Tag: union and intersection of sets

Questions Related to union and intersection of sets

If X=(multiples of $2$ ), Y = ( multiples of $5$) , Z= (multiples of $10$), then $ \displaystyle X \cap(Y\cap Z)    $ is equal to 

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. Multiples of $10$

  2. Multiples of $5$

  3. Multiples of $2$

  4. Multiples of $7$

Show answer
Correct Option: A
Explanation:
$x= (multiples\ of\ 2\ is\ 2, 4, 6, 8, 10.............)$
$y=(multiples\ of\ 5\ is\ 5, 10,15,20,25.........)$
$z=(multiples\ of\ 10\ is\ 10,20,30,40...........)$
Then, $ X\cap(Y\cap Z)$

Apply the value
$=(2, 4, 6, 8, 10,.......)\cap[ (5, 10, 15, 20, 25,........)\cap (10, 20, 30, 40,.......)]$
$=(2, 4, 6, 8, 10,.......)\cap (10, 20, 30,........)$
$=(10, 20, 30,........)$

Hence, this is multiple of $10$.


Hence, this is the answer.

If $A={x:x^2-3x+2=0}$ and $B={x:x^2+4x-5=0}$ then the value of $A-B$ is

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. ${1,2}$

  2. ${2}$

  3. ${1}$

  4. ${5,2}$

Show answer
Correct Option: B
Explanation:

$A={x:x^2-3x+2=0}\implies A={1,2}$
$B={x:x^2+4x-5=0}\implies A={1,-5}$
$\therefore A-B={2}$

There are $19$ hockey players in a club. On a particular day $14$ were wearing the prescribed hockey shirts, while $11$ were wearing the prescribed hockey pants. None of them was without hockey pant or hockey shirt. How many of them were in complete hockey uniform?

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. $8$

  2. $6$

  3. $9$

  4. $7$

Show answer
Correct Option: B
Explanation:

$n(S\cup P) = 19$

$n(S)=14 $
$ n(P)= 11 $
$n(S\cup P)= n(S)+n(P)-n(S\cap P)$
$19=14+11-n(S\cap P)$
$n(S\cap P)= 25-19 = 6$

Out of $450$ students in a school, $193$ students read Science Today, $200$ students read Junior Statesman, while $80$ students read neither. How many students read both the magazines?

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. $137$

  2. $80$

  3. $57$

  4. $23$

Show answer
Correct Option: D
Explanation:

$n(U)=450$

n(Students read Science Today) $=193 =n(S) $

n(Students read Junior Statesman) $=200=n(J) $

n(students read neither) $=80$

$n(S \cup J) $= n(U)-n(students read neither) $= 450-80=370$

Also, $n(S \cup J) = n(S) + n(J)-n(S\cap J) $

$370 = 193+200-n(S\cap J) $

$n(S\cap J)=393-370 =23 $

In a community of $175$ persons, $40$ read the Times, $50$ reads the Samachar and $100$ do not read any. How many persons read both the papers?

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. $10$

  2. $15$

  3. $20$

  4. $25$

Show answer
Correct Option: B
Explanation:

$n(U)=175$

n(read Times) $=40 =n(T) $

n(read the samachar) $=50 = n(S) $

n(do not read any) $= 100$

$n(T\cup S)=  n(U)-$ n(do not read any) $= 175-100 =75$

$\therefore n(T \cup S)= n(T)+ n(S)- n(T\cap S) $

$n(T\cap S) =90-75 =15$

In a group of $15, 7$ have studied, German, $8$ have studied French, and $3$ have not studied either. How many of these have studied both German and French?

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. $0$

  2. $3$

  3. $4$

  4. $5$

Show answer
Correct Option: B
Explanation:

$n(U)=15$

$n(German) =7 =n(G) $

$n(French) =8 =n(F) $

n(students who have studied neither) $=3$

$n(G \cup F) = n(U)-$ n(students who studied neither) $= 15-3=12 $

$n(G \cup F) = n(G) + n(F)-n(G\cap F) $

$12 = 7+8-n(G\cap F) $

$n(G\cap F)=15-12=3 $

In a class consisting of $100$ students, $20$ know English and $20$ do not know Hindi and $10$ know neither English nor Hindi. The number of students knowing both Hindi and English is

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. $5$

  2. $10$

  3. $15$

  4. $20$

Show answer
Correct Option: B
Explanation:

$n(U)=100$

$n(English) =20 =n(E) $

$n(Hindi) =100- 20= 80 =n(H) $

n(students who have studied neither Hindi nor English) $=10$

$n(E \cup H) = n(U)-$ n(students who studied neither) $= 100-10 =90 $

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

$90 = 20+80-n(E\cap H) $

$n(E\cap H)= 10$

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 $A \cap B$

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

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

Show answer
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?

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. $30$

  2. $22$

  3. $12$

  4. $8$

Show answer
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?

mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

  1. $55$

  2. $44$

  3. $36$

  4. $28$

Show answer
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) $= 28+16 =44$