Sample term test 2
- Page ID
- 247832
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These mock exams are provided to help you prepare for Term/Final tests. The best way to use these practice tests is to try the problems as if you were taking the test. Please don't look at the solution until you have attempted the question(s). Only reading through the answers or studying them, will typically not be helpful in preparing since it is too easy to convince yourself that you understand them.
You can try timing yourself for \(40\) minutes.
Let the universal set be \(U = \{ x \in \mathbb{Z} \mid -7 \le x \le 7 \}\).
Define the following subsets of \(U\):
\(
A = \{ x \in U \mid x \text{ is an odd integer} \}, \quad
B = \{ x \in U \mid x^2 - 49= 0 \}, \quad
C = \{ 1, 2, 3,4,5,6,7 \}\).
- Write \(A\) and \(B\) explicitly using roster notation.
- Find \(A \cap C\) and \(A \cup B\).
- Find the set difference \(C \setminus B\).
- Determine the complement \(A^c\).
Decide whether the following statements are true or false.
- \(7 \in \{6, 7, 8, 9\}\)
- \(5 \notin \{6, 7, 8, 9\}\)
- \(\{2\} \nsubseteq \{1, 2\}\)
- \(\emptyset \nsubseteq \{\alpha, \beta, x\}\)
- \(\emptyset = \{\emptyset\}\)
- \(\emptyset\subseteq A\), for any set \(A\)
- \( \emptyset \in A,\) for any set \(A\)
- \(A\subseteq A,\) for any set \(A\)
A survey of 80 students gives:
\(
|A|=42,
\qquad
|B|=37,
\qquad
|C|=31,
\)
\(
|A\cap B|=18,
\qquad
|A\cap C|=15,
\qquad
|B\cap C|=13,
\)
and
\(
|A\cap B\cap C|=7.
\)
1. Find the number of students in:
- \(A\cap B\) only;
- \(A\cap C\) only;
- \(B\cap C\) only;
- \(A\) only;
- \(B\) only;
- \(C\) only.
2. Determine how many students belong to none of
the three sets.
Given non-empty subsets \(A,B,C\) of a universal set \(U\), prove the true statements and give counterexamples for the false statements.
- \( A-(B \cap C)=(A-B) \cup(A-C).\)
- If \( A \cap B= A \cap C\) then \(B= C\).
- If \( A \cup B= A \cup C\) then \(B= C\).
Let \(A, B, C, D, E\) be sets. Use set properties to show the following:
- \((A\cup B)\setminus C=(A\setminus C)\cup(B \setminus C).\)
- \((A\cup B)\setminus (C \setminus A)=A\cup(B \setminus C).\)