Sets Test

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Sets Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

If the domain of $$h(x)$$ is $$R-A$$ , where $$'R'$$ is set of real number then the product of all elements in set $$'A'$$ is

A
$$0$$

B
$$e^{2}$$

C
$$\dfrac {1}{e^{2}}$$

D
$$e^{4}$$

Solution

product of all elements in set
$$A'=e\times e=e^{2}$$
where demain of $$h(x)=R-A$$ & $$R$$ is a set of real numbers
Question 2
1 / -0

The set which begins with additive identity is

A
W

B
N

C
Q

D
Z

Solution

Additive identity is $$0$$.

The set that begins with additive identity is whole numbers which us denoted by $$W$$.
Question 3
1 / -0

If $$A$$ and $$B$$ are events such that
$$P(A\cup B)=\cfrac{3}{4},P(A\cap B)=\cfrac{1}{4},P(\overline { A } )=\cfrac{2}{3}$$, then $$P(\overline { A } \cap B)$$ is

A
$$\cfrac{5}{12}$$

B
$$\cfrac{3}{8}$$

C
$$\cfrac{5}{8}$$

D
$$\cfrac{1}{4}$$

Solution

Given, $$P(A \cup B)=\dfrac {3}{4}, P(A \cap B)=\dfrac {1}{4}, P(\bar A)=\dfrac {2}{3}$$
We need to find $$P(\bar A\cap B)$$.
Therefore, $$P(B)=\dfrac {2}{3}$$
We know $$P(B)=P(A\cup B)+P(A \cap B)-P(A)$$
Therefore, $$P(\bar A\cap B)=\dfrac {2}{3}-\dfrac {1}{4}=\dfrac {5}{12}$$
Question 4
1 / -0

$$A - ( A - B ) $$ is equivalent to which expression

A
$$B$$

B
$$A \cup B$$

C
$$A \cap B$$

D
$$B-A$$

Solution

$$A-\left(A-B\right)$$
$$=A-\left(A\cap{B}^{C}\right)$$
$$=A\cap{\left(A\cap{B}^{C}\right)}^{c}$$
$$=A\cap\left[{A}^{c}\cup{\left({B}^{c}\right)}^{c}\right]$$
$$=A\cap\left[{A}^{c}\cup\,B\right]$$ since $${\left({B}^{c}\right)}^{c}=B$$
$$=\left(A\cap{A}^{c}\right)\cup\,\left(A\cap\,B\right)$$
$$=\phi\,\cup\,\left(A\cap\,B\right)$$ since $$A\cap{A}^{c}=\phi$$
$$=A\cap\,B$$
Question 5
1 / -0

If set 's' contains all the real values of x for which $$log_ {(2x+3)^{x^2}}<1$$ is true, then set 'S' contain:

A
$$(log_25,log_2 7)$$

B
$$[log_34,log_3 8]$$

C
$$\left(\dfrac{-3}{2},1\right)$$

D
$$(-1,0)$$
Question 6
1 / -0

Let $$A=\left\{ 1,2,3,4 \right\} ,B=\left\{ 2,4,6 \right\} $$. Then the number of sets $$C$$ such that $$A\cap B\subseteq C\subseteq A\cup B$$ is

A
$$6$$

B
$$9$$

C
$$8$$

D
$$10$$

Solution

$$A=\left\{1,2,3,4\right\}, B=\left\{2,4,6\right\}$$

$$A\cap B=\left\{2,4\right\}\ \ \ A\cup B=\left\{1,2,3,4,6\right\}$$

$$A\cap B\subseteq C\subseteq A\cup B\Rightarrow C=\left\{2,4,D\right\}$$ where $$D$$ is subsets of $$\left\{1,3,6\right\}$$

$$\Rightarrow$$ No.of sets $$C$$= no.of subsets of $$\left\{1,3,6\right\}=8$$
Question 7
1 / -0

In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

A
10

B
11

C
12

D
13

Solution

Let $$A$$, $$B$$, and $$C$$ be the set of people who like $$product\ A$$, $$product\ B,\ and\ product\ C$$ respectively.

Accordingly,

$$ n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12$$,

$$n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8$$

The Venn diagram for the given problem can be drawn as above.

It can be seen that number of people who like product C only is
$$={29 – (4 + 8 + 6)} = 11$$
Question 8
1 / -0

Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set $$A \times B$$, each having at least three elements is............

A
$$275$$

B
$$510$$

C
$$219$$

D
$$256$$

Solution

Set A has $$4$$ elements
Set B has $$2$$ elements
Number of elements in set $$\left( A\times B\right)=4\times 2=8$$
Total number of subsets of $$\left( A\times B\right) =2^8=256$$
Number of subsets having $$0$$ elements $$=^{ 8 }{ { C }_{ 0 } }=1$$
Number of subsets having $$1$$ elements each $$=^{ 8 }{ { C }_{ 1 } }=8$$
Number of subsets having $$2$$ elements each $$=^{ 8 }{ { C }_{ 2 } }=\dfrac{8!}{2!6!}=28$$
Number of subsets having atleast $$3$$ elements $$=256-1-8-28$$
$$=256-37$$
$$=219$$
Hence, the answer is $$219.$$
Question 9
1 / -0

The number of subsets $$ R$$ of $$P=(1,2,3,....,9)$$ which satisfies the property "There exit integers a<b<c with a$$\in $$R, b$$\in $$R,c$$\in $$R" is

A
$$512$$

B
$$466$$

C
$$467$$

D
None of these

Solution

Every subset of P containing at least 3 elements from P will always satisfy the required property since each number is distinct in the set P.

So, Total no. of ways = $$^{9}C_3 + ^{9}C_4 + .... + ^{9}C_9 $$
$$= (^{9}C_0 + ^{9}C_1 + ..... + ^{9}C_9) - ^{9}C_0 - ^{9}C_1 - ^{9}C_2$$
$$=2^9 - 46$$
$$=512 -46$$
$$=466$$
Question 10
1 / -0

If the number of $$5$$ elements subsets of the set $$A\left\{\ a_{1},a_{2}.....a_{20}\right\}$$ of $$20$$ distinct elements is $$k$$ times the number of $$5$$ elements subsets containing $$a_{4}$$, then $$k$$ is

A
$$5$$

B
$$\dfrac{20}{7}$$

C
$$4$$

D
$$\dfrac{10}{3}$$

Solution

No of $$5$$ elements subset $$=^{20}C_{5}=\dfrac {20!}{15! \, 15!}=\dfrac {20 \times 19\times 18\times 17\times 16\times 15!}{15! \, 5\times 4\times 3\times 2\times 1}=15504$$

No of $$5$$ elements subset containing $$a_{4}$$ $$=^{19}C_{4}=\dfrac {19!}{4! \, 15!}$$

$$\Rightarrow \dfrac {20!}{15! \, 15!}=k\dfrac {19!}{4!\, 15!}$$

$$\Rightarrow \dfrac {20\times 19!}{5\times 4!}=k\dfrac {19!}{4!}$$

$$\Rightarrow k=4$$

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Re-Attempt Weekly Quiz Competition

Sets Test - 39

Result

Sets Test - 39

Score

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Rank

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SHARING IS CARING

Question 1

$$0$$

$$e^{2}$$

$$\dfrac {1}{e^{2}}$$

$$e^{4}$$

Solution

Question 2

W

N

Q

Z

Solution

Question 3

$$\cfrac{5}{12}$$

$$\cfrac{3}{8}$$

$$\cfrac{5}{8}$$

$$\cfrac{1}{4}$$

Solution

Question 4

$$B$$

$$A \cup B$$

$$A \cap B$$

$$B-A$$

Solution

Question 5

$$(log_25,log_2 7)$$

$$[log_34,log_3 8]$$

$$\left(\dfrac{-3}{2},1\right)$$

$$(-1,0)$$

Question 6

$$6$$

$$9$$

$$8$$

$$10$$

Solution

Question 7

10

11

12

13

Solution

Question 8

$$275$$

$$510$$

$$219$$

$$256$$

Solution

Question 9

$$512$$

$$466$$

$$467$$

None of these

Solution

Question 10

$$5$$

$$\dfrac{20}{7}$$

$$4$$

$$\dfrac{10}{3}$$

Solution

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