MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 13

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?

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.

Suppose

${ A }_{ 1 },{ A }_{ 2 },....{ A }_{ 30 },$ are thirty sets each with five elements and

${ B }_{ 1 },{ B }_{ 2 },....B_{ A },$ are n sets each with three elements such

$\overset { 30 }{ \underset { i-1 }{ U } } \quad { A }_{ 1 }=\overset { n }{ \underset { j-1 }{ U } } \quad =s$ If each element of belongs to exactly ten of the

${ A }_{ 1 }$ exactly 9 of the

${ A }_{ 1 }$ ,then value of n is:

0%
15
0%
135
0%
45
0%
90

The value of

$\left( {A \cup B \cup C} \right) \cap \left( {A \cap {B^c} \cap {C^c}} \right) \cap {C^c}$ is

0%
$B \cap {C^c}$
0%
${B^c} \cap {C^c}$
0%
$B \cap {C}$
0%
$A \cap {B^c} \cap {C^c}$

If A and B are any two sets, then

$A\bigcup (A\bigcap B)$ is equal to

0%
$A$
0%
$B$
0%
$A'$
0%
$B'$

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

$n((A \cup B)^C)=$

0%
$17$
0%
$9$
0%
$11$
0%
$3$

Let x= {1, 2, 3, 4, 5} The number of different ordered pairs (y, z) that can be formed such ordered pairs (y , z) that can be formed such that

$y\sqsubseteq x,z\sqsubseteq x$ and

$y\cap z$ is empty is

0%
${ 2 }^{ 5 }$
0%
$5^{ 3 }$
0%
$5^{ 2 }$
0%
$3^{ 5 }$

If n(A)=115, n(B)=326, n(A-B)=47, then

$n(A\cup B)$ is equal to

0%
373
0%
165
0%
370
0%
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.

Let S = {x

$\epsilon$

$R$ : x 0 and 2

$\left | \sqrt{x} 3 \right |$ +

$\sqrt{x}$ (

$\sqrt{x}$ 6) + 6 = 0} .
Then S :

0%
Contains exactly four elements
0%
Is an empty set
0%
Contains exactly one elements
0%
Contains exactly two elements

Explanation

Given,

$S=\{x \in R: x \geqslant 0$ , and

$2|\sqrt{x}-3|+\sqrt{x}|\sqrt{x}-b|+b=0$

Let

$\sqrt{x}=t$

Then,

$2|t-3|+t|t-6|+6=0$

Now,

taking Case I, where t<3

$-2 t+6+t^{2} 4 b+6=0$

$t^{2}-8 t+12=0$

$t=2,6$

$\because$ we have taken, t

$<3$

$\begin{aligned} \therefore \quad t &=2, \\ \sqrt{x} &=2 \\ x &=4 \end{aligned}$

Case II where

$t \geqslant 3$

$2 t-6+t^{2}-6 t+6=0$

$t^{2}-4 t=0$

$t=4,0$

$\because$ We have taken,

$t \geqslant 3$

$\begin{aligned} \therefore t &=4 \\ \sqrt{x} &=4 \\ x &=16 \\ \therefore x &=40 or 16 \end{aligned}$

Option D exactly two elements.

Let N= {1, 2, 3, ........., 10}, then sum of the product of number taken two at a time from the set N is

0%
$55\times 24$
0%
$45\times 24$
0%
$512\times 45$
0%
$55\times 12$

Let

$A$ be a set containing

$32$ elements. Let

$x_k$ denotes the number of '

$k$ ' element sub-sets of

$A$ . Then

$\displaystyle \sum^{8}_{i = 0} (x_{4i}) =$

0%
$2^{30} + 2^8$
0%
$2^{29} + 2^8$
0%
$2^{15}(2^{15} + 1)$
0%
$2^{14}(2^{16} + 1)$

The set

$(A\cap B')'\cup (B\cap C)$ is equal to?

0%
$A'\cup B\cup C$
0%
$A'\cup B$
0%
$A'\cup C'$
0%
$A'\cap B$

Explanation

$\left( A\cap B' \right) '\cup \left( B\cap C \right)$

$=\left( A\cap \bar B \right)\cup \left( B\cap C \right)$

$=\left( \bar A\cap B\right)\cup \left( B\cap C \right)$

$=\bar A\cup B\cup C$

Hence, the answer is

$A'\cup B\cup C.$

Tell whether set A is a subset of set B.

0%
set A : whole numbers less than 8
set B : whole numbers less than 10
0%
set A : prime numbers
set B : odd numbers
0%
set A : numbers divisible by 6
set B : numbers divisible by 3
0%
set A : set of letters in the word 'FLAT'
set B : set of letters in the word 'PLATE'

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

0%
S=0
0%
$S=ax+91-1){ x }^{ 2 }\forall a\varepsilon (0,\infty )$
0%
$S=ax+(1-a){ x }^{ 2 }\forall a\varepsilon R$
0%
$S=ax+(1-a){ x }^{ 2 }\forall a\varepsilon R(0,\quad 2)$

The solution set of

$x^{2}+5x+6=0$ is ........

0%
{2,3}
0%
{-2,-3}
0%
{2,-3}
0%
{-2,3}

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

0%
$26$
0%
$27$
0%
$28$
0%
Any of the three values $26, 27, 28$ is possible

$P = \{ x:x < 3,x \in N\}$ 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

0%
${(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)}$
0%
$(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)$
0%
$(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)$
0%
$(1,1),(1,2),(2,1),(2,2)$

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

0%
A $\cup$ B $\cup$ C
0%
A $\cup$ (B $\cap$ C)
0%
A $\cap$ B $\cap$ C
0%
A $\cap$ (B $\cup$ C)

$If\,\,A = \left\{ {x:{x^2} = 1} \right\}\,\,\,and\,B = \left\{ {x:{x^4} = 1} \right\},\,then\,A\Delta B\,\,\,is\,\,equal\,to\,$

0%
$\left\{ {i, - i} \right\}\,$
0%
$\,\left\{ { - 1,1} \right\}\,$
0%
$\left\{ { - 1,1,i - i} \right\}$
0%
$\,\left\{ {1,i} \right\}$

Let

$A$ and

$B$ be two sets. The

$(A\cup B)'\cup(A'\cap B)$ =

0%
$A'$
0%
$A$
0%
$B'$
0%
$none\ of\ these$

Examine whether the following statements are true or false:

$\left\{a\right\}\subset \left\{\{a\},\, b \right\}$

0%
True
0%
False

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 13 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 13 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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