Tag: basic operations on sets

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

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

  1. Multiples of $10$
  2. Multiples of $5$
  3. Multiples of $2$
  4. Multiples of $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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?

  1. $8$
  2. $6$
  3. $9$
  4. $7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$n(S\cup P) = 19$

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

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

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?

  1. $137$
  2. $80$
  3. $57$
  4. $23$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $

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

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?

  1. $10$
  2. $15$
  3. $20$
  4. $25$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

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

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

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?

  1. $0$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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

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

  1. $5$
  2. $10$
  3. $15$
  4. $20$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

$n(E\cap H)= 10$

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

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. $\left \{1, 3, 5, 8\right \}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

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

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. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

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

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. $28$
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $= 28+16 =44$