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CBSE Questions for Class 11 Engineering Maths Sets Quiz 4 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sets
Quiz 4
Examine whether the following statements are true or false:
$$(1,2,3)\subset (1,3,5)$$
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0%
True
0%
False
Explanation
False. $$2\epsilon (1,2,3)$$; however, $$2\not{\subset}(1,3,5)$$
Examine whether the following statements are true or false:
$$(a)\epsilon (a,b,c)$$
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0%
True
0%
False
Explanation
False. The element of $$(a, b, c)$$ are $$a, b, c$$. Therefore, $$(a)\subset (a,b,c)$$.
Examine whether the following statements are true or false:$$\left\{a \right\} \subset \left\{a,b,c \right\}$$
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0%
True
0%
False
Explanation
a
∈
{
a
,
b
,
c
}
State true or false for each of the following. Correct the wrong statement.
If T = {a, l, a, h, b, d, h}, then $$n (T) = 5$$
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0%
True
0%
False
Explanation
Given
If $$T = \{$$a, l, a, h, b, d, h$$\}$$, then $$n (T) = 5$$
$$T = \{$$a, l, h, b, d$$\}$$
i.e. $$n (T) = 5 $$
Hence, the given statement is true
For sets $$\phi , A = \left \{ 1,3 \right \} = B = \left \{ 1, 5, 9 \right \} , C = \left \{ 1, 5, 7, 9 \right \}$$ True option is
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0%
$$ A \subset B $$
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$$ B \subset C $$
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$$ C \subset B $$
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$$ A \subset C $$
Explanation
Option (B) is correct , because all the elements of set $$B = \left \{ 1, 5, 9 \right \}$$ is present in set $$ C = (1, 5, 7, 9) $$ Hence , $$ B \subset C $$
Examine the following statements:
$$\left \{ a, b \right \} \subset $$ { b, a, c]
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0%
True
0%
False
Explanation
$$ \left \{ a, b \right \} \subset $$ { b, a, c]
Elements of first set (a,b) is present in second set .
Hence , the statement is true
If $$ A = \left \{ 2 , 4, 6, 8 \right \} $$ and $$ B = \left \{ 1, 4, 7, 8 \right \}$$ then A - B and B - A will be respectively:
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0%
$$\left \{ 2 ,6 \right \} ; \left \{ 1, 7 \right \}$$
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$$\left \{ 1, 7 \right \} ; \left \{ 4 , 8 \right \}$$
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$$\left \{ 1, 7 \right \} ; \left \{ 2, 6 \right \}$$
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$$\left \{ 4, 8 \right \} ; \left \{ 1, 7 \right \}$$
Explanation
$$ A = \left \{ 2 , 4, 6, 8 \right \}$$ and $$ B = \left \{ 1, 4, 7, 8 \right \}$$
$$ A - B = \left \{ 2, 6 \right \} $$ and $$ B - A = \left \{ 1 , 7 \right \}$$
Hence , option (A) is correct
If $$ A = \left \{ 1, 2, 3, 4, 5, 6 \right \} , B = \left \{ 2, 4, 6, 8 \right \}$$, then A - B will be :
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0%
$$\left \{1, 3, 5, 8 \right \}$$
0%
$$\left \{ 1, 3, 5 \right \}$$
0%
$$\left \{1, 2, 3, 4, 5, 6, 8 \right \}$$
0%
= { }
Explanation
Given , $$ A = \left \{ 1, 2, 3, 4, 5, 6 \right \}$$ and $$ B = \left \{2, 4, 6, 8 \right \}$$
A - B means A contains the element which is not present in B.
Thus, $$A - B = \left \{ 1, 3, 5 \right \}$$
Hence , option (B) is correct.
Find the equivalent set for
$$A - B $$.
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$$A \cup (A \cap B)$$
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B - A
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$$A - (A \cap B)$$
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$$A \cap B$$
Explanation
Hence By this graph we see that $$A-B = A - (A \cap B)$$
The equation x + cosx $$=$$ a has exactly one positive root. Complete set of values of 'a' is
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0%
(0, 1)
0%
$$(-\infty, 1)$$
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(-1,1)
0%
$$(1,\infty)$$
0%
$$(0,\infty)$$
Explanation
Let f(x) = x + cosx a
$$\Rightarrow f'(x)=1-sinx\geq 0\forall x \epsilon R.$$
Thus f(x) is increasing in $$\left ( -\infty ,\infty \right )$$, as zero of f'(x) don't for an interval. f(0) = 1 a
For a positive root, $$1-a< 0$$
$$\Rightarrow a> 1$$
Let $$S$$ be a non-empty subset of $$R$$. Consider the following statement:
$$p$$ : There is a rational number $$x$$ such that $$x > 0$$.
which of the following statements is the negation of the statement P ?
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There is no rational number $$\mathrm{x}\in \mathrm{S}$$ such that $$\mathrm{x}\leq 0$$
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Every rational number $$\mathrm{x}\in \mathrm{S}$$ satisfies $$\mathrm{x}\leq 0$$
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$$\mathrm{x}\in \mathrm{S}$$ and $$\mathrm{x}\leq 0= \mathrm{x}$$ is not rational
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There is a rational number $$\mathrm{x}\in \mathrm{S}$$ such that $$\mathrm{x}\leq 0$$
Explanation
$$\mathrm{P}$$: there is a rational number $$\mathrm{x}\in \mathrm{S}$$ such that $$\mathrm{x}>0$$
$$\sim \mathrm{P}$$: Every rational number $$\mathrm{x}\in \mathrm{S}$$ satisfies $$\mathrm{x}\leq 0$$
Hence, option 'B' is correct.
If $$A= $${1,2,3};$$ B=$${4,5}. then $$A-B=$${1,2,3} number cancellation of numbers in A and B
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0%
{4,5}
0%
{1,2,3,4,5}
0%
$$\phi $$
0%
{1,2,3}
The smallest set $$A$$ such that $$A\cup \left\{ 1,2 \right\} =\left\{ 1,2,3,5,9 \right\} $$ is
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0%
$$\left\{ 2,3,5 \right\} $$
0%
$$\left\{ 3,5,9 \right\} $$
0%
$$\left\{ 1,2,5,9 \right\} $$
0%
$$\left\{ 1,2 \right\} $$
Explanation
$$A\cup \{1,2\}=\{1,2,3,5,9\}$$
Thus
$$A=\{1,2,3,5,9\}-\{1,2\}$$
Hence
$$A=\{3,5,9\}$$
If A = {1, 2, 3} B = {4, 5}, then find A - B.
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0%
{1, 4, 5}
0%
{1, 4, 3}
0%
{1, 2, 3}
0%
{4, 5}
Explanation
Given, $$A=\{1,2,3\}$$ and $$B=\{4,5\}$$.
Since $$A$$ and $$B$$ are two disjoint sets i.e. $$A\cap B=\phi$$ then we've,
$$A-B$$
$$=\{1,2,3\}$$.
In an examination $$80\%$$ passed in English, $$85\%$$ in Maths, $$75\%$$ in both and $$40$$ students failed in both subjects. Then the number of students appeared are
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0%
$$300$$
0%
$$400$$
0%
$$500$$
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$$600$$
Explanation
$$n(E)=80$$
$$n(M)=85$$
$$n(E \cap M)=75$$
$$n(E \cup M)=n(E)+n(M)-n(E \cap M)=80+85-75=90$$
$$n(E \cup M)'=10$$
Let n be the total number of students appeared
$$\dfrac{10}{100} \times n=40$$
$$\therefore n=400$$
$$p\cap (q\cup r)=?$$
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0%
$$p\cup (q\cap r)$$
0%
$$(p\cap q)\cap (q\cap p)$$
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$$(p\cap q)\cup r$$
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$$(p\cap q)\cup (p\cap r)$$
In an examination, $$34\%$$ of the candidates fail in Arithmetic and $$42\%$$ in English. If $$20\%$$ fail in Arithmetic and English, the percentage of those passing in both subjects is :
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0%
$$44$$
0%
$$45$$
0%
$$46$$
0%
$$47$$
Explanation
Let $$A$$ denote students fail in Arithmetic,
$$B$$ denote students fail in English
$$n(A)=34$$
$$n(B)=42$$
$$n(A \cap B)=20$$
$$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 34+42-20=56$$
$$n(A \cup B)' = 100 - n(A \cup B) = 100-56=44$$
In a science talent examination, 50% of the candidates fail in Mathematics and 50% fail in Physics. If 20% fail in both these subjects, then the percentage who pass in both Mathematics and Physics is:
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0%
0%
0%
20%
0%
25%
0%
50%
Explanation
By set theory
$$n(M\cup P) = n(M) + n(P)-n(M\cap P)$$ where M and P are sets of students failing in respective subjects.
$$=0.5 + 0.5 - 0.2 = 0.8$$
This indicates $$80\%$$ of the class fails in at least one of the given subjects while $$20\%$$ pass in both.
If $$A - B = \phi$$ and $$B - A = \phi$$ then A and B are
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0%
Overlapping
0%
Equivalent
0%
Disjoint
0%
Equal
Explanation
if $$A-B = \phi$$ and $$B-A = \phi$$
hence this relation contains $$A-B$$ and $$B-A$$ doesn't contains any element,
so, the elements in $$A$$ & $$B$$ are same.
All the students of a batch opted Psychology, Business, or both. 73% of the students opted Psychology and 62% opted Business. If there are 220 students, how many of them opted for both Psychology and business?
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0%
$$60$$
0%
$$100$$
0%
$$77$$
0%
$$35$$
Explanation
By set theory
$$n(P\cap B) = n(P) + n(B)-n(P\cup B)$$
$$=0.73+0.62-1.00=0.35$$
$$35\%$$ of $$220 = 77$$
If $$A$$ and $$B$$ have some elements in common, then $$n(A \cup B)$$ is:
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0%
$$= n(A) + n (B)$$
0%
$$ > n (A) + n(B)$$
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$$< n(A) + n(B)$$
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$$\geq n (A)+ n (B)$$
Explanation
We know that $$n(A\cup B)+n(A\cap B) = n(A)+n(B)$$
But $$n(A\cap B) >0$$ always
$$\Rightarrow n(A)+n(B)=n(A\cup B)+n(A\cap B)>n(A\cup B)$$
Hence, $$n(A\cup B)<n(A)+n(B)$$.
If $$A - B = \emptyset $$, then relation between A and B is :
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$$A \neq B$$
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$$B \subset A$$
0%
$$A \subset B$$
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$$A = B$$
Explanation
If A and B are disjoint it would mean A is a null set. Otherwise A and B must be equal to $$A\cap B$$ AT LEAST.
Hence option C is correct.
If $$A \cap B = \phi$$, then $$A \Delta B$$ =
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$$A \cap B$$
0%
$$A \cup B$$
0%
$$A - B$$
0%
$$B - A$$
Explanation
We know that $$A\Delta B=A\cup B-A\cap B $$
Given $$A\cap B=\phi $$
$$\therefore A\Delta B=A\cup B-\phi=A\cup B $$
Hence, the answer is $$A\cup B$$.
In a group of 15 women, 7 have nose studs, 8 have ear rings and 3 have neither. How many of these have both nose studs and ear rings?
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0%
$$0$$
0%
$$2$$
0%
$$3$$
0%
$$7$$
Explanation
Since 3 women have neither nose studs nor earrings
$$n(N\cup E) = 15-3 = 12$$
By set theory
$$n(N\cap E) = n(N) + n(E)-n(N\cup E)$$
$$=7+8-12 = 3$$
There are 19 hockey players in a club. On a particular day 14 were wearing the prescribed hockey shirts, while 11 were wearing the prescribed hockey pants. None of then was without hockey pant or hockey shirt. How many of them were in complete hockey uniform?
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0%
8
0%
6
0%
9
0%
7
Explanation
We can look at it in 2 ways
First by set theory
$$n(A\cap B) = n(A) + n(B)-n(A\cup B)$$
$$=14+11-19=6$$
Qualitatively, we know that 14 people are wearing prescribed hockey shirts,which leaves us with 5 players who must be wearing hockey pants. So out of 11 players who are wearing hockey pants, 5 are not wearing hockey shirts while the other 6 are in complete uniform.
In a class, 20 opted for Physics, 17 for Maths, 5 for both and 10 for other subjects. The class contains how many students?
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$$35$$
0%
$$42$$
0%
$$52$$
0%
$$60$$
Explanation
By set theory
$$n(P\cup M) = n(P) + n(M)-n(P\cap M)$$
$$ =20+17-5=32$$
So total no. of students $$= 32+10 =42$$
$$32$$ opted for at least one subject from Physics and maths while $$10$$ opted for other.
In a community of 175 persons, 40 read the Times, 50 read the Samachar and 100 do not read any. How many persons read both the papers?
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0%
10
0%
15
0%
20
0%
25
Explanation
Since 100 do not read any
$$n(T\cup S)=175-100=75$$
By set theory
$$n(T\cap S) = n(T) + n(S)-n(T\cup S)$$
$$=40+50-75 = 15$$
In a party, 70 guests were to be served tea or coffee after dinner. There were 52 guests who preferred tea while 37 preferred coffee. Each of the guests liked one or the other beverage. How many guests liked both tea and coffee?
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0%
15
0%
18
0%
19
0%
33
Explanation
By set theory
$$n(T\cap C) = n(T) + n(C)-n(T\cup C)$$
$$=52+37-70 = 19$$
In a group of $$15, 7$$ have studied German, $$8$$ have studied French, and $$3$$ have not studied either. How many of these have studied both German and French?
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0%
$$0$$
0%
$$3$$
0%
$$4$$
0%
$$5$$
Explanation
Since $$3$$ have neither studied German nor French
$$n(G\cup F) = 15-3 = 12$$
By set theory
$$n(G\cap F) = n(G) + n(F)-n(G\cup F)$$
$$=7+8-12 = 3$$
In a certain group of 75 students, 16 students are taking physics, geography and English; 24 students are taking physics and geography, 30 students are taking physics and English; and 22 students are taking geography and English. However, 7 students are taking only physics, 10 students are taking only geography and 5 students are taking only English.
How many of these students are taking physics?
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n(P) = 45
0%
n(P) = 35
0%
n(P) = 65
0%
n(P) = 55
Explanation
7 students are taking Physics only.
So
$$7=n(P)-n(P\cap G)-n(P\cap E) +n(P\cap G \cap E)$$
$$\Rightarrow n(P)=7$$
$$+ n (P\cap G)+n(P\cap E) -n(P\cap G \cap E)$$
So $$n(P)=7+24+30-16=45$$
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