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CBSE Questions for Class 11 Engineering Maths Sets Quiz 13 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sets
Quiz 13
Let R = {(1, 3), (4, 2), (2, 3), (3, 1)} be a relation on the set A = (1, 2, 3, 4). The relation R is
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Transitive
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Symmetric
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Reflexive
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None of these
$$13$$ Planar Surfaces, each of area $$1\ m^{2}$$ are such that their total area is $$7\ m^{2}$$. If no three of more of these surfaces have any region in common; Then, the overlap of some Two of These surfaces has an Area.
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Not less than $$\dfrac{7}{13} m^{2}$$
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Not less than $$\dfrac{2}{13} m^{2}$$
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Not less than $$\dfrac{1}{13} m^{2}$$
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Not Less than $$\dfrac{1}{17} m^{2}$$
Explanation
Assume are of overlap of any $$2$$ surfaces is less than $$\dfrac{1}{13}m^2$$
Let us ignore the unit of area in our notation.
We know that, $$Area(A_{1}\cup A_{2}) = Area(A_{1}) + Area(A_{2}) - Area(A_{1}\cap A_{2})$$
But, $$Area (A_{1}) = Area (A_{2}) = 1.$$ (as given in question)
Also, $$Area(A_{1}\cap A_{2}) < \dfrac {1}{13}$$ (as per our assumption)
Hence, $$Area(A_{1}\cup A_{2}) > 1 + 1 - \dfrac {1}{13}$$
$$Area(A_{1}\cup A_{2}) > 2 - \dfrac {1}{13}$$
Similarly,
$$Area (A_{1}\cup A_{2}\cup A_{3}) = Area (A_{1}) + Area(A_{2}) + Area(A_{3})$$
$$ - Area(A_{1}\cap A_{2})- Area (A_{1}\cap A_{3}) - Area(A_{2} \cap A_{3})$$
$$ + Area (A_{1}\cap A_{2}\cap A_{3})$$
But $$Area(A_{1}\cap A_{2}\cap A_{3}) = 0$$ (As per the condition in Question)
Thus, $$Area(A_{1}\cup A_{2}\cup A_{3}) > 1 + 1 + 1 - \dfrac {1}{13} - \dfrac {1}{13} - \dfrac {1}{13} + 0$$.
i.e. $$Area(A_{1} \cup A_{2} \cup A_{3}) > 3 - \dfrac {3}{13}$$.
Similarly,
$$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) = Area(A_{1}) + Area (A_{2}) + Area(A_{3}) + Area(A_{4})$$
$$- Area(A_{1}\cap A_{2}) - Area(A_{1}\cap A_{3}) - Area (A_{1} \cap A_{4})$$
$$- Area(A_{2} \cap A_{3}) - Area(A_{2} \cap A_{4}) - Area (A_{3} \cap A_{4})$$
$$ + Area(A_{1}\cap A_{2}\cap A_{3})+ Area(A_{1}\cap A_{2}\cap A_{4})$$
$$ + Area (A_{2}\cap A_{3}\cap A_{4}) - Area (A_{1}\cap A_{2}\cap A_{3}\cap A_{4})$$
But $$Area(A_{1}\cap A_{2}\cap A_{3}\cap A_{4}) = 0$$
$$Area (A_{1} \cap A_{2}\cap A_{3}) = 0$$, etc. (as per the condition in Question)
So, $$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) > 1 + 1 + 1 + 1 - 6\left (\dfrac {1}{13}\right ) + 3(0) - 0$$.
i.e. $$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) > 4 - \dfrac {6}{13}$$
In general,
$$Area(A_{1}\cup A_{2}\cup A_{3}\cup ...\cup A_{n}) > n - \dfrac {^{n}C_{2}}{13}$$
Put $$n = 13$$;
$$Area(A_{1}\cup A_{2}\cup A_{3}\cup ....\cup A_{13}) > 13 - \dfrac {^{13}C_{2}}{13}$$
That is $$Area(A_{1}\cup A_{2}\cup A_{3}\cup ....\cup A_{13}) > 7$$
Hence, our assumption was wrong.
So, Area of overlap of any $$2$$ surfaces is not less than $$\dfrac {1}{13}$$.
Consider that $$n(S)$$ represented the number of elements in set S. If $$n(A\cup B\cup C)=40, n(A\cap B'\cap C')=5, n(B\cap A'\cap C')=10, n(C\cap B' \cap A')=6$$ then number of element which belongs to at least two of the set is
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Less than $$19$$
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More than $$21$$ but less than $$40$$
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$$19$$
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$$20$$
Explanation
Given,
$$n(A\cup B\cup C)=40,n(A'\cap B'\cap C')=5$$
$$n(B\cap A'\cap C')=10,n(C\cap B'\cap A')=6$$
Then number of element Which belongs to at least two of the sets is-
$$\le n(A\cup B\cup C)- n(A'\cap B'\cap C')-n(B\cap A'\cap C')-n(C\cap B'\cap A')$$
$$\le 40-5-10-6$$
$$\le 19$$
If n(A)=115, n(B)=326, n(A-B)=47, then $$n(A\cup B)$$ is equal to
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373
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165
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370
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none of these
Explanation
$$n(A)=115$$
$$n(B)=326$$
$$n(A-B)=47$$
$$\therefore$$ $$n\left( A\cap B \right) =n\left( A \right) -n\left( A-B \right) $$
$$=115-47=68$$
$$\therefore$$ $$n\left( A\cup B \right) =n\left( A \right) +n\left( B \right) -n\left( A\cap B \right) $$
$$=115+326-68$$
$$=326+47$$
$$=373$$
Option A.
For 3 sets A,B,C if A$$\subset B,B\subset C$$ then
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A$$\cup B\subset C$$
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C$$\subset A\cup B$$
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A-B=C
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None of these
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
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15
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3
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45
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35
Explanation
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,
$$S=\cfrac{3n}{9}=15=\cfrac{3n}{9}$$
$$n=45$$
Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set $$A \times B$$, each having at least three elements is............
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$$275$$
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$$510$$
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$$219$$
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$$256$$
Explanation
Set A has $$4$$ elements
Set B has $$2$$ elements
Number of elements in set $$\left( A\times B\right)=4\times 2=8$$
Total number of subsets of $$\left( A\times B\right) =2^8=256$$
Number of subsets having $$0$$ elements $$=^{ 8 }{ { C }_{ 0 } }=1$$
Number of subsets having $$1$$ elements each $$=^{ 8 }{ { C }_{ 1 } }=8$$
Number of subsets having $$2$$ elements each $$=^{ 8 }{ { C }_{ 2 } }=\dfrac{8!}{2!6!}=28$$
Number of subsets having atleast $$3$$ elements $$=256-1-8-28$$
$$=256-37$$
$$=219$$
Hence, the answer is $$219.$$
The value of $$\left( {A \cup B \cup C} \right) \cap \left( {A \cap {B^c} \cap {C^c}} \right) \cap {C^c}$$ is
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$$B \cap {C^c}$$
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$${B^c} \cap {C^c}$$
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$$B \cap {C}$$
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$$A \cap {B^c} \cap {C^c}$$
If A =[a,b],B={c,d} and C={d,e}, then { (a,c),(a,d),(a,e),(b,c),(b,d),(b,e) is equal to
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$$A\cap (B\cup C)$$
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$$A\cup (B\cap C)$$
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$$A\times (B\cup C)$$
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$$A\times (B\cap C)$$
The number of subsets $$R\,of\,P\, = \left\{ {1,2,3,....8} \right\}$$ which satisfies the property 'There exist integers $$a < \,b < \,c$$ with $$a\, \in \,R,b\, \notin \,R,c\, \in \,R''\,$$ is
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215
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219
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222
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223
Given $$n(U) =20, n(A) =12, n(B) =9, n(A \cap B) =4$$, where U is the universal set, A and B are subset of U, then $$n((A \cup B)^C)=$$
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$$17$$
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$$9$$
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$$11$$
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$$3$$
if S is a set of p(x) is polynomial of degree <2 such that p(0)=, P(1)=1, p(x)>0 $$\forall \quad x\quad \varepsilon $$ (0, 1) then
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S=0
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$$S=ax+91-1){ x }^{ 2 }\forall a\varepsilon (0,\infty )$$
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$$S=ax+(1-a){ x }^{ 2 }\forall a\varepsilon R$$
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$$S=ax+(1-a){ x }^{ 2 }\forall a\varepsilon R(0,\quad 2)$$
Let $$A,B,C$$ finite sets. Suppose then $$n(A)=10, n(B)=15, n(C)=20, n(A\cap B)=8$$ and $$n(B\cap C)=9$$. Then the possible value of $$n(A\cup B\cup C)$$ is
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$$26$$
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$$27$$
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$$28$$
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Any of the three values $$26, 27, 28$$ is possible
X and Y are two sets and $$f:X\rightarrow Y$$. If $$f(c)=\left\{ y;c\subset X,y\subset Y \right\} $$ and $${ f }^{ 1 }(d)=\left\{ x;d\subset Y,x\subset X \right\} $$, then the true statement is
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$$f({ f }^{ 1 }(b))=b$$
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$$f({ f }^{ 1 }(a))=a$$
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$$f({ f }^{ 1 }(b))=b,b\subset y$$
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$$f({ f }^{ 1 }(a))=a,a\subset x$$
A = {n/n is a digits in the number 33591} and $$B=\left\{ n/n\in N,n<10 \right\} ,$$ then $$B-A = $$
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$$\left\{ 2,4,6,8 \right\} $$
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$$\left\{ 7,2,4,8,6 \right\} $$
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$$\left\{ 1,3,5,7 \right\} $$
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$$\left\{ \left( 1,2 \right) ,\left( 1,3 \right) ,\left( 2,3 \right) \right\} $$
Let A, B, C be three seta such that A $$\cup$$ B $$\cup$$ C = $$U$$ , where $$U$$ is the universal set then ,
[(A B) $$\cup$$ (B C) $$\cup$$ (C A)] is equal to
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A $$\cup$$ B $$\cup$$ C
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A $$\cup$$ (B $$\cap$$ C)
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A $$\cap$$ B $$\cap$$ C
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A $$\cap$$ (B $$\cup$$ C)
Let $$n\left( U \right) =700,n\left( A \right) =200,n\left( B \right) =300$$ and $$n\left( A\cap B \right) =100,$$ then $$n\left( { A }^{ c }\cap { B }^{ c } \right) =$$
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400
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600
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300
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200
Which set is most like the given set?
(8, 18, 37)
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(4, 9, 20)
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(16, 33, 67)
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(13, 30, 67)
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(5, 12, 25)
If $$A = \{ 4,5,8,12 \} , B = \{ 1,4,6,9 \} \text { and } C = \{ 1,2,3,4 \}$$ then $$A - ( C - B ) =$$
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$$A$$
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$$B$$
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$$C$$
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$$\phi$$
Tell whether set A is a subset of set B.
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set A : whole numbers less than 8
set B : whole numbers less than 10
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set A : prime numbers
set B : odd numbers
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set A : numbers divisible by 6
set B : numbers divisible by 3
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set A : set of letters in the word 'FLAT'
set B : set of letters in the word 'PLATE'
The solution set of $$x^{2}+5x+6=0 $$ is ........
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{2,3}
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{-2,-3}
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{2,-3}
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{-2,3}
If $$P = \{ x:x < 3,x \in N\} $$ and $$Q = \{ x:x \le 2,x \in W\} $$, where W is the set of whole numbers then the set of whole numbers then the set $$(P \cup Q) \times (P \cap Q)$$ is
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$${(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)}$$
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$$(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)$$
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$$(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)$$
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$$(1,1),(1,2),(2,1),(2,2)$$
If $$\text{A} = \{ 1,2,3,4,5\} \,,\text{B} = \{ 2,3,5,6\} $$ and $$\text{C} = \{ 4,5,6,7\} $$
Then, which of the following is true?
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$$A \cup \left( {B \cap C} \right) = \left( {A \cup B} \right) \cap (A \cup C)$$
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$$ {A \cup B} =\{1,2,3,4,5\}$$
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$$ {A \cup B} =\{1,2,3,4,5,6\}$$
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Both $$\text{A}$$ and $$\text{C}$$
If $$n\left( {P \cap Q} \right) = 23,\,\,n\left( {P \cup Q} \right) = 57,$$ and $$n\left( {Q - P} \right) = 26,$$ then $$n\left( {P - Q} \right) = \_\_\_\_\_\_\_\_.$$
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24
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54
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8
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14
if P and Q are two sets having 5 elements in com- mon, then how many elenents do P $$ \times $$ Q and Q $$ \times $$ P have in common?
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5
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10
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25
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20
$$If\,\,A = \left\{ {x:{x^2} = 1} \right\}\,\,\,and\,B = \left\{ {x:{x^4} = 1} \right\},\,then\,A\Delta B\,\,\,is\,\,equal\,to\,$$
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$$\left\{ {i, - i} \right\}\,$$
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$$\,\left\{ { - 1,1} \right\}\,$$
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$$\left\{ { - 1,1,i - i} \right\}$$
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$$\,\left\{ {1,i} \right\}$$
Let $$A$$ and $$B$$ be two sets. The $$(A\cup B)'\cup(A'\cap B)$$=
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$$A'$$
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$$A$$
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$$B'$$
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$$none\ of\ these$$
Examine whether the following statements are true or false:
$$\left\{a\right\}\subset \left\{\{a\},\, b \right\}$$
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True
0%
False
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