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CBSE Questions for Class 11 Engineering Maths Sets Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sets
Quiz 12
State whether the following statements are true(T) or false(F):
A collection of books is a set.
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0%
True
0%
False
Explanation
A collection of books is a set.(False)
Correct:
A collection of different books is a set.
State true or false for each of the following. Correct the wrong statement If $$A = \{0\}$$, then $$n (A) = 0$$
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0%
True
0%
False
Explanation
Given
If $$A = \{0\}$$, then $$n (A) = 0$$
The statement given here is false
Correct statement: If $$A = \{0\}$$, then $$n (A) = 1$$
State true or false for each of the following. Correct the wrong statement
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0%
True
0%
False
Explanation
Given
$$n (φ) = 1 $$
The statement given here is false
Correct statement: $$n (φ) = 0$$
In $$n(P)= n(M)$$, then $$P \rightarrow M$$
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0%
True
0%
False
Explanation
$$n(P)= n(M)$$ means the number of elements of set P = Number of elements of set M.
$$\therefore$$ P and M are equivalent.
Hence, the given statement is true.
M $$\cup$$ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4} then which of the following represent set N?
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0%
{1, 2, 3}
0%
{3, 4, 5, 6}
0%
{2, 5, 6}
0%
{4, 5, 6}
Explanation
We have, M $$\cup$$ N = {1, 2, 3, 4, 5, 6} and M = {1, 2, 4}
Since, {1, 2, 3} $$\cup$$ {1, 2, 4} = {1, 2, 3, 4} $$\neq$$ M $$\cup$$ N = {1, 2, 3, 4, 5, 6};
{3, 4, 5, 6} $$\cup$$ {1, 2, 4} = {1, 2, 3, 4, 5, 6} = M $$\cup$$ N = {1, 2, 3, 4, 5, 6};
{2, 5, 6} $$\cup$$ {1, 2, 4} = {1, 2, 4, 5, 6} $$\neq$$ M $$\cup$$ N = {1, 2, 3, 4, 5, 6}; and
{4, 5, 6} $$\cup$$ { 1, 2, 4} = {1, 2, 4, 5, 6} $$\neq$$ M $$\cup$$ N = {1, 2, 3, 4, 5, 6}
If $$P \ \subseteq \ M$$, then Which of the following set represent $$P \ \cap \ (P \ \cup \ M)$$ ?
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0%
P
0%
M
0%
P $$\cup$$ M
0%
P' $$\cap$$ M
Find the correct option for the given question.
Which of the following collections is a set?
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0%
Colours of the rainbow
0%
Tall trees in the school campus
0%
Rich people in the village
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Easy examples in the book
Explanation
Since, the colors of the rainbow are well defined such as Violet, Indigo, Blue, Green, Yellow, Orange and Red.
So, the collection of colors of the rainbow is a set.
Examine whether the following statements are true or false:
$$\left\{a,e \right\} \subset \left\{x : x \ is\ a\ vowel\ in\ the\ English\ alphabet \right\}$$
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0%
True
0%
False
Explanation
a
,
e
∈
{
a
,
e
,
i
,
o
,
u
}
Examine whether the following statements are true or false:$$\left\{x : x\ is\ an\ even\ natural\ number\ less\ than\ 6 \right\} \subset \left\{x : x is\ a\ natural\ number\ which\ divides\ 36 \right\}$$
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0%
True
0%
False
Explanation
{
2
,
4
}
∵{2,4}
is a set but no in
{
1
,
2
,
3
,
4
,
6
,
9
,
12
,
18
,
36
}
Examine whether the following statements are true or false:$$\left\{1,2,3 \right\} \subset \left\{1,3,5 \right\}$$
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0%
True
0%
False
Explanation
2
∉
{
1
,
3
,
5
}
Examine whether the following statements are true or false:
$$\left\{a,b \right\} \not \subset \left\{b,c,a \right\}$$
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0%
True
0%
False
Explanation
a
,
e
∈
{
a
,
e
,
i
,
o
,
u
}
In each of the following, determine whether the statements is true or false if it is true prove it if it false given an example.
If $$A \subset B$$ and $$B \subset C$$, then $$A \subset C$$
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0%
True
0%
False
Explanation
Given
A
⊂
B
A⊂B
and
Now we have to prove
A
⊂
C
A⊂C
∴
x
∈
C
∵
B
∈
C
∴x∈C∵B∈C
∀
X
∈
A
⇒
X
∈
C
∴
A
∈
C
Examine the following statements:
{x : x is an even natural number less then 6} $$\subset $$ { x : x is natural number which divide 36 }
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0%
True
0%
False
Explanation
{x : x is an even natural number less then 6} $$\subset $$ { x : x is natural number which divide 36 }
$$ \left \{ 2, 4\right \} \subset \left \{ 1, 2, 3, 4 , 6, 9, 12, 18, 36\right \} $$
Hence, the statement is true
If $$ \cap = \left \{ 1, 2, 3, 4, 5, 6 \right \}, A = \left \{ 2, 3 \right \} $$ and $$B = \left \{ 3, 4, 5 \right \}$$ then :
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0%
$$ (A \cap B )' = \left \{ 2, 3, 4, 5 \right \}$$
0%
$$B - A = \left \{ 4, 5 \right \}$$
0%
$$A - B = \left \{ 2, 4, 5 \right \}$$
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$$( A \cap B ) = \left \{ 3 \right \}$$
Explanation
Option (B) is correct because ,
B -A means B contain the element which is not present in A
Thus , $$ B - A = \left \{4,5 \right \}$$
If $$A=\left [ \frac{5}{111} \frac{-3}{336}\right ]$$ and det $$(-3A^{2013}+A^{2014})=\alpha ^{\alpha }\beta ^{2}(1+\gamma +\gamma ^{2})$$ then, where $$\alpha ,\beta ,\gamma $$ are integers
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0%
$$\alpha = 2013$$
0%
$$\beta = 3$$
0%
$$\gamma = 10$$
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none of these
If A = {1, 2, 3}, B = {3, 4, 5}, C = {4, 6}, then
A x $$( B \cup C)$$ =
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0%
{(1, 3) (1, 4) (1, 5) (1, 6) (2, 3) (2, 4) (2, 5) (2, 6) (3,3) (3, 4) (3,5) (3, 6)}
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A x $$(B \cap C)$$
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B x $$(A \cap C)$$
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all the above
Explanation
If A = {1 , 2 , 3} , B = {3 , 4 , 5} , C = {4 , 6}
then, to find A X (B U C)
B U C = combination of unique elements of set B and C.
B U C = {3, 4 , 5, 6}
Using associative property of set:
A X (
B U C) = (A X B ) U ( A X C)
To find cartesian
products:
(i) A X B = {1 , 2 , 3 } X {3 , 4 , 5}
= {(1 , 3), (1 , 4), (1 , 5), (2 , 3), (2 , 4), (2 , 5), (3 , 3), (3 , 4), (3 , 5)}
(i) A X C = {1 , 2 , 3 } X {4 , 6}
= {(1 , 4), (1 , 6), (2 , 4), (2 , 6), (3 , 4), (3 , 6)}
(A X B ) U ( A X C) =
{(1 , 3), (1 , 4), (1 , 5), (2 , 3), (2 , 4), (2 , 5), (3 , 3), (3 , 4), (3 , 5)} U
{(1 , 4), (1 , 6), (2 , 4), (2 , 6), (3 , 4), (3 , 6)}
Ans =
{(1 , 3), (1 , 4), (1 , 5),
(1 , 6),
(2 , 3), (2 , 4), (2 , 5)
, (2 , 6)
, (3 , 3), (3 , 4), (3 , 5),
(3 , 6)
}
$$A - (B \cup C)= $$
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0%
$$(A - B) \cap (A - C)$$
0%
$$(A - B) \cup (A - C)$$
0%
$$(A - B) \cup C$$
0%
$$(A - B) \cap C$$
Which of the following statements is true (if N, W and I are sets of Natural, Whole and
Integer numbers respectively ?
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0%
$$N\, \subset \, W\, \subset \, I$$
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$$I\, \subset \, N\, \subset \, W$$
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$$W\, \subset \, N\, \subset \, I$$
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$$I\, \subset \, W\, \subset \, N$$
S = {1, 2, 3, 5, 8, 13, 21, 34 }. Find $$\displaystyle \sum $$ max (A), where the sum is taken over all 28 elements subsets A to S.
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0%
844
0%
480
0%
484
0%
488
There are 6 boxes numbered 1, 2, ...Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:
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0%
5
0%
21
0%
33
0%
60
The set $$\{x/| x L|< K\}$$ is the same for all $$K > 0$$ and for all L, as
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0%
$$\{x/0 < x < L + K\}$$
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$$\{x/L K < x < L + K\}$$
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$$\{x/|L K| < x <|L + K|\}$$
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$$\{x/|L x| > K\}$$
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$$\{x/ K < x < L\}$$
The dual of $$-p\wedge (q\vee \sim r)$$ is
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$$p\vee (\sim q \wedge r)$$
0%
$$\sim p\vee (q\wedge r)$$
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$$p\wedge (q\wedge r)$$
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$$\sim q\vee(q\wedge \sim r)$$
If $$S$$ represents the set of all real numbers $$x$$ such that $$1\le x \le 3$$ and $$T$$ represents the set of all real numbers $$x$$ such that $$2 \le x \le 5$$, the set represented by $$S \cap T$$ is
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$$2 \le x \le 3$$
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$$1 \le x \le 5$$
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$$x \le 5 $$
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$$x\ge 5$$
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none of these
Explanation
$$S\cap T$$(common part of S and T) = $$2\leq x\leq 3$$
In a town of $$10,000$$ families it was found that $$40\%$$ families buy newspaper $$A$$, $$20\%$$ families buy newspaper $$B$$ and $$10\%$$ families buy newspaper $$C$$. $$5\%$$ families buy $$A$$ and $$B$$, $$3\%$$ buy $$B$$ and $$C$$ and $$4\%$$ buy $$A$$ and $$C$$. If $$2\%$$ families buy all the three newspaper, find the number of families which buy (i) $$A$$ only (ii) $$B$$ only (iii) none of $$A, B$$ and $$C$$
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(i) $$3000$$ (ii) $$1800$$ (iii) $$4600$$
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(i) $$3300$$ (ii) $$1400$$ (iii) $$4000$$
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(i) $$3500$$ (ii) $$1600$$ (iii) $$3800$$
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none
Union set is defined as
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a collection of sets is the set of all elements in the collection
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It is one of the fundamental operations through which sets can be combined and related to each other.
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Set whole each element is an element of all the present set
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None of these
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
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0%
$$5$$
0%
$$6$$$
0%
$$0$$
0%
$$4$$
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
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0%
$$15$$
0%
$$16$$
0%
$$17$$
0%
$$18$$
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
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0%
$$15$$
0%
$$3$$
0%
$$45$$
0%
$$None\ of\ these$$
Explanation
$$(c)$$.
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 $$n=45$$.
All the permissible values of $$b$$, if $$a=0$$ and $${S}_{2}$$ is a subset of $$\left( 0,\pi \right) $$
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0%
$$b\in \left( -n\pi ,2n\pi \right) ;\in Z$$
0%
$$b\in \left( -n\pi ,2\pi -n\pi \right) ;\in Z$$
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$$b\in \left( -n\pi ,n\pi \right) ;\in Z$$
0%
none of these
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?
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0%
$$20$$
0%
$$30$$
0%
$$36$$
0%
$$42$$
Explanation
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.
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