) + n(C) - n(A ∩ B ) - n( B∩C ) - n(A ∩ C) + n ( A∩ B∩ C)
= 25 + 26 + 26 - 11 – 8 - 9 + 3 = 52
(ii) The number of people who read exactly one newspaper is all the people who read H but
not I and T or who read I but not H and T who read T but not Hand I .
We will first work out
The number of people who read H and T but not I
= n ( A ∩B ∩ C’) = n(A ∩B ) – n(A∩B∩C) = 11- 3 = 8
The number of people who read H and I but not T
= n ( A ∩B ‘ ∩ C) = n (A∩C) – n(A∩B∩C) = 9 – 3 = 6
The number of people who read T and I but not H
= n ( A’ ∩B ∩ C) = n (B∩C) – n(A∩B∩C) = 8 – 3 = 5
Now ,
The number of people who read H but not I and T
n(A∩B’∩C’) = n( A) - n(A∩B∩C’) - n(A∩B ‘∩C) - n(A∩B∩C)
= 25 – 8 – 6 – 3 = 8
The number of people who read T but not I and H
n(A’∩B∩C’) = n(B) - n(A’∩B∩C) - n(A∩B∩C’) - n(A∩B∩C)
=26 – 5 – 8 – 3 = 10
The number of people who read I but not H and T
n(A’∩B’∩C) = n(C) - n(A’∩B∩C)- n(A∩B’∩C)- n(A∩B∩C)
= 26 – 5 – 6 – 3 = 12
Therefore ,
The number of people who read exactly one newspaper
= 8 + 10 + 12 = 30
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