Set Theory Test 18

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Set Theory Test 18

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Weekly Quiz Competition

Question 1
1 / -0

For any three sets , A B and C, $$B\setminus (A \cup C)$$ is:

A
$$(A\setminus B)\cap (A\setminus C)$$

B
$$(B\setminus A)\cap (B \setminus C)$$

C
$$(B \setminus A)\cap (A\setminus C)$$

D
$$(A\setminus B)\cap (B \setminus C)$$

Solution

In the problem statement we are taking union of $$A$$ and $$C$$ and then taking its difference from $$B$$
This means $$B$$ \ $$ (A\cup C)$$ will contain elements that are in $$B$$ but not in both $$B$$ $$\textbf{and}$$ $$C$$.
$$\therefore B$$ \ $$ (A\cup C) $$ is equivalent to taking difference of $$A,B$$ and $$A,C$$ and then taking there intersection i.e. $$(B$$ \ $$ A)\cap (B$$ \ $$ C)$$
Hence option ($$b$$) is correct.
Question 2
1 / -0

Which one of the following is not true?

A
$$A\setminus B=A\cap B'$$

B
$$A\setminus B=A\cap B$$

C
$$A\setminus B=(A\cup B)\cap B'$$

D
$$A\setminus B=(A\cup B)\setminus B$$

Solution

$$A$$ \ $$B$$ contains elements that are in set $$A$$ and not in set $$B$$
$$A\cap B$$ contains elements that are both in $$A$$ and $$B$$
Therefore option ($$b$$) contains the incorrect statement.
Question 3
1 / -0

If A and B are finite sets and $$A \subset B$$, then

A
$$n(A \cap B) = \phi$$

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

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

D
$$n(A \cup B) = n (A)$$

Solution

A and B are finite set and $$A\subset B$$
$$A\subset B$$ means A is a subset of B
If A is subset of B, then all elements of A are present in B
$$\therefore n\left( A\bigcup { B } \right) =n\left( B \right) $$
Question 4
1 / -0

Let X be a set of $$5$$ elements. The number d of ordered pairs (A, B) of subsets of X such that $$A\neq \Phi, B\neq \Phi, A\cap B=\Phi$$ satisfies.

A
$$50\leq d \leq 100$$

B
$$101\leq d \leq 150$$

C
$$151 \leq d \leq 200$$

D
$$201 \leq d$$

Solution

Total no. of ordered pairs is
$${ =C }_{ 2 }^{ 5 }\times 2!+{ C }_{ 3 }^{ 5 }\left( \frac { 3! }{ 1!\times 2! } \times 2! \right) +{ C }_{ 4 }^{ 5 }\left( \frac { 4! }{ 1!\times 3! } \times 2!+\frac { 4! }{ 2!\times 2! } \times \frac { 2! }{ 2! } \right) +{ C }_{ 5 }^{ 5 }\left( \frac { 5! }{ 1!\times 4! } \times 2!+\frac { 5! }{ 2!\times 3! } \times 2! \right) \\ =10(2)+10(6)+5(8+6)(10+20)\\ =180\\ $$
So option $$C$$ is correct
Question 5
1 / -0

Consider the following:
1. $$A\cup \left( B\cap C \right) =\left( A\cap B \right) \cup \left( A\cap C \right) $$
2. $$A\cap \left( B\cup C \right) =\left( A\cup B \right) \cap \left( A\cup C \right) $$
Which of the above is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Figure $$1$$ represents $$A\cup (B\cap C)$$ and figure $$2$$ represents $$(A\cap B)\cup (A\cap C)$$. We ca see both of them are not equal.
Figure $$3$$ represents $$A\cap (B\cup C)$$ and figure $$4$$ represents $$(A\cup B)\cap (A\cup C)$$. We can see both of them are not equal.
So, both the equations are incorrect.
Question 6
1 / -0

Directions For Questions

In a city, three daily newspapers A, B, C are published, $$42\%$$ read A; $$51\%$$ read B; $$68\%$$ read C; $$30\%$$ read A and B; $$28\%$$ read B and C; $$36\%$$ read A and C; $$8\%$$ do not read any of the three newspapers.

...view full instructions

What is the percentage of persons who read only two papers ?

A
$$19\%$$

B
$$31\%$$

C
$$44\%$$

D
None of the above

Solution

Let total numbers of people be 100, So for newspapers
$$n(A)=42, n(B)=51, n(C)=68, n(A\cap B)=30$$
$$ n(B\cap C)=28$$
$$ n(A\cap C)=36$$

Also given, $$ n(\overline { A\cup B\cup C) } =8$$
$$ n(A\cup B\cup C)=100-n(\overline { A\cup B\cup C) } =100-8=92$$
Also,
$$ n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$$
$$92=42+51+68-30-28-36+n(A\cap B\cap C)$$
$$n(A\cap B\cap C)=25$$

Now, for people who read only two papers$$=n(A\cap B)+n(B\cap C)+n(A\cap C)-3n(A\cap B\cap C)$$
$$=30+28+36-75$$
$$=19$$
Thus, $$19\%$$ people read only two newspapers.
Hence, A is correct.
Question 7
1 / -0

Let U = { $$x \in N : 1 \le x \le 10 $$ } be the universal set, $$N$$ being the set of natural numbers. If $$A = \{1, 2, 3, 4\}$$ and $$B = \{2, 3, 6, 10\} $$ then what is the complement of $$(A - B)$$ ?

A
$$\{6, 10\}$$

B
$$\{1, 4\}$$

C
$$\{2, 3, 5, 6, 7, 8, 9, 10\}$$

D
$$\{5, 6, 7, 8, 9, 10\}$$

Solution

Now $$U=\{ 1,2,3,4,5,6,7,8,9,10\} $$
and $$ (A-B)=\{ 1,4\}$$
So the compliment of $$(A-B)$$ considering $$U$$ as the universal set is
$$\{ 2,3,5,6,7,8,9,10\}$$
Question 8
1 / -0

Out of 500 first year students, 260 passed in the first semester and 21 0 passed in the second semester. If 170 did not pass in either semester, how many passed in both semesters ?

A
$$30$$

B
$$40$$

C
$$70$$

D
$$140$$

Solution

Let A be the set of students who passed first semester so $$n(A)=260$$
and B be the set of students who passed second semester so $$n(B)=210$$.
Now $$170$$ did not passed any semester
So, $$(500-170=330)$$ students passed atleast one of the semesters
$$\therefore n(A\cup B)=330$$
Now $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$
$$330=260+210-n(A\cap B)$$
$$n(A\cap B)=140$$
Question 9
1 / -0

A market research group conducted a survey of $$1000$$ consumers and reported that $$720$$ consumers like product A and $$420$$ consumers like product B. Then, the least number of consumers that must have liked both the products is.

A
$$140$$

B
$$180$$

C
$$210$$

D
$$190$$

Solution

Total consumers $$=1000$$
Like product $$A=n(A)=720$$
Like product $$B=n(B)420$$
$$n(A\cap { B })$$ (Both the products) $$=n(A)+n(B)-n(A\cup { B }) $$
$$=720+420-1000$$
$$=140$$
Question 10
1 / -0

In a group of $$50$$ people, two tests were conducted, one for diabetes and one for blood pressure. $$30$$ people were diagnosed with diabetes and $$40$$ people were diagnosed with high blood pressure. What is the minimum number of people who were having diabetes and high blood pressure?

A
$$0$$

B
$$10$$

C
$$20$$

D
$$30$$

Solution

Let $$A$$ be the set containing people diagnosed with diabetes and $$B$$ be the set containing people diagnosed with Blood pressure.
We know set formula
$$n(A\cup B)+n(A\cap B)=n(A)+n(B)$$
$$n(A\cap B)=x$$ are people having both
$$n(A\cup B)$$ are total people
$$n(A)$$ is people having diabetes and $$n(B)$$ is having blood pressure.
$$\Rightarrow 50+x=40+30\\ x=20$$
Thus, there should be minimum of 20 people having both.

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Set Theory Test 18

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Set Theory Test 18

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SHARING IS CARING

Question 1

$$(A\setminus B)\cap (A\setminus C)$$

$$(B\setminus A)\cap (B \setminus C)$$

$$(B \setminus A)\cap (A\setminus C)$$

$$(A\setminus B)\cap (B \setminus C)$$

Solution

Question 2

$$A\setminus B=A\cap B'$$

$$A\setminus B=A\cap B$$

$$A\setminus B=(A\cup B)\cap B'$$

$$A\setminus B=(A\cup B)\setminus B$$

Solution

Question 3

$$n(A \cap B) = \phi$$

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

$$n(A \cap B) = n (B)$$

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

Solution

Question 4

$$50\leq d \leq 100$$

$$101\leq d \leq 150$$

$$151 \leq d \leq 200$$

$$201 \leq d$$

Solution

Question 5

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 6

$$19\%$$

$$31\%$$

$$44\%$$

None of the above

Solution

Question 7

$$\{6, 10\}$$

$$\{1, 4\}$$

$$\{2, 3, 5, 6, 7, 8, 9, 10\}$$

$$\{5, 6, 7, 8, 9, 10\}$$

Solution

Question 8

$$30$$

$$40$$

$$70$$

$$140$$

Solution

Question 9

$$140$$

$$180$$

$$210$$

$$190$$

Solution

Question 10

$$0$$

$$10$$

$$20$$

$$30$$

Solution

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