MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 4

Examine whether the following statements are true or false:

$(a)\epsilon (a,b,c)$

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:

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

0%
True
0%
False

State true or false for each of the following. Correct the wrong statement.
If T = {a, l, a, h, b, d, h}, then

$n (T) = 5$

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

0%
$A \subset B$
0%
$B \subset C$
0%
$C \subset B$
0%
$A \subset C$

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 ,

$B \subset C$

The equation x + cosx

$=$ a has exactly one positive root. Complete set of values of 'a' is

0%
(0, 1)
0%
$(-\infty, 1)$
0%
(-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 ?

0%
There is no rational number $\mathrm{x}\in \mathrm{S}$ such that $\mathrm{x}\leq 0$
0%
Every rational number $\mathrm{x}\in \mathrm{S}$ satisfies $\mathrm{x}\leq 0$
0%
$\mathrm{x}\in \mathrm{S}$ and $\mathrm{x}\leq 0= \mathrm{x}$ is not rational
0%
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.

The smallest set

$A$ such that

$A\cup \left\{ 1,2 \right\} =\left\{ 1,2,3,5,9 \right\}$ is

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

$A=\{3,5,9\}$

$A$ and

$B$ have some elements in common, then

$n(A \cup B)$ is:

0%
$= n(A) + n (B)$
0%
$> n (A) + n(B)$
0%
$< n(A) + n(B)$
0%
$\geq n (A)+ n (B)$

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,

$n(A\cup B)<n(A)+n(B)$ .

$A \cap B = \phi$ , then

$A \Delta B$ =

0%
$A \cap B$
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 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

0%
$300$
0%
$400$
0%
$500$
0%
$600$

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

$\dfrac{10}{100} \times n=40$

$\therefore n=400$

$p\cap (q\cup r)=?$

0%
$p\cup (q\cap r)$
0%
$(p\cap q)\cap (q\cap p)$
0%
$(p\cap q)\cup r$
0%
$(p\cap q)\cup (p\cap r)$

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?

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 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 :

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:

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.

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?

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

$220 = 77$

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?

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

$n(G\cap F) = n(G) + n(F)-n(G\cup F)$

$=7+8-12 = 3$

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 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

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

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

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

$\text{It is given that , }$

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

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

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

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

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

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

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

$=100-10$

$=90$

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

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

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

$=100-20$

$=80$

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

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

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

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

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

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

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

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

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

State true or false:

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

0%
True
0%
False

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 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?

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

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

$=7+8-12 = 3$

In a class, 20 opted for Physics, 17 for Maths, 5 for both and 10 for other subjects. The class contains how many students?

0%
$35$
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?

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

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

$=40+50-75 = 15$

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

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

$=200+193-370 = 23$

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?

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 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?

0%
n(P) = 45
0%
n(P) = 35
0%
n(P) = 65
0%
n(P) = 55

Explanation

7 students are taking Physics only.

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

$\Rightarrow n(P)=7$

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

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

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 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.

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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