Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

If among natural numbers $A={5,6,7}$ and $B={8,9,10}$ , then 

  1. $A \cap B =$ null
  2. $(A$ $\cup$ $B)' = A'\cap B'$
  3. $A$ $\cap$ $B = \{2,3,4\}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Since, no elements of A and B are the same.
Therefore, the Intersection is empty.
i.e. $A \cap B = $ null
We know, $U = $  { $ N $ }
$A = $  { $ 5,6,7 $ }
$B = $  { $ 8,9,10 $ } 
($A \cup B$) $=$ {$5,6,7$) $ \cup $ {$8,9,10$}
               $=$ {$5,6,7,8,9,10$}
Therefore,  ($A \cup B$)'$= $ All natural numbers except $ {5,6,7,8,9,10} \dots (i)$
Now, $A'= $ All Natural numbers except ${5,6,7}$
and $B'=$ All Natural numbers except ${8,9,10}$
Therefore, $A' \cap B' = $ All Natural numbers except ${5,6,7,8,9,10}\dots (ii)$
By comparing $(i) $ and $(ii)$ we get,
$(A \cup B) = A' \cap B'$

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

Given $A={x\in N :x<6} ,B={3,6,9}$ and $C={x \in N: 2x-5\le 8}$

  1. $A \cup $ (B $ \cap $C)=(A $ \cap $B) $\cap $(A $ \cap$ C)
  2. (A $\cup$B)'=A'$\cap$B'
  3. A $\cup$B=null set

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

$A = { x \in N : x < 6 }$

$\therefore A = { 1,2,3,4,5 }$
$B = { 3,6,9 }$
$C = { x \in N : 2x - 5 \leq 8 }$
$\Rightarrow C = { 1,2,3,4,5,6 }$
Option A. $A \cup (B \cap C) = A \cup {3, 6 } = { 1,2,3,4,5,6 }$
RHS $= (A \cap B) \cap ( A \cup C) = { 3 } \cap {1,2,3,4,5,6} = {3}$
LHS $\neq$ RHS

Option B. $(A \cup B)' = A' \cap B'$ is always true by De Morgan's Law  . 
$A\cup B={1,2,3,4,5,6,9}$
$(A\cup B)'={7,8,10,11,...}$
$A'={x\in N:x\geq 7}$ and $B'={1,2,4,5,7,8,10,11,12...}$
$A'\cap B'={7,8,10,11,12...}$
$\therefore (A\cup B)'=A'\cap B'$

Option C. $A \cap B = { 3 } \neq$ null set
Hence, only B is correct.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

If $U = {3, 4, 5, 6, 7, 8, 9}, X = {3, 4}, Y = {5, 6}$ and $Z = {7, 8, 9}$, then $\displaystyle Y'\cap \left ( X\cap Z \right )'$ is equal to 

  1. $\displaystyle X\cup Y$
  2. $\displaystyle Y\cup Z$
  3. $\displaystyle X'\cap Y'$
  4. $X\cup Z$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ { Y ' } = U - Y = { 3, 4, 7, 8, 9} $

$ X\cap Z = {0} $ as there is no element common in sets X and Z

$\Rightarrow (X\cap Z) ' = U - (X\cap Z) =  {3, 4, 5, 6, 7, 8, 9} $

Now, $Y' \cap (X\cap Z)'  = { 3, 4, 7, 8, 9} \cap {3, 4, 5, 6, 7, 8, 9} ={ 3, 4, 7, 8, 9} $

This is equal to $ X \cup Z = {3,4,7,8,9} $

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

If $U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$, $A = {0, 3, 4, 7}$ ,$B = {1, 2, 8, 9}$
then $(A U B)'$ is

  1. $\{2, 5\}$
  2. $\{5, 6\}$
  3. $\{8, 9\}$
  4. $\{6, 7\}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$, $A = {0, 3, 4, 7}$ ,$B = {1, 2, 8, 9}$


$A^c={1,2,5,6,8,9}, B^c= {0,3,4,5,6,7} $

By De Morgan's law

$(A\cup B)^c = A^c\cap B^c$

$=\{1,2,5,6,8,9\}\cap \{0,3,4,5,6,7\}$

$=\{5,6\}$

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

Out of 800 boys in a school 224 played cricket, 240 played hockey and 236 played basketball. Of the total 64 played both basketball and hockey, 80 played cricket and basketball and 40 played cricket and hockey, 24 players all the three games. The number of boys who did not play any game is

  1. $128$
  2. $216$
  3. $240$
  4. $260$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

No. of players who played at least one game is:


By set theory

$n(C\cup H\cup B)= n(C) +n(H) +n(B)-n(B\cap H)-n(C\cap B) -n(C\cap H) + n(C\cap H \cap B)$

$=224+240+236-64-80-40+24=540$

Hence $260$ players do not play any game.

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

For any two sets A and B, $\left { (A\setminus B)\cup (B\setminus A) \right }\cap (A\cap B)$ is:

  1. $\phi $
  2. $ A \cup B$
  3. $A \cap B$
  4. $A' \cap B'$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$(A$ \ $B)\cup(B$ \ $A)$ contains that elements form set $A$ and $B$ that are not contained in the other set.

$(A$ \ $B)\cup(B$ \ $A)=(A\cap B)$ \ $(A\cap B)$
$\therefore (A$ \ $B)\cup(B$ \ $A)$ does not contain elements of $A\cap B$
Hence {$(A$ \ $B)\cup(B$ \ $A)$}$\cap (A\cap B)=${$\phi$}

Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

In order to draw a graph of $f(x) = ax^{2} + bx + c$, a table of values was constructed. These values of the function for a set of equally spaced increasing values of $x$ were $3844, 4096, 4227, 4356, 4489, 4624$, and $4761$. The one which is incorrect is

  1. $4096$
  2. $4356$
  3. $4489$
  4. $4761$

  5. None of these
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

We are told that the values of $f(x)$ listed correspond to
7$f(x), f(x + h), f(x + 2h), ...., f(x + 7h)$.
Observe that the difference between successive values is given by
$f(x + h) - f(x) = a(x + h)^{2} + b(x + h) + c - (ax^{2} + bx + c)$
$= 2ahx + ah^{2} + bh$.
Since the difference is a linear function of $x$, it must change by the same amount whenever $x$ is increased by $h$. But the successive differences of the listed values
$3844\ 3969\ 4096\ 4227\ 4356\ 4489\ 4624\ 4761$
are $125\ 127\ 131\ 129\ 133\ 135\ 137$
so that, if only one value is incorrect; $4227$ is the value.