Tag: union and intersections

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Questions Related to union and intersections

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

Let $S=\left{ \left( x,y \right) :\dfrac { y\left( 3x-1 \right)  }{ x\left( 3x-2 \right)  } <0 \right}$ and $S'=\left{ \left( x,y \right) \in A\times B;\ -1\le A\le 1,-1\le B\le 1 \right} $ There area of $S\cap S'$ is

  1. $1$
  2. $3$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The region S is defined by the inequality involving fractions, and S prime is a bounded square region. Solving the inequality for x and y and intersecting with the square yields a region of area 2. Careful sketching of the sign scheme for the rational expression reveals the valid domain.

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

Let A={1, 2, 3, 4), B={2, 3, 4, 5}, then $n{ (A\times B)\cap (B\times A)} =$?

  1. 13

  2. 16

  3. 9

  4. 10
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sets A and B share three common elements, namely 2, 3, and 4. The intersection of A cross B and B cross A is equivalent to the Cartesian product of the intersection of A and B with itself. Since the intersection has 3 elements, its square has 3 times 3, or 9 elements.

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

Let $P={ \theta :sin\theta -cos\theta =\sqrt { 2 } cos\theta } $ and $Q={ sin\theta + cos\theta =\sqrt { 2 } sin\theta } $ be two sets. Then:

  1. $P\subset Q\quad and\quad Q-P\neq \emptyset $
  2. $Q\subset P$
  3. $P\subset Q$
  4. $P=Q$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$P = \left{ {\theta :\sin \theta  - \cos \theta  = \sqrt 2 \cos \theta } \right}$

$Q = \left{ {\theta :\sin \theta  + \cos \theta  = \sqrt 2 \sin \theta } \right}$
From $P$
$\sin \theta  - \cos \theta  = \sqrt 2 \cos \theta $
$\sin \theta  = \left( {\sqrt 2  + 1} \right)\cos \theta $
$\frac{{\sin \theta }}{{\cos \theta }} = \left( {\sqrt 2  + 1} \right)$
$\tan \theta  = \left( {\sqrt 2  + 1} \right)$
from $Q$
$\sin \theta  + \cos \theta  = \sqrt 2 \sin \theta $
$\sin \theta \left( {\sqrt 2  - 1} \right) = \cos \theta $
$\frac{{\sin \theta }}{{\cos \theta }} = \left( {\sqrt 2  - 1} \right)$
$\tan \theta  = \left( {\sqrt 2  - 1} \right)$
$\tan \theta  = \frac{1}{{\sqrt 2  - 1}} \times \frac{{\sqrt 2  + 1}}{{\sqrt 2  + 1}} = \sqrt 2  + 1$
$\therefore P = Q$
Hence,
option $(D)$ is correct answer.

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

If $A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8}$ and C= ${3,4,5,6}$, 

then verify : $A - (B \cup C) = (A - B) \cap (A - C)$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6, 8\}$ and $C=\{3,4,5,6\}$

For the LHS:

Union of two sets will have the elements of both sets.

So, $ B \cup C = \{2,3,4,5,6,8 \}$ 

$ A - (B \cup C) $ will have elements of $A$ which are not in $ (B \cup C) $

So, $ A - (B \cup C) = \{ 1 \}$ ..... $(1)$

For the RHS:

$ A - B $ will have elements of $A$ which are not in $B$.

So, $ A - B = \{ 1,3,5 \}$  

$ A - C $ will have elements of $A$ which are not in $C$.

So, $ A - C = \{ 1,2 \}$  

Intersection of two sets has the common elements of both the sets. 

$\Rightarrow (A - B) \cap (A - C) = \{1\}$ ..... $(2)$

From $(1)$ and $(2),$ we have

$ A - (B \cup C) =(A - B) \cap (A - C) $

Hence, the given expression is true.
Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

Let $A = {$ multiples of $3$ less than $20 }$
      $B = {$ multiples of $5$ less than $20}$
Then  $A$ $\displaystyle\cap$ $B$ is

  1. $\{3, 5\}$
  2. $\{5, 9\}$
  3. ${15}$
  4. $\displaystyle\phi $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$A = {$ multiples of $ 3 $ less than $20}$

    $= {3,6,9,12,15,18}$
$B={ $ multiples of $5$ less than $20}$
    $= {5,10,15}$

$A \cap B = 15$

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

Let $P _1$ be the set of all prime numbers, i.e., $P _1=\left {2, 3, 5, 7, 11, ....\right }$, Let $Pn=\left {np|p\epsilon P _1|\right }$, i.e., the set of all prime multiples of n. Then which of the following sets is non empty?

  1. $P _1\cap P _{23}$
  2. $P _7\cap P _{21}$
  3. $P _{12}\cap P _{20}$
  4. $P _{20}\cap P _{24}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Check by option
$P _{12}=\left {24, 36, 60, 84, ....\right }$
$P _{20}=\left {40, 60, 100, .....\right }$
$P _{12}\cap P _{20}$ has common element.

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

In a group of $760$ persons, $510$ can speak Hindi and $360$ can speak English. Find how many can speak Hindi only.

  1. $250$
  2. $400$
  3. $1270$
  4. $150$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
People who can speak both Hindi and English $= n (H ∩ E)$

$n (P) = n(E) + n(H) – n (H ∩ E)$

$n(E\cap H)=510+360-760=110$

We can see that, $H$ is disjoint union of $n(H–E)$ and $n (H ∩ E).$


(If $A$ and $B$ are disjoint then $n (A ∪ B) = n(A) + n(B))$

$∴ H = n(H–E) ∪ n (H ∩ E).$

$⇒ n(H) = n(H–E) + n (H ∩ E).$

$⇒ 510 = n (H – E)+ 110$

$⇒ n(H–E) = 400.$

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

There are $25$ trays on a table in the cafeteria. Each tray contains a cup only, a plate only, or both a cup and a plate. If $15$ of the trays contain cups and $21$ of the trays contain plates, how many contain both a cup and a plate?

  1. $10$
  2. $11$
  3. $12$
  4. $13$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $'A'$ denote the set with a cup
$'B'$ denote the set with a plate
$'A$ $\cap$ $B'$ denote the set with both a cup and a plate
$'A$ $\cup$ $B'$ denote the set with the trays either a cup or a plate
$n($ $X)$ denote the number of elements in the set $'X'$

Given, total number of trays $n(A$ $\cup$ $B)$ $=$ $25$
Number of trays that contain cups $n($$A)$ $=$ $15$
Number of trays that contain cups $n($$B)$ $=$ $21$

To find the trays with both a cup and a plate $n(A$ $\cap$ $B)$,

We know that
$n(A$ $\cup$ $B)$ $=$ $n($$A)$$+$ $n($$B)$ $-$ $n(A$ $\cap$ $B)$
Rearranging the terms, we get
$n(A$ $\cap$ $B)$ $=$ $n($$A)$$+$ $n($$B)$ $-$ $n(A$ $\cup$ $B)$
From the above,
$n(A$ $\cap$ $B)$ $=$ $15$ $+$ $21$ $-$ $25$
$=$ $11$
$n(A$ $\cap$ $B)$ $=$ $11$

Therefore, number of trays with both a cup and a plate is $'11'$.
Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

Let A = {x : x is a square of a natural number and x is less than 100} and B is a set of even natural numbers. What is the cardinality of $ A \cap B$ ?

  1. 4

  2. 5

  3. 9

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$A = {1, 4. 9. 16, 25, 36, 49, 64, 81 }$

$B = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24, 26, 28, 30, 32, 34, 36, 38,...}$
$A \cap B = {4, 16, 36, 64}$
Hence, $n(a \cap B) = 4$.