Tag: mathematics and statistics

Questions Related to mathematics and statistics

Find the angle measure of $4$ radians.

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $114.591^{\circ}$

  2. $141.372^{\circ}$

  3. $229.183^{\circ}$

  4. $282.743^{\circ}$

  5. $458.366^{\circ}$

Show answer
Correct Option: C
Explanation:

We know, $\pi$ radians is equal to $180$ degrees
$\therefore 4$ radians is equal to $4 \times \dfrac {180 }{\pi} = 180 \times 1.2733 = 229.183$ degrees

1 radian =

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $58^0.17$

  2. $65^0.17$

  3. $57^0.30$

  4. $66^0.71$

Show answer
Correct Option: C
Explanation:
We know that, $\pi$ radian$=180^o$

so $1$ radian$=\left(\dfrac{180}{\pi}\right)^o$

$=\left(\dfrac{180}{22/7}\right)^o$

$=(57.30)^0$

If $\displaystyle\cot  \theta+\left ( \frac{1}{\sqrt{3}} \right )\sin \theta =\frac{2}{\sqrt{3}} $ then find $\displaystyle \theta $ in circular measure

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $\displaystyle \frac{\pi ^{2}}{2} $

  2. $\displaystyle \frac{\pi ^{2}}{3} $

  3. $\displaystyle \frac{\pi ^{2}}{4} $

  4. $\displaystyle \frac{\pi ^{2}}{6} $

Show answer
Correct Option: D

The degree measure of 1 radian (taking $\pi =\dfrac { 22 }{ 7 }$ ) is

mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

  1. $55^o{ 61 }^{ ' }{ 22 }^{ " }$ (approx.)

  2. $57^o{ 16 }^{ ' }{ 22 }^{ " }$ (approx.)

  3. $57^o{ 22 }^{ ' }{ 16 }^{ " }$ (approx.)

  4. $57^o{ 22 }^{ ' }{ 22 }^{ " }$ (approx.)

Show answer
Correct Option: B
Explanation:

$\pi\ radians = 180^{\circ}$

$1\ radian=\frac { 180 }{ \pi  } = \frac { 180 }{ \frac { 22 }{ 7 }  } $
$1\ radian=57.272727$
The integer part constitutes the degree part. The mantissa is converted to minutes by multiplying with ${60}'$
Minutes = $0.272727*{60}' = {16.3636}'$
The integer part constitutes the minutes. The mantissa is converted to seconds by multiplying with ${60}''$
Seconds = $0.3636*{60}''\approx {22}''$
Hence, the degree measure of 1 radian is $57^{\circ}{16}'{22}''$

Let $\ast$ be a binary operation on the set $Q$ of rational numbers as follows:
(i) $a\ast b = a - b$ (ii) $a\ast b = a^{2} + b^{2}$
(iii) $a\ast b = a + ab$ (iv) $a\ast b = (a - b)^{2}$
(v) $a\ast b = \dfrac {ab}{4}$ (vi) $a\ast b = ab^{2}$
Find which of the binary operations are commutative and which are associative

mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

  1. $ii, iv, v$ are commutative and $v$ associative

  2. $ii, iv, v$ are not commutative and $v$ associative

  3. $iii, iv, v$are commutative and $v$ associative

  4. $vi, iv, v$are commutative and $v$ associative

Show answer
Correct Option: A
Explanation:

$(i)$  $a\ast b=a-b$

Check commutative is
$a\ast b=b\ast a$
$a\ast b=a-b$
$b\ast a=b-a$
Since, $a\ast b\neq b\ast a$
$\ast$ is not commutative.
Check associative
$\ast$ is associative if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c={ (a-b) }^{ \ast  }c=(a-b)-c=a-b-c\ a\ast (b\ast c)=a\ast (b-c)=a-(b-c)=a-b+c$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(ii)$  $a\ast b={ a }^{ 2 }+{ b }^{ 2 }$
Check commutative
$\ast$ is commutative if $a\ast b=b\ast a$
$a\ast b={ a }^{ 2 }+{ b }^{ 2 }\ b\ast a={ b }^{ 2 }+{ a }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative
$\ast$ is associative if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c=({ a }^{ 2 }+{ b }^{ 2 })\ast c={ ({ a }^{ 2 }+{ b }^{ 2 }) }^{ 2 }+{ c }^{ 2 }\ a\ast (b\ast c)=a\ast ({ b }^{ 2 }+{ c }^{ 2 })={ a }^{ 2 }+{ ({ b }^{ 2 }+{ c }^{ 2 }) }^{ 2 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(iii)$ $a\ast b=a+b$
Check commutative
$\ast$ is commutative is $a\ast b=b\ast a$
$ a\ast b=a+ab;\quad b\ast a=b+ba$
Since $ a\ast b\neq b\ast a$
$\ast$ is not commutative.
$(iv)$ $a\ast b={ (a-b) }^{ 2 }$
Check commutative
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b={ (a-b) }^{ 2 }\quad ;\quad b\ast a={ (b-a) }^{ 2 }={ (a-b) }^{ 2 }$
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative
$\ast$ if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c={ (a-b) }^{ 2 }\ast c={ [{ (a-b) }^{ 2 }-c] }^{ 2 }\ a\ast (b\ast c)=a\ast { (b-c) }^{ 2 }={ [a-{ (b-c) }^{ 2 }] }^{ 2 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(v)$ $a\ast b=\cfrac { ab }{ 4 } $
Check commutative.
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b=\cfrac { ab }{ 4 } \quad ;\quad b\ast a=\cfrac { ba }{ 4 } =\cfrac { ab }{ 4 } $
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative.
$\ast$ is association if $(a\ast b)\ast c=a\ast (b\ast c)$
$(a\ast b)\ast c=(\cfrac { \cfrac { ab }{ 4 } \ast c }{ 4 } )=\cfrac { abc }{ 16 } \ a\ast (b\ast c)=a\ast (\cfrac { bc }{ 4 } )=\cfrac { a\times \cfrac { bc }{ 4 }  }{ 4 } =\cfrac { abc }{ 16 } $
Since $ (a\ast b)\ast c=a\ast (b\ast c)\quad \forall\quad a,b,c\epsilon Q$
$\ast$ is an associative binary operation.
$(vi)$ $a\ast b={ ab }^{ 2 }$
check commutative.
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b={ ab }^{ 2 }\quad ;\quad b\ast a={ ba }^{ 2 }$
Since $ a\ast b\neq b\ast a$
$\ast$ is not commutative.
Check associative 
$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$
$(a\ast b)\ast c={ ab }^{ 2 }\ast c=({ ab }^{ 2 }){ c }^{ 2 }=a{ b }^{ 2 }{ c }^{ 2 }.\ a\ast (b\ast c)=a\ast { bc }^{ 2 }=a{ ({ bc }^{ 2 }) }^{ 2 }=a{ b }^{ 2 }{ c }^{ 4 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associate binary operation.

State whether the following statements are true of false. Justify.
(i) For an arbitrary binary operation $\ast$ on as set $N, a\ast a = a\forall a \epsilon N$
(ii) If $\ast$ is a commutative binary operation on $N$, then $a\ast (b\ast c) = (c\ast b) \ast a$

mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

i)Given $a\ast a=a\quad \forall a\epsilon N$ on a set $N$ for an arbitary binary operation $\ast$.

The binary operation can be defined as,$a\times b={ a }^{ 2 }$
Clearly ${ a }^{ 2 }\neq a\quad \forall\quad a\epsilon A$.Therefore the statement is false.
ii)Given binary operator $\ast$ on set $N$.We need to prove that $a\ast (b\ast a)=(c\ast b)\ast a$
An operation $\ast$ on $A$  is commutative 
if $a\ast b=b\ast a\quad \forall\quad a,b\epsilon A$
$\therefore a\ast (b\ast a)=(c\ast b)\ast a$
 since $\ast$ is commutative.
$\therefore $the statement is true.

Consider a binary operation $\ast$ on $N$ defined as $a\ast b = a^{3} + b^{3}$. Choose the correct answer

mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

  1. Is $\ast$ both associative and commutative?

  2. Is $\ast$ commutative but not associative?

  3. s $\ast$ associative but not commutative?

  4. Is $\ast$ neither commutative nor associative?

Show answer
Correct Option: B
Explanation:

Check commutative 

$\ast$ is commuatative if
$a \ast b=b\ast a$
$a\ast b={ a }^{ 3 }+{ b }^{ 3 }\quad ;\quad b\ast a={ b }^{ 3 }+{ a }^{ 3 }={ a }^{ 3 }+{ b }^{ 3 }$
Since ,$ a\ast b=b\ast a,\forall\quad a,b\epsilon N$
$\ast$ is commutative
Check associative
$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$
$(a\ast b)\ast c={ ({ a }^{ 3 }+{ b }^{ 3 }) }^{ \ast  }c={ ({ a }^{ 3 }+{ b }^{ 3 }) }^{ 3 }+{ c }^{ 3 }\ a\ast (b\ast c)={ a }^{ 3 }\ast ({ b }^{ 3 }+{ c }^{ 3 })={ { a }^{ 3 } }+{ ({ b }^{ 3 }+{ c }^{ 3 }) }^{ 3 }$
Since, $ (a\ast b)\ast c\neq a\ast (b\ast c),\forall\quad a,b,c\epsilon N$
$\ast$ is not associative.
$\therefore $Is $\ast$ commutative but not associative.

Let $$ be a binary operation defined on the set of rational numbers $Q$ defined by $a * b= ab + 1,$ in this statement  $$ is a commutative.

mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
$a\text{*}\,b=ab+1$
$b\text{*}\,a=ba+1=ab+1$
$\therefore\,a\text{*}\,b=b\text{*}\,a$
Hence the given statement is true.

Consider the following statements for non empty sets A, B and C
1 $\displaystyle A-\left ( B-C \right )=\left ( A-B \right )\cup C $
2 $\displaystyle A-\left ( B\cup C \right )=\left ( A-B \right )- C $
which of the statements given above is/are correct?

mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

  1. 1 only

  2. 2 only

  3. Both 1 and 2

  4. Neither 1 nor 2

Show answer
Correct Option: B
Explanation:

1. $A- (B - C) = \displaystyle A-(B\cap C')$
$\displaystyle =A\cap (B\cap C')'$
$\displaystyle =A\cap (B'\cup (C'))$
$\displaystyle = A\cap (B'\cup C)$
Thus, $\displaystyle A-(B-C)\neq (A-B)\cup C$


2. A- $\displaystyle (B\cup C)=A\cap (B\cap C)'$
$\displaystyle = A\cap (B'\cap C)'$
$\displaystyle (A-B)-C=(A\cap B')-C$
$\displaystyle =A\cap B'\cap C'$
$\displaystyle \Rightarrow A-(B\cup C)=(A-B)-C$
Associative property. 

If A, B, C are any three sets, $A-(B\cup C)$ will be

mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

  1. $(A-B)\cap(B-C)$

  2. $(A-B)\cup(B-C)$

  3. $(A-B)\cup(A-C)$

  4. $(A-B)\cap(A-C)$

Show answer
Correct Option: D
Explanation:

Formula for $ A - (B \cup C) = (A-B) \cap (A-C) $