Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

If $n(A) = n(B)$ then

  1. $n(A - B) = n(B - A)$
  2. $n(AB) = n(A) + n(B)$
  3. $n(A - B) =\phi$
  4. $n(AB) = n(B) - n(A - B)$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

If, $n(A)=n(B)$

a.  $n(A-B) = n(B-A)$. As, the no. of elements are same, if we subtract A from B or B from A, we will get the same no. of elements

b.  $n(AB)\neq n(A)+n(B)$. It is $n(A\cup B)=n(A)+n(B)$

c.  $n(A-B)= \phi$

d.  $n(AB)\neq n(B) - n(A-B)$. It is $n(AB) = n^2$ where n is the no. of elements of these sets

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

If $n(A) = n(B)$ then:

  1. $n(A- B) = n(B- A)$
  2. $n(AB)= n(A) + n(B)$
  3. $n(A- B)=n(A)-n(B)$
  4. $n(AB) = n(B) - n(A-B)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given: $n(A) = n(B)$
$n(A)$ is the cardinal no. of set $A$ and same for the set $B$

Thus, the number of elements are always same no matter what type of operation we are performing.

Hence, $n(A-B)=n(B-A)$.
Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

Let $U$ be the universal set for sets $A$ and $B$ such that $n(A)=200 , n(B)=300$ and $n(A\cap B)=100$, then $n(A'\cap B')$ is equal to $300$ provided that $n(U)$ is equal to

  1. $600$
  2. $700$
  3. $800$
  4. $900$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$=200+300-100$
$=400$
$n(A'\cap B')=n(A\cup B)'$
                    $=n(U) - n(A\cup B)$
$300=n(U)-400$
$n(U)=700$

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

A market research group conducted a survey of $2500$ consumers and reported that $1620$ consumers like product $p _{1}$ and $1500$ consumers like product $p _{2}$ then (Note $A$ and $B$ denotes the set of products $p _{1}$ and $p _{2}$ respectively)

  1. $\displaystyle n\left ( A \cup B \right )\geq 620$
  2. $\displaystyle n\left ( A \cap B \right )\leq 1500$
  3. $\displaystyle 620 \leq n\left ( A \cap B \right )\leq 1500$

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$n(A)  =1620, n(B) = 1500$

$\Rightarrow n(A\cap B) \leq min{n(A), n(B)} $

$\Rightarrow n(A\cap B) \leq 1500$

Also $n(A\cup B) \geq  n(A)+n(B) -n(U) =620$

 $n(A\cup B) \geq620$

Hence all options are correct.

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

In a community it is found that $52$% people like coffee and $73$% like tea. If $x\%$ like both coffee and tea then

  1. $\displaystyle x\geq 25$
  2. $\displaystyle x\leq 52$
  3. $\displaystyle 25\leq x\leq 52 $

  4. all of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $A=$ number of people like coffee, $B=$ number of people like tea.


$\therefore \ n(A)=52$%  $ \ n(B) = 73$%


$\displaystyle n\left ( A\cap B \right ) =x$%

Let total people in the community =$100 \displaystyle = n\left ( U \right )$
 
$\displaystyle \therefore n\left ( A\cup B \right )\leq 100$ 

$\displaystyle n\left ( A \right )+n\left ( B \right )-n\left ( A\cap B \right )\leq 100$ 

$\displaystyle 52+73-x\leq 100$ 

$\Rightarrow \displaystyle 125-x\leq 100$ 

$\displaystyle\Rightarrow  \therefore x\geq 25$...(i)

Again $\displaystyle A\cap B\subseteq A$

$\displaystyle \Rightarrow n\left ( A\cap B \right )\leq n\left ( A \right )$ 

$\displaystyle x \leq 52$ ...(ii)

$\displaystyle

\therefore $ By (i) and (ii)$\displaystyle 25\leq x\leq 52$