Tag: introduction to set

Quizzes (6) Questions (82)

Questions Related to introduction to set

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

Americans like at least one of cheese or apples. A survey shows that $63$% of the Americans like cheese while $76$% like apples. If $x$ % of the Americans like both cheese and apples, then

  1. $x = 39$
  2. $x= 63$
  3. $3 \leq x \leq 63$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $A$ be the percent of americans like cheese
           $B$ be the percent of americans like apples

$p(A)=\dfrac{63}{100}=0.63$

$p(B)=\dfrac{76}{100}=0.76$

Let $p(A\cap B)=x$

$p(A\cup B)=1$ as every american likes either cheese or apples

$p(A\cup B)=p(A)+p(B)-p(A\cap B)$

$1=0.63+0.76-x$

$x=1.39-1$

$x=0.39\Rightarrow x=0.39\times 100=39\%$
Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

Let the sets $A={2,4,6, 8, ...}$ and $B={3, 6, 9, 12, ...}$, and $n(A)=200, n(B)=250$. Then

  1. $n\left ( A\cap B \right )=67$
  2. $n\left ( A\cup B \right )=450$
  3. $n\left ( A\cap B \right )=66$
  4. $n\left ( A\cup B \right )=384$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

In A, last term will be $400$.

In B, the terms are also in A.P having a common difference of $3$.

Hence 

$a _{n}=a _1+(n-1)d$.

Now $n=250$ for the last term.

Hence

$a _{250}=3+(250-1).3$
$=3(1+250-1)$
$=750.$

Now $A\cap B$ will have elements which are multiples of $6$.

Last term will be $400-4=396$.

Hence
$a _{n}=a+(n-1).d$
$d=6,n=?,a=6$ and $a _{n}=396$

Hence
$396=6+(n-1).6$
Or 
$66=n$.

Hence
$n(A\cap B)=66$.

Now 
$n(A \cup B)=n(A)+n(B)-n(A\cap B)$
$=200+250-66$
$=384$.

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

In a group of children $35$ play football out of which $20$ play football only, $22$ play hockey; $25$ play cricket out of which $11$ play cricket only. Out of these $7$ play cricket and football but not hockey, $3$ play football and hockey but not cricket and $12$ play football and cricket both. How many play all three games?

  1. $5$
  2. $2$
  3. $12$
  4. $60$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $F,H$ and $C$ denote the no. of children who play Football, Hockey and Cricket respectively.
Given $n(F)=35,n(H)=22,n(C)=25$ and $n(F\cap C\cap H^{\prime})=7,n(F\cap H\cap C^{\prime})=3,n(F\cap C)=12$
but $n(F\cap C\cap H^{\prime})=n(F\cap C)-n(F\cap C\cap H)$
$\Rightarrow 7=12-n(F\cap C\cap H)$
$\Rightarrow n(F\cap C\cap H)=5$
$\therefore$ No of children who play all three games is $5$
Hence, option A

Multiple choice maths introduction to set different sets de morgan's law de morgan's law for set theory
Given, universal set = {$x \,\,\epsilon\,\, Z$ : $- 6 < x \leq 6$}, N = {$n$ : $n$ is a non-negative number} and P = {$x$ : $x$ is a nonpositive number}. Find :$P'$
  1. $\{-1,-2,-3,-4,-5,-6\}$
  2. $\{1,2,3,4,5,6\}$
  3. $\{0,1,2,3,4,5,6\}$
  4. $\{0,-1,-2,-3,-4,-5,-6\}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Universal set includes ${-5,-4,-3,-2,-1,0,1,2,3,4,5,6}$


P=${-5.-4.-3,-2,-1,0}$.

Hence P' only has the elements in option B. It does not contain 0.
P'=${1,2,3,4,5,6}$.

Multiple choice maths introduction to set different sets de morgan's law de morgan's law for set theory

If the universal set ${x\in W ,3<x≤12} ,A={5,7,9}$, then $A'=$

  1. $\{3,6,8,10,11,12\}$
  2. $\{4,6,8,10,11,12\}$
  3. $\{6,8,10,11,12\}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given universal set is $\{x\in W ,3<x≤12\}$
It can be written as a set $\{4,5,6,7,8,9,10,11,12\}$
$A$ is given as $A=\left\{5,7,9\right\}$
$\therefore A'=W-A=$ All the elements in the universal set but not in set $A$
         $ =\{4,6,8,10,11,12\}$
Hence, option B is correct
Multiple choice maths introduction to set different sets de morgan's law de morgan's law for set theory

Let  $U={x: \in, W: 3<x< 12} $, $B={4,6,8,10}$ . $B'$

  1. $\{6,7,9,11,12\}$
  2. $\{5,7,9,11\}$
  3. $\{5,7,9,10,11,12 \}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given universal set is $\{x\in W ,3<x<12\}$
It can be written as a set $\{4,5,6,7,8,9,10,11\}$
$B$ is given as $B=\left\{4,6,8,10\right\}$
$\therefore B'=W-B=$ All the elements in the universal set but not in set $B$
         $ =\{5,7,9,11\}$
Hence, option B is correct