Sets Test - 48

The given problem is,

\(A=\{x: x\) is a multiple of 3\(\}\)

\(B=\{x: x\) is a multiple of 4\(\}\)

\(C=\{x: x\) is a multiple of 12\(\}\)

If we look closely at the set \(C\), we will observe that

\(C=A \cap B\)

This is because L.C.M of \( (3,4)\) is 12.

Now, let's come back to the question

\(A \cap(B \cap C)=A \cap(B \cap(A \cap B\), (since \(C=A \cap B\))

\(=A \cap(A \cap B)\)

\(=A \cap B\)

\(=C\)

Therefore,

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

Let,

Set A of students who like mathematics, so:

\(n(A)=20\)

Set \(B\) of students who like science, so:

\(n(B)=15\)

Given:

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

Set of students who like at least one of the 2 given subjects is \(A \cup B\), then:

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

\(n(A \cup B)=20+15-5=30\)

Students who do not like either of the subjects = Total students - \(n(A \cup B)\)

\(=50-30\)

\(=20\)

Given:

\(B=\{1,2,4,7,14,28\}\)

As we can see that, all the element of \(B\) are factors of 28.

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

\(\{x: x \in N\) and \(x\) is a factor of 28\(\}\)

Given:

\(A=\{x: x\) is a natural number and \(1

Therefore, \(A=\{2,3,4\}\)

\(B=\{x: x\) is a natural number and \(4

Therefore, \(\mathrm{B}=\{5,6,7\}\)

Here, not a single element is common in both sets.

So, \((A \cap B)=\phi\)

Let \(\mathrm{A}=\{0,1,2, \ldots, 9\}\)

Number of elements in set \(\mathrm{A}=10\)

We know that:

The number of elements in the power set of any set \(\mathrm{A}\) is given by: 

\(=2^{n}\) 

where, \(n\) is the number of elements of the set \(A\).

\(\therefore\) The number of element in the power set \(P(A)\) will be \(=2^{10} =1024\)

Therefore, the power set of \(\{0,1,2, \ldots, 9\}\) is \(1024\).

Given:

\(A=\) The set of vowels in English dictionary

and, \(B=\) The set of even prime numbers

As we know that, there are 5 vowels in English dictionary i.e., \(A\) \(=\{5\}\)

\(\Rightarrow n(A)=5\)

Similarly we also know that, 2 is the only even prime number i.e., \(B\) \(=\{2\}\)

\(\Rightarrow n(B)=1\)

As we know that, any set which contains finite number of elements is called a finite set.

So, both the sets \(A\) and \(B\) are finite sets.

Given:

Total consumers \(=1000\)

Consumers who like product \(\mathrm{A}=720\)

Consumers who like product \(\mathrm{B}=450\)

We know that:

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

Least number that like both the products are \(=n(A \cap B)\)

\(\Rightarrow 1000=720+450-\mathrm{n}(\mathrm{A} \cap \mathrm{B})\)

\(\Rightarrow 1000=1170-\mathrm{n}(\mathrm{A} \cap \mathrm{B})\)

\(\Rightarrow \mathrm{n}(\mathrm{A} \cap \mathrm{B})=170\)

\(\therefore 170\) consumers like both the products \(A\) and \(B\).

Given:

\(A=\left\{4 x^{2}-x-5=0: x \in I\right\}\)

\(4 x^{2}-x-5=0\)

\(\Rightarrow(x+1)(4 x-5)=0\)

\(\Rightarrow x=-1,1.25\)

For set \(A, x \in I\)

\(\therefore\) Roster form of set \(A\) will be \(\{-1\}\).

Let \(n(A)=\) Number of students opted Mathematics \(=70\),

\(n(B)=\) Number of students opted Physics $=46\), and

\(n(C)=\) Number of students opted Chemistry $=28\).

Then, \(n(A \cap B)=23, n(B \cap C)=9, n(A \cap C)=14\) and \(n(A \cap B \cap C)=4\).

Now, \(n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)\)

\(-n(B \cap C)-n(A \cap C)+n(A \cap B \cap C)\)

\(=70+46+28-23-9-14+4=102\)

Therefore, the number of students who did not opted for any of the three courses \(=140-102=38\)

We know that:

Distributive law in sets:

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

Given:

\(A\) and \(B\) are two sets.

\(A \cap(B \cup A)^{c}\)

As we know from De morgan's law:

\((A \cup B)^{c}=A^{c} \cap B^{c}\)

Therefore,

\((B \cup A)^{c}=B^{c} \cap A^{c}\)

Now,

\(A \cap(B \cup A)^{c}=A \cap\left(B^{c} \cap A^{c}\right)\)

\(=\left(A \cap B^{c}\right) \cap\left(A \cap A^{c}\right)\) (From distributive law)

\(=\left(A \cap B^{c}\right) \cap \phi \quad\) \((\because x \cap \phi=\phi)\)

\(=\phi\)