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Let be the set of all numbers, which are perfect squares, $\mathrm{B}$ be the set of all numbers which are multiples of 5 and $C$ be the set of all numbers, which are divisible by 2 and 3 . Consider
the following statements :
I. $A, B, C$ are mutually exdusive
II. $A, B, C$ are mutually exhaustive. III. The number of elements in the complement set of $\mathrm{A} \cup \mathrm{B}$ is 12
Which of the statements given above are the correct?
Options:
the following statements :
I. $A, B, C$ are mutually exdusive
II. $A, B, C$ are mutually exhaustive. III. The number of elements in the complement set of $\mathrm{A} \cup \mathrm{B}$ is 12
Which of the statements given above are the correct?
Solution:
2113 Upvotes
Verified Answer
The correct answer is:
I and III only
Let, $\mathrm{U}=\{1,2,3, \ldots, 20\}$
$\mathrm{A}=$ Set of all numbers which are perfect square $=\{1,4,9,16\}$
$\mathrm{B}=$ Set of all numbers which are multiples of $5=\{5,10,15,20\}$
$\mathrm{C}=$ Set of all numbers which are divisible by 2 and 3 $=\{6,12,18\}$
$\mathrm{A} \cup \mathrm{B}=\{1,4,9,16,5,10,15,20\}$
$\Rightarrow \mathrm{n}(\mathrm{A} \cup \mathrm{B})=8$
$\Rightarrow \mathrm{n}(\mathrm{A} \cup \mathrm{B})^{\prime}=20-8=12$
Also, $\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})$
$\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\{1,4,9,16,5,10,15,20,6,12,18\}$
$\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=11$
and $\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})=4+4+3=11$
$\mathrm{A}=$ Set of all numbers which are perfect square $=\{1,4,9,16\}$
$\mathrm{B}=$ Set of all numbers which are multiples of $5=\{5,10,15,20\}$
$\mathrm{C}=$ Set of all numbers which are divisible by 2 and 3 $=\{6,12,18\}$
$\mathrm{A} \cup \mathrm{B}=\{1,4,9,16,5,10,15,20\}$
$\Rightarrow \mathrm{n}(\mathrm{A} \cup \mathrm{B})=8$
$\Rightarrow \mathrm{n}(\mathrm{A} \cup \mathrm{B})^{\prime}=20-8=12$
Also, $\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})$
$\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\{1,4,9,16,5,10,15,20,6,12,18\}$
$\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=11$
and $\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})=4+4+3=11$
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