Sets Test

Result

Sets Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

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 ?

A
$$30$$

B
$$22$$

C
$$12$$

D
$$8$$

Solution

Since 6 people are wearing neither hat nor sweater

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

By set theory

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

$$=18+24-30 = 12$$
Question 2
1 / -0

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.

A
$$15%$$

B
$$20%$$

C
$$30%$$

D
$$35%$$

Solution

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.
Question 3
1 / -0

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?

A
$$137$$

B
$$80$$

C
$$57$$

D
$$23$$

Solution

Since 80 do not read any

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

By set theory

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

$$=200+193-370 = 23$$
Question 4
1 / -0

Directions For Questions

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.

...view full instructions

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

A
8

B
9

C
10

D
11

Solution

7 students are taking Physics only.

So

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

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

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

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.
Question 5
1 / -0

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

A
Only one.

B
Many.

C
Three.

D
None of the options.

Solution

$$ We\quad know\quad that\quad the\quad common\quad chord\quad of\quad two\quad intersecting\quad circles\quad is\\ perpendicular\quad to\quad the\quad line\quad joining\quad the\quad centres\quad of\quad the\quad circles.\\ Here\quad are\quad two\quad intersecting\quad circles\quad with\quad centres\quad O\quad \& \quad O'.\\ The\quad common\quad chord\quad is\quad AB\quad which\quad is\quad perpendicular\quad to\quad OO'\quad at\quad P.\\ Now\quad if\quad the\quad circles\quad intersect\quad in\quad more\quad than\quad two\quad points\quad then\\ those\quad common\quad chords\quad will\quad have\quad different\quad incinations\quad with\quad the\quad \\ line\quad OO'\quad resulting\quad in\quad many\quad perpendiculars\quad on\quad OO'\quad with\quad \\ different\quad inclinations.\quad This\quad assumption\quad is\quad impossible.\\ So\quad AB\quad is\quad the\quad only\quad common\quad chord.\\ Ans-\quad Option\quad A. $$
Question 6
1 / -0

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

A
equal in counting

B
can not be equal in counting

C
$$24$$

D
$$\displaystyle 2^{6}\times 2^{4} $$

Solution

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
Question 7
1 / -0

The set of natural number is subset of set of real numbers.
State true or false:

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

$$
x-5<0\\ \Rightarrow \quad x\quad <\quad 5\\ Since\quad x\quad is\quad a\quad natural\quad no\\ x\quad =\quad \{ 1,2,3,4\} \\
$$
Question 8
1 / -0

Given the universal set $$B = \{-7,-3,-1,0,5,6,8,9\}$$, find :
$$B = \{x : -4 < x < 6\}$$

A
$$\{ -7, 0, 5,6\}$$

B
$$\{5,6,8,9\}$$

C
$$\{-3, -1, 0, 5\}$$

D
$$\{0,5\}$$

Solution

The only 4 no.s that lie in the given range are -3, 0 . -1 and 5.
Question 9
1 / -0

If $$A = (6, 7, 8, 9), B = (4, 6, 8, 10)$$ and $$C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$$ ; find :$$B - C$$

A
$$\{4, 6\}$$

B
$$\{4,6,8\}$$

C
$$\{6, 8, 10\}$$

D
$$\{8, 10\}$$

Solution

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

$$B-C=\{4,8,10\}$$
Question 10
1 / -0

If A = (6, 7, 8, 9), B = (4, 6, 8, 10) and C = {$$x$$ : $$x \,\,\epsilon\,\,N$$ : $$2 < x \leq 7$$} ; find :
A - B

A
$$\{6,8\}$$

B
$$\{7,9\}$$

C
$$\{6,9\}$$

D
$$\{6,7,9,10\}$$

Solution

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

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Re-Attempt Weekly Quiz Competition

Sets Test - 24

Result

Sets Test - 24

Score

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Rank

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SHARING IS CARING

Question 1

$$30$$

$$22$$

$$12$$

$$8$$

Solution

Question 2

$$15%$$

$$20%$$

$$30%$$

$$35%$$

Solution

Question 3

$$137$$

$$80$$

$$57$$

$$23$$

Solution

Question 4

8

9

10

11

Solution

Question 5

Only one.

Many.

Three.

None of the options.

Solution

Question 6

equal in counting

can not be equal in counting

$$24$$

$$\displaystyle 2^{6}\times 2^{4} $$

Solution

Question 7

True

False

Ambiguous

Data insufficient

Solution

Question 8

$$\{ -7, 0, 5,6\}$$

$$\{5,6,8,9\}$$

$$\{-3, -1, 0, 5\}$$

$$\{0,5\}$$

Solution

Question 9

$$\{4, 6\}$$

$$\{4,6,8\}$$

$$\{6, 8, 10\}$$

$$\{8, 10\}$$

Solution

Question 10

$$\{6,8\}$$

$$\{7,9\}$$

$$\{6,9\}$$

$$\{6,7,9,10\}$$

Solution

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