MCQ Questions for Class 11 Engineering Maths Sets Quiz 5

One Hundred Twenty-five

$(125)$ aliens descended on a set of film as Extra Terrestrial Beings.

$40$ had two noses,

$30$ had three legs,

$20$ had four ears,

$10$ had two noses and three legs,

$12$ had three legs and four ears,

$5$ had two noses and four ears and

$3$ had all the three unusual features. How many were there without any of these unusual features?

0%
$5$
0%
$35$
0%
$80$
0%
None of these

Explanation

Let

$A=$ set of aliens having two noses

$B=$ set of aliens having three legs

$C=$ set of aliens having four ears

$n(A)=40,n(B)=30,n(C)=20,n(A\cap B)=10,n(B\cap C)=12,n(C\cap A)=5,n(A\cap B\cap C)=3$

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

$n(A\cup B\cup C)=40+30+20-(10+12+5)+3$

$n(A\cup B\cup C)=66$

Aliens having any of the deformities

$=66$

So, aliens without any deformities

$=125-66=59$

In a class of 80 children, 35% children can play only cricket, 45% children can play only table-tennis and the remaining children can play both the games. In all, how many children can play cricket?

0%
$55$
0%
$44$
0%
$36$
0%
$28$

Explanation

Clearly

$35\%$ children can play cricket. Also

$20\%$ can play both.

$55\%$ children can play cricket

Total no. of kids =

$0.55\times 80 =44$

In a class consisting of 100 students, 20 know English and 20 do not know Hindi and 10 know neither English nor Hindi. The number of students knowing both Hindi and English is:

0%
5
0%
10
0%
15
0%
20

Explanation

$\textbf{Step -1: Write all the given data in symbols.}$

$\text{Let }E\text{ be the set of students who know English.}$

$H\text{ be the set of students who know Hindi.}$

$\text{It is given that , }$

$n\left(U\right)=100$

$n\left(E\right)=20, n\left(\overline{H}\right)=20$

$\text{and } n\left(\overline{E\cup H}\right)=10$

$\textbf{Step -2: Draw Venn Diagram}$

$\textbf{Step -3: Find the number of students who know either English or Hindi.}$

$\text{Number of students who know either English or Hindi are:}$

$n(E\cup H)=n(U)-n(\overline{E\cup H})$

$=100-10$

$=90$

$\textbf{Step -4: Find the number of students who Hindi.}$

$\text{Number of students who know Hindi are:}$

$n(H)=n(U)-n(\overline{H})$

$=100-20$

$=80$

$\textbf{Step -5: Find the number of students who know both English or Hindi.}$

$\text{Number of students who know both Hindi and English are }n(E\cap H).$

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

$\therefore n(E\cup H)=n(E)+n(H)-n(E\cap H)$

$\Rightarrow 90=20+80-n(E\cap H)$

$\Rightarrow n(E\cap H)=10$

$\textbf{Hence, the correct option is B.}$

In a survey of brand preference for toothpastes,

$82$ of the population (number of people covered for the survey is

$100$ ) liked at least one of the brands: I, II and III.

$40$ of those liked brand I,

$25$ liked brand II and

$35$ liked brand III. If

$8$ of those asked, showed liking for all the three brands, then what percentage of those liked more than one of the three brands?

0%
$13$
0%
$10$
0%
$8$
0%
$5$

Explanation

By set theory

$n(I\cup II \cup III) = n(I) + n(II)+n(III)-n(I\cap II)-n(II\cap III)-n(I\cap III) +n(I\cap II\cap III)$

$82=40+25+35-x+8$

$\Rightarrow x = 26$

But

$n(I\cap II)+n(II\cap III)+n(I\cap III)$ counts

$n(I\cap II\cap III)$ thrice whereas it needs to be counted only once.

Hence answer

$=26-2\times8=10$

In a certain group of 36 people, 18 are wearing hats and 24 are wearing sweaters. If six people are wearing neither a hat nor a sweater, then how many people are wearing both a hat and a sweater ?

0%
$30$
0%
$22$
0%
$12$
0%
$8$

Explanation

Since 6 people are wearing neither hat nor sweater

$n(H\cup S) = 36-6 = 30$

By set theory

$n(H\cap S) = n(H) + n(S)-n(H\cup S)$

$=18+24-30 = 12$

While preparing the progress reports of the students, the class teacher found that 70% of the students passed in Hindi, 80% passed in English and only 65% passed in both the subjects. Find out the percentage of students who failed in both the subjects.

0%
$15%$
0%
$20%$
0%
$30%$
0%
$35%$

Explanation

The sets E and H represent the students failing in the respective subjects.

$n(H\cup E) = 1-0.65 = 0.35$

By set theory

$n(H\cap E) = n(H) + n(E)-n(N\cup E)$

$=0.3+0.2-0.35 = 0.15$

Hence

$15\%$ of students failed in both subjects.

Out of 450 students in a school, 193 students read Science Today, 200 students read Junior Statesman, while 80 students read neither. How many students read both the magazines?

0%
$137$
0%
$80$
0%
$57$
0%
$23$

Explanation

Since 80 do not read any

$n(S\cup J)=450-80=370$ ....................(S = Science Today; J= Junior Statesman)

By set theory

$n(J\cap S) = n(J) + n(S)-n(J\cup S)$

$=200+193-370 = 23$

If $Y\cup \left\{ 1,2 \right\} =\left\{ 1,2,3,5,9 \right\}$ , then

0%
The smallest set of $Y$ is $\left\{ 3,5,9 \right\}$
0%
The smallest set of $Y$ is $\left\{ 2,3,5,9 \right\}$
0%
The largest set of $Y$ is $\left\{ 1,2,3,5,9 \right\}$
0%
The largest set of $Y$ is $\left\{ 2,3,4,9 \right\}$

Explanation

Since the set of the right hand side has

$5$ elements,

$\therefore$ smallest set of

$Y$ has three elements and largest set of

$Y$ has five elements,

$\therefore$ smallest set of

$Y$ is

$\left\{ 3,5,9 \right\}$

and largest set of

$Y$ is

$\left\{ 1,2,3,4,9 \right\}$

How many students in this group are not taking any of the three subjects?

0%
8
0%
9
0%
10
0%
11

Explanation

7 students are taking Physics only.

$7=n(P)-n(P\cap G)-n(P\cap E) +n(P\cap G \cap E)$

$\Rightarrow n(P)=7$

$+ n (P\cap G)+n(P\cap E) -n(P\cap G \cap E)$

$n(P)=7+24+30-16=45$

Similarly

$n(G)= 10+24+22-16=40$

$n(E)=5+22+30-16=41$

$n(P\cup G \cup E)= n(P) +n(G) +n(E)$

$-n(G\cap E)-n(P\cap G)-n(P\cap E) +n(P\cap G \cap E)$

$=45+40+41-22-24-30+16=66$

$75-66=9$ students are not taking any of the 3 subjects.

State true or false:

A set of rational number is a subset of a set of real numbers.

0%
True
0%
False

If two intersecting circles with two points in common are drawn then how many common chords can be drawn? Draw the figure and write the answer

0%
Only one.
0%
Many.
0%
Three.
0%
None of the options.

Explanation

$We\quad know\quad that\quad the\quad common\quad chord\quad of\quad two\quad intersecting\quad circles\quad is\\ perpendicular\quad to\quad the\quad line\quad joining\quad the\quad centres\quad of\quad the\quad circles.\\ Here\quad are\quad two\quad intersecting\quad circles\quad with\quad centres\quad O\quad \& \quad O'.\\ The\quad common\quad chord\quad is\quad AB\quad which\quad is\quad perpendicular\quad to\quad OO'\quad at\quad P.\\ Now\quad if\quad the\quad circles\quad intersect\quad in\quad more\quad than\quad two\quad points\quad then\\ those\quad common\quad chords\quad will\quad have\quad different\quad incinations\quad with\quad the\quad \\ line\quad OO'\quad resulting\quad in\quad many\quad perpendiculars\quad on\quad OO'\quad with\quad \\ different\quad inclinations.\quad This\quad assumption\quad is\quad impossible.\\ So\quad AB\quad is\quad the\quad only\quad common\quad chord.\\ Ans-\quad Option\quad A.$

Which of the following sets is/are empty?

0%
$\displaystyle \left \{ x : x \in R ,x^{2} -4=0\right \}$
0%
$\displaystyle \left \{ x : x \in R ,x^{4} +4=0\right \}$
0%
$\displaystyle \left \{ x : x \in R ,x^{3} =1\right \}$
0%
$\displaystyle \left \{ x : x \in R ,x^{8}+x^{4}+1=0 \right \}$

Explanation

$x=2,-2$ is a solution

$x^{4}$ is positive always and 4 is also positive.The sum of two positive numbers is always positive.Hence x doesn't have a solution.

$x^{3}$ =1

$\Rightarrow x=1$ .

$x^{8}+x^{4}+1$ is always positive because the individual terms are positive.Hence x here doesn't have a solution.

The set of natural number is subset of set of real numbers.

State true or false:

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

$x-5<0\\ \Rightarrow \quad x\quad <\quad 5\\ Since\quad x\quad is\quad a\quad natural\quad no\\ x\quad =\quad \{ 1,2,3,4\} \\$

Let

$\displaystyle A= \left \{ 7,8,9,a,b,c \right \}$ and

$\displaystyle B= \left \{ 1,2,3,4 \right \}$ then number of universal relation from the set

$A$ to set

$B$ and set

$B$ to set

$A$ are

0%
equal in counting
0%
can not be equal in counting
0%
$24$
0%
$\displaystyle 2^{6}\times 2^{4}$

Explanation

Relation

$R$ is said to be universal relation from set

$A$ to set

$B$ or set

$B$ to set

$A$ if

$\displaystyle R= A\times B$

$\displaystyle \therefore$ Number of relations from

$A$ to

$B$ for

$\displaystyle R= A\times B$

$\displaystyle = n(A)\times n(B)= 6\times 4= 24$

Ans: C

$\displaystyle \left \{ \left ( 1, 2 \right ) \right \}, and B=\left \{ 1, 3 \right \}, then\left ( A\times B \right )\cup \left ( B\times A \right )$

$\displaystyle =\left \{ \left ( 1, 3 \right ), \left ( 2, 3 \right ), \left ( 3, 1 \right ), \left ( 3, 2 \right ), \left ( 1, 1 \right ), \left ( 1, 2 \right ), \left ( 2, 1 \right ),\right \}$

0%
True
0%
False

Explanation

$A=\left\{ 1,2 \right\} \\ B=\left\{ 1,3 \right\} \\ A\times B=\left\{ \left( 1,1 \right) ,\left( 1,3 \right) ,\left( 2,1 \right) ,\left( 2,3 \right) \right\} \\ B\times A=\left\{ \left( 1,1 \right) ,\left( 1,2 \right) ,\left( 3,1 \right) ,\left( 3,2 \right) \right\} \\ \therefore \left( A\times B \right) \cup \left( B\times A \right) =\left\{ \left( 1,1 \right) ,\left( 1,3 \right) ,\left( 1,2 \right) ,\left( 2,1 \right) ,\left( 2,3 \right) ,\left( 3,1 \right) \left( 3,2 \right) \right\}$

True or false :

$\displaystyle A\times \left ( B\cap C \right )=\left ( A\times B \right )\cap \left ( A\times C \right )$

0%
True
0%
False

Explanation

$A\times (B\cap C)=(A\times B)\cap (A\times C)$

Let

$(a, b)$ been arbitrary elements of

$A\times (B\cap C)$ , then

$(a, b)\in A\times (B\cap C)$

$\Rightarrow a\in A$ and

$b\in (B\cap C)$

$\Rightarrow a\in A$ and

$(b\in B$ and

$b\in C)$

$\Rightarrow (a\in A$ and

$b\in B)$ and

$(a\in A$ and

$b\in C)$

$\Rightarrow (a, b)\in A\times B$ and

$(a, b)\in A\times C$

$\Rightarrow (a, b)\in (A\times B)\cap (A\times C)$

$\therefore A\times (B\cap C)\subset(A\times B)\cap (A\times C)$ .....

$(1)$

Now, let

$(x, y)$ bean arbitrary elements of

$(A\times B)\cap (A\times C)$ , then

$(x, y)\in (A\times B)\cap (A\times C)$

$\Rightarrow (x, y)\in (A\times B)$ and

$(x, y)\in (A\times C)$

$\Rightarrow (x\in A$ and

$y\in B)$ and

$(x\in A$ and

$y\in C)$

$\Rightarrow x\in A$ and

$(y\in B$ and

$y\in C)$

$\Rightarrow x\in A$ and

$y\in(B\cap C)$

$\Rightarrow (x, y)\in A\times (B\cap C)$

$\therefore (A\times B)\cap (A\times C)\subset A\times (B\cap C)$ .......

$(2)$

Hence from equation

$(1)$ and

$(2)$

we get

$A\times (B\cap C)=(A\times B)\cap (A\times C)$

so this equation is true.

1123081_183268_ans_1fd838363bba4979bd27f99082a272f6.jpg

Given the universal set

$B = \{-7,-3,-1,0,5,6,8,9\}$ , find :

$B = \{x : -4 < x < 6\}$

0%
$\{ -7, 0, 5,6\}$
0%
$\{5,6,8,9\}$
0%
$\{-3, -1, 0, 5\}$
0%
$\{0,5\}$

If universal set

$\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$ and

$C = \{c, d, e, f, g\}$ , then find

$B - A$

0%
$\{b,c,e,f\}$
0%
$\{a, b, f, h\}$
0%
$\{a, g, h\}$
0%
$\{a,c,e,g\}$

Explanation

Given :

$\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$ and

$C = \{c, d, e, f, g\}$

$B - A$ will have elements of

$B$ which are not in

$A$

So,

$B - A =\{a,g,h \}$ .

If universal set

$\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$ and

$C = \{c, d, e, f, g\}$ find

$(B - C)'$

0%
$\{d, e, f, g\}$
0%
$\{c, d, e, f, g\}$
0%
$\{c, d, f, g\}$
0%
$\{a, c, d, e, f, g\}$

Explanation

Given :

$\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$ and

$C = \{c, d, e, f, g\}$

$B - C$ will have elements of

$B$ which are not in

$C$

So,

$B - C = \{ a, b, h\}$

And

${(B-C)}'= \xi - (B-C) =$ {

$c,d,e,f,g$ }

If A = (6, 7, 8, 9), B = (4, 6, 8, 10) and C = {

$x$ :

$x \,\,\epsilon\,\,N$ :

$2 < x \leq 7$ } ; find :

A - B

0%
$\{6,8\}$
0%
$\{7,9\}$
0%
$\{6,9\}$
0%
$\{6,7,9,10\}$

Explanation

$A-B = A-A\cap B= \{7,9\}$

$A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and

$C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$ ; find :

$B - C$

0%
$\{4, 6\}$
0%
$\{4,6,8\}$
0%
$\{6, 8, 10\}$
0%
$\{8, 10\}$

Explanation

$C=\{3,4,5,6,7\}$

$B-C=\{4,8,10\}$

$A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and

$C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$ ; find :

$B - B$

0%
$\phi$
0%
$\{0\}$
0%
$\{6,7\}$
0%
$\{4\}$

Explanation

Given :

$A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and

$C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$

$B - B$ will always be a null set it will contain elements of

$B$ which are not in

$B$ i.e. no elements.

So,

$B - B = \phi$

Verify:

$A'\cap B = B - (A \cap B)$ .

0%
True
0%
False

Explanation

Let

$U$ be the universal set

Whole numbers whose Squares are less than or equal to

$16$ are

$0, 1, 2, 3, 4$ .
So, set

$A = \{ 0, 1, 2, 3, 4 \}$
And

$\dfrac {x}{3} - 2 < 3$

$\Rightarrow \dfrac {x}{3} < 5$

$\Rightarrow x < 15$
Whole numbers which are less than

$15$ are

$0, 1, 2,3,4,5,6.... 14$
So, set

$B = \{ 0, 1, 2,3,4,5,6.... 14 \}$
For the LHS

${A}' = U - A = \{ 5,6,7,8,9,10........... \}$
Intersection of two sets has the common elements of both the sets.
So,

${A}'\cap B = \{ 5,6,7,8,9,10,11,12,13,14 \}$ .....(1)
For the RHS

$(A \cap B) = \{0,1,2,3,4 \}$

$B - (A \cap B)$ will have elements of B which are not in

$(A \cap B)$
So,

$B - (A \cap B) = \{ 5,6, 7, 8,9, 10, 11, 12,13,14 \}$ .....(2)
From (1) and (2), we have

${A}' \cap B =B - (A \cap B)$

Hence, the given expression is correct.

$A=\left\{ x:x\neq x \right\}$ represents-

0%
$\left\{ 0 \right\}$
0%
$\left\{ \right\}$
0%
$\left\{ 1 \right\}$
0%
$\left\{ x \right\}$

Explanation

It is fundamental concept that a void set is defined as,

$A =\{x : x\neq x\}=\{\}$

Which of the following statements is false?

0%
$2$ $\in$ {first five counting numbers.}
0%
$f$ ${\notin }$ {consonants}
0%
Rose ${\notin }$ {the set of all fruits.}
0%
Asia ${\in }$ {Continents}

Explanation

$\textbf{Step-1: Observing the Option 1}$

$\text{2 belongs to the sets of first five counting numbers}$

$\text{so option (A) is True.}$

$\textbf{Step-2: Observing other option}$

$\text{f belongs to the set of consonants so option B is false}$

$\text{Rose belongs to the set of all fruits so option C is True}$

$\text{Asia belongs to the sets of continents so option D is also True}$

$\textbf{Hence, the false statement is option B}$

Let

$A = \{2,3,5,7,8,11\}$ then which among the following is true?

0%
${7\notin A}$
0%
$\{2,3\} \not\subset \ of \ A$
0%
$2 \in A$
0%
None of the above

Explanation

$2\in A$ means 2 is element of A which is true.

Which of the following statements is true

0%
$3\subseteq \left\{ 1,3,5 \right\}$
0%
$3\in \left\{ 1,3,5 \right\}$
0%
$\left\{ 3 \right\} \in \left\{ 1,3,5 \right\}$
0%
$\left\{ 3,5 \right\} \in \left\{ 1,3,5 \right\}$

Explanation

Element/ number three is denoted by

$3$ and not

$\{3\}$ .

$\{3\}$ denotes a subset containing element 3. Hence

$\{3\}\subseteq \{1,3,5\}$
And

$3\in \{1,3,5\}$

Let

$P = \{ x | x$ is a multiple of

$3$ and less than

$100$ ,

$x$

$\displaystyle \in$

$N \}$

$Q = \{ x | x$ is a multiple of

$10$ and less than

$100$ ,

$x$

$\displaystyle \in$

$N\}$

0%
$\displaystyle Q\subset P$
0%
$\displaystyle P\cup Q=$ $\{ x | x$ is multiple of 30 $;$ $\displaystyle x \in N$ $\}$
0%
$\displaystyle P\cap Q=\phi$
0%
$\displaystyle P\cap Q=$ $\{ x | x$ is a multiple of 30 $;$ $\displaystyle x\in N$ $\}$

Explanation

P =

$\{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 ..., 60, ...,90, ....,99\}$
Q =

$\{10, 20, 30, 40, 50, 60, 70, 80, 90 \}$
Then looking at the multiple choices we get

$P \displaystyle \cap Q$ =

$\{30, 60, 90\}$

$\displaystyle \Rightarrow$ P

$\displaystyle \cap$ Q = {x

$\displaystyle \mid$ x is a multiple of 30 x

$\displaystyle \epsilon$ N}

Find the set of all solutions of the equation

$2^{\left | y \right |}-\left | 2^{y-1}-1 \right |=2^{y-1}+1$ , the solution includes

0%
$y =-1$
0%
$y>1$
0%
$y=1$
0%
$y<1$

Explanation

Here,

$2^{\left | y \right |}-\left | 2^{y-1}-1 \right |=2^{y-1}+1$
We know to define modulus, we have three cases as
Case I: y< 0

$\Rightarrow$

$2^{-y}+\left ( 2^{y-1}-1 \right )=2^{y-1}+1$

$\Rightarrow$

$2^{-y}=2^{1}$ (as when y< 0 |y|=-y and

$\left | 2^{y-1}-1 \right |=-\left ( 2^{y-1}-1 \right )$ )
Hence, y=-1, which is true when y< 0 (i)
Case II:

$0\leq y< 1$

$\Rightarrow$

$2^{y}+\left ( 2^{y-1}-1 \right )=2^{y-1}+1$

$\Rightarrow$

$2^{y}=2$ (as when

$0\leq y< 1$ |y|=-y and

$\left | 2^{y-1}-1 \right |=-\left ( 2^{y-1}-1 \right )$ )

$\Rightarrow$

$y=1$ , which shows no solution as,

$0\leq y< 1$ (ii)
Case III:

$y\geq 1$

$\Rightarrow$

$2^{y}-\left ( 2^{y-1}-1 \right )=2^{y-1}+1$

$\Rightarrow$

$2^{y}=2^{y-1}+2^{y-1}$

$\Rightarrow$

$2^{y}=2.2^{y-1}$ (as when

$y\geq 0$ |y|=y and

$\left | 2^{y-1}-1 \right |=\left ( 2^{y-1}-1 \right )$ )

$\Rightarrow$

$2^{y}=2^{y}$ , which is an identity therefor, it is true

$\forall y\geq 1$ (iii)
Hence, from Eqs. (i), (ii), (iii) the solution of set is

$\left \{ y: y\geq 1\cup y=-1 \right \}$ .

$Q=\{ x:x=\cfrac { 1 }{ y },$ where

$y\in N \}$ , then

0%
$0\in Q$
0%
$1\in Q$
0%
$2\in Q$
0%
$\cfrac{2}{3} \in Q$

Explanation

Since

$\cfrac{1}{y}\ne 0$ ,

$\cfrac{1}{y}\ne 2$ ,

$\cfrac{1}{y}\ne \cfrac{-2}{3}$ [

$\because y\in N$ ]

$\therefore \cfrac{1}{y}$ can be

$1$

Ans: B

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CBSE Questions for Class 11 Engineering Maths Sets Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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