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CBSE Questions for Class 11 Engineering Maths Sets Quiz 6 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sets
Quiz 6
If X and Y are two sets then $$\displaystyle X\cap (Y\cup X)'$$ equals:
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$$X$$
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$$Y$$
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$$\displaystyle \phi $$
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$$\{ 0 \}$$
Explanation
We have $$\displaystyle X\cap \left ( Y\cup X \right )'=X\cap \left ( Y'\cap X' \right )$$
= $$\displaystyle X\cap Y'\cap X'=X\cap X'\cap Y'$$
= $$\displaystyle \phi \cap Y'= \phi $$
reflexive, symmetric and transitive.
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$$\displaystyle R_{3}= \left \{ \left ( 1,1 \right ), \left ( 2,2 \right ), \left ( 3,3 \right ), \left ( 4,4 \right ), \left ( 1,2 \right ), \left ( 2,1 \right ) \right \}$$
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$$\displaystyle R_{3}= \left \{ \left ( 1,1 \right ), \left ( 2,2 \right ), \left ( 3,3 \right ), \left ( 4,4 \right ), \left ( 1,2 \right ), \left ( 2,1 \right ),\left ( 1,3 \right ),\left ( 3,1 \right ),\left ( 4,1 \right ),\left ( 1,4 \right ) \right \}$$
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$$\displaystyle R_{3}= \left \{ \left ( 1,1 \right ), \left ( 2,2 \right ), \left ( 3,3 \right ), \left ( 4,4 \right ) \right \}$$
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none of these
Explanation
We define $$\displaystyle R_{3}$$ as follows:
$$\displaystyle R_{3}=
\left \{ \left ( 1,1 \right ), \left ( 2,2 \right ), \left ( 3,3 \right
), \left ( 4,4 \right ), \left ( 1,2 \right ), \left ( 2,1 \right )
\right \}.$$
Then evidently $$\displaystyle R_{3}$$ is reflexive, symmetric and transitive, that is, $$\displaystyle R_{3}$$ is an equivalence relation on A.
$$\displaystyle \left ( 1,2 \right )
\epsilon R_{3}, \left ( 2,1 \right ) \epsilon R_{3}\Rightarrow \left (
1,1 \right ) \epsilon R_{3}$$
Given $$\displaystyle \xi $$ = {x : x is a natural number}
A = {x : x is an even number x $$\displaystyle \in $$ N}
B = {x : x is an odd number, x $$\displaystyle \in $$ N}
Then $$\displaystyle (B\cap A)-(x-A)=....$$
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$$\displaystyle \phi $$
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$$A$$
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$$B$$
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$$A-B$$
Explanation
$$A = \{2, 4, 6, 8, ........\}$$
$$B = \{1, 3, 5, 7, .......\}$$
$$\displaystyle B\cap A=\left \{ 2,4,6,8,..... \right \}\cap \left \{ 1,3,5,7,.... \right \}=\phi $$
$$\displaystyle \xi -A=\left \{ 1,2,3,4,5,6,.... \right \}-\left \{ 2,4,6,8,.... \right \}=\left \{ 1,3,5,7,.... \right \}$$
$$\displaystyle \therefore B\cap A-(\xi -A)=\phi -\left \{ 1,3,5,7,.... \right \}=\phi $$
Which of the following statements is true ?
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$$\displaystyle 3\quad \subseteq \quad \left\{ 1,3,5 \right\} $$
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$$\displaystyle 3\quad \in \quad \left\{ 1,3,5 \right\} $$
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$$\displaystyle \{ 3\} \quad \in \quad \left\{ 1,3,5 \right\} $$
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$$\displaystyle \{ 3,5\} \quad \in \quad \left\{ 1,3,5 \right\} $$
Explanation
$$\displaystyle 3\quad \subseteq \quad \left\{ 1,3,5 \right\} $$
An element cannot be a subset of another set. Hence,false.
$$\displaystyle 3\quad \in \quad \left\{ 1,3,5 \right\} $$
$$3$$ lies in the set. Hence, true.
$$\displaystyle \{ 3\} \quad \in \quad \left\{ 1,3,5 \right\} $$
$$\{ 3\}$$ doesnot lie in the set. Hence, false.
$$\displaystyle \{ 3,5\} \quad \in \quad \left\{ 1,3,5 \right\} $$
$$\{ 3,5\}$$ doesnot lie in the set. Hence, false.
Hence, option B.
$$\displaystyle A=\left\{ x:x\neq x \right\} $$ represents:
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$${0}$$
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$$\phi$$
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$${1}$$
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$${x}$$
Explanation
$$x=x$$ for all $$x$$
Hence, there is no element in A.
A is an empty set .
$$A=\{ \}$$
If n (A) = 120, N(B) = 250 and n (A - B) = 52, then find $$\displaystyle n(A\cup B)$$
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$$302$$
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$$250$$
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$$368$$
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None of the above
Explanation
$$\displaystyle n(A-B)=n(A)-n(A\cap B)$$
$$\displaystyle \Rightarrow 52=120-n(A\cap B)$$
$$\displaystyle \Rightarrow n(A\cap B)=120-52=68$$
Now $$\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B)$$
$$\displaystyle =120+250-68$$
$$=302$$
The solution set of $$x+2<9$$ over a set of positive even integers is
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$$\displaystyle \left \{ 8,10,12,... \right \}$$
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$$\displaystyle \left \{ 2,4,6 \right \}$$
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$$\displaystyle \left \{ 1,2,3,4,5,6 \right \}$$
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$$\displaystyle \left \{ 2,4,6,8 \right \}$$
Explanation
$$x+2<9$$
$$x<7$$
Since $$x$$ is even positive integer then $$x\in [2,7)$$.
Hence the set of positive even integers which fall in the above set is
$$\{2,4,6\}$$.
The solution set of $$3 x - 4 < 8$$ over the set of non-negative square numbers is
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$$\displaystyle \left \{ 1,2,3 \right \}$$
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$$\displaystyle \left \{ 1,4 \right \}$$
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$$\displaystyle \left \{ 1 \right \}$$
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$$\displaystyle \left \{ 16 \right \}$$
Explanation
$$3x-4<8$$
$$3x<12$$
$$x<4$$
Hence set of non-negative square numbers belonging to the above set is
$$\{1\}$$.
Let $$P$$ and $$Q$$ be two sets then what is $$\displaystyle (P\cap Q')\cup (P\cup Q)'$$ equal to ?
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$$\displaystyle (P\cap Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi $$
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$$\displaystyle (P\cup Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi $$
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$$\displaystyle (P\cap Q')\cup (P\cap Q)'=\xi \cap Q'=\xi \cap Q'=\xi $$
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none of the above
Explanation
$$\displaystyle (P\cap Q')\cup (P\cup Q)'=(P\cap Q')\cup (P'\cap Q') $$
$$=$$$$\displaystyle (P\cup P')\cap (P\cup Q')\cap (Q'\cup P')\cap (Q'\cup Q')$$
$$=$$$$\displaystyle \xi \cap \left \{ Q'\cup (P\cap P') \right \}\cap Q'$$
$$=\displaystyle \xi \cap \{Q'\cup \xi )\cap Q'$$
$$=$$$$\displaystyle \xi \cap Q'\cap Q'=\xi \cap Q'=\xi$$
If $$A$$ and $$B$$ are finite sets which of the following is the correct statement?
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$$n(A - B) = n(A) - n(B)$$
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$$n(A - B) = n(B - A)$$
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$$n(A - B) = n(A) -$$ $$\displaystyle n\left ( A\cap B \right )$$
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$$n(A - B) = n(B) - $$$$\displaystyle n\left ( A\cap B \right )$$
U is a universal set and n(U) =A, B and C are subset of U. If n(A) = 50, n(B) = 70, $$\displaystyle n\left ( B\cup C \right )=\Phi $$, $$\displaystyle n\left ( B\cap C \right )=15 $$ and $$\displaystyle A\cup B\cup C=U $$. then n(C) equals
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40
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50
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55
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60
If $$n (A) = 115, n(B) = 326, n(A - B) = 47$$, then $$\displaystyle n(A+B)$$ is equal to
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373
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165
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370
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None
Explanation
$$n(A-B)=47\Rightarrow n(A)-n(A\cap B) = 47\Rightarrow n(A\cap B)=115-47=68$$
Hence $$n(A+B)=115+326-68=373$$
In a town of 10000 families, it was found that 40% families buy a newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy both A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then the number of families which buy A only.
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$$3300$$
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$$3500$$
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$$3600$$
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$$3700$$
Explanation
$$n(A) = 40% of 10,000 = 4,000$$
$$n(B) = 20% of 10,000 = 2,000$$
$$n(C) = 10% of 10,000 = 1,000$$
$$n (A \cap B) = 5% of 10,000 = 500$$
$$n (B \cap C) = 3% of 10,000 = 300$$
$$n(C \cap A) = 4% of 10,000 = 400,$$
$$n(A \cap B \cap C) = 2% of 10,000 = 200$$
We want to find $$n(A\ only) = n(A) – [n(A\cap B) + n(A \cap C)] + n(A \cap B \cap C)$$
$$ n(A\ only) = 4000 – [500 + 400] + 200 = 4000 – 700 = 3300$$
If A and B are two disjoint sets and N is the universal set then $$\displaystyle A^{c}\cup \left [ \left ( A\cup B \right )\cap B^{c} \right ]$$ is
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$$\displaystyle \phi $$
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A
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B
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N
Explanation
Since $$A$$ and $$B$$ are disjoint sets $$B^c\cap(A\cup B)$$ will be only $$A$$ as there is no intersection between $$A$$ and $$B$$.
Of course $$A^c\cup A=N$$
Out of 800 boys in a school 224 played cricket, 240 played hockey and 236 played basketball. Of the total 64 played both basketball and hockey, 80 played cricket and basketball and 40 played cricket and hockey, 24 players all the three games. The number of boys who did not play any game is
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$$128$$
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$$216$$
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$$240$$
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$$260$$
Explanation
No. of players who played at least one game is:
By set theory
$$n(C\cup H\cup B)= n(C) +n(H) +n(B)-n(B\cap H)-n(C\cap B) -n(C\cap H) + n(C\cap H \cap B)$$
$$=224+240+236-64-80-40+24=540$$
Hence $$260$$ players do not play any game.
Suppose $$\displaystyle A_{1},A_{2}.....A_{30}$$ are thirty sets having 5 elements and $$\displaystyle B_{1},B_{2}....B_{n}$$ are n sets each with 3 elements. Let $$\displaystyle \bigcup_{i=1}^{30}Ai=\bigcup_{i=1}^{n}Bj=S$$ and each elements of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n is equal to
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35
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45
0%
55
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65
Explanation
Since each element of S belongs to $$10$$ $$ A_{i}s$$, there are only $$15 $$elements in S $$\left[\dfrac{30\times 5}{10}\right]$$
Since each of the $$15$$ elements belongs to 9 $$B_j s$$ there are 135 elements in the second union.
No. of sets $$=\dfrac{135}{3}=45$$
S = {1, 2, 3, 5, 8, 13, 21, 34}. Find $$\displaystyle \sum max\left ( A \right )$$ where the sum is taken over all 28 two elements subsets A to S
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844
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480
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484
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488
Given n(A) = 11, n(B) = 13, n(C) = 16, $$\displaystyle n\left ( A\cap B \right )=3,n\left ( B\cap C \right )=6,n\left ( A\cap C \right )=5\: \: and\: \: n\left ( A\cap B\cap C \right )=2$$ then the value of $$\displaystyle n [ A\cup B \cup C ]=$$
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$$24$$
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$$27$$
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$$25$$
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$$28$$
Explanation
We know,
$$n(A\cup B\cup C)= n(A) +n(B) +n(C)-n(A\cap B)-n(B\cap C) -n(C\cap A) + n(A\cap B \cap C)$$
$$=11+13+16-3-6-5+2=28$$
In a group, if 60% of people drink tea and 70% drink coffee. What is the maximum possible percentage of people drinking either tea or coffee but not both?
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$$100\%$$
0%
$$70\%$$
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$$30\%$$
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$$10\%$$
Explanation
To find maximum possible percentage of people drinking either coffee or tea, we can assume everyone drinks at least either of the options.
Hence
$$a+b=100$$
$$a+2b=60+70=130$$
$$b=30$$
hence $$a=70$$ which is the required area of the venn diagram.
If out of 150 students who read at least one newspaper The Times of India, The Hindustan Times and The Hindu. There are 65 who read The Times of India, 41 who read The Hindu and 50 who read The Hindustan Times. What is the maximum possible number of students who read all the three newspaper?
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$$7$$
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$$42$$
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$$3$$
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Cannot be determined
Explanation
From venn diagram
$$a+b+c=150$$
$$a+2b+3c=156$$
Hence $$b+2c=6$$
To maximise c we take minimum value of b that is 0.
Hence $$ c=3$$
If A and B are two disjoint sets and N is universal set, then $$\displaystyle A^{\circ}\cup \left [ \left ( A\cup B \right )\cap B^{\circ} \right ]$$ is
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$$\displaystyle \phi $$
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$$A$$
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$$B$$
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$$N$$
Explanation
In the Venn Diagram, we can see that $$ (A \cup B) \cap B' $$ covers the pink coloured region inside $$N$$ except $$B$$.
And $$ A' $$ is the region which is shown using striking lines , that is the region outside $$A$$
And if we perform $$ A' \cup (A \cup B) \cap B' $$, it will cover the pink region as well s the striken region, which is nothing but the complete set $$N$$.
Suppose $$\displaystyle A_{1},A_{2},....,A_{30}$$ are thirty sets each having 5 elements and $$\displaystyle B_{1},B_{2},....,B_{n}$$ are n sets each with 3 elements. Let $$\displaystyle \bigcup_{i=1}^{30}A_{i} = \bigcup_{j=1}^{n}B_{j}=S $$ and each elements of S belongs to exactly 10 of the $$\displaystyle A_{i}$$ and exactly 9 of the $$\displaystyle B_{j}$$. Then n is equal to-
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35
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45
0%
55
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65
Explanation
$$A_0,A_1,.............,A_{30}\implies$$ each of 5 elements
$$B_1,B_2,B_3...........n\implies$$ each of 3 elements
The number of elements in the union of the A sets is $$5(30)-r$$ where 'r' is the number repeats likewise the number of elements in the B sets $$3n-rB$$
Each element in the union (in5) is repeated 10 times in A which means if x was the real number of elements in A (not counting repeats) then q out of those 10 should be thrown away or 9x .likewise on the B side 8x of those elements should be thrown away So, $$\implies 150-9x=3n-8x$$
$$n=50-3x$$
Now in figure out what x is we need to use the fact that the union of a group of sets contains every member of each sets . If every element in 'S' is repeated 10 times that means every element in the union of the n's is repeated 10 times .
This means that $$10/10\implies 15$$ is the number of in the A's without repeats counted (same for the B's aswell ) So now
$$\cfrac{50-15}{3}=n$$
$$n=45$$
Subset:- A proper subset is nothing but it contain atleast one more element of main set .
Ex:$$\{3,4,5\}$$ is a set then the possible subsets are
$$\{3\},\{4\},\{5\},\{1,5\},\{3,4\}$$
Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in $$A\cup B$$ ?
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$$9$$
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$$6$$
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$$3$$
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$$18$$
Explanation
Let $$A$$ be the left circle and $$B$$ be the right circle.There are $$3$$ elements in $$A$$ and $$6$$ elements in $$B$$.
The union of $$A$$ and $$B$$ contains elements that are in either circle.Thus,the union of $$A$$ and $$B$$ will be all of the elements in $$A$$ along with all of the element $$B$$.However,you have to subtract the elements that are in the overlapping area because you are counting twice.
If $$A$$ and $$B$$ don't overlap at all,then the union will ontain $$9$$ elements.If $$A$$ is completely inside $$B$$ then the union will only contain $$6$$ elements,which is the minimum no. of elements in the union of $$A$$ and $$B$$.
let $$A=1,2, B=2,3$$
$$\therefore A\cup B=1,2,3$$ which is $$3$$ elements.
$$\therefore A $$ has $$2$$ elements ,$$B$$ has $$2$$ elements,and there is $$1$$ element overlapping.
$$\therefore 2+2-1=3!=3\times 2\times 1=6$$
If $$A=\left \{1, 2, 3, .....9\right \}$$ and $$B=\left \{2, 3, 4, 5, 7, 8\right \}$$, then A-B is given by
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$$\left \{1, 6, 7, 8\right \}$$
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$$\left \{1, 6, 9\right \}$$
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$$\left \{1, 9\right \}$$
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$$\left \{6, 9\right \}$$
Explanation
$$A=\left\{1,2,3,4,5,6,7,8,9 \right\}$$
$$B=\left\{2,3,4,5,7,8 \right\}$$
A-B is a set of all those elements of A which are not in B.
$$\therefore A-B=\left\{1,6,9 \right\}$$
The set of real numbers r satisfying
$$\displaystyle \frac{3 r^2- 8r+5}{4r^2-3r+7}>0$$ is
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the set of all real numbers
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the set of all positive real numbers
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the set of all real numbers strictly between 1 and 5/3
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the set of all real numbers which are either less than 1 or greater than 5/3
Explanation
$$\displaystyle \frac{3.r^2 - 8.r + 5 }{4. r^2 - 3.r + 7} > 0$$
$$\Rightarrow 3r^2 -3r -5r + 5 > 0$$ and $$4r^2 - 3r + 7 > 0 $$ as D < 0
$$\Rightarrow 3r(r-1) -5(r -1) > 0$$
$$\Rightarrow (3r - 5) (r - 1) > 0$$
$$(3r - 5) (r - 1) > 0$$
$$r \varepsilon (- \infty, 1) \cup (5/3, \infty)$$
Let $$ S=\left\{1,2,3,.....40\right\} $$ and let $$A$$ be a subset of $$S$$ such that no two elements in $$A$$ have their sum divisible by $$5$$ What is the maximum number of elements possible in $$A$$?
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$$10$$
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$$13$$
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$$17$$
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$$20$$
Explanation
There are $$20$$ maximum number of elements possible in $$A$$
$$A=(1,2,5,6,7,11,12,15,16,17,21,22,25,26,27,31,32,35,36,37)$$
Look at the set of numbers below.
Set : $$\left \{6, 12, 30, 48\right\}$$
Which statement about all the numbers in this set is NOT true?
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They are all multiples of $$3$$
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They are all even numbers
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They are all factors of $$48$$
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They are all divisible by $$2$$
Explanation
Here above statement C is not true because 30 is not
factors of
48.
So, Answer (C)
They are all factors of
48
Let $$A$$ and $$B$$ be two sets such that $$n(A)=16$$, $$n(B)=12$$, and $$n(A\cap B)=8$$. Then $$n(A\cup B)$$ equals
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$$28$$
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$$20$$
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$$36$$
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$$12$$
Explanation
$$n(A\cup B)=n(A)+n(B)-n(A\cap B)=16+12-8=20$$
If A={a,b,c,d,e}, B={a,c,e,g} and C={b,d,e,g} then which of the following is true?
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$$\displaystyle C\subset \left ( A\cup B \right )$$
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$$\displaystyle C\subset \left ( A\cap B \right )$$
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$$\displaystyle A\cup B=A\cup C$$
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Both(1) and (3)
Explanation
For the given sets,
$$ (A \cup B ) = $$ { $$ a,b,c,d,e,g $$} (Combination of all elements of both sets)
Clearly, elements of C are a part of $$ A \cup B $$ as well,
So, $$ C\subset (A\cup B) $$
And, $$ (A \cap B ) = $$ { $$ a,c,e $$} (Common elements of both sets )
As the elements of C are not completely a part of $$ A \cap B $$ , Option B is not True.
Also,
$$ (A \cup C ) = $$ { $$ a,b,c,d,e,g $$}
Clearly, $$ (A \cup B ) = (A \cup C ) $$
Hence, both first option and third options are True.
Let $$A=\{1,2,3,4,5,6\}$$. How many subsets of $$A$$ can be formed with just two elements, one even and one odd?
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$$6$$
0%
$$8$$
0%
$$9$$
0%
$$10$$
Explanation
$$\{1,2\}$$, $$\{1,4\}$$, $$\{1,6\}$$, $$\{2,3\}$$, $$\{2,5\}$$, $$\{3,4\}$$, $$\{3,6\}$$, $$\{4,5\}$$, $$\{5,6\}$$,
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