Set Theory Test 13

Result

Set Theory Test 13

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

The solution set of $3 x - 4 < 8$ over the set of non-negative square numbers is

A
$\displaystyle \left \{ 1,2,3 \right \}$

B
$\displaystyle \left \{ 1,4 \right \}$

C
$\displaystyle \left \{ 1 \right \}$

D
$\displaystyle \left \{ 16 \right \}$

Solution

$3x-4<8$
$3x<12$
$x<4$
Hence set of non-negative square numbers belonging to the above set is
$\{1\}$ .
Question 2
1 / -0

Let $P$ and $Q$ be two sets then what is $\displaystyle (P\cap Q')\cup (P\cup Q)'$ equal to ?

A
$\displaystyle (P\cap Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi$

B
$\displaystyle (P\cup Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi$

C
$\displaystyle (P\cap Q')\cup (P\cap Q)'=\xi \cap Q'=\xi \cap Q'=\xi$

D
none of the above

Solution

$\displaystyle (P\cap Q')\cup (P\cup Q)'=(P\cap Q')\cup (P'\cap Q')$
$=$ $\displaystyle (P\cup P')\cap (P\cup Q')\cap (Q'\cup P')\cap (Q'\cup Q')$
$=$ $\displaystyle \xi \cap \left \{ Q'\cup (P\cap P') \right \}\cap Q'$
$=\displaystyle \xi \cap \{Q'\cup \xi )\cap Q'$
$=$ $\displaystyle \xi \cap Q'\cap Q'=\xi \cap Q'=\xi$
Question 3
1 / -0

If $A$ and $B$ are finite sets which of the following is the correct statement?

A
$n(A - B) = n(A) - n(B)$

B
$n(A - B) = n(B - A)$

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

D
$n(A - B) = n(B) -$ $\displaystyle n\left ( A\cap B \right )$

Solution
Question 4
1 / -0

U is a universal set and n(U) = 160. A, B and C are subset of U. If n(A) = 50, n(B) = 70, $\displaystyle n\left ( B\cup C \right )=\Phi$ , $\displaystyle n\left ( B\cap C \right )=15$ and $\displaystyle A\cup B\cup C=U$ . then n(C) equals

A
40

B
50

C
55

D
60
Question 5
1 / -0

If $n (A) = 115, n(B) = 326, n(A - B) = 47$ , then $\displaystyle n(A+B)$ is equal to

A
373

B
165

C
370

D
None

Solution

$n(A-B)=47\Rightarrow n(A)-n(A\cap B) = 47\Rightarrow n(A\cap B)=115-47=68$

Hence $n(A+B)=115+326-68=373$
Question 6
1 / -0

If A and B are two disjoint sets and N is the universal set then $\displaystyle A^{c}\cup \left [ \left ( A\cup B \right )\cap B^{c} \right ]$ is

A
$\displaystyle \phi$

B
A

C
B

D
N

Solution

Since $A$ and $B$ are disjoint sets $B^c\cap(A\cup B)$ will be only $A$ as there is no intersection between $A$ and $B$ .

Of course $A^c\cup A=N$
Question 7
1 / -0

Suppose $\displaystyle A_{1},A_{2},....,A_{30}$ are thirty sets each having 5 elements and $\displaystyle B_{1},B_{2},....,B_{n}$ are n sets each with 3 elements. Let $\displaystyle \bigcup_{i=1}^{30}A_{i} = \bigcup_{j=1}^{n}B_{j}=S$ and each elements of S belongs to exactly 10 of the $\displaystyle A_{i}$ and exactly 9 of the $\displaystyle B_{j}$ . Then n is equal to-

A
35

B
45

C
55

D
65

Solution

$A_0,A_1,.............,A_{30}\implies$ each of 5 elements
$B_1,B_2,B_3...........n\implies$ each of 3 elements
The number of elements in the union of the A sets is $5(30)-r$ where 'r' is the number repeats likewise the number of elements in the B sets $3n-rB$
Each element in the union (in5) is repeated 10 times in A which means if x was the real number of elements in A (not counting repeats) then q out of those 10 should be thrown away or 9x .likewise on the B side 8x of those elements should be thrown away So, $\implies 150-9x=3n-8x$
$n=50-3x$
Now in figure out what x is we need to use the fact that the union of a group of sets contains every member of each sets . If every element in 'S' is repeated 10 times that means every element in the union of the n's is repeated 10 times .
This means that $10/10\implies 15$ is the number of in the A's without repeats counted (same for the B's aswell ) So now
$\cfrac{50-15}{3}=n$
$n=45$
Subset:- A proper subset is nothing but it contain atleast one more element of main set .
Ex: $\{3,4,5\}$ is a set then the possible subsets are
$\{3\},\{4\},\{5\},\{1,5\},\{3,4\}$
Question 8
1 / -0

Suppose $\displaystyle A_{1},A_{2}.....A_{30}$ are thirty sets having 5 elements and $\displaystyle B_{1},B_{2}....B_{n}$ are n sets each with 3 elements. Let $\displaystyle \bigcup_{i=1}^{30}Ai=\bigcup_{i=1}^{n}Bj=S$ and each elements of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n is equal to

A
35

B
45

C
55

D
65

Solution

Since each element of S belongs to $10$ $A_{i}s$ , there are only $15$ elements in S $\left[\dfrac{30\times 5}{10}\right]$

Since each of the $15$ elements belongs to 9 $B_j s$ there are 135 elements in the second union.

No. of sets $=\dfrac{135}{3}=45$
Question 9
1 / -0

S = {1, 2, 3, 5, 8, 13, 21, 34}. Find $\displaystyle \sum max\left ( A \right )$ where the sum is taken over all 28 two elements subsets A to S

A
844

B
480

C
484

D
488

Solution
Question 10
1 / -0

Given n(A) = 11, n(B) = 13, n(C) = 16, $\displaystyle n\left ( A\cap B \right )=3,n\left ( B\cap C \right )=6,n\left ( A\cap C \right )=5\: \: and\: \: n\left ( A\cap B\cap C \right )=2$ then the value of $\displaystyle n [ A\cup B \cup C ]=$

A
$24$

B
$27$

C
$25$

D
$28$

Solution

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

$=11+13+16-3-6-5+2=28$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Set Theory Test 13

Result

Set Theory Test 13

Score

-

Rank

-

SHARING IS CARING

Question 1

{1,2,3}\displaystyle \left \{ 1,2,3 \right \}{1,2,3}

{1,4}\displaystyle \left \{ 1,4 \right \}{1,4}

{1}\displaystyle \left \{ 1 \right \}{1}

{16}\displaystyle \left \{ 16 \right \}{16}

Solution

Question 2

(P∩Q′)∪(P∪Q)′=ξ∩Q′=ξ∩Q′=ξ\displaystyle (P\cap Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi (P∩Q′)∪(P∪Q)′=ξ∩Q′=ξ∩Q′=ξ

(P∪Q′)∪(P∪Q)′=ξ∩Q′=ξ∩Q′=ξ\displaystyle (P\cup Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi (P∪Q′)∪(P∪Q)′=ξ∩Q′=ξ∩Q′=ξ

(P∩Q′)∪(P∩Q)′=ξ∩Q′=ξ∩Q′=ξ\displaystyle (P\cap Q')\cup (P\cap Q)'=\xi \cap Q'=\xi \cap Q'=\xi (P∩Q′)∪(P∩Q)′=ξ∩Q′=ξ∩Q′=ξ

none of the above

Solution

Question 3

n(A−B)=n(A)−n(B)n(A - B) = n(A) - n(B)n(A−B)=n(A)−n(B)

n(A−B)=n(B−A)n(A - B) = n(B - A)n(A−B)=n(B−A)

n(A−B)=n(A)−n(A - B) = n(A) -n(A−B)=n(A)− n(A∩B)\displaystyle n\left ( A\cap B \right )n(A∩B)

n(A−B)=n(B)−n(A - B) = n(B) - n(A−B)=n(B)−n(A∩B)\displaystyle n\left ( A\cap B \right )n(A∩B)

Solution

Question 4

40

50

55

60

Question 5

373

165

370

None

Solution

Question 6

ϕ\displaystyle \phi ϕ

A

B

N

Solution

Question 7

35

45

55

65

Solution

Question 8

35

45

55

65

Solution

Question 9

844

480

484

488

Solution

Question 10

242424

272727

252525

282828

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.

$\displaystyle \left \{ 1,2,3 \right \}$

$\displaystyle \left \{ 1,4 \right \}$

$\displaystyle \left \{ 1 \right \}$

$\displaystyle \left \{ 16 \right \}$

$\displaystyle (P\cap Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi$

$\displaystyle (P\cup Q')\cup (P\cup Q)'=\xi \cap Q'=\xi \cap Q'=\xi$

$\displaystyle (P\cap Q')\cup (P\cap Q)'=\xi \cap Q'=\xi \cap Q'=\xi$

$n(A - B) = n(A) - n(B)$

$n(A - B) = n(B - A)$

$n(A - B) = n(A) -$ $\displaystyle n\left ( A\cap B \right )$

$n(A - B) = n(B) -$ $\displaystyle n\left ( A\cap B \right )$

$\displaystyle \phi$

$24$

$27$

$25$

$28$