Tag: introduction to set

Questions Related to introduction to set

If $A = \left {1, 2, 3, 4, 5, 6, 7, 8\right }$ and $B \left {1, 3, 5, 7\right }$, then find $A - B$ and $A \cap B$

  1. $\left {3, 5\right }$ and $\left {2, 4, 6\right }$

  2. $\left {2, 4, 6\right }$ and $\left {1, 5\right }$

  3. $\left {2, 4, 6, 8\right }$ and $\left {1, 3, 5, 7\right }$

  4. $\left {1, 3, 5, 8\right }$


Correct Option: C
Explanation:

$A=\{1,2,3,4,5,6,7,8\}$

$B=\{1,3,5,7\}$

$A-B=\{1,2,3,4,5,6,7,8\} - \{1,3,5,7\} = \{2,4,6,8\}$

$A \cap B = \{1,2,3,4,5,6,7,8\} \cap \{1,3,5,7\} =\{1,3,5,7\}$

In a certain group of $36$ people, $18$ are wearing hats and $24$ are wearing sweaters. If six people are wearing neither a hat nor a sweater, then how many people are wearing both a hat and a sweater?

  1. $30$

  2. $22$

  3. $12$

  4. $8$


Correct Option: C
Explanation:

$n(U)=36$

$n(Hats) =18 =n(H) $

$n(Sweaters) =24 =n(S) $

n(Wearing neither hat nor Sweater) =6

$n(S \cup H) = n(U)-$ n(Wearing neither hat nor sweater) $= 36-6 = 30 $

$n(S \cup H) = n(S) + n(H)-n(S\cap H) $

$30 = 24+18-n(S\cap H) $

$n(S\cap H)=42-30 = 12 $

In a class of $80$ children, $35$% children can play only cricket, $45$% children can play only table-tennis and the remaining children can play both the games. In all, how many children can play cricket?

  1. $55$

  2. $44$

  3. $36$

  4. $28$


Correct Option: B
Explanation:

n(Children who can play only cricket) $ = 35\% = 80\times \dfrac{35}{100} = 28$

n(Children who can play only table tennis) $ = 45\% = 80\times \dfrac{45}{100} = 36$

$n(C \cap T) = 80-28-36 = 16$

n(Children can play cricket) $= 28+16 =44$

If x belongs to set of integers, A is the solution set of $2(x-1)< 3x-1$ and B is the solution set of $4x-3\leq 8+x$, find A$\cap$B.

  1. $\left{0, 1, 2\right}$

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

  3. $\left{0, 1, 2, 3\right}$

  4. $\left{0, 2, 4\right}$


Correct Option: C
Explanation:
$2(x-1)< 3x-1$
$\Rightarrow 2x-2< 3x-1$
$\Rightarrow 2x-3x< -1+2$
$\Rightarrow -x< 1$
$\Rightarrow x> -1$

$4x-3\leq 8+x$
$4x-3-x+3\leq 8+x-x+3$
$\Rightarrow 4x-x\leq 8+3$
$\Rightarrow 3x\leq 11$
$x\leq \dfrac{11}{3}\Rightarrow x\le 3.67$
Then $-1< x\leq 3.67$
Then, $A\cap B$ is $ \{0,1,2,3 \}$

If $A = {1, 3, 5, 7, 8, 6}$, $B = {2, 4, 6, 8, 9}$ .Find $A\cap B$

  1. ${8}$

  2. ${9}$

  3. ${10}$

  4. None of these


Correct Option: A

If $A = \left {2, 3, 4, 8, 10\right }, B = \left (3, 4, 5, 10, 12\right }, C = \left {4, 5, 6, 12, 14\right }$, then $(A\cap B)\cup (A\cap C)$ is equal to

  1. $\left {3, 4, 10\right }$

  2. $\left {2, 9, 10\right }$

  3. $\left {4, 5, 6\right }$

  4. $\left {3, 5, 14\right }$


Correct Option: A
Explanation:

$ A={2,3,4,8,10} , B={3,4,5,10,12} , C={4,5,6,12,14}$


To find : $(A\cap B)$ $\cup  (A\cap C)$

$A\cap B$ = elements common in sets $A$ and $B$. 
$A\cap C$ = elements common in sets $A$ and $C$ 

$A\cap C = \{4\}$ 
$(A\cap B)\cup (A\cap C)=$elements in $(A\cap B)$ and $(A\cap C)$
$(A\cap B) \cup  (A\cap C) = \{3,4,10\} \cup \{4\} $
$(A\cap B) \cup  (A\cap C)={3,4,10}$

If U = {1, 2, 3, 4, 5, 6}; A = {3, 5}; B = {2, 3, 4} C = {4, 5}, find then A $\cap$ (B $\cup$ C).

  1. {1, 2, 3, 4, 5, 6}

  2. {3, 5}

  3. {2, 3, 4}

  4. {4, 5, 6}


Correct Option: B
Explanation:
Given, 
$U=\{1, 2, 3, 4, 5, 6\}$
$A=\{3, 5\}$
$B=\{2, 3, 4\}$
$C=\{4, 5\}$

$B\cup C=\{2, 3, 4, 5\}$

$A\cap (B\cup C)=\{3, 5\}$.

Which is the simplified representation of 
$\left( {{A^/} \cap \,{B^/} \cap \,C} \right) \cup \left( {B\, \cap \,C} \right) \cup \left( {A \cap \,C} \right)$  

where A,B,C are subsets of X

  1. A

  2. B

  3. C

  4. $X \cap \,\left( {A \cup B \cup C} \right)$


Correct Option: D

If $aN=\left{ ax:x\epsilon N \right}$, then the set $3N\cap 7N$ is

  1. $21\ N$

  2. $10\ N$

  3. $4\ N$

  4. None of these


Correct Option: A

If $n(U)= 60, n(A)= 35, n(B)= 24$ and $n(A \cup B)' = 10$ ,then $n(A \cap B)$ is

  1. $9$

  2. $8$

  3. $6$

  4. $7$


Correct Option: A
Explanation:
we have formula,

$n(A\cup B)=n(U)-n(A\cup B)'$

$n(A\cup B)=60-10=50$

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$50=35+24-n(A\cap B)$

$\therefore n(A\cap B)=9$