MATHS M.C.Q (1 Marks)

3. 45.

Solution:

It is given that each set $\text{A}_\text{j}(1\leq\text{i}\leq30)$ contains 5 elements and $\bigcup\limits^{30}_\text{i = 1}\text{A}_\text{i}=\text{S}.$

$\therefore\text{n(S)}=30\times5=150$

But, it is given that each element of S belong to exactly 10 of the A_i^'s.

$\therefore$ Number of distinct elements in $\text{S}=\frac{150}{10}=15......(1)$

It is also given that each set $\text{B}_\text{j}(1\leq\text{j}\leq\text{n})$ contains 3 elements and $\bigcup\limits^{\text{n}}_\text{j = 1}\text{B}_\text{j}=\text{S}.$

$\therefore\text{ n(S)}=\text{n}\times3=\text{3n}$

Also, each element of S belong to eactly 9 of B_j^'s.

$\therefore$ Number of distinct elements in $\text{S}=\frac{\text{3n}}{9}......(2)$

From (1) and (2), we have

$\frac{\text{3n}}{9}=15$

$\Rightarrow\text{n} = 45.$

Hence, the correct answer is option (c).

4. 20.

Solution:

We have:

$\text{n(A}\cap\text{B) = n(A) + n(B)} - \text{n(A}\cup\text{B)}$

$=70+60-110$

$=20.$

1. A.

Solution:

$\text{A}\cap\text{(A}\cup\text{B)}=\text{(A}\cap\text{A)}\cup\text{(A}\cap\text{B)}=\text{A}\cup\text{(A}\cap\text{B)}=\\\text{AA}\cap\text{(A}\cup\text{B)}=\text{(A}\cap\text{A)}\cup\text{(A}\cap\text{B)}=\text{A}\cup\text{(A}\cap\text{B)}=\text{A.}$

4. None of these.

Solution:

$(4\not\in\text{A) }(4\not\in\text{A})$

$\{4\}\not\subset\text{A}$

$\text{B}\not\subset\text{}A$

Thus, we can say that none of these options satisfy the given relation.

3. 5.

Solution:

We know:

$\text{n(A}\cup\text{B) = n(A) + n(B)} - \text{n(A}\cap\text{B)}$

Now,

$\text{n(A}\cap\text{B) = n(A) + n(B)} -\text{n(A}\cup\text{B)}$

$=16+14-25$

$=5.$

3. $\text{A}- \text{B = A} - \text{B}'.$

Solution:

It includes all those elements of A which do not belongs to complement of B which is equal to $\text{A}\cap\text{B}$ but not equal to A - B.

$\therefore$ (c) ic false.

2. $\text{B}\subset\text{A}.$

Solution:

Only this case is possible.

2. 7, 4.

Solution:

We know that if a set X contains k elements, then the number of subsets of X are 2^k.

It is given that the number of subsets of a set containing m elements is 112 more than the number of subsets of set containing n elements.

$\therefore 2^\text{m}-2^\text{n}=112$

$\Rightarrow2^\text{n}(2^\text{m - n}-1)=2\times2\times2\times2\times7$

$\Rightarrow2^\text{n}(2^{\text{m}-\text{n}}-1)=2^4(2^3-1)$

$\Rightarrow\text{n}=4$ and $\text{m}-\text{n}=3$

$\therefore\text{ m}-4=3$

$\Rightarrow\text{m}=7$

Thus, the values of m and n are 7 and 4, respectively.

Hence, the correct answer is option (b).

1. $\text{B}\subseteq\text{A}.$

Solution:

The union of two sets is a set of all those elements that belong to A or to B or to both A and B.

If $\text{A}\cup\text{B = A},$ then $\text{B}\subseteq\text{A}.$

3. $\phi.$

Solution:

$\text{A}\cap\text{(A}\cup\text{B)}'$

$=\text{A}\cap\text{(A}'\cup\text{B}')$ (De Morgen Law)

$=\text{(A}\cap\text{A}')\cap\text{B}'$

$=\phi\cap\text{B}'$

$=\phi$

Hence, the correct answer is option (c).

3. $\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}.$

Solution:

$\text{(A}-\text{B)}\cup\text{(B}-\text{A)}=\text{(A}\cap\text{B}')\cup\text{(B}\cap\text{A}')$

$=[\text{A}\cup\text{(B}\cup\text{A}')]\cap[\text{B}'\cup\text{(B}\cap\text{A}')]$ [Using distribution law]

$=[\text{(A}\cup\text{B})\cap\text{(A}\cup\text{A}')]\cap[\text{(B}'\cup\text{B})\cap\text{(B}'\cup\text{A}')]$ [Using distribution law]

$=[\text{(A}\cup\text{B)}\cup\text{(U)}]\cap[\text{(U)}\cap\text{(B}'\cup\text{A}')]$ $[\text{A}\cup\text{A'= U = B}'\cup\text{B}]$

$=[\text{A}\cup\text{B}]\cap[\text{B}'\cup\text{A}']$ $\begin{bmatrix}\text{(A}\cup\text{B)}\cap\text{(U)}=\text{(A}\cup\text{B)}\\\text{ and (U)}\cap\text{(B}'\cup\text{A)}'=\text{(B}'\cup\text{A}')]\end{bmatrix}$

$=[\text{A}\cup\text{B}]\cap[\text{(A}\cap\text{B)}']$ $[\text{(A}\cap\text{B)}'=\text{B}'\cup\text{A}']$

$=[\text{A}\cup\text{B}]\cap[\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}]$

$=[\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}].$

2. $\text{(A}- \text{B)}\cup\text{(B} - \text{A)}.$

Solution:

The symmetric difference of A and B is given by:-

$\text{(A} - \text{B)}\cup\text{(B}- \text{A)}.$

3. $\{\text{x : x}\not=\text{x}\}.$

4. $\text{F}_2\cup\text{F}_3\cup\text{F}_4\cup\text{F}_1.$

Solution:

We know that every rectangle, rhombus and square in a plane is a parallelogram but every trapezium is not a parallelogram.

So, F₁ is either of F₁ or F₂ or F₃ or F₄.

$\therefore\text{F}_1=\text{F}_1\cup\text{F}_2\cup\text{F}_3\cup\text{F}_4$

Hence, the correct answer is option (d).

2. {1, 2, 4, 5}.

Solution:

Here,

$\text{A} = \{1, 2, 3\}$ and

$\text{B} = \{3, 4, 5\}$

The symmetric difference of A and B is given by:-

$\text{(A} - \text{B)}\cup\text{(B} -\text{A)}$

Now, are have:

$\text{(A} - \text{B)}= \{1, 2\}$

$\text{(B} - \text{A)}=\{4, 5\}$

$\text{(A}-\text{B)}\cup\text{(B}-\text{A)}=\{1, 2, 4, 5\}.$

2. $\text{A}\cap\overline{\text{B}}.$

Solution:

A = {x : x is a multiple of 3}

A = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, .....

B = {x : x is a multiple of 5}

B = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ......

Now, we have:

A - B = 3, 6, 9, 12, 18, 21, 24, 27, 30, 33,36, 39, 42, ....

$=\text{A}\cap\overline{\text{B}}.$

3. 31.

Solution:

The number of proper subsets of any set is given by the formula 2n - 1, where n is the number of elements in the set.

Here,

n = 5

$\therefore$ Number of proper subsets of A = 25 - 1 = 31.

4. $\text{A}\cap\text{B}^\text{c}.$

Solution:

A and B are two sets.

$\text{A}\cap\text{B}$ is the common region in both the sets.

$\text{A}\cap\text{B}^\text{c}$ is all the region in the universal set except $\text{A}\cap\text{B}.$

Now,

$\text{(A}\cap\text{A}\cap\text{B)}^\text{c}=\text{(A}\cap\text{B)}^\text{c}.$

3. 300.

Solution:

$\text{n(A}'\cap\text{B}')=\text{n(A}\cup\text{B}')$

$=\text{n(U)}-\text{n(A}\cup\text{B})$

$=700 - 200 + 300 - 100 = 300.$

3. $\text{A}\cap\text{B}'.$

Solution:

A - B belongs to those elements of A that do not belong to B.

$\therefore\text{A} - \text{B = A}\cap\text{B}'.$

3. [4, 5).

Solution:

$\text{A} = \{\text{x : x} \in \text{R}, \text{x > 4}\}$ and 

$\text{B}= \{\text{x}\in\text{R : x} < 5\}$

$\text{A}\cap\text{B}=[4, 5).$

3. 60.

Solution:

Let M, P and C denote the sets of students who have opted for mathematics, physics, and chemistry, respectively.

Here,

$\text{n(M)}= 100, \text{ n( P)} = 70, \text{ n(C)} = 40$

Now,

$\text{n(M}\cap\text{P)}=30,\text{n(M}\cap\text{C)}=28,\\\text{n(P}\cap\text{C)}=23,\text{n(M}\cap\text{P}\cap\text{C)}=18$

Number of students who opted for only mathematics:

$\text{n(M}\cap\text{P}'\cap\text{C)}'=\{\text{M}\cap\text{(P}\cap\text{C})'\}$

$=\text{n(M)}-\text{n}\{\text{M}\cap\text{(P}\cap\text{C})\}$

$=\text{n(M)}-\text{n}\{\text{(M}\cap\text{P)}\cup\text{(M}\cap\text{C})\}$

$=\text{n(M)}-\{\text{n(M}\cap\text{P)}+\text{n(M}\cap\text{C})-\text{n(M}\cap\text{P}\cap\text{C}\}$

$=100-(30+28-18)$

$=60$

$\therefore$  the number of students who opted for mathematics alone is 60.

1. $\text{A}\cap\text{(B}-\text{C)}=\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}$

2. $\text{A}\cap\text{(B - C)}=\text{(A}\cap\text{B)} - \text{C}$

3. $\text{A}\cup\text{(B} - \text{C)}=\text{(A}\cup\text{B)}\cap\text{(A}\cup\text{C}').$

Solution:

Let x be any arbitrary element of $\text{A}\cap\text{B}-\text{C.}$

Thus, we have,

$\text{x}\in\text{A}\cap\text{(B - C)}\Rightarrow\text{x}\in\text{A}$ and $\text{x}\in\text{B}-\text{C}$

$\Rightarrow\text{x}\in\text{A}$ and $\text{(x}\in\text{B and x}\not\in\text{C)}$

$\Rightarrow\text{x}\in\text{A and x}\in\text{B}$ and $\Rightarrow\text{X}\in\text{A and x}\not\in\text{C}$

$\Rightarrow\text{x(A}\cap\text{B)}$ and $\text{x}\not\in\text{(A}\cap\text{C)}$

$\Rightarrow\text{x}\in[\text{(A}\cap\text{B)}-\text{(A}\cap\text{C)}]$

$\Rightarrow\text{A}\cap\text{(B}-\text{C)}\subseteq\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}$

Similarly, $\text{(A}\cap\text{B)}-\text{(A} - \text{C)}\subseteq\text{(A}\cap\text{(B}-\text{C)}$

Hence, $\text{A}\cap\text{(B} - \text{C)}=\text{(A}\cap\text{B)} - \text{(A}\cap\text{C)}.$

1. $\text{n(A) + n(B)}.$

Solution:

Two sets are disjoint if they do not  have a common element in them, i.e., $\text{A}\cap\text{B}=\phi.$

$\therefore\text{n(A}\cup\text{B) = n(A) + n(B)}.$

3. 60%.

Solution:

Suppose C and B represents the population travels by car and bus respectively.

$\text{n(C}\cup\text{B) = n(C) + n(B)} -\text{n(B}\cap\text{C)}$

$=0.20+0.50-0.10$

$=0.6\text{ or }60\%.$

2. $\text{A}'\cup\text{B}.$

Solution:

$\text{(A}\cup\text{B}')'\cup\text{(B}\cap\text{C})$

$=[\text{A}\cap\text{(B}')']\cup\text{(B}\cap\text{C})$ (De Morgen law)

$=\text{(A}'\cap\text{B})\cup\text{(B}\cap\text{C})$

$=\text{(A}'\cup\text{C})\cup\text{B}$ (Distributive law)

Disclimer: The question seems to be incorrect or there is some printing mistake in the question. The options given in the question does not match with the answer.