Set Theory Test 15

Result

Set Theory Test 15

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Look at the set of numbers below.
Set : $\left \{6, 12, 30, 48\right\}$
Which statement about all the numbers in this set is NOT true?

A
They are all multiples of $3$

B
They are all even numbers

C
They are all factors of $48$

D
They are all divisible by $2$

Solution

Here above statement C is not true because 30 is not factors of 48.
So, Answer (C) They are all factors of 48
Question 2
1 / -0

In a class of 55 students, the number of students studying different subjects are 23 in Mathematics and 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects, The number of students who have taken exactly one subject is

A
20

B
24

C
23

D
22

Solution

Total no. of student=55
Let n(M)=student who studying mathematics
n(C)=student who studying chemistry
n(P)=student who studying Physics
$\therefore n(M)=23,n(P)=24,n(C)=19,n(M\cap P)=12,n(M\cap C)=9,n(P\cap C)=7,n(M\cap P\cap C)=4$
Number of student who studying mathematics but not physic and chemistry
$\Rightarrow n(M)-[(n(M\cap C)+n(M\cap P)]+n(M\cap P\cap C)$
$\Rightarrow 23-[9+12]+4$
$\Rightarrow 23-21+4=6$

Number of student who studying chemistry but not physic and matehematics
$\Rightarrow n(C)-[(n(M\cap C)+n(P\cap C)]+n(M\cap P\cap C)$
$\Rightarrow 19-[9+7]+4$
$\Rightarrow 19-16+4=7$

Number of student who studying physics but not mathematics and chemistry
$\Rightarrow n(P)-[(n(M\cap P)+n(P\cap C)]+n(M\cap P\cap C)$
$\Rightarrow 24-[12+7]+4$
$\Rightarrow 24-19+4=9$

$\therefore$ no. of student studying exactly one subject $=6+7+9=22$
Question 3
1 / -0

Let $A$ and $B$ be two sets such that $n(A)=16$ , $n(B)=12$ , and $n(A\cap B)=8$ . Then $n(A\cup B)$ equals

A
$28$

B
$20$

C
$36$

D
$12$

Solution

$n(A\cup B)=n(A)+n(B)-n(A\cap B)=16+12-8=20$
Question 4
1 / -0

Let $A=\{1,2,3,4,5,6\}$ . How many subsets of $A$ can be formed with just two elements, one even and one odd?

A
$6$

B
$8$

C
$9$

D
$10$

Solution

$\{1,2\}$ , $\{1,4\}$ , $\{1,6\}$ , $\{2,3\}$ , $\{2,5\}$ , $\{3,4\}$ , $\{3,6\}$ , $\{4,5\}$ , $\{5,6\}$ ,
Question 5
1 / -0

In a class 60% of the students were boys and 30% of them had I class. If 50%of the students in the class had I class, find the fraction of the girls in the class who did not have a I class.

A
$1/5$

B
$4/5$

C
$1/4$

D
$1/3$

Solution

$Let\quad the\quad number\quad of\quad students\quad be\quad x,\\ boys\quad with\quad first\quad class=0.3*0.6x=0.18x,\\ number\quad of\quad girls=0.4x,\\ students\quad with\quad first\quad class=0.5x,\\ girls\quad with\quad first\quad class=0.5x-0.18x=0.32x,\\ fraction\quad of\quad girls\quad without\quad first\quad class=\dfrac { 0.4-0.32x }{ 0.4x } =\dfrac { 1 }{ 5 }$
Question 6
1 / -0

If $A=\{\dots,-6,-4,-2,0,2,4,6,\dots\}$ , then

A
$10\notin A$

B
$-10\notin A$

C
$5\in A$

D
$50\in A$

Solution

50 is even.
Question 7
1 / -0

The sets $\displaystyle S_{x}$ are defined to be $(x, x + 1, x + 2, x + 3, x + 4)$ where $x=1, 2, 3,.....80$ . How many of these sets contain $6$ or its multiple?

A
$65$

B
$66$

C
$59$

D
$60$

Solution

There are $14$ multiples of $6$ till $84$ .

Since $5$ consecutive no.s are chosen only one set in $6$ consecutive sets will not have a multiple of $6$ . So till $78$ sets there are

$78-\dfrac{78}6=78-13=65$ sets containing 6 or multiples of 6.

$S_{79}$ does not contain any multiple of 6

Hence $S_{80}$ must contain a multiple of 6.

Answer $=66$ sets
Question 8
1 / -0

If $\displaystyle Q=\left((x|x=\frac{1}{y}\:\ \text{wher} \ e\:y\in N\right)$ , then

A
$0\notin Q$

B
$1\in Q$

C
$2\in Q$

D
$\displaystyle\frac{2}{3}\in Q$

Solution

When $y=1$ , $\displaystyle x=\frac{1}{1}=1$ .
Question 9
1 / -0

If A={a,b,c,d,e}, B={a,c,e,g} and C={b,d,e,g} then which of the following is true?

A
$\displaystyle C\subset \left ( A\cup B \right )$

B
$\displaystyle C\subset \left ( A\cap B \right )$

C
$\displaystyle A\cup B=A\cup C$

D
Both(1) and (3)

Solution

For the given sets,
$(A \cup B ) =$ { $a,b,c,d,e,g$ } (Combination of all elements of both sets)

Clearly, elements of C are a part of $A \cup B$ as well,
So, $C\subset (A\cup B)$

And, $(A \cap B ) =$ { $a,c,e$ } (Common elements of both sets )
As the elements of C are not completely a part of $A \cap B$ , Option B is not True.

Also,
$(A \cup C ) =$ { $a,b,c,d,e,g$ }

Clearly, $(A \cup B ) = (A \cup C )$

Hence, both first option and third options are True.
Question 10
1 / -0

If the universal set is U = $\displaystyle \left \{ 1^{2},2^{2},3^{2},4^{2},5^{2},6^{2} \right \}$ What is the complement of the intersection of set A = $\displaystyle \left \{ 2^{2},4^{2},6^{2} \right \}$ and set B= $\displaystyle \left \{ 2^{2},3^{2},4^{2} \right \}$ ?

A
$\displaystyle \left \{ 2^{2},4^{2} \right \}$

B
$\displaystyle \left \{ 1^{2},5^{2} \right \}$

C
$\displaystyle \left \{ 1^{2},5^{2},6 ^{2} \right \}$

D
$\displaystyle \left \{ 1^{2},3^{2},5^{2},6^{2} \right \}$

E
Answer required

Solution

$A\cap B=\{2^2,4^2\}$
$\bar{A\cap B}=U-(A\cap B)=\{1^2,3^2,5^2,6^2\}$
Option D is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Set Theory Test 15

Result

Set Theory Test 15

Score

-

Rank

-

SHARING IS CARING

Question 1

They are all multiples of 333

They are all even numbers

They are all factors of 484848

They are all divisible by 222

Solution

Question 2

20

24

23

22

Solution

Question 3

282828

202020

363636

121212

Solution

Question 4

666

888

999

101010

Solution

Question 5

1/51/51/5

4/54/54/5

1/41/41/4

1/31/31/3

Solution

Question 6

10∉A10\notin A10∈/A

−10∉A-10\notin A−10∈/A

5∈A5\in A5∈A

50∈A50\in A50∈A

Solution

Question 7

656565

666666

595959

606060

Solution

Question 8

0∉Q0\notin Q0∈/Q

1∈Q1\in Q1∈Q

2∈Q2\in Q2∈Q

23∈Q\displaystyle\frac{2}{3}\in Q32​∈Q

Solution

Question 9

C⊂(A∪B)\displaystyle C\subset \left ( A\cup B \right )C⊂(A∪B)

C⊂(A∩B)\displaystyle C\subset \left ( A\cap B \right )C⊂(A∩B)

A∪B=A∪C\displaystyle A\cup B=A\cup CA∪B=A∪C

Both(1) and (3)

Solution

Question 10

{22,42} \displaystyle \left \{ 2^{2},4^{2} \right \} {22,42}

{12,52} \displaystyle \left \{ 1^{2},5^{2} \right \} {12,52}

{12,52,62} \displaystyle \left \{ 1^{2},5^{2},6 ^{2} \right \} {12,52,62}

{12,32,52,62} \displaystyle \left \{ 1^{2},3^{2},5^{2},6^{2} \right \} {12,32,52,62}

Answer required

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

They are all multiples of $3$

They are all factors of $48$

They are all divisible by $2$

$28$

$20$

$36$

$12$

$6$

$8$

$9$

$10$

$1/5$

$4/5$

$1/4$

$1/3$

$10\notin A$

$-10\notin A$

$5\in A$

$50\in A$

$65$

$66$

$59$

$60$

$0\notin Q$

$1\in Q$

$2\in Q$

$\displaystyle\frac{2}{3}\in Q$

$\displaystyle C\subset \left ( A\cup B \right )$

$\displaystyle C\subset \left ( A\cap B \right )$

$\displaystyle A\cup B=A\cup C$

$\displaystyle \left \{ 2^{2},4^{2} \right \}$

$\displaystyle \left \{ 1^{2},5^{2} \right \}$

$\displaystyle \left \{ 1^{2},5^{2},6 ^{2} \right \}$

$\displaystyle \left \{ 1^{2},3^{2},5^{2},6^{2} \right \}$