MATHS M.C.Q (1 Marks)

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).

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

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

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.

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

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).

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).

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).

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}.

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 $=\text{n}(\text{H}'\cap\text{F}')$

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

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

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

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. $\text{A}$

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. $\phi.$

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

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).

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).