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CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 10 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Set Theory
Quiz 10
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Report Question
0%
10
0%
11
0%
12
0%
13
Explanation
Let
A
,
B
, and
C
be the set of people who like
p
r
o
d
u
c
t
A
,
p
r
o
d
u
c
t
B
,
a
n
d
p
r
o
d
u
c
t
C
respectively.
Accordingly,
n
(
A
)
=
21
,
n
(
B
)
=
26
,
n
(
C
)
=
29
,
n
(
A
∩
B
)
=
14
,
n
(
C
∩
A
)
=
12
,
n
(
B
∩
C
)
=
14
,
n
(
A
∩
B
∩
C
)
=
8
The Venn diagram for the given problem can be drawn as above.
It can be seen that number of people who like product C only is
=
29
–
(
4
+
8
+
6
)
=
11
In a battle
70
%
of the combatants lost one eye,
80
%
an ear,
75
%
an arm,
85
%
a leg and
x
%
lost all the four limbs the minimum value of
x
is
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0%
10
0%
12
0%
15
0%
n
o
n
e
o
f
t
h
e
s
e
The number of subsets
R
of
P
=
(
1
,
2
,
3
,
.
.
.
.
,
9
)
which satisfies the property "There exit integers a<b<c with a
∈
R, b
∈
R,c
∈
R" is
Report Question
0%
512
0%
466
0%
467
0%
None of these
Explanation
Every subset of P containing at least 3 elements from P will always satisfy the required property since each number is distinct in the set P.
So, Total no. of ways =
9
C
3
+
9
C
4
+
.
.
.
.
+
9
C
9
=
(
9
C
0
+
9
C
1
+
.
.
.
.
.
+
9
C
9
)
−
9
C
0
−
9
C
1
−
9
C
2
=
2
9
−
46
=
512
−
46
=
466
If
A
=
{
3
,
{
4
,
5
}
,
6
}
, then the statement is true or false
{
3
}
⊆
A
Report Question
0%
True
0%
False
Explanation
A
=
{
3
,
{
4
,
5
}
,
6
}
We have to find out if
{
3
}
is subset of
A
or not.
Since,
3
is present in elements of
A
, then
{
3
}
⊆
{
A
}
∴
Given statement is true.
In a universal set x,
n\left( x \right) = 50
,
n\left( A \right) = 35
,
n\left( B \right) = 20
,
n\left( {A' \cap B'} \right) = 5
,then
n\left( {A \cup B} \right),n\left( {A \cap B} \right)
are repsectively
Report Question
0%
45,10
0%
10,45
0%
25,30
0%
15,25
Explanation
Given
n\left(X\right) = 50
,
n\left(A\right) = 35
,
n\left(B\right) = 20
n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = 5
(i)\,n\left(A \cup B\right)
n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = n\left(X\right) - n\left(A\cup\,B\right)
n\left(A\cup\, B\right) = n\left(X\right) - n\left({A}^{\prime}\cap\,{B}^{\prime}\right)
n\left(A\cup\, B\right) = 50 - 5= 45
(ii) n\left(A \cap B\right)
n\left(A\cup B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cap B\right)
n\left(A\cap\, B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cup\, B\right)
n\left(A\cap\, B\right) = 35 + 20 - 45= 10
If
p
and
q
are two proposition, then
\sim(p\leftrightarrow q)
is
Report Question
0%
\sim p \wedge \sim q
0%
\sim p \vee \sim q
0%
(p \wedge \sim q) \vee (\sim p \wedge q)
0%
none of these
If
A
and
B
are two non empty sets then
( A \cup B ) ^ { C } = ?
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0%
A ^ { C } \cup B ^ { C }
0%
A ^ { C } \cap B ^ { C }
0%
A \cup B ^ { C }
0%
A ^ { C } \cap B
Explanation
If A and B are two no empty sets then
{ \left( A\cup B \right) }^{ C }={ A }^{ C }\cap { B }^{ C }
Let
S={1,2,3,.....10}
and
P={1,2,3,4,5}
The number of subsets
'Q'
of
S
such that
p \cup Q=S
, are.....
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0%
128
0%
256
0%
32
0%
64
If 'p' is true and 'q' is false, then which of the following statement is not true?
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0%
p\vee q
0%
p\Rightarrow q
0%
p\wedge (\sim q)
0%
q\Rightarrow p
If the number of
5
elements subsets of the set
A\left\{\ a_{1},a_{2}.....a_{20}\right\}
of
20
distinct elements is
k
times the number of
5
elements subsets containing
a_{4}
, then
k
is
Report Question
0%
5
0%
\dfrac{20}{7}
0%
4
0%
\dfrac{10}{3}
Explanation
No of
5
elements subset
=^{20}C_{5}=\dfrac {20!}{15! \, 15!}=\dfrac {20 \times 19\times 18\times 17\times 16\times 15!}{15! \, 5\times 4\times 3\times 2\times 1}=15504
No of
5
elements subset containing
a_{4}
=^{19}C_{4}=\dfrac {19!}{4! \, 15!}
\Rightarrow \dfrac {20!}{15! \, 15!}=k\dfrac {19!}{4!\, 15!}
\Rightarrow \dfrac {20\times 19!}{5\times 4!}=k\dfrac {19!}{4!}
\Rightarrow k=4
Let
A
and
B
be two sets.
A \cap B = \{1, 2\}, A \cup B = \{1, 2, 3, 4, 5\}
and if
n_1
is the maximum number of function from
A
to
B
and
n_2
is the minimum number of function form
A
to
B
then
n_1 - n_2
equals
Report Question
0%
32
0%
56
0%
64
0%
81
The value of set
(A\cup B\cup C)\cap(A\cap B^1\cap C^1)^1\cap C^1
is equal to
Report Question
0%
B\cap C^1
0%
A\cap C
0%
B\cap C^1
0%
A\cap C^1
If two sets
P
and
Q,n\left(P\right)=5,n\left(Q\right)=4
then
n\left(P\times Q\right)=
Report Question
0%
20
0%
9
0%
25
0%
5/4
Explanation
\begin{aligned} n(P \times Q) &=n(P) \times n(Q) \\ &=5 \times 4 \\ &=20 \end{aligned}
\therefore
option
A
is correct.
The statement that is true among the following is
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0%
p\Longrightarrow q
is equivalent to
p\wedge -q
0%
p\vee q
and
p\wedge q
have the same truth value
0%
The converse of
tanx=0\Longrightarrow x=0
is
x\neq 0\Longrightarrow tanx=0
0%
The contrapositive of
3x+2=8\Longrightarrow x=2
is
x\neq 2\Longrightarrow 3x+2\neq 8
Let
Q
be a non empty subset of
N
and
q
is a
statement as given below :
q:
There exists an even number
a \in Q
Negation of the statement
q
will be :
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0%
There is no even number in the set
Q
0%
Every
a \in Q
is an odd number.
0%
(a)
and
(b)
both
0%
None of these
Explanation
Let
Q
be a non empty subset of
N
and
q.
There exists even number
a\in Q
negation of statement
q.
There does not exist a even number in set Q.
Hence, the answer is there is no even number in the set
Q.
If A= {1, 2, 5} and B= {3, 4, 5, 9}, then
A \bigcup B
is equal to :
Report Question
0%
\{1, 2, 5, 9\}
0%
\{1, 2, 3, 4, 9\}
0%
\{1, 2, 3, 4, 5, 9\}
0%
None of these
Explanation
A=\left\{ 1,2,5 \right\} \quad B=\left\{ 3,4,5,9 \right\}
A\cup B=\left\{ 1,2,3,4,5,9 \right\}
Option C.
In a battle
70
% of the combatants lost one eye,
80
% an ear,
75
% an arm,
85
% a leg,
x
% lost all the four limbs the minimum value of
x
is
Report Question
0%
10
0%
12
0%
15
0%
5
Explanation
70\%
of the combatants lost one eye,
80\%
an ear,
75\%
an arm and
85\%
a leg.
Now,
\Rightarrow
The combatants who lost one eye and one ear
=(70+80-100)\%
=50\%
\Rightarrow
The combatants who lost one eye, one ear and one erm
=(50+75-100)\%
=25\%
\Rightarrow
The combatants who lost one eye, one ear one arm and one leg
=(25-85-100)\%
=10\%
\therefore
Combatant who lost all the four limbs
=10\%
\therefore
x=10\%
Two sets A and B are defined as follows
A=\left\{ \left( x,y \right) :y={ e }^{ 2x },x\in R \right\}
and
B=\left\{ \left( x,y \right) :y={ x }^{ 2 },x\in R \right\}
, then
Report Question
0%
A\subset B
0%
B\subset A
0%
A\bigcup B
0%
A\cap B=\phi
Explanation
\begin{array}{l} A=\left\{ { \left( { x,\, y } \right) \, :\, y={ e^{ 2x } },\, x\in R } \right\} \, and \\ B=\left\{ { \left( { x,\, y } \right) \, :\, y={ x^{ 2 } },\, x\in R } \right\} \\ then, \\ A=\left( { { e^{ \circ } },\, { e^{ 1 } },\, { e^{ 2 } },....,\, { e^{ \infty } } } \right) \\ =\left( { 1,\, { e^{ 1 } },\, { e^{ 2 } },....... } \right) \\ and, \\ B=\left( { 0,\, 1,\, 4,\, ..... } \right) \\ \therefore A\subset B \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}
If
A=\left\{1,2,3,4\right\}; B=\left\{2,4,6,8\right\}; C=\left\{3,4,5,8\right\}
then
A\cap B\cap C=
Report Question
0%
\phi
0%
{4}
0%
\mu
0%
{2,4}
Explanation
\begin{matrix} A=\left\{ { 1,2,3,4 } \right\} \\ B=\left\{ { 2,4,6,8 } \right\} \\ C=\left\{ { 3,4,5,8 } \right\} \\ A\cap B\cap C=\left\{ { 4 } \right\} . \\ \end{matrix}
Examine whether the following statements are true or false:
\left\{b, c\right\}\subset \left\{a, \left\{b, c\right\}\right\}
Report Question
0%
True
0%
False
In a group of 800 people, 550 can speak Hindi and 450 can speak English. How many can speak both Hindi and English ?
Report Question
0%
100
0%
150
0%
200
0%
None of these
Explanation
Let H denote the set of people speaking Hindi and E denote the set of people speaking English. We are given :
n(H)=550, n(E)=450, n(H\cup E)=800
Now,
n(H\cup E)=n(H)+n(E)-n(H\cap E)
n(H \cap E)=550+450-800=200
Hence, 200 persons can speak both Hindi and English.
Let A,B are two sets such that n(A)=4 and n(B)=Then the least possible number of elements in the power set of
(A\cup B)
is
Report Question
0%
16
0%
64
0%
256
0%
1024
Explanation
Given,
n(A)=4,n(B)=6
Then the least number of possible elements in
n(A\cup B)=2^{n(A)}.2^{n(B)}=2^4.2^6=2^{10}=1024
Left A = {1, 2, 3, 4, 5, 6} and B= {1, 2, 3, 4} be two sets, then the number of functions that can be defined from A to B such that the element '2' in B has exactly 3 pre - images i A, is equal ot
Report Question
0%
540
0%
440
0%
810
0%
620
if universal set
\sum { =\left\{ a,b,c,d,e,f,g,h, \right\} A=\left\{ b,c,d,e,f \right\} } and\quad C=\left\{ c,d,e,f,g \right\} ,\quad then\quad find\quad B-A
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0%
\left\{ b,c,e,f \right\}
0%
\left\{ a,b,f,h \right\}
0%
\left\{ a,g,h \right\}
0%
\left\{ a,c,e,g \right\}
Explanation
Given,
B=
{
a,b,c,d,e,f,g,h
}
A=
{
b,c,d,e,f
}
,
C=
{
c,d,e,f,g
}
\therefore B-A=
{
a,b,c,d,e,f,g,h
}-{
b,c,d,e,f
}
\implies B-A=
{
a,g,h
}
If P(S) denotes the set of all subsets of a given set S, then the number of one to one function from the set s={1,2,3} to the set of P(S) is
Report Question
0%
336
0%
8
0%
36
0%
320
Explanation
\begin{array}{l} Given \\ P\left( s \right) =\left\{ { 1,\, 2,\, 3 } \right\} \\ P\left( s \right) =all\, subset\, of\, 5 \\ =\left\{ { \phi ,\left\{ 1 \right\} ,\left\{ 3 \right\} ,\left\{ { 1,2 } \right\} ,\left\{ { 2,3 } \right\} ,\left\{ { 3,1 } \right\} ,\left\{ { 1,2,3 } \right\} } \right\} \\ No.\, of\, element\, of\, 5=0\left( 5 \right) =3 \\ similarly \\ O\left( { p\left( s \right) } \right) =8 \\ Total\, no.\, of\, one-one\, function \\ =\frac { { n! } }{ { \left( { n-m } \right) ! } } \, ,\, where\, n=o\left( { p\left( s \right) } \right) \, and\, m=o\left( 5 \right) \\ =\frac { { 8! } }{ { \left( { 8-3 } \right) ! } } =\frac { { 8! } }{ { 5! } } \\ =336\, \, \\ No.\, of\, one-one\, function\, is\, 336 \\ Hence,\, \, option\, \, A\, \, is\, correct, \end{array}
The function
f(x)
satisfies the condition
(x-2)f(x)+2f\left(\dfrac{1}{x}\right)=2
for all
x\neq 0
. Then the value of
f(2)
is
Report Question
0%
\dfrac{1}{2}
0%
1
0%
\dfrac{7}{4}
0%
\dfrac{-3}{2}
Explanation
Putting
x=2
in
\left( x-2 \right) f\left( x \right) +2f\left( \dfrac { 1 }{ x } \right) =2
\left( 2-2 \right) f\left( 2 \right) +2f\left( \dfrac { 1 }{ 2 } \right) =2
2f\left( \dfrac { 1 }{ 2 } \right) =2
f\left( \dfrac { 1 }{ 2 } \right) =1
If
A=\left\{1, 2, 3, 4\right\}
, then the number of subsets of
A
that contain the element
2
but not
3
, is
Report Question
0%
16
0%
4
0%
8
0%
24
Explanation
The subsets are be
\left\{1, 2, 4\right\},\left\{1, 2\right\}, \left\{2, 4\right\}, \left\{2\right\}
Number of subsets of
A
that contain the element
2
but not
3
is
4
Let
p
and
q
be two statements. amongst the following, the statement is equivalent to
p\rightarrow q
is
Report Question
0%
\displaystyle p\wedge \sim q
0%
\displaystyle \sim p\vee q
0%
\displaystyle \sim p\wedge q
0%
\displaystyle p\vee \sim q
Explanation
\begin{array}{l} \text { Given, } \quad P \longrightarrow q \\ \text { ne trinow that, } \rightarrow \text { - if then } \\ \Rightarrow \text { if } P \text { then } q \end{array}
\begin{array}{lll} p & q & p \rightarrow q \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}
\begin{array}{c} \Rightarrow \text { Evaluating option } A(p \wedge \sim q) \\ P \text { iq } p \wedge \sim q \\ 111 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \text { Incorrect assumption } \end{array}
\begin{array}{c} \Rightarrow \text { Eualuating } B(\sim p \wedge q) \\ \sim p \quad q \quad \sim p \wedge q \\ 0 \\ 0 \\ 1 \\ 1 \quad 0 \\ \text { Incorrect assumption } \end{array}
\begin{array}{rlrl} \Rightarrow \text { Evaluating } & c(\sim p \vee q) \\ & \sim p & q & \sim p v q \\ & 0 & 1 & & \\ 0 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ & \text { correct assumption } \end{array}
Let
\displaystyle S=\left\{a\in N,a\le100\right\}
if the equation
\displaystyle[{\tan}^{2}x]-\tan x-a=0
has real roots, then the number of elements
S
is (when
[.]
is greatest integer function)
Report Question
0%
10
0%
1
0%
9
0%
0
Explanation
Given equation is,
[tan^2x]-tanx-a=0
\therefore D=b^2-4ac=(-1)^2-4(1)(-a)
=1+4a
So D must be a odd perfect square
\implies \sqrt{1+4a}=2\lambda+1
\implies 1+4a=4\lambda^2+1+4\lambda
\implies a=\lambda(\lambda+1)
For different values og=f
\lambda
we get
a=0,2,6,12,20,30,42,56,72,90
Hence there are
10
values of
a
.
If
A=\left\{ 1,3,5,7,9,11,13,15,17 \right\} ,B=\left\{ 2,4,....,18 \right\}
and N is the universal set, then
A'\cup \left( \left( A\cup B \right) \cap B' \right)
Report Question
0%
A
0%
N
0%
B
0%
R
Explanation
A=\left\{1,3,5,7,9,11,13,15,17\right\}
{A}^{\prime}=N-A=N-\left\{1,3,5,7,9,11,13,15,17\right\}=\left\{2,4,6,8,10,12,14,16,18\right\}
A\cup B=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}
{B}^{\prime}=N-B=N-\left\{2,4,6,8,10,12,14,16,18\right\}=\left\{1,3,5,7,9,11,13,15,17\right\}
\left(A\cup B\right)\cap {B}^{\prime}
=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}\cap\left\{1,3,5,7,9,11,13,15,17\right\}
=\left\{1,3,5,7,9,11,13,15,17\right\}
{A}^{\prime}\cup \left(\left(A\cup B\right)\cap {B}^{\prime}\right)
=\left\{2,4,6,8,10,12,14,16,18\right\}\cup\left\{1,3,5,7,9,11,13,15,17\right\}
=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\right\}=N
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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