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Set Theory Problem

Mathematics

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1. Set Theory Problem
Given:
Total students, n(U)=120n(U) = 120
Students who liked cricket, n(C)=60n(C) = 60
Students who liked basketball, n(B)=55n(B) = 55
Students who did not like any of the two games, n(C∪B)′=20n(C \cup B)' = 20

a) If C and B denote the sets of students who liked cricket and basketball game respectively, write the cardinality of n(B∪C)n(B \cup C).
We know that n(U)=n(C∪B)+n(C∪B)′n(U) = n(C \cup B) + n(C \cup B)'.

120=n(C∪B)+20120 = n(C \cup B) + 20 n(C∪B)=120−20n(C \cup B) = 120 - 20 n(C∪B)=100n(C \cup B) = 100

The cardinality of n(B∪C)n(B \cup C) is 100\boxed{100}.

b) Present the above information in a Venn-diagram.
First, find the number of students who liked both games, n(C∩B)n(C \cap B).
We use the formula: n(C∪B)=n(C)+n(B)−n(C∩B)n(C \cup B) = n(C) + n(B) - n(C \cap B).

100=60+55−n(C∩B)100 = 60 + 55 - n(C \cap B) 100=115−n(C∩B)100 = 115 - n(C \cap B) n(C∩B)=115−100n(C \cap B) = 115 - 100 n(C∩B)=15n(C \cap B) = 15

Now, find the number of students who liked cricket only, n0(C)n_0(C).

n0(C)=n(C)−n(C∩B)=60−15=45n_0(C) = n(C) - n(C \cap B) = 60 - 15 = 45

And the number of students who liked basketball only, n0(B)n_0(B).

n0(B)=n(B)−n(C∩B)=55−15=40n_0(B) = n(B) - n(C \cap B) = 55 - 15 = 40

The Venn diagram is as follows:
(Imagine two overlapping circles. The left circle represents Cricket (C), the right circle represents Basketball (B). The overlapping region is C∩BC \cap B. The rectangle enclosing both circles represents the Universal set (U).)

  • In the region for C only: 45
  • In the region for B only: 40
  • In the intersection region (C∩BC \cap B): 15
  • Outside both circles (neither liked): 20

c) Find the number of students who liked cricket only.
The number of students who liked cricket only is n0(C)n_0(C).

n0(C)=n(C)−n(C∩B)=60−15=45n_0(C) = n(C) - n(C \cap B) = 60 - 15 = 45

The number of students who liked cricket only is 45\boxed{45}.

d) Compare the number of students who liked both cricket and basketball and who liked except these two games.
Number of students who liked both cricket and basketball, n(C∩B)=15n(C \cap B) = 15.
Number of students who liked neither of the two games, n(C∪B)′=20n(C \cup B)' = 20.
Comparing these two values: 20>1520 > 15.
Therefore, the number of students who liked neither game is greater than the number of students who liked both games.

2. Simple and Compound Interest Problem
Given:
Principal, P=Rs. 60,000P = \text{Rs. } 60,000
Time, T=2T = 2 years
Rate, R=10%R = 10\% p.a.

a) Define compound interest.
Compound interest is the interest calculated on the initial principal, which also includes all of the accumulated interest from previous periods on a deposit or loan. It is often referred to as "interest on interest."

b) Find the compound interest received by Bijaya after 2 years.

Step 1: Calculate the amount (AA) using the compound interest formula.

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