Set Theory Test 31

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Set Theory Test 31

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Weekly Quiz Competition

Question 1
1 / -0

If $$S_k=\begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix}$$, k$$\in N^+$$, where N is the set of all natural numbers, then $$(S_2)^n(S_k)^{-1}$$ for n$$\in$$N is?

A
$$S_{2n-k}$$

B
$$S_{2n+k-1}$$

C
$$S_{2^n+k-1}$$

D
$$S_{2^n+k+1}$$

Solution
Question 2
1 / -0

Directions For Questions

Let $${S}_{1}$$ be the set of all those solutions of the equation $$\left( 1+a \right) \cos { \theta } \cos { \left( 2\theta -b \right) } =\left( 1+a\cos { 2\theta } \right) \cos { \left( \theta -b \right) } $$ which are independent of $$a$$ and $$b$$ and $${S}_{2}$$ be the set of all such solutions which are dependent on $$a$$ and $$b$$, then

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All the permissible values of $$b$$, if $$a=0$$ and $${S}_{2}$$ is a subset of $$\left( 0,\pi \right) $$

A
$$b\in \left( -n\pi ,2n\pi \right) ;\in Z$$

B
$$b\in \left( -n\pi ,2\pi -n\pi \right) ;\in Z$$

C
$$b\in \left( -n\pi ,n\pi \right) ;\in Z$$

D
none of these
Question 3
1 / -0

If B= { $$y_{1}, y_{2}, y_{3} $$} and A = {$$x_{1}, x_{2}, x_{3}, .......x_{8}$$} having 3 elements in set B and 8 elements in set A. Then if the number of ways that $$ f: A\rightarrow B$$ is onto such that exactly 4 elements of $$ x \epsilon A$$ are matched to $$ y \epsilon B$$ such that $$ f(x) = y_{3} $$ is $$(_{r}^{n}\textrm{C}).m$$, then the value of $$m + r - n$$ is

A
12

B
11

C
9

D
10

Solution
Question 4
1 / -0

State which of the following is total number of reflexive relations form set $$A = \left \{a, b, c\right \}$$ to set $$B = \left \{d, e\right \}$$ is

A
$$2^{6}$$

B
$$2^{8}$$

C
$$2^{4}$$

D
$$2^{15}$$

Solution

Number of reflexive relation from$$A=[a,b,c]$$ and $$B=[d,e]$$is

$$=2^{n^2-n}=2^{3^2-3}=2^{9-3}=2^6$$
Question 5
1 / -0

Sets $$A$$ and $$B$$ have $$5$$ and $$6$$ elements respectively and $$\left( A\triangle B \right) =C$$ then the number of elements in set $$\left( A-\left( B\triangle C \right) \right)$$ is

A
$$5$$

B
$$6$$$

C
$$0$$

D
$$4$$

Solution
Question 6
1 / -0

Suppose $${ A }_{ 1 },{ A }_{ 2 },,{A }_{ 30 }$$ are thirty sets each having $$5$$ elements and $${ B }_{ 1 },{ B }_{ 2 },..,{B}_{ n }$$ are $$n$$ sets each with $$3$$ elements, let $$\displaystyle \bigcup _{ i=1 }^{ 30 }{ { A }_{ i } } =\bigcup _{ j=1 }^{ n }{ { B }_{ j } =S}$$ and each element of $$S$$ belongs to exactly $$10$$ of the $${A}_{i}s$$ and exactly $$9$$ of the $${B}_{j}s.$$ Then $$n$$ is equal to

A
$$15$$

B
$$3$$

C
$$45$$

D
$$None\ of\ these$$

Solution

$$(c)$$.
Since each $$A_{ i }$$ has $$5$$ elements, we have
$$\overset { 30 }{ \underset { i=1 }{ \Sigma } } \ n\left( { A }_{ i } \right) =5\times 30=150$$. . . $$(1)$$
Let $$S$$ consist of $$m$$ distinct elements. Since each elements of $$S$$ belongs to exactly $$12$$ of the $$A_{ i }$$ $$s$$ we also have
$$\overset { 30 }{ \underset { i=1 }{ \Sigma } } \quad n\left( { A }_{ i } \right) =10m$$. . . $$(2)$$
Hence from $$(1)$$ and $$(2)$$, $$10m=150$$ or $$m=15$$. Again since each $$B_{ i }$$ has $$3$$ elements and each element of $$S$$ belongs to exactly $$9$$ of the $$B_{ j }$$ $$s$$ we have
$$\overset { 30 }{ \underset { j=1 }{ \Sigma } } \quad n\left( { B }_{ j } \right) =3n$$ and $$\overset { 30 }{ \underset { j=1 }{ \Sigma } } \quad n\left( { B }_{ j } \right) =9m$$
It follows that $$3n=9m=9\times 15$$
. . . $$[\therefore m=15]$$
This gives $$n=45$$.
Question 7
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Suppose $$A_1 , A_2,... A_{30}$$ are thirty sets each having 5 elements and $$B_1, B_2,..., B_n$$ are n sets each with 3 elements , let $$\underset{i = 1}{\overset{30}{\cup}} A_i = \underset{j = 1}{\overset{n}{\cup}} B_j = S$$ and each element of S belongs to exactly 10 of the $$A_i's$$ and exactly 9 of the $$B_j'S$$. then n is equal to

A
15

B
3

C
45

D
35

Solution

If elements are not repeated then the number of elements in $$A_1\cup A_2\cup A_3\cup A_{30}$$ is $$30\times 5$$
But each elements is used 10 times
so, $$S=\cfrac{30\times 5}{10}=15$$
If the elements in $$B_1,B_2,.....,B_n$$ are not repeated.
then total number of element in
$$B_1\cup B_2\cup B_3.....\cup B_n$$ is 3n
but each elements is repeated 9 times
so,
$$S=\cfrac{3n}{9}=15=\cfrac{3n}{9}$$
$$n=45$$
Question 8
1 / -0

An investigator interviewed $$100$$ students to determine their preferences for the three drinks: milk (M), coffee(C) and tea (T). He reported the following: $$10$$ students had all the three drinks M, C, T; $$20$$ had M and C only; $$30$$ had C and T; $$25$$ had M and T; $$12$$ had M only; $$5$$ had C only; $$8$$ had T only. Then how many did not take any of the three drinks is?

A
$$20$$

B
$$30$$

C
$$36$$

D
$$42$$

Solution

Let $$N_M$$ be the number of students who had Milk(M) only, $$N_T$$ be the number of students who had Tea(T) only, $$N_C$$ be the number of students who had Coffee(C) only, $$N_{MC}$$ is the number of students who had Milk(M)&Coffee(C) but no Tea(T), $$N_{MT}$$ is the number of students who had Milk(M)&Tea(T) but no Coffee(C), $$N_{TC}$$ is the number of students who had Tea(T)&Coffee(C) but no Milk(M) and $$N_{MCT}$$ is the number of students who had all the three drinks Milk(M), Coffee(C), Tea(T).

To find the number of students who did not take any of the drink we have to take away students who take any of the drink from $$100$$ students.

Students who take any of the drink are as follows:

$$N_M=12$$, $$N_C=5$$, $$N_T=8$$, $$N_{MCT}=10$$.

$$N_{MC}= 20 −N_{MCT} = 20 − 10 = 10$$.

$$N_{MT}= 25 − N_{MCT} = 25 − 10 = 15$$.

$$N_{TC}= 30 −N_{MCT} = 30 − 10 = 20$$.

Now, number of students who take any of the drink will be:

$$N_M + N_C + N_T + N_{MC} + N_{MT} + N_{TC}+ N_{MCT} =12 + 5 + 8 + 10 + 15 + 20 + 10 = 80$$.

Finally, the number of students who did not take any of the drink is $$100 − 80 = 20$$.

Hence, $$20$$ students did not take any of the three drinks.
Question 9
1 / -0

Suppose $${ A }_{ 1 },{ A }_{ 2 },....{ A }_{ 30 },$$ are thirty sets each with five elements and $${ B }_{ 1 },{ B }_{ 2 },....B_{ A },$$ are n sets each with three elements such $$\overset { 30 }{ \underset { i-1 }{ U } } \quad { A }_{ 1 }=\overset { n }{ \underset { j-1 }{ U } } \quad =s$$ If each element of belongs to exactly ten of the $${ A }_{ 1 }$$ exactly 9 of the $${ A }_{ 1 }$$,then value of n is:

A
15

B
135

C
45

D
90
Question 10
1 / -0

The value of $$\left( {A \cup B \cup C} \right) \cap \left( {A \cap {B^c} \cap {C^c}} \right) \cap {C^c}$$ is

A
$$B \cap {C^c}$$

B
$${B^c} \cap {C^c}$$

C
$$B \cap {C}$$

D
$$A \cap {B^c} \cap {C^c}$$

Solution

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Re-Attempt Weekly Quiz Competition

Set Theory Test 31

Result

Set Theory Test 31

Score

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Rank

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SHARING IS CARING

Question 1

$$S_{2n-k}$$

$$S_{2n+k-1}$$

$$S_{2^n+k-1}$$

$$S_{2^n+k+1}$$

Solution

Question 2

$$b\in \left( -n\pi ,2n\pi \right) ;\in Z$$

$$b\in \left( -n\pi ,2\pi -n\pi \right) ;\in Z$$

$$b\in \left( -n\pi ,n\pi \right) ;\in Z$$

none of these

Question 3

12

11

9

10

Solution

Question 4

$$2^{6}$$

$$2^{8}$$

$$2^{4}$$

$$2^{15}$$

Solution

Question 5

$$5$$

$$6$$$

$$0$$

$$4$$

Solution

Question 6

$$15$$

$$3$$

$$45$$

$$None\ of\ these$$

Solution

Question 7

15

3

45

35

Solution

Question 8

$$20$$

$$30$$

$$36$$

$$42$$

Solution

Question 9

15

135

45

90

Question 10

$$B \cap {C^c}$$

$${B^c} \cap {C^c}$$

$$B \cap {C}$$

$$A \cap {B^c} \cap {C^c}$$

Solution

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