Tag: introduction to set

Questions Related to introduction to set

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. $150$


Correct Option: B
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 $n (A ∪ B) = n(A) + n(B))$

$∴ H = n(H–E) ∪ n (H ∩ E).$

$⇒ n(H) = n(H–E) + n (H ∩ E).$

$⇒ 510 = n (H – E)+ 110$

$⇒ n(H–E) = 400.$

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. $13$


Correct Option: B
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 $'11'$.

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


Correct Option: A
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, $n(a \cap B) = 4$.

If $A = ${1,4,6}, $B = ${3,6}, then find $(A \cap B)$

  1. $({1,3,4,6})$

  2. $\phi$

  3. ${6}$

  4. none of these


Correct Option: C
Explanation:
Given,
$A=(1,4,6)$, $B=(3,6)$

$(A\cap B)=common \  of \  A \  and \ B=6  $

$\therefore (A \cap B)=6$

$If\ A = {1, 2, 3}, B = {4, 5}, then\ find\ A \cap B$

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

  2. $\phi $

  3. ${4,5}$

  4. ${1, 2, 3}$


Correct Option: B

A survey shows that $63\%$ of Indians like mangoes whereas $76\%$ like apple. If $x%$ of the Indians like both mangoes and apples, then

  1. $x=39$

  2. $x=63$

  3. $39\le x\le 63$

  4. $None\ of\ these$


Correct Option: C
Explanation:
Let $100$ denote the population of India, then
$n(A)=63$, $n(B)=76$, $n(A\cap B)=x$
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$\therefore x=63+76-n(A\cap B)$ ..$(1)$
If should be noted that $n(A\cup B)\neq 100$ but $n(A\cup B)\le 100$ because there are Indians who may like other fruits besides mangoes and apples.
Hence from $(1)$, $x=139-n(A\cup B)$
$\therefore x\ge 39$ or $39\le x$
Again $A\cap B\subset A$, $A\cap B\subset B$
$\therefore n(A\cap B)\le n(A)=63$, $n(A\cap B)\le n(B)=76$
$\therefore x\le 63$.
Hence $39\le x\le 63$.

Two set A and B are as under 
A = {(a,b) $\epsilon$ R $\times$ R : $\mid a - 5\mid$ < $1$ and  $\mid b - 5\mid$ < $1$};
B = {(a,b) $\epsilon$ R $\times$ R : $4(a-6)^2 + 9(b-5)^2$ $\leq 36$. Then, 

  1. B $\subset$ A

  2. A $\subset$ B

  3. A $\bigcap$ B = $\phi$ (an empty set)

  4. nither A $\subset$ B nor B $\subset$ A


Correct Option: B

$R$ is the set of all positive odd integers less than $20$; $S$ is the set of all multiples of $3$ that are less than $20$. How many elements are in the set $R$ $\cap$ $S$?

  1. 0

  2. 1

  3. 2

  4. 3

  5. 4


Correct Option: D
Explanation:

Given, $R=$ {$1,3,5,7,9,11,13,15,17,19$} , $S=$ {$3,6,9,12,15,18$}
Therefore the intersection of $S$ and $R$ is {$3,9,15$}.
So the number of elements which are common to both is $3$.

Let $Z$ denotes the set of all integers and $A=\left{ \left( a,b \right) :{ a }^{ 2 }+3{ b }^{ 2 }=28,a,b\in Z \right} $ and $B=\left{ \left( a,b \right) :a < b,a,b\in Z \right} $. Then, the number of elements in $A\cap B$ is

  1. $2$

  2. $4$

  3. $6$

  4. $5$


Correct Option: C
Explanation:

$\because A=\left{ \left( a,b \right) :{ a }^{ 2 }+3{ b }^{ 2 }=28,a,b\in Z \right} $
   $=\left{ \left( 5,1 \right) ,\left( -5,-1 \right) ,\left( 5,-1 \right) ,\left( -5,1 \right) ,\left( 4,2 \right) ,\left( -4,-2 \right) ,\left( 4,-2 \right) ,\left( -4,2 \right) ,\left( 1,3 \right) ,\left( -1,-3 \right) ,\left( 1,-3 \right) ,\left( -1,3 \right)  \right}$
and $B=\left{ \left( a,b \right) :a<b,a,b\in Z \right} $
$\therefore A\cap B=\left{ \left( 1,3 \right) ,\left( -1,3 \right) ,\left( -4,-2 \right) ,\left( -4,2 \right) ,\left( -5,-1 \right) ,\left( -5,1 \right)  \right} $
$\therefore $ The number of elements in $A\cap B$ is $6$.

Let $A,B$ be two sets such that $A\cap B=\phi$, then $A=\phi$ and $B=\phi$?

  1. True

  2. False


Correct Option: B
Explanation:

For two sets $A,B$ if $A\cap B=\phi$ then it is not necessary that $A=\phi$ and $B=\phi$.

For example,
Let $A={1,2,3}$ and $B={4,5,6}$.
Now $A\cap B=\phi$ still $A\ne \phi $ and $B\ne \phi$.