Sets Test - 46

Given:

\(n(A)=20, n(B)=35\) and \(n(A \cup B)=45\)

We know that,

\(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)

\(\Rightarrow n(A \cap B)=n(A)+n(B)-n(A \cup B)\)

\(\Rightarrow n(A \cap B)=20+35-45\)

\(\therefore n(A \cap B)=10\)

Given:

\(A=\{2,3,4,5,6,7\}, B=\{6,7,8\} \text { and } C=\{1,5,8,9\}\)

Let, \(P=(B \cup C)\)

\(\Rightarrow P=(B \cup C)\)

\(=\{6,7,8\} \cup\{1,5,8,9\}\)

\(=\{1,5,6,7,8,9\}\)

\(\Rightarrow A \cap P=\{2,3,4,5,6,7\} \cap\{1,5,6,7,8,9\}\)

\(\Rightarrow A \cap(B \cup C)=\{5,6,7\}\)

Therefore, number of elements \(=3\)

Let \(A\) be the set of people who like coffee and \(B\) be the set of people who like tea.

Given that:

\(\mathrm{n}(\mathrm{A})=37, \mathrm{n}(\mathrm{B})=52\) and \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=70\)

Since, every person likes at least one drink (0 elements outside \(\mathrm{A}\) and \(\mathrm{B}\) ), we have:

\(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)

\(\Rightarrow 70=37+52-n(A \cap B)\)

\(\Rightarrow n(A \cap B)=89-70\)

\(=19\)

People who like coffee and NOT tea is given by:

\(n(A-B)=n(A)-n(A \cap B)\)

\(=37-19\)

\(=18\)

Given:

\(D=\{-6,-4,-2,0,2,4,6\}\)

As we can see that, all the element of \(D\) are even integers from \(-6\) to \(6\).

So, the required set-builder form of set\(D\) will be:

\(\{x: x=2 n\), where \(n \in Z\) and \(-3 \leq n \leq 3\}\).

Given:

\(A=\{a, b, c, d, e\}, B=\{a, c, e, g\}\) and \(C=\{b, e, f, g\}\)

As we know that,

\(A \cup(B \cap C)=(A \cup B) \cap(A \cup C)\)

Here, we have to find the value of \((A \cup B) \cap(A \cup C)\)

i.e., we have to find the value of \(A \cup(B \cap C)\)

\(\Rightarrow B \cap C=\{e, g\}\)

\(\Rightarrow A \cup(B \cap C)=\{a, b, c, d, e\} \cup\{e, g\}=\{a, b, c, d, e, g\}\)

Therefore, \( (A \cup B) \cap(A \cup C)=\{a, b, c, d, e, g\}\)

It is given that \(Y \subseteq X, Z \subseteq X\). Let \(a \in X\), then we have following chances that

1. \(a \in Y, \quad a \in Z\)

2. \(a \notin Y, \quad a \in Z\)

3. \(a \in Y, \quad a \notin Z\)

4. \(a \notin Y, \quad a \notin Z\)

It is required that \(Y \cap Z=\phi\). Hence, the items (2), (3), (4) above are chances for '\(a\)' to satisfy \(Y \cap Z=\phi\). Therefore, \(Y \cap Z=\phi\) has 3 chances for \(a\). Thus, for five elements of \(X\), the number of required chances is \(3 \times 3 \times 3 \times 3 \times 3=3^5\)

Given: 

\(\mathrm{n}(\mathrm{U})=42, \mathrm{n}(\mathrm{X})=15, \mathrm{n}(\mathrm{Y})=12\) and \(\mathrm{n}(\mathrm{X} \cup \mathrm{Y})=22\)

The shaded region represents:

\(\mathrm{n(X \cup Y)^{\prime}+n(X \cap Y)}\)

We know that:

\(\mathrm{n(X \cup Y)^{\prime}=n(U)-n(X \cup Y)}\)

\(\Rightarrow \mathrm{n(X \cup Y)^{\prime}=42-22=20}\)

As we know that, 

\(\mathrm{n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)}\)

\(\Rightarrow 22=15+12-\mathrm{n}(\mathrm{X} \cap \mathrm{Y})\)

\(\Rightarrow \mathrm{n}(\mathrm{X} \cap \mathrm{Y})=5\)

So, 

\(\mathrm{n(X \cup Y)^{\prime}+n(X \cap Y)}\)

\(=20+5\)

\(=25\)

Given \(S=\{1,2,3, \ldots \ldots, 100\}\)

Total non-empty subsets are \(2^{100}-1\).

Subsets whose product of elements is odd are \(2^{50}-1\)

Required number of non-empty subsets

= Total non-empty subsets - Subsets whose product of elements is odd

= \(2^{100}-1-\left(2^{50}-1\right) \)

= \( 2^{100}-2^{50} \)

= \(2^{50}\left(2^{50}-1\right)\)

Given:

\(B=\{1000,1331,1728,2197,2744,3375, \ldots, 8000\}\)

As we can see that, all the element of \(B\) are cube of natural numbers from 10 to 20.

So, the required set-builder form of set \(B\) will be:

\(\left\{x^{3}: 10 \leq x \in N \leq 20\right\}\)

\(n(A)=5, n(B)=7\)

Therefore,

\(n(A \cup B) \)

\(=  n(A)+n(B)-n(A \cap B) \)

\(=  5+7-n(A \cap B)\)

Maximum number of elements in \(A \cap B\) is 5. That is,

\(n(A \cap B)=5\)

Therefore, minimum number in

\(n(A \cup B)=5+7-5=7\)