MATHS M.C.Q (1 Marks)

2. $\phi.$

Solution:

Let $\text{x}\in\text{X}\cap(\text{X}\cup\text{Y})'$

$\Rightarrow \text{x}\in\text{X}\cap(\text{X}'\cup\text{Y})'$

$\Rightarrow \text{x}\in(\text{X}\cap\text{X})\cap(\text{X}\cap\text{Y}')$

$\Rightarrow \text{x}\in\phi\cap(\text{x}\cap\text{Y}')\ \big[\because \text{A}\cap\text{A}'=\phi\big]$

$\Rightarrow \text{x}\in\phi$

Hence, the correct option is (c).

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.

4. $\text{A}$

Solution:

(A’)’ = A

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.

4. $(1,\infty)$

Solution:

Let $\text{f(x) = x} + \cos \text{x a}$

$\Rightarrow \text{f}'(\text{x})=1-\sin\geq0\forall\text{x}\in\text{R}.$

Thus f(x) is increasing in $(-\infty,\infty),$ as zero of f'(x) don't for an interval. f(0) = 1a

For a positive root, 1 - a < 0

⇒ a > 1

3. $\text{A}\cap\text{B}=\phi$

Solution:

Given that: $\text{A}=\Big\{(\text{x},\text{y}|\text{y}=\frac{1}{\text{x}},0\neq\text{x}\in\text{R}\Big\},$

and $\text{B}=\big\{(\text{x},\text{y}|\text{y}=\text{x},\text{x}\in\text{R}\big\}$

It is very clear that $\text{y}=\frac{1}{\text{x}}$ and y = -x

$\because \frac{1}{\text{x}}\neq -\text{x}$

$\therefore \text{A}\cap\text{B}=\phi$

Hence, the correct option is (c).

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

4. 25

Solution:

Total number of students = 60

Number of students who play cricket = 25

Number of students who play tennis = 20

Number of students who play cricket and tennis both = 10

$\Rightarrow \text{n}(\text{C}\cap\text{T})=10$

$\therefore \text{n}(\text{C}\cap\text{T}) =\text{n(C)}+\text{n(T)}-\text{n}(\text{C}\cap\text{T})$

$=25+20-10=45-10=35$

$\therefore \text{n}(\text{C}'\cap\text{T}') = \text{n(U)}-\text{n}(\text{C}\cap\text{T})$

$=60-35=25$

Hence, the correct option is (b).

3. $39 \leq \text{x} \leq 63$

Solution:

Let p% of the people watch a channel and q% of the people watch another channel

$\because \text{n}(\text{p}\cap\text{q})=\text{x}\% \text{ and n}(\text{p}\cup\text{q})\leq100$

So, $\text{n}(\text{p}\cap\text{q})\geq\text{n(p)}+\text{n(q)}-\text{n}(\text{p}\cap\text{q})$

$100\geq63+76-\text{x}$

$100\geq139-\text{x}\Rightarrow\text{x}\geq139-100\Rightarrow \text{x}\geq39$

Now n(p) = 63

$\therefore \text{n}(\text{p}\cap\text{q})\leq\text{n(p)}\Rightarrow \text{x}\geq63$

So $39\leq\text{x}\geq63.$ Hence, the correcr option is (c).

3. 60%.

Solution:

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

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

$=0.20+0.50-0.10$

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

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

Solution:

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

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

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

4. $(\text{A } \cup \text{ B})\cup\text{C}=\text{A} \cup (\text{B }\cup\text{ C})$

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.

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

2. 290

Solution:

Let H be the set of persons who read Hindi and E be the ser of persons who read English.

Then, n(U) = 840, n(H) = 450, n(E) 300, $\text{n}(\text{H}\cap\text{E})=200$

Number of persons who read neither $=\text{n}(\text{H}'\cap\text{F}')$

$=\text{n}(\text{H}\cup\text{E})'=\text{n(U)}-\text{n}(\text{H}\cup\text{E})$

$=840-\big[\text{n(H)}+\text{n(E)}-\text{n}(\text{H}\cap\text{E})\big]$

$=840-(450+300-200)=290$

3. 5

Solution:

A subset containing 3 elements = {1, 2, 3};{1, 3, 4};{1, 2, 4} and {2, 3, 4}

A subset containing 4 elements ={1, 2, 3, 4}

$\therefore$ there are five subsets containing at least 3 elements.

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

Solution:

We konw that: $(\text{A}\cap\text{B})'=\text{A}'\cup\text{B}'$ [De Morgan's law]

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

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

$=\text{A}'\cup\text{B}$

Hence, the correct optiom is (b).

1. $\frac{1}{5}$

1. $\text{X} \subset \text{Y}$

Solution:

X = {8n - 7n - 1| n $\in$ N} = {0, 49, 490, .....}

Y = {49n - 49 | n $\in$ N} = {0, 49, 147, ....., 490, .....}

Clearlut, every element of X is in Y but every element of Y is not in X.

$\therefore \text{X}\subset\text{Y}$

1. 373

Solution:

n(A) = 115, n(B) = 326

n(A - B) = 47

n(A) = n(A - B) + n(A ∩ B)

n(A ∩ B) = n(A) -n(A - B)

$\therefore$ n(A ∩ B) = 115 - 47 = 68

$\therefore$ n(A ∪ B) = n(A) + n(B) -n(A ∩ B)

⇒ n(A ∪ B) = 115 + 326 - 68

⇒ n A ∪ B) = 373

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

Solution:

Every rectangel, rhombus, square in a plane is a parallelogram but every trapezium is not a parallelogram.

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

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).

3. 45

Solution:

Number of elements in $\text{A}_1\cup\text{A}_2\cup\text{A}_3\ ..... \cup \text{A}_{30}=30\times5=150$ (When repetition is not allowed)

But each element is repeated 10 times

$\therefore \text{n(S)}=\frac{30\times5}{10}=\frac{150}{10}=15\ .....\text{(i)}$

Number of elements in $\text{B}_1\cup\text{B}_2\cup\text{B}_3\ ...... \text{B}_\text{n}=3\text{n}$ (when repetitiom is not allowed)

But each element is repeated 09 times

$\therefore \text{n(S)}=\frac{3\text{n}}{9}=\frac{\text{n}}{3}\ .....\text{(ii)}$

From (i) and (ii) we get

$\frac{\text{n}}{3}=15\Rightarrow \text{n}=15\times3=45$

Hence, the corrrect option is (c).

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

1. $\text{A}$

Solution:

Given that: $\text{A}\cap(\text{A}\cup\text{B})$

Let $\text{x}\in\text{A}\cap(\text{A}\cup\text{B})$

$\Rightarrow \text{x}\in\text{A}$ and $\text{x}\in(\text{A}\cup\text{B})$

$\Rightarrow \text{x}\in\text{A}$ and $(\text{x}\in\text{A}\text{ or x}\in\text{B})$

$\Rightarrow (\text{x}\in\text{A and x}\in\text{A})$ or $(\text{x}\in\text{A and x}\in\text{B})$

$\Rightarrow \text{x}\in\text{A}$ or $\text{x}\in(\text{A}\cap\text{B})$

$\Rightarrow \text{x}\in\text{A}$

Hence, the correct option is (a).

2. 6

Solution:

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

Now A has 3 elements and B has 6 elements. If they are disjoint, then n(A ∩ B) = 0.

$\therefore$ n(A ∪ B) = 6 + 3 = 9

If A ⊂ B then A ∪ B = B

$\therefore$ (A ∪ B) = n(B) = 6

B cannot be a subset of A and hence the other possibility of A ∪ B = A is ruled out.

4. $\text{f}$

4. 7

Solution:

A = {2, 3, 4, 5, 7}

n(A) = 5

B = {7, 8, 9}

n(B) = 3

n(A ∩ B) = 1

$\therefore$ (A ∪ B) = n(A) + n(B) − n(A ∩ B)

(A ∪ B) = 5 + 3 − 1 = 7

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. 5.

Solution:

We know:

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

Now,

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

$=16+14-25$

$=5.$

2. [5, 6, 7, 8, 9]

Solution:

Given A = [5, 6, 7] and B = [7, 8, 9]

then $\text{A } \cup \text{ B}=[5, 6, 7, 8, 9]$

2. 41

Solution:

The number of students who took at least one of the three subjects can be found by finding out $\text{A }\cup \text{ B}\cup \text{ C},$ 

where A is the set of those who took Physics, B the set of those who took Chemistry and C the set of those who opted for Math.

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

A is the set of those who opted for Physics $=\frac{120}{2}=60\text{ Students}$

B is the set of those who opted for Chemistry $=\frac{120}{5}=24$

C is the set of those who opted for Math $=\frac{120}{7}=17$

The 10th, 20th, 30th….. numbered students would have opted for both Physics and Chemistry.

Therefore $\text{A }\cap \text{ B}=\frac{120}{10}=12$

The 14th, 28th, 42^nd…… Numbered students would have opted for Physics and Math.

Therefore, $\text{C }\cap \text{ A}=\frac{120}{14}=8$

The 35th, 70th…. numbered students would have opted for Chemistry and Math.

Therefore, $\text{B }\cap \text{ C}=\frac{120}{35}=3$

And the 70th numbered student would have opted for all three subjects.

Therefore, $\text{A }\cup \text{ B }\cup \text{ C}= 60 + 24 + 17 - (12 + 8 + 3) + 1 = 79$

Number of students who opted for none of the three subjects = 120 – 79 = 41

2. $\text{A } – (\text{B} \cap \text{C}) = ({1, 2, 4})$

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

Solution:

If A and B are disjoint it would mean A is a null set. Otherwise A and B must be equal to A ∩ B AT LEAST.

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.

1. ϕ

Solution:

Given: A = (6, 7, 8, 9), B = (4, 6, 8, 10) and C = {x : x $\in$ N : 2 < x ≤ 7}

B − B will always be a null set it will contain elements of B which are not in B i.e. no elements.

So, B − B = ϕ

4. 20.

Solution:

We have:

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

$=70+60-110$

$=20.$

3. 300.

Solution:

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

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

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

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,

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

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. 13

Solution:

Here n(M ∪ B) = 50, n(M) = 35, n(B) = 37

$\therefore$ n(M ∩ B) = n(M) + n(B) -n(M ∪ B)

= 35 + 37 - 50 = 22

⇒ 22 student have opted for both Mathematics and Biology. 

Now the number of students who have opted for Mathematics only

= n(M) -n(M ∩ B)

= 35 - 22 = 13

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:

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

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

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

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

2. 66

Solution:

Since 5 consecutive no.s are chosen only one set in 6 consecutive sets will not have a multiple of 6. So till 78 sets there are

$78-\frac{78}{6}=78-13=65$ sets containing 6 or multiples of 6.

S_79​ does not contain any multiple of 6

Hence S_80​ must contain a multiple of 6.

Answer = 66 sets

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

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]

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

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

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

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

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

3. $\text{A}$

Solution:

We are given that A is the subset of B

⇒ Every element of A is an element of B.

Therefore, the intersection elements of sets A and B are A ∩ B = 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. $\text{S}\cup\text{T}\cup\text{C}=\text{S}$

Solution:

The given conditions of the question may be represented by the following Venn diagram. From the given Venn diagram, we conclude thta

$\text{S}\cup\text{T}\cup\text{C}=\text{S}$

Hence, the correct option is (c).

4. R = {(x, y) : 0 < x < a, 0 < y < b}.

Solution:

Since, R be the set of points inside the rectangle.

$\therefore$ R = {(x, y) : 0 < x < a, 0 < y < b}.

1. $(\text{A}\times\text{B}) \cup (\text{A}\times\text{C})$

Solution:

Given A, B and C are any three sets.

Now, $\text{A}\times(\text{B }\cup \text{C})=(\text{A}\times\text{B}) \cup (\text{A}\times\text{C})$

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 $\text{B}\subseteq\text{A}.$

4. 69

Solution:

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

$\therefore​​​(\text{A}\triangle\text{B})=\text{ (A - B)} ∪ \text{(B - A)}$

$​​\Rightarrow (\text{A}\triangle\text{B}) = \text{n(A)} \text{ -n(A ∩ B)} +\text{ n(B)} − \text{n(A ∩ B)}$ 

$​​\Rightarrow (\text{A}\triangle\text{B})=\text{ n(A)} +\text{ n(B)} \text{ -2n} (A∩B) = 65 + 32 - 2 × 14 = 69$ 

1. [5, 6, 7, 8, 9]

Solution:

Given A = [5, 6, 7] and B = [7, 8, 9]

then $\text{A }\cup \text{ B}$ = [5, 6, 7, 8, 9]

1. The smallest set of Y is {3, 5, 9}

3. The largest set of Y is {1, 2, 3, 5, 9}

Solution:

Since the set of the right hand side has 5 elements,

$\therefore$ smallest set of Y has three elements and largest set of

Y has five elements,

$\therefore$ smallest set of Y is {3, 5, 9}

and largest set of Y is {1, 2, 3, 4, 9}

2. 400

Solution:

$\text{n(E)}=80$

$\text{n(M)}=85$

$\text{n(E}\cap\text{M})=75$

$\text{n(E}\cup\text{M})=\text{n(E)}+\text{n(M)}-\text{n(E}\cap\text{m})$

$=80+85-75=90$

$\text{n(E}\cup\text{M})'=10$

Let n be the total number of students appeared

$\frac{10}{100}\times\text{n}=40$

$\therefore\text{n}=400$

2. 600

Solution:

Student passed in atleast one subject

= n (P ∪ M) = n(P) + n(M) -n (P ∪ M)

= 80 + 85 - 70 = 95

$\therefore$ 5% student failed in both the subjects

⇒ 5% of total students = 30

⇒ Total students $=\frac{30\times100}{5}=600$

2. $2 \text{ and }3$

Solution:

2 and 3 is the null set.

2. 400

Solution:

n(E) = 80

n(M) = 85

n(E ∩ M) = 75

n(E ∪ M) = n(E) + n(M) − n(E ∩ M) = 80 + 85 − 75 = 90

n(E∪M)′ = 10

Let n be the total number of students appeared

$\frac{100}{100}\times\text{n}=40$

∴ n = 400

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

Solution:

Only this case is possible.

1. 228

Solution:

Total number of students = 950

Number of students who are members of lion club = 646

Number of students who are members of country club = 532

Number of students who are members of both club = (646 + 532) - 950 = 1178 - 950 = 228

$\therefore$ There are 228 students who joined both clubs.

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

Solution:

Given A, B and C are any three sets.

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

2. 31

Solution:

Given that: S = {x | x is a positive multiple of 3 < 100}

$\therefore$ S = {3, 6, 9, 12, 15 18, ....., 99}

n(S) = 33

T = (x | x is a prime number < 20)

$\therefore$ T = {2, 3, 5, 7, 11, 13, 17, 19}

n(T) = 8

So, n(S) + n(T) = 33 + 8 = 41

Hence, the correct option is (b).

1. B = C

Solution:

Given A, B, C be three sets such that $\text{A } \cup \text{ B}=\text{A } \cup \text{ C}$ and $\text{A } \cap \text{ B}=\text{A } \cap \text{ C},$ then, B = C

2. N

Solution:

Given that:

A = {1, 3, 5, 7, 9, 11, 13, 15, 17}

B = {2, 4, ...., 18}

U = N = {1, 2, 3, 4, 5, .....}

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

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

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

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

$=\text{N}\cup(\text{A}'\cup\text{B}')\ \big[\because \text{A}'\cup\text{A}=\text{N}\big]$

$=\text{A}'\cup\text{B}'$

$=(\text{A}\cup\text{B}')=(\phi)'=\text{N} \ \big[\because \text{A}\cap\text{B}=\phi\big]$

Hence, the correct option is (b).

1. A

Solution:

(A ∩ A) ∪ (A ∩ B) = A ∪ (A ∩ B) = A

$\therefore\text{A}\cap\text{B}\subset\text{A}$

1. A.

Solution:

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

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

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).

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

Solution:

Hence By this graph we see that $\text{A}-\text{B}=\text{A}-(\text{A}\cap\text{B})$

1. $(-\infty,\infty)$

Solution:

Let the given function is

y = 3x - 2

⇒ y + 2 = 3x

$\Rightarrow \text{x} =\frac{(\text{y}+2)}{3}$

Now x is satisfied by all values.

So, Range {f(x)} = R $=(-\infty,\infty)$

3. 45

Solution:

Number of elements in A = 10 

Number of subsets of A containing exactly two elements 

= Number of ways we can select 2 elements from 10 elements 

$10\text{c}_2=\frac{10\times9}{2}=45$

$\therefore$ Number of subsets of A containing exactly two elements = 45

2. $\text{E}\subset\text{I}$

Solution:

Given, I is the set of isosceles triangle and E is the equilateral triangles.

We know that every equilateral triangle is an isosceles triangle but the converse is not true.

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).