Sets Test

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Sets Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Suman is given an aptitude test containing 80 problems, each carrying I mark to be tackled in 60 minutes. The problems are of 2 types; the easy ones and the difficult ones. Suman can solve the easy problems in half a minute each and the difficult ones in 2 minutes each. (The two type of problems alternate in the test). Before solving a problem, Suman must spend one-fourth of a minute for reading it. What is the maximum score that Suman can get if he solves all the problems that he attempts?

A
30

B
40

C
53

D
60

Solution

$$Assuming\quad that\quad all\quad the\quad problems\quad suman\quad solves\quad are\quad right,\\ marks\quad earned\quad per\quad minute\quad solving\quad easy\quad problem=\dfrac { 1 }{ 3/4 } =\dfrac { 4 }{ 3 } \\ marks\quad earned\quad per\quad minute\quad solving\quad difficult\quad problem=\frac { 1 }{ 9/4 } =\frac { 4 }{ 9 } \\ since\quad we\quad have\quad to\quad maximize\quad the\quad marks,\quad we\quad will\quad solve\quad all\quad the\quad easy\quad problems\quad first.\\ marks\quad earned\quad by\quad solving\quad easy\quad problems=40,\\ time\quad spent=\frac { 3 }{ 4 } *40=30\quad minutes\\ time\quad left\quad for\quad difficult\quad problems=30\quad minutes,\\ problems\quad solved=\frac { 30 }{ 9/4 } =13,\\ marks\quad earned=13,\\ total\quad marks=40+13=53$$
Question 2
1 / -0

Which one of the following is correct?

A
$$A \times (B - C) = (A - B) \times (A - C)$$

B
$$A \times (B - C) = (A \times B) - (A \times C)$$

C
$$\displaystyle A\cap \left ( B\cup C \right )=\left ( A\cap B \right )\cup C$$

D
$$\displaystyle A\cup \left ( B\cap C \right )=\left ( A\cup B \right )\cap C$$

Solution

$$(A) A\times(B-C) = (A\times B) - (A-C)$$
$$ A\times(B-C) \ne (A\times B) - (A-C)$$

$$(B) A\times(B-C) = (A\times B) - (A\times C)$$

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

$$(D) A\cup (B\cap C) = (A\cup B) \cap (A\cup C)$$
$$ (A\cup B)\cap C \ne (A\cup B) \cap (A\cup C)$$

So , B is correct.
Question 3
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 4
1 / -0

If $$20$$% of three subsets (i.e., subsets containing exactly three elements) of the set $$A = \left \{a_{1}, a_{2}, ...., a_{n}\right \}$$ contain $$a_{2}$$, then the value of $$n$$ is

A
$$15$$

B
$$16$$

C
$$17$$

D
$$18$$

Solution
Question 5
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 6
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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 $$\displaystyle \bigcup _{ i=1 }^{ 30 }{ { A }_{ i } } =\bigcup _{ j=1 }^{ n }{ { B }_{ j } =S}$$ 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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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

...view full instructions

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 8
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 9
1 / -0

If n(A)=115, n(B)=326, n(A-B)=47, then $$n(A\cup B)$$ is equal to

A
373

B
165

C
370

D
none of these

Solution

$$n(A)=115$$
$$n(B)=326$$
$$n(A-B)=47$$
$$\therefore$$ $$n\left( A\cap B \right) =n\left( A \right) -n\left( A-B \right) $$
$$=115-47=68$$
$$\therefore$$ $$n\left( A\cup B \right) =n\left( A \right) +n\left( B \right) -n\left( A\cap B \right) $$
$$=115+326-68$$
$$=326+47$$
$$=373$$
Option A.
Question 10
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For 3 sets A,B,C if A$$\subset B,B\subset C$$ then

A
A$$\cup B\subset C$$

B
C$$\subset A\cup B$$

C
A-B=C

D
None of these

Solution

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Re-Attempt Weekly Quiz Competition

Sets Test - 44

Result

Sets Test - 44

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SHARING IS CARING

Question 1

30

40

53

60

Solution

Question 2

$$A \times (B - C) = (A - B) \times (A - C)$$

$$A \times (B - C) = (A \times B) - (A \times C)$$

$$\displaystyle A\cap \left ( B\cup C \right )=\left ( A\cap B \right )\cup C$$

$$\displaystyle A\cup \left ( B\cap C \right )=\left ( A\cup B \right )\cap C$$

Solution

Question 3

$$5$$

$$6$$$

$$0$$

$$4$$

Solution

Question 4

$$15$$

$$16$$

$$17$$

$$18$$

Solution

Question 5

$$20$$

$$30$$

$$36$$

$$42$$

Solution

Question 6

$$15$$

$$3$$

$$45$$

$$None\ of\ these$$

Solution

Question 7

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

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

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

none of these

Question 8

$$2^{6}$$

$$2^{8}$$

$$2^{4}$$

$$2^{15}$$

Solution

Question 9

373

165

370

none of these

Solution

Question 10

A$$\cup B\subset C$$

C$$\subset A\cup B$$

A-B=C

None of these

Solution

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