Tag: mathematics and statistics

Questions Related to mathematics and statistics

If $A = {2, 3, 5}$ and $B = {5, 7}$, find the set with highest number of elements:

  1. $A \times B$

  2. $ B \times A$

  3. $A \times A$

  4. $B \times B$


Correct Option: C
Explanation:
$A=\left \{ 2,3,5 \right \}$

$B=\left \{ 5,7 \right \}$

$A\times B=\left \{ (2,5),(2,7),(3,5),(3,7),(5,5),(5,7) \right \}$

$B\times A=\left \{ (5,2),(5,3),(5,5),(7,2),(7,3),(7,5) \right \}$

$A\times A=\left \{ (2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5) \right \}$

$B\times B=\left \{ (5,5),(5,7),(7,5),(7,7) \right \}$

$\therefore A\times A$ has the highest number of elements

For two sets $A$ and $B$, $A\times B=B\times A$.

  1. True

  2. False


Correct Option: B
Explanation:

The given statement is false.

Example:-
Let us consider $A={1,2}$ and $B={3,4}$.

Now $A\times B={(1,3),(1,4),(2,3),(2,4)}$......(1).

And $B\times A={(3,1),(3,2),(4,1),(4,2)}$..........(2).

Form (1) and (2) it's evident that $A\times B\ne B\times A$.

Let A and B be sets containing 2 and 4 elements respecetively. The number of subsets $A \times B$ having 3 or more elements is 

  1. $219$

  2. $211$

  3. $256$

  4. $220$


Correct Option: A
Explanation:
Let $A=\left\{x,y\right\}$
$B=\left\{a,bc,d\right\}$
$A\times B$ has $2\times 4=8$ elements
Total substance of $A\times B={2}^{8}=256$
$\therefore\,$Total number of subsets of $A\times B$ having $3$ or more elements
$=256-\left(1\,null \,set+8\,single\,ton\,set-^{8}C _{2}\,having\,2\,elements\right)$
$=256-1-8-\dfrac{8!}{6!2!}$
$=256-1-8-\dfrac{8\times 7\times 6!}{6!2!}$
$=256-1-8-28=219$

If $A$ and $B$ are independent event such that $P(A \cap B')=\dfrac {3}{25}$ and $P(A' \cap B)=\dfrac {8}{25}$, then $P(A)=$

  1. $1/5$

  2. $3/8$

  3. $2/5$

  4. $4/5$


Correct Option: C
Explanation:
$A$ & $B$ are independent
$P(A\cap B)=\dfrac {3}{25}$ and $P(A'\cap B)=\dfrac {8}{25}\quad P(A)=(?)$
$\rightarrow \ P(A)+P(B)=1---(i)$ ($A$& $B$ are independent )
$\rightarrow \ P(A\cap B')=P(A)-P(A\cap B)$
$P(A)-P(A\cap B)=\dfrac {3}{25}----(ii)$
$\rightarrow \ P(A' \cap B)=P(B)-P(A\cap B)$
$P(B)-P(A\cap B)=\dfrac {8}{25}-----(iii)$
$\rightarrow \ $ solving equation $(ii)$ and $(iii)$
$\dfrac {\,\,\, P\left( A \right) -P\left( A\cap B \right) =\dfrac { 3 }{ 25 } \\\,\,\, P\left( B \right) -P\left( A\cap B \right) =\dfrac { 8 }{ 25 } \\ -\quad \,\,\,\,\,\,\,+\quad \quad\quad\quad- }{ P\left( A \right) -P\left( B \right) =\dfrac { 3 }{ 25 } -\dfrac { 8 }{ 25 } \\ P\left( A \right) -P\left( B \right) =\dfrac { 3-8 }{ 25 }  } $
                           $=\dfrac {-5}{25}$
$\rightarrow \ P(A)-P(B)=\dfrac {-1}{5}$
$P(A)-(1-P(A))=\dfrac {-1}{5}$
$P(A)-I+P(A)=\dfrac {-1}{5}$
$\therefore \ 2P(A)=\dfrac {-1}{5}+1$
$\therefore \ 2P(A)=\dfrac {4}{5}$
$\therefore \ P(A)=\dfrac {4}{5\times 2}$
$\therefore \ P(A)=\dfrac {2}{5}$


The number of ordered pairs (x, y) of natural numbers satisfy the equation $x^2+y^2+2xy-2018x-2018y-2019^o=0$ is?

  1. $0$

  2. $1009$

  3. $2018$

  4. $2019$


Correct Option: A

The number of ordered triplet $\left(x,y,z\right),x,y,z$ are positive integers satisfying $xyz=105$

  1. $15$

  2. $27$

  3. $54$

  4. $35$


Correct Option: A

If $\int\dfrac{2\cos x-\sin x+\lambda}{\cos x-\sin x-2}dx=A In\left|\cos x+\sin x-2\right|+Bx+C$. Then the ordered triplet $\left(A,B,\lambda\right)$, is 

  1. $\left(\dfrac{1}{2},\dfrac{3}{2},-1\right)$

  2. $\left(\dfrac{3}{2},\dfrac{1}{2},-1\right)$

  3. $\left(\dfrac{1}{2},-1, \dfrac{3}{2}\right)$

  4. $\left(\dfrac{3}{2},-1, \dfrac{1}{2}\right)$


Correct Option: A

If $A={1, 2, 3}$ and $B={3, 8}$, then $(A\cup B)\times (A\cap B)$ is

  1. ${(3, 1), (3, 2), (3, 3), (3, 8)}$

  2. ${(1, 3), (2, 3), (3, 3), (8, 3)}$

  3. ${(1, 2), (2, 2), (3, 3), (8, 8)}$

  4. ${(8, 3), (8, 2), (8, 1), (8, 8)}$


Correct Option: B
Explanation:

$A\cup B={1,2,3,8}$
$A\cap B={3}$
$\therefore (A\cup B)\times (A\cap B)$
$={(1,3),(2,3),(3,3),(8,3)}$

Let $A=\left { 1,2,3 \right }$ and $B=\left { a,b \right }$.Which of the following subsets of $A\times B$ is a mapping from $A$ to $B$

  1. $\left { \left ( 1,a \right ),\left ( 3,b \right ),\left ( 2,a \right ),\left ( 2,b \right ) \right }$

  2. $\left { \left ( 1,b \right ),\left ( 2,a \right ),\left ( 3,a \right ) \right }$

  3. $\left { \left ( 1,a \right ),\left ( 2,b \right ) \right }$

  4. none of these


Correct Option: A,B,C
Explanation:

$A=\left{ 1,2,3 \right} \ B=\left{ a,b \right} \ A\times B=\left{ \left( 1,a \right) ,\left( 2,a \right) ,\left( 3,a \right) ,\left( 1,b \right) ,\left( 2,b \right) ,\left( 3,b \right)  \right} .$


$ \left{ \left( 1,a \right) ,\left( 3,b \right) ,\left( 2,a, \right) \left( 2,b \right)  \right} \subset A\times B$

$ \left{ \left( 1,b \right) ,\left( 2,a \right) ,\left( 3,a \right)  \right} \subset A\times B$

$ \left{ \left( 1,a \right) ,\left( 2,b \right)  \right} \subset A\times B$

Let $ A= { 1,2,3,.......50} $ and $B={2,4,6.......100}$ .The number of elements $\left ( x, y \right )\in A\times B$ such that $x+y=50$

  1. $24$

  2. $25$

  3. $50$

  4. $75$


Correct Option: A
Explanation:

The elements will be
$(2,48),(48,2)$
$(4,46), (46,4)$
:
:
$(2n,50-2n), (50-2n,2n)$
Now we have
$2,4,6,8...$ upto $48$
This forms an A.P
The number terms is $24$.