MCQ Questions for Class 11 Engineering Maths Sets Quiz 8

If A and B are finite sets and

$A \subset B$ , then

0%
$n(A \cap B) = \phi$
0%
$n(A \cup B) = n(B)$
0%
$n(A \cap B) = n (B)$
0%
$n(A \cup B) = n (A)$

Explanation

A and B are finite set and

$A\subset B$

$A\subset B$ means A is a subset of B
If A is subset of B, then all elements of A are present in B

$\therefore n\left( A\bigcup { B } \right) =n\left( B \right)$

The set

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

0%
$B\cap C'$
0%
$A\cap C$
0%
$B' \cap C'$
0%
None of these

Explanation

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

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

$= [(A\cap A')\cup (B\cup C)]\cap C'$

$= (\phi \cup B\cup C)\cap C' = (B\cup C)\cap C'$

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

$= (B\cap C') \cup \phi = B\cap C'$ .

$A=\begin{Bmatrix}x\in R : x^2+6x-7<0\end{Bmatrix}$ and

$B=\begin{Bmatrix}x\in R : x^2+9x+14<0\end{Bmatrix}$ , then which of the following is/are correct?

$(A\cap B)=(-2, 1)$

$(A \setminus B)=(-7, -2)$
Select the correct answer using the code given below :

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

Solution:

We have,

$A=\{x\in R:x^2+6x-7<0\}$ and

$B=\{x\in R:x^2+9x+14<0\}$

$\therefore A=\{x\in R:(x-1)(x+7)<0\}$ and

$B=\{x\in R:(x+2)(x+7)<0\}$

$A=(-7,1)$ and

$B=(-7,-2)$

$(A\cap B)=(-7,-2)$

$(A\setminus B)=[-2,1)$

Hence, D is the correct option.

Let X be a set of

$5$ elements. The number d of ordered pairs (A, B) of subsets of X such that

$A\neq \Phi, B\neq \Phi, A\cap B=\Phi$ satisfies.

0%
$50\leq d \leq 100$
0%
$101\leq d \leq 150$
0%
$151 \leq d \leq 200$
0%
$201 \leq d$

Explanation

Total no. of ordered pairs is

${ =C }_{ 2 }^{ 5 }\times 2!+{ C }_{ 3 }^{ 5 }\left( \frac { 3! }{ 1!\times 2! } \times 2! \right) +{ C }_{ 4 }^{ 5 }\left( \frac { 4! }{ 1!\times 3! } \times 2!+\frac { 4! }{ 2!\times 2! } \times \frac { 2! }{ 2! } \right) +{ C }_{ 5 }^{ 5 }\left( \frac { 5! }{ 1!\times 4! } \times 2!+\frac { 5! }{ 2!\times 3! } \times 2! \right) \\ =10(2)+10(6)+5(8+6)(10+20)\\ =180\\$

So option

$C$ is correct

$A = \left \{(x, y) : x^{2} + y^{2} \leq 1; x, y \epsilon R\right \}$ and

$B = \left \{(x, y) : x^{2} + y^{2} \geq 4; x, y \epsilon R\right \}$ , then

0%
$A - B = \phi$
0%
$B - A = \phi$
0%
$A\cap B \neq \phi$
0%
$A\cap B = \phi$

Explanation

$A$ is the set of all points on the inner circle

$x^{2} + y^{2} = 1. B$ is the set of all points on the outer circle

$x^{2} + y^{2} = 4$ .

$\therefore A - B = A, B - A = B, A\cap B = \phi$ .

Consider the following:

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

$A\cap \left( B\cup C \right) =\left( A\cup B \right) \cap \left( A\cup C \right)$
Which of the above is/are correct?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

Figure

$1$ represents

$A\cup (B\cap C)$ and figure

$2$ represents

$(A\cap B)\cup (A\cap C)$ . We ca see both of them are not equal.

Figure

$3$ represents

$A\cap (B\cup C)$ and figure

$4$ represents

$(A\cup B)\cap (A\cup C)$ . We can see both of them are not equal.

So, both the equations are incorrect.

What is the percentage of persons who read only two papers ?

0%
$19\%$
0%
$31\%$
0%
$44\%$
0%
None of the above

Explanation

Let total numbers of people be 100, So for newspapers

$n(A)=42, n(B)=51, n(C)=68, n(A\cap B)=30$

$n(B\cap C)=28$

$n(A\cap C)=36$

Also given,

$n(\overline { A\cup B\cup C) } =8$

$n(A\cup B\cup C)=100-n(\overline { A\cup B\cup C) } =100-8=92$

Also,

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$

$92=42+51+68-30-28-36+n(A\cap B\cap C)$

$n(A\cap B\cap C)=25$

Now, for people who read only two papers

$=n(A\cap B)+n(B\cap C)+n(A\cap C)-3n(A\cap B\cap C)$

$=30+28+36-75$

$=19$

Thus,

$19\%$ people read only two newspapers.

Hence, A is correct.

A market research group conducted a survey of

$1000$ consumers and reported that

$720$ consumers like product A and

$420$ consumers like product B. Then, the least number of consumers that must have liked both the products is.

0%
$140$
0%
$180$
0%
$210$
0%
$190$

Explanation

Total consumers

$=1000$
Like product

$A=n(A)=720$
Like product

$B=n(B)420$

$n(A\cap { B })$ (Both the products)

$=n(A)+n(B)-n(A\cup { B })$

$=720+420-1000$

$=140$

In a group of

$50$ people, two tests were conducted, one for diabetes and one for blood pressure.

$30$ people were diagnosed with diabetes and

$40$ people were diagnosed with high blood pressure. What is the minimum number of people who were having diabetes and high blood pressure?

0%
$0$
0%
$10$
0%
$20$
0%
$30$

Explanation

Let

$A$ be the set containing people diagnosed with diabetes and

$B$ be the set containing people diagnosed with Blood pressure.

We know set formula

$n(A\cup B)+n(A\cap B)=n(A)+n(B)$

$n(A\cap B)=x$ are people having both

$n(A\cup B)$ are total people

$n(A)$ is people having diabetes and

$n(B)$ is having blood pressure.

$\Rightarrow 50+x=40+30\\ x=20$

Thus, there should be minimum of 20 people having both.

In a city, three daily newspapers A, B, C are published,

$42\%$ read A;

$51\%$ read B;

$68\%$ read C;

$30\%$ read A and B;

$28\%$ read B and C;

$36\%$ read A and C;

$8\%$ do not read any of the three newspapers.

What is the percentage of persons who read only one paper ?

0%
$38\%$
0%
$48\%$
0%
$51\%$
0%
None of the above.

Explanation

Let total numbers of people be 100, So for newspapers

$n(A)=42, n(B)=51, n(C)=68, n(A\cap B)=30$

$n(B\cap C)=28$

$n(A\cap C)=36$

Also given,

$n(\overline { A\cup B\cup C) } =8$

$n(A\cup B\cup C)=100-n(\overline { A\cup B\cup C) } =100-8=92$

Also,

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$

$92=42+51+68-30-28-36+n(A\cap B\cap C)$

$n(A\cap B\cap C)=25$

Now,

People who read only one paper

$=n(A)+n(B)+n(C)-2n(A\cap B)-2n(B\cap C)-2n(A\cap C)+3n(A\cap B\cap C)$

$=42+51+68-60-56-72+75$

$=48$

Hence, B is correct.

$a\mathbb{N}=(an:n\in \mathbb{N})$ and

$b\mathbb{N}\cap c\mathbb{N}=d\mathbb{N}$ , where

$a,b,c\in \mathbb{N}$ and

$b,c$ are coprime, then

0%
$b=cd$
0%
$c=bd$
0%
$d=bc$
0%
None of these

Explanation

Given :

$a\mathbb{N}=(an:n\in \mathbb{N})$ and

$b\mathbb{N}\cap c\mathbb{N}=d\mathbb{N}$ , where

$a,b,c\in \mathbb{N}$ and

$b,c$ are coprime

We have,

$b\mathbb{N}=(bn:n\in \mathbb{N})$ ,

$c\mathbb{N}=(cn:n\in \mathbb{N})$

and

$d\mathbb{N}=(dn:n\in \mathbb{N})$

Also,

$bc\in b\mathbb{N} \cap c\mathbb{N}=d\mathbb{N}$

$\therefore$

$bc=d$ .....

$[\because b,c$ are coprime

$]$ .

A college awarded

$38$ medals in football,

$15$ in basketball and

$20$ in cricket. If these medals went to a total of

$58$ men and only three men got medals in all the three sports. Then the number of students who received medals in exactly two of the three sports, is

0%
$18$
0%
$15$
0%
$9$
0%
$6$

Explanation

$\textbf{Step – 1: Formulating the information to use Inclusion-Exclusion Rule.}$

$\text{Let number of men who have got medals in football, basketball, and cricket be represented as}$

$n(F),n(B)$ and

$n(C)$ .

$\text{Therefore the number of men who won in all the three sports will be}$ $n(F \cap B \cap C)$

$\text{and the number of people who got medals in two of the three sports is}$

$n(F \cap B) + n(B \cap C) + n(C \cap F)$

$\text{and the total number of men is represented as}$ $n(F \cup B \cup C)$ .

$\text{From the given data we have:}$

$n(F) = 38$

$n(B) = 15$

$n(C) = 20$

$n(F \cup B \cup C) = 58$

$n(F \cap B \cap C) = 3$

$\textbf{Step – 2: Apply the Inclusion-Exclusion Rule}$ .

$\text{Using the Inclusion-Exclusion Rule we get:}$

$n(F \cup B \cup C) = n(F) + n(B) + n(C) - n(F \cap B) - n(B \cap C) - n(C \cap F) + n(F \cap B \cap C)$

$\Rightarrow 58 = 38 + 15 + 20 - n(F \cap B) - n(B \cap C) - n(C \cap F) + 3$

$\Rightarrow n(F \cap B) + n(B \cap C) + n(C \cap F) = 38 + 15 + 20 + 3 - 58$

$\Rightarrow n(F \cap B) + n(B \cap C) + n(C \cap F) = 18$

$\therefore\text{ the number of students who have received medals in exactly two of the three sports is}$

${= n(F \cap B) + n(B \cap C) + n(C \cap F) - 3 \times n(F \cap B \cap C) }$

${= 18 - 3 \times 3 }$

${= 18 - 9}$

${= 9}$

$\textbf{Hence, the number of students who have received medals in exactly two of the three sports is 9}$

$S$ is a set with

$10$ elements and

$A = \left \{(x, y) : x, y\epsilon S, x\neq y\right \}$ , then number of elements in

$A$ is

0%
$100$
0%
$90$
0%
$50$
0%
$45$

Explanation

Given number of elements in

$S = 10$
and

$A = \left \{(x, y) : x, y\epsilon S, x\neq y\right \}$

$\therefore$ Number of elements in

$A = 10\times 9 = 90$ .

$A=\left\{ a,b,c \right\} ,$

$B=\left\{ b,c,d \right\}$ and

$C=\left\{ a,d,c \right\}$ , then

$\left( A-B \right) \times \left( B\cap C \right)$ is equal to

0%
$\left\{ \left( a,c \right) ,\left( a,d \right) \right\}$
0%
$\left\{ \left( a,b \right) ,\left( c,d \right) \right\}$
0%
$\left\{ \left( c,a \right) ,\left( d,a \right) \right\}$
0%
$\left\{ \left( a,c \right) ,\left( a,d \right) ,\left( b,d \right) \right\}$

Explanation

Given that;

$A=\left\{ a,b,c \right\} ,$

$B=\left\{ b,c,d \right\}$ and

$C=\left\{ a,d,c \right\}$

Now,

$A-B=\left\{ a,b,c \right\} -\left\{ b,c,d \right\} =\left\{ a \right\}$
and

$B\cap C=\left\{ b,c,d \right\} \cap \left\{ a,d,c \right\}$

$=\left\{ c,d \right\}$

Therefore,

$\left( A-B \right) \times \left( B\cap C \right) =\left\{ a \right\} \times \left\{ c,d \right\}$

$=\left\{ \left( a,c \right) ,\left( a,d \right) \right\}$

$A/B=\left\{ a,b \right\} ,B\setminus A=\left\{ c,d \right\}$ and

$A\cap B=\left\{ e,f \right\}$ then the set

$B$ is equal to

0%
$\left\{ a,b,c,d \right\}$
0%
$\left\{ e,f,c,d \right\}$
0%
$\left\{ a,b,e,f \right\}$
0%
$\left\{ c,d,a,e \right\}$
0%
$\left\{ d,e,a,b \right\}$

Explanation

Given,

$A/B=\left\{ a,b \right\} ,B\setminus A=\left\{ c,d \right\}$

and

$A\cap B=\left\{ e,f \right\}$

$\therefore A-B=\left\{ a,b \right\} ...(i)\quad$

$B-A=\left\{ c,d \right\} ...(ii)$

$A\cap B=\left\{ e,f \right\} ...(iii)$
From Eqs. (i),(ii),(iii) we get

$B=\left\{ c,d,e,f \right\}$

Let

$A$ and

$B$ be two events such that

$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } ,$

$P\left( A\cap B \right) =\dfrac { 1 }{ 4 }$ and

$P\left( \overline { A } \right) =\dfrac { 1 }{ 4 }$ , where,

$\overline { A }$ stands for complement of event

$A$ . Then, event

$A$ and

$B$ are

0%
Mutually exclusive and independent
0%
Independent but not equally likely
0%
Equally likely but not independent
0%
Equally likely and mutually exclusive

Explanation

Given that,

$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } ,$

$P\left( A\cap B \right) =\dfrac { 1 }{ 4 } ,$

$P\left( \overline { A } \right) =\dfrac { 1 }{ 4 }$

$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 }$

$\Rightarrow 1-P\left( A\cup B \right) =\dfrac { 1 }{ 6 }$

$\Rightarrow \left[ 1-P\left( A \right) \right] -P\left( B \right) +P\left( A\cap B \right) =\dfrac { 1 }{ 6 }$

$\Rightarrow P\left( \overline { A } \right) -P\left( B \right) +P\left( A\cap B \right) =\dfrac { 1 }{ 6 }$

$\Rightarrow \dfrac { 1 }{ 4 } -P\left( B \right) +\dfrac { 1 }{ 4 } =\dfrac { 1 }{ 6 }$

$\Rightarrow P\left( B \right) =\dfrac { 2 }{ 4 } -\dfrac { 1 }{ 6 }$

$\Rightarrow P\left( B \right) =\dfrac { 1 }{ 3 }$
and

$P\left( A \right) =1-P\left( \overline { A } \right) =1-\dfrac { 1 }{ 4 } =\dfrac { 3 }{ 4 }$

Since,

$P\left( A\cap B \right) =P\left( A \right) \cdot P\left( B \right)$ , so the events

$A$ and

$B$ are independent events but not equal likely.

Let

$X$ and

$Y$ be two non-empty sets such that

$X\cap A=Y\cap A=\phi$ and

$X\cup A=Y\cup A$ for some non-empty set

$A$ . Then which of the following is true?

0%
$X$ is a proper subset of $Y$
0%
$Y$ is a proper subset of $X$
0%
$X=Y$
0%
$X$ and $Y$ are disjoint sets
0%
${ X }/{ A }=\phi$

Explanation

We have,

$X\cup A=Y\cup A$

$\Rightarrow X\cap \left( X\cup A \right) =X\cap \left( Y\cup A \right)$

$\Rightarrow X=\left( X\cap Y \right) \cup \left( X\cap A \right)$

$\left[ \because X\cap \left( X\cup A \right) =X \right]$

$\Rightarrow X=\left( X\cap Y \right) \cup \phi$

$\left[\because X\cap A=\phi \right]$

$\Rightarrow X=X\cap Y$ .....(i)
Again,

$X\cup A=Y\cup A$

$\Rightarrow Y\cap \left( X\cup A \right) =Y\cap \left( Y\cup A \right)$

$\Rightarrow \left( Y\cap X \right) \cup \left( Y\cap A \right) =Y$

$\Rightarrow \left( Y\cap X \right) \cup \phi =Y$

$\Rightarrow Y\cap X=Y$

$\Rightarrow X\cap Y=Y$ ....(ii)
From equations (i) and (ii), we get

$X=Y$

$X = \left \{1, 2, 3, ..., 10\right \}$ and

$A = \left \{1, 2, 3, 4, 5\right \}$ . Then, the number of subsets

$B$ of

$X$ such that

$A - B = \left \{4\right \}$ is

0%
$2^{5}$
0%
$2^{4}$
0%
$2^{5} - 1$
0%
$1$
0%
$2^{4} - 1$

Explanation

Given,

$X = \left \{1, 2, 3, ...., 10\right \}$
and

$A = \left \{1, 2, 3, 4, 5\right \}$
Now,

$A - B = \left \{4\right \}$

$\Rightarrow B = A - \left \{4\right \} = \left \{1, 2, 3, 4, 5\right \} - \left \{4\right \}$

$= \left \{1, 2, 3, 5\right \}$

$\therefore X - \left \{4\right \} - B$

$= \left \{1, 2, 3, ..., 10\right \} - \left \{4\right \} - \left \{1, 2, 3, 5\right \}$

$= \left \{6, 7, 8, 9, 10\right \}$
Therefore,number of subsets

$B$ of

$X$ i.e.,

$\left \{6, 7, 8, 9, 10\right \}$

$= 2^{5}$ .

If the sets

$A$ and

$B$ are as follows:

$A=\left\{ 1,2,3,4 \right\} ,B=\left\{ 3,4,5,6 \right\}$ , then

0%
$A-B=\left\{ 1,2 \right\}$
0%
$B-A=\left\{ 5,6 \right\}$
0%
$\left[ \left( A-B \right) -\left( B-A \right) \right] \cap A=\left\{ 1,2 \right\}$
0%
$\left[ \left( A-B \right) -\left( B-A \right) \right] \cup A=\left\{ 3,4 \right\}$

Explanation

Given,

$A=\left\{ 1,2,3,4 \right\} ,B=\left\{ 3,4,5,6 \right\}$

$\therefore A-B=\left\{ 1,2 \right\} ;\quad B-A=\left\{ 5,6 \right\} \quad$

$\left[ (A-B)-(B-A) \right] \cap A=\left[ \left\{ 1,2 \right\} -\left\{ 5,6 \right\} \right] \cap \left\{ 12,3,4, \right\} =\left\{ 1,2 \right\}$

$n(A)=10,n(B)=6$ and

$(C)=5$ for three disjoint sets

$A,B,C$ then

$n(A\cup B\cup C)$ equals

0%
$21$
0%
$11$
0%
$1$
0%
$9$

Explanation

Since,

$A,B,C$ are disjoint sets

$\therefore n(A\cup B\cup C)=n(A)+n(B)+n(C)$

$=10+6+5=21$

The total number of subsets of {1, 2, 6, 7} are?

0%
$16$
0%
$8$
0%
$64$
0%
$32$

Explanation

We have to find the total number of subsets of

$\{1,2,6,7\}$ .

We know that, for a set containing

$n$ elements, the total number of subsets is

$2^n$ .

Consider

$\{1,2,6,7\}$ , wich has

$4$ elements.

$\therefore$ here

$n=4$

Hence total number of subsets is

$2^4=16$ .

Thus the total number of subsets of

$\{1,2,6,7\}$ is

$16$ .

If U = {1, 3, 5, 7, 9, 11, 13}, then which of the following are subsets of U.
B = {2, 4}
A = {0}
C = {1, 9, 5, 13}
D = {5, 11, 1}
E = {13, 7, 9, 11, 5, 3, 1}
F = {2, 3, 4, 5}

0%
$A$ and $B$
0%
$C$ , $D$ and $E$
0%
$A$ and $D$
0%
only $A$

Explanation

Given

$U=\{1,3,5,7,9,11,13\}, A=\{0\}, B=\{2,4\}, C=\{1,9,5,13\}, \\ D=\{5,11,1\}, E=\{13,7,9,11,5,3,1\}, F=\{2,3,4,5\}$

Now we have to find the subsets of

$U$

Consider

$A=\{0\}$
Since

$0\notin U, A\nsubseteq U$

Consider

$B=\{2,4\}$
Since

$2,4\notin U, B\nsubseteq U$

Consider

$C=\{1,9,5,13\}$
Since

$1,9,5,13\in U, C\subset U$

Consider

$D=\{5,11,1\}$
Since

$5,11,1\in U, D\subset U$

Consider

$E=\{13,7,9,11,5,3,1\}$
Since

$13,7,9,11,5,3,1\in U, E\subseteq U$

Consider

$F=\{2,3,4,5\}$
Since

$2,3,4\notin U, F\nsubseteq U$

Therefore

$C,D,E$ are subsets of

$U$ .

Let

$S = \left \{(a, b): a, b\epsilon Z, 0\leq a, b\leq 18\right \}$ . The number of elements

$(x, y)$ in

$S$ such that

$3x + 4y + 5$ is divided by

$19$ is

0%
$38$
0%
$19$
0%
$18$
0%
$1$

Explanation

Maximum value of

$3x+4y+5$ will occur at

$(18,18)$

$\Rightarrow$ Maximum value

$=3\times18+4\times18+5=131$

And, the minimum value will occur at

$0,0$

$\Rightarrow$ Minimum value

$=3\times0+4\times0+5=5$

Now, the multiples of

$19$ that lie between minimum and maximum value of

$3x+4y+5$ are

$19,38,57,76,95,114$

$\mathbf{Case 1-}$ When

$3x+4y+5=19$

By trial and error the solutions for this case is-

$(2,2)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case is

$1$

$\mathbf{Case 2-}$ When

$3x+4y+5=38$

By trial and error the solutions for this case are-

$(11,0);(7,3);(3,6)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$3$

$\mathbf{Case 3-}$ When

$3x+4y+5=57$

By trial and error the solutions for this case are-

$(16,1);(12,4);(8,7);(4,10);(0,13)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$5$

$\mathbf{Case 4-}$ When

$3x+4y+5=76$

By trial and error the solutions for this case are-

$(17,5);(13,8);(9,11);(5,14);(1,17)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$5$

$\mathbf{Case 5-}$ When

$3x+4y+5=95$

By trial and error the solutions for this case are-

$(18,9);(14,12);(10,15);(6,18)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$4$

$\mathbf{Case 6-}$ When

$3x+4y+5=114$

By trial and error the solutions for this case are-

$(15,16)$

$\Rightarrow$ The number of elements

$(x,y)$ in

$S$ for this case are

$1$

Therefore, the total number of elements

$(x,y)$ in

$S$ which are divisible by 19

$=1+3+5+5+4+1=19$

Hence the correct answer is option

$B$

State whether true or false.
Quadrilateral

$\subseteq$ polygon

0%
True
0%
False

Explanation

We have to state whether the statement "

$Quadrilateral \subseteq polygon$ " is true or false.

We know that, "A polygon is any 2-dimensional shape formed with any number of straight lines. "

Also we know that "A quadrilateral is a closed figure with four straight lines. "

Hence

$Quadrilateral \subseteq polygon$

Therefore the given statement is true.

The relation R defined on set A = {1, 2, 3, 4, 5} is defined by R =

${(x, y): |x^2 - y^2| > 4}$ . Which of the following could be the range of relation R?

0%
{4, 5}
0%
{3, 4, 5 }
0%
{1, 2, 3, 4, 5 }
0%
{1, 2}

$A={1 , 11 , 21 , 31 ,....... , 541 , 551}$ . B is a subset of A such that

$x+y\neq552$ , for any

$x , y \epsilon B.$ The maximum number of elements in B is

0%
$26$
0%
$30$
0%
$29$
0%
$28$

State whether the following statement is true or false.
Whole numbers

$\subseteq$ natural numbers

0%
True
0%
False

Explanation

We have to state whether the statement "

$whole \:numbers \subseteq natural \:numbers$ " is true or false.

We know that, whole numbers start with

$0$ . That is

$0,1,2,3,...$ and natural numbers start with

$1$ . That is

$1,2,3,...$

Thus

$0 \notin natural \: numbers$ .

$\Rightarrow whole \:numbers \nsubseteq natural \:numbers$ .

Hence the given statement is false.

If $a.N = \left\{ ax\thinspace :\thinspace x\in N \right\}$ then $3N\cap 7N=$

0%
$21\ N$
0%
$10\ N$
0%
$4\ N$
0%
$none$

Explanation

Clearly the set

$aN$ will include all multiples of

$a$ .

Hence

$3N$ and

$7N$ contain multiples of 3 and 7 respectively.

Their intersection must have the multiples of their L.C.M,i.e,21

Hence

$3N\cap 7N=21N$

$a, b, c, d$ are four distinct numbers chosen from the set

$\left \{1, 2, 3, ..., 9\right \}$ , then the minimum value of

$\dfrac {a}{b} + \dfrac {c}{d}$ is

0%
$\dfrac {3}{8}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {13}{36}$
0%
$\dfrac {25}{72}$

Explanation

To minimize the value of

$\dfrac{a}{b}+\dfrac{c}{d}$ ,

$b$ and

$d$ must highest numbers chosen from the set and

$a$ and

$c$ must be the lowest numbers of given set.

Therefore,

$b=9,\, d=8,\, a=2$ and

$c=1$ .

Thus, the minimum value for

$\dfrac{a}{b}+\dfrac{c}{d}$

$=\dfrac{2}{9}+\dfrac{1}{8}=\dfrac{25}{72}$

Let

$A,B$ and

$C$ be three events such that

$P(A)=0.3,P(B)=0.4,P(C)=0.8,P(A\cup B)=0.08,\quad P(A\cap C)=0.28,P(A\cap B\cap C)=0.09$ . If

$P(A\cup B\cup C)\ge 0.75$ , then

$P(B\cap C)$ satisfies

0%
$P(B\cap C)\le 0.23$
0%
$P(B\cap C)\le 0.48\quad$
0%
$0.23\le P(B\cap C)\le 0.48$
0%
$0.23\le P(B\cap C)\ge 0.48$

Explanation

Since

$P(A\cup B\cup C)\ge 0.75\quad$

$\quad \therefore 0.75\le P(A\cup B\cup C)\le 1$

$\Rightarrow 0.75\le P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\le 1\quad \quad$

$\Rightarrow 0.75\le 1.23-P(B\cap C)\le 1\Rightarrow -0.48\le -P(B\cap C)\le -0.23$

$\Rightarrow 0.23\le P(B\cap C)\le 0.48\quad$

CBSE Questions for Class 11 Engineering Maths Sets Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sets Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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