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ljafxwvp | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let $R=\{(1,3),(4,2),(2,4),(2,3),(3,1)\}$ be a relation on the set $A=\{1,2,3,4\}$. The relation $R$ is : | [{"identifier": "A", "content": "a function"}, {"identifier": "B", "content": "transitive"}, {"identifier": "C", "content": "not symmetric"}, {"identifier": "D", "content": "reflexive"}] | ["C"] | null | <p>Let's evaluate each of the properties for the relation $R$.</p>
<p><b>Relation R :</b> $R=\{(1,3),(4,2),(2,4),(2,3),(3,1)\}$</p>
<ul>
<li><p>A relation is a function if each element in the domain is related to exactly one element in the codomain. In this case, for example, 2 is related to both 4 and 3, so $R$ is not a function.</p>
</li>
<br/><li><p>A relation is transitive if for every pair of elements $(x,y)$ and $(y,z)$ in the relation, $(x,z)$ is also in the relation. In this case, for example, $(1,3)$ and $(3,1)$ are in the relation but $(1,1)$ is not, so $R$ is not transitive.</p>
</li>
<br/><li><p>A relation is symmetric if for every pair $(x,y)$ in the relation, $(y,x)$ is also in the relation. In this case, for example, $(1,3)$ is in the relation but $(3,1)$ is not, so $R$ is not symmetric.</p>
</li>
<br/><li><p>A relation is reflexive if every element is related to itself, i.e., if all pairs of the form $(x,x)$ are in the relation for all $x$ in the set. In this case, none of the pairs are of the form $(x,x)$, so $R$ is not reflexive.</p>
</li>
</ul>
<p>Therefore, the correct answer is <b>Option C :</b> $R$ is not symmetric.</p>
| mcq | aieee-2004 |
ljaeehms | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let $R=\{(3,3),(6,6),(9,9),(12,12),(6,12)$, $(3,9),(3,12),(3,6)\}$ be a relation on the set $A=\{3,6,9,12\}$. The relation is : | [{"identifier": "A", "content": "reflexive and symmetric only"}, {"identifier": "B", "content": "an equivalence relation"}, {"identifier": "C", "content": "reflexive only"}, {"identifier": "D", "content": "reflexive and transitive only"}] | ["D"] | null | <p>We have to examine whether the relation $R$ satisfies the properties of reflexivity, symmetry, and transitivity.</p>
<p><b>Relation R :</b> $R=\{(3,3),(6,6),(9,9),(12,12),(6,12)$, $(3,9),(3,12),(3,6)\}$ on set $A=\{3,6,9,12\}$.</p>
<p>We will evaluate each of the three properties :</p>
<ul>
<li><p>Reflexivity : A relation is reflexive if all elements in the set are related to themselves. Here, (3,3), (6,6), (9,9), and (12,12) are all present in $R$, so $R$ is reflexive.</p>
</li>
<br/><li><p>Symmetry : A relation is symmetric if for every pair (a,b) in the relation, the pair (b,a) is also in the relation. In $R$, we have pairs such as (6,12) and (3,6), but their corresponding reverse pairs (12,6) and (6,3) are not in $R$, so $R$ is not symmetric.</p>
</li>
<br/><li><p>Transitivity : A relation is transitive if for every pair of pairs ((a,b), (b,c)) in the relation, the pair (a,c) is also in the relation. For the pairs (3,6) and (6,12) in $R$, we have (3,12) in $R$, and for the pairs (3,6) and (6,6) in $R$, we also have (3,6) in $R$. Therefore, $R$ is transitive.</p>
</li>
</ul>
<p>Therefore, the relation $R$ is reflexive and transitive, but not symmetric.</p>
<p>So, the correct answer is <b>Option D :</b> $R$ is reflexive and transitive only.</p>
| mcq | aieee-2005 |
ljadq9q3 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let $W$ denote the words in the English dictionary. Define the relation $R$ by
<br/><br/>$R=\{(x, y) \in W \times W \mid$ the words $x$ and $y$ have at least one letter in common}. Then, $R$ is | [{"identifier": "A", "content": "reflexive, symmetric and not transitive"}, {"identifier": "B", "content": "reflexive, symmetric and transitive"}, {"identifier": "C", "content": "reflexive, not symmetric and transitive"}, {"identifier": "D", "content": "not reflexive, symmetric and transitive"}] | ["A"] | null | <p>Let's evaluate the relation $R$ for the properties of reflexivity, symmetry, and transitivity.</p>
<p><b>Relation R :</b> $R={(x, y) \in W \times W \mid}$ the words $x$ and $y$ have at least one letter in common}.</p>
<ul>
<li>Reflexivity : Each word in English obviously has at least one letter in common with itself, so the relation is reflexive.</li>
<br/><li>Symmetry : If a word $x$ has at least one letter in common with a word $y$, then $y$ necessarily has that same letter in common with $x$. So, the relation is symmetric.</li>
<br/><li>Transitivity : This property is not always satisfied. For instance, consider the three words 'cat', 'bat', and 'bee'. 'Cat' and 'bat' share a letter (the 'a'), and 'bat' and 'bee' share a letter (the 'b'), but 'cat' and 'bee' do not share any letters. Therefore, even though 'cat' is related to 'bat' and 'bat' is related to 'bee', 'cat' is not related to 'bee', so the relation is not transitive.</li>
</ul>
<p>In conclusion, the correct answer is <b>Option A :</b> $R$ is reflexive, symmetric and not transitive.</p> | mcq | aieee-2006 |
jRE7pdboFFzdk15T | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let R be the real line. Consider the following subsets of the plane $$R \times R$$ :
<br/> $$S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}$$
<br/> $$T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm int}} eger\,} \right\}$$,
<p> Which one of the following is true ? </p> | [{"identifier": "A", "content": "Neither S nor T is an equivalence relation on R"}, {"identifier": "B", "content": "Both S and T are equivalence relation on R"}, {"identifier": "C", "content": "S is an equivalence relation on R but T is not"}, {"identifier": "D", "content": "T is an equivalence relation on R but S is not"}] | ["D"] | null | Given $$S = \left\{ {\left( {x,y} \right):y = x + 1\,\,} \right.\,$$
<br><br>and $$\,\,\,\left. {0 < x < 2} \right\}$$
<br><br>As $$\,\,\,\,x \ne x + 1\,\,\,$$
<br><br>for any $$\,\,\,x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S$$
<br><br>$$\therefore$$ $$S$$ is not reflexive.
<br><br>Hence $$S$$ in not an equivalence relation.
<br><br>Also $$\,\,\,T = \left\{ {x,\left. y \right)} \right.:x - y$$ is an integer $$\left. {} \right\}$$
<br><br>as $$x - x = 0$$ is an integer $$\forall x \in R$$
<br><br>$$\therefore$$ $$T$$ is reflexive.
<br><br>If $$x-y$$ is an integer then $$y-x$$ is also an integer
<br><br>$$\therefore$$ $$T$$ is symmetric
<br><br>If $$x-y$$ is an integer and $$y - z$$ is an integer then
<br><br>$$(x-y)+(y-z)=x-z$$ is also an integer.
<br><br>$$\therefore$$ $$T$$ is transitive
<br><br>Hence $$T$$ is an equivalence relation | mcq | aieee-2008 |
ljad974k | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Consider the following relations
<br/><br/>$R=\{(x, y) \mid x, y$ are real numbers and $x=w y$ for some rational number $w\}$;
<br/><br/>$S=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p\right.$ and $q$ are integers such that $n, q \neq 0$ and $q m=p m\}$. Then | [{"identifier": "A", "content": "$R$ is an equivalence relation but $S$ is not an equivalence relation\n"}, {"identifier": "B", "content": "Neither $R$ nor $S$ is an equivalence relation"}, {"identifier": "C", "content": "$S$ is an equivalence relation but $R$ is not an equivalence relation"}, {"identifier": "D", "content": "$R$ and $S$ both are equivalence relations"}] | ["C"] | null | <p>Let's evaluate each relation for the properties of an equivalence relation: reflexivity, symmetry, and transitivity.</p>
<p><b>Relation R :</b> $R=(x, y) \mid x, y$ are real numbers and $x=w y$ for some rational number $w$.</p>
<ul>
<li>Reflexivity : For all $x$ in $R$, $x = 1x$. Since 1 is a rational number, every element is related to itself.</li>
<br/><li>Symmetry : For all $x, y$ in $R$, if $x = w y$ for some rational $w$, then $y = \frac{1}{w}x$. However, if $w = 0$, then $\frac{1}{w}$ is undefined, and therefore, $R$ doesn't satisfy symmetry.</li>
<br/><li>Transitivity : If $x = w y$ and $y = v z$ for some rational numbers $w$ and $v$, then $x = (w v)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.</li>
</ul>
<p>Therefore, $R$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.</p>
<p><b>Relation S :</b> $S=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p\right.$ and $q$ are integers such that $n, q \neq 0$ and $q m=p m\}$</p>
<ul>
<li>Reflexivity : For all $\frac{m}{n}$, $\frac{m}{n} = \frac{m}{n}$. Since $n \neq 0$ and $m = m$, every element is related to itself.</li>
<br/><li>Symmetry : For all $\frac{m}{n}$, $\frac{p}{q}$, if $q m = p n$, then $n p = m n$. So if $\frac{m}{n}$ is related to $\frac{p}{q}$, then $\frac{p}{q}$ is related to $\frac{m}{n}$.</li>
<br/><li>Transitivity :
<br/><br/>$$
\begin{array}{rlr}
\frac{m}{n} R \frac{p}{q} \text { and } \frac{p}{q} R \frac{r}{s} \\\\
\Rightarrow m q=n p \text { and } p s=r q \\\\
\Rightarrow m q \cdot p s=n p \cdot r q \\\\
\Rightarrow \quad m s=n r \\\\
\Rightarrow \frac{m}{n}=\frac{r}{s} \Rightarrow \frac{m}{n} R \frac{r}{s}
\end{array}
$$
<br/><br/> So if $\frac{m}{n}$ is related to $\frac{p}{q}$ and $\frac{p}{q}$ is related to $\frac{r}{s}$, then $\frac{m}{n}$ is related to $\frac{r}{s}$. The relation $S$ is transitive.</li>
</ul>
<p>Therefore, $S$ is an equivalence relation on the set of all fractions where denominator is not zero.</p>
<p>In conclusion, the correct answer is
<br/><br/><b>Option C :</b> $S$ is an equivalence relation but $R$ is not an equivalence relation.</p>
| mcq | aieee-2010 |
ljaclkwr | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let $R$ be the set of real numbers.
<br/><br/><b>Statement I :</b> $A=\{(x, y) \in R \times R: y-x$ is an integer $\}$ is an equivalence relation on $R$.
<br/><br/><b>Statement II :</b> $ B=\{(x, y) \in R \times R: x=\alpha y$ for some rational number $\alpha\}$ is an equivalence relation on $R$. | [{"identifier": "A", "content": "Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I."}, {"identifier": "B", "content": "Statement I is true, Statement II is false."}, {"identifier": "C", "content": "Statement I is false, Statement II is true."}, {"identifier": "D", "content": "Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I."}] | ["B"] | null | <p>An equivalence relation on a set must satisfy three properties: reflexivity (every element is related to itself), symmetry (if an element is related to a second, the second is related to the first), and transitivity (if a first element is related to a second, and the second is related to a third, then the first is related to the third).</p>
<p><b>Statement I :</b> $A={(x, y) \in R \times R: }$ y-x is an integer .</p>
<ul>
<li>Reflexivity : For all $x$ in $R$, $x - x = 0$ which is an integer. So, every element is related to itself.</li>
<br/><li>Symmetry : For all $x, y$ in $R$, if $y - x$ is an integer, then $x - y = - (y - x)$ is also an integer. So, if $x$ is related to $y$, then $y$ is related to $x$.</li>
<br/><li>Transitivity : For all $x, y, z$ in $R$, if $y - x$ and $z - y$ are integers, then $(z - y) + (y - x) = z - x$ is also an integer. So, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.</li>
</ul>
<p>Therefore, $A$ is an equivalence relation on $R$.</p>
<p><b>Statement II :</b> $ B={(x, y) \in R \times R: x=\alpha y}$ for some rational number $\alpha$.</p>
<ul>
<li>Reflexivity : For all $x$ in $R$, $x = 1x$. Since 1 is a rational number, every element is related to itself.</li>
<br/><li>Symmetry : For all $x, y$ in $R$, if $x = \alpha y$ for some rational $\alpha$, then $y = \frac{1}{\alpha}x$. However, if $\alpha = 0$, then $\frac{1}{\alpha}$ is undefined, and therefore, $B$ doesn't satisfy symmetry.</li>
<br/><li>Transitivity : If $x = \alpha y$ and $y = \beta z$ for some rational numbers $\alpha$ and $\beta$, then $x = (\alpha \beta)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.</li>
</ul>
<p>Therefore, $B$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.</p>
<p>In conclusion, the correct answer is <br/><br/><b>Option B :</b> Statement I is true, Statement II is false.</p>
| mcq | aieee-2011 |
CJSlXy9fvMO64kYVWmKoJ | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Consider the following two binary relations on the set A = {a, b, c} :
<br/>R<sub>1</sub> = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and
<br/>R<sub>2</sub> = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}.
<br/>Then :
| [{"identifier": "A", "content": "both R<sub>1</sub> and R<sub>2</sub> are not symmetric."}, {"identifier": "B", "content": "R<sub>1</sub> is not symmetric but it is transitive."}, {"identifier": "C", "content": "R<sub>2</sub> is symmetric but it is not transitive. "}, {"identifier": "D", "content": "both R<sub>1</sub> and R<sub>2</sub> are transitive."}] | ["C"] | null | Here both R<sub>1</sub> and R<sub>2</sub> are symmetric as for any (x, y) $$ \in $$ R<sub>1</sub>, we have (y, x) $$ \in $$ R<sub>1</sub> and similarly for any (x, y) $$ \in $$ R<sub>2</sub>, we have (y, x) $$ \in $$ R<sub>2</sub><br><br>
In R<sub>1</sub>, (b, c) $$ \in $$ R<sub>1</sub>, (c, a) $$ \in $$ R<sub>1</sub> but (b,a) $$ \notin $$ R<sub>1</sub><br><br>
Similarly in R<sub>2</sub>, (b, a) $$ \in $$ R<sub>2</sub>, (a, c) $$ \in $$ R<sub>2</sub> but (b, c) $$ \notin $$ R<sub>2</sub><br><br>
$$ \therefore $$ R<sub>1</sub> and R<sub>2</sub> are not transitive. | mcq | jee-main-2018-online-15th-april-morning-slot |
DOhFcGHeiIM3tbQSkae3L | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let <b>N</b> denote the set of all natural numbers. Define two binary relations on <b>N</b> as <sub>R</sub> = {(x, y) $$ \in $$ <b>N $$ \times $$ N</b> : 2x + y = 10} and R<sub>2</sub> = {(x, y) $$ \in $$ <b>N $$ \times $$ N</b> : x + 2y = 10}. Then : | [{"identifier": "A", "content": "Range of R<sub>1</sub> is {2, 4, 8)."}, {"identifier": "B", "content": "Range of R<sub>2</sub> is {1, 2, 3, 4}."}, {"identifier": "C", "content": "Both R<sub>1</sub> and R<sub>2</sub> are symmetric relations."}, {"identifier": "D", "content": "Both R<sub>1</sub> and R<sub>2</sub> are transitive relations."}] | ["B"] | null | For R<sub>1</sub>; 2x + y = 10 and x, y $$ \in $$ N possible values for x and y are :
<br><br>x = 1, y = 8 i.e. (1, 8);
<br><br>x = 2, y = 6 i.e (2, 6);
<br><br>x = 3, y = 4 i.e (3, 4);
<br><br>x = 4, y = 2 i.e (4, 2)
<br><br>$$\therefore\,\,\,$$ R<sub>1</sub> = { (1, 8), (2, 6), (3, 4), (4, 2) }
<br><br>$$\therefore\,\,\,$$ Range of R<sub>1</sub> is {2, 4, 6, 8}
<br><br>R<sub>1</sub> is not symmetric.
<br><br>R<sub>1</sub> is not transitive also as
<br><br>(3, 4), (4, 2) $$ \in $$ R , but (3, 2) $$ \notin $$ R<sub>1</sub>
<br><br>For R<sub>2</sub> : x + 2y = 10 and x, y $$ \in $$ N
<br><br>Possible values of x, and y are :
<br><br>x = 8, y= 1 i.e (8, 1)
<br><br>x = 6, y = 2 i.e (6, 2)
<br><br>x = 4, y = 3 i.e (4, 3) and
<br><br>x = 2, y = 4 i.e (2, 4)
<br><br>$$\therefore\,\,\,$$ R<sub>2</sub> = {(8, 1) (6, 2) (4, 3) (2, 4)}
<br><br>$$\therefore\,\,\,$$ Range of R<sub>2</sub> = $$\left\{ {1,2,3,4} \right\}$$
<br><br>R<sub>2</sub> is not symmetric and transitive | mcq | jee-main-2018-online-16th-april-morning-slot |
Ped5VSnU29ThaR3uOOjgy2xukf3yias0 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let R<sub>1</sub>
and R<sub>2</sub>
be two relation defined as
follows :
<br/>R<sub>1</sub>
= {(a, b) $$ \in $$ R<sup>2</sup>
: a<sup>2</sup>
+ b<sup>2</sup> $$ \in $$ Q} and
<br/>R<sub>2</sub>
= {(a, b) $$ \in $$ R<sup>2</sup>
: a<sup>2</sup>
+ b<sup>2</sup> $$ \notin $$ Q},
<br/>where Q is the
set of all rational numbers. Then : | [{"identifier": "A", "content": "Neither R<sub>1</sub>\n nor R<sub>2</sub>\n is transitive."}, {"identifier": "B", "content": "R<sub>2</sub>\n is transitive but R<sub>1</sub>\n is not transitive."}, {"identifier": "C", "content": "R<sub>1</sub>\n and R<sub>2</sub>\n are both transitive."}, {"identifier": "D", "content": "R<sub>1</sub>\n is transitive but R<sub>2</sub>\n is not transitive."}] | ["A"] | null | For R<sub>1</sub> :<br><br>Let a = 1 + $$\sqrt 2 $$, b = 1 $$-$$ $$\sqrt 2 $$, c = $${8^{{1 \over 4}}}$$<br><br>aR<sub>1</sub>b : a<sup>2</sup> + b<sup>2</sup> = 6 $$ \in $$ Q<br><br>bR<sub>1</sub>c : b<sup>2</sup> + c<sup>2</sup> = 3 $$-$$ 2$$\sqrt 2 $$ + 2$$\sqrt 2 $$ = 3 $$ \in $$ Q<br><br>aR<sub>1</sub>c : a<sup>2</sup> + c<sup>2</sup> = 3 + 2$$\sqrt 2 $$ + 2$$\sqrt 2 $$ $$ \notin $$ Q<br><br>$$ \therefore $$ R<sub>1</sub> is not transitive.<br><br>For R<sub>2</sub> : <br><br>Let a = 1 + $$\sqrt 2 $$, b = $$\sqrt 2 $$, c = 1 $$-$$ $$\sqrt 2 $$<br><br>aR<sub>2</sub>b : a<sup>2</sup> + b<sup>2</sup> = 5 + 2$$\sqrt 2 $$ $$ \notin $$ Q<br><br>bR<sub>2</sub>c : b<sup>2</sup> + c<sup>2</sup> = 5 $$-$$ 2$$\sqrt 2 $$ $$ \notin $$ Q<br><br>aR<sub>2</sub>c : a<sup>2</sup> + c<sup>2</sup> = 3 + 2$$\sqrt 2 $$ + 3 $$-$$ 2$$\sqrt 2 $$ = 6 $$ \in $$ Q<br><br>$$ \therefore $$ R<sub>2</sub> is not transitive. | mcq | jee-main-2020-online-3rd-september-evening-slot |
cS4oUt1tfL1cKkH1tC1klughovj | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1, $$-$$1) is the set : | [{"identifier": "A", "content": "$$S = \\{ (x,y)|{x^2} + {y^2} = \\sqrt 2 \\} $$"}, {"identifier": "B", "content": "$$S = \\{ (x,y)|{x^2} + {y^2} = 2\\} $$"}, {"identifier": "C", "content": "$$S = \\{ (x,y)|{x^2} + {y^2} = 1\\} $$"}, {"identifier": "D", "content": "$$S = \\{ (x,y)|{x^2} + {y^2} = 4\\} $$"}] | ["B"] | null | Given R = {(P, Q) | P and Q are at the same distance from the origin}.<br><br>Then equivalence class of (1, $$-$$1) will contain al such points which lies on circumference of the circle of centre at origin and passing through point (1, $$-$$1).<br><br>i.e., radius of circle = $$\sqrt {{1^2} + {1^2}} = \sqrt 2 $$<br><br>$$ \therefore $$ Required equivalence class of (S)<br><br>$$ = \{ (x,y)|{x^2} + {y^2} = 2\} $$. | mcq | jee-main-2021-online-26th-february-morning-slot |
UWZpD3f8zFp3ctRl2Z1kmm3f6jr | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Define a relation R over a class of n $$\times$$ n real matrices A and B as <br/><br/>"ARB iff there exists a non-singular matrix P such that PAP<sup>$$-$$1</sup> = B". <br/><br/>Then which of the following is true? | [{"identifier": "A", "content": "R is reflexive, transitive but not symmetric"}, {"identifier": "B", "content": "R is symmetric, transitive but not reflexive."}, {"identifier": "C", "content": "R is reflexive, symmetric but not transitive"}, {"identifier": "D", "content": "R is an equivalence relation"}] | ["D"] | null | For reflexive relation,<br/><br/> $\forall(A, A) \in R$ for matrix $P$.<br/><br/>
$\Rightarrow A=P A P^{-1}$ is true for $P=1$<br/><br/>
So, $R$ is reflexive relation.<br/><br/>
For symmetric relation,<br/><br/>
Let $(A, B) \in R$ for matrix $P$.<br/><br/>
$$
\Rightarrow \quad A=P B P^{-1}
$$<br/><br/>
After pre-multiply by $P^{-1}$ and post-multiply by $P$, we get<br/><br/>
$$
P^{-1} A P=B
$$<br/><br/>
So, $(B, A) \in R$ for matrix $P^{-1}$.<br/><br/>
So, $R$ is a symmetric relation.<br/><br/>
For transitive relation,<br/><br/>
Let $A R B$ and $B R C$<br/><br/>
So, $A=P B P^{-1}$ and $B=P C P^{-1}$<br/><br/>
Now, $A=P\left(P C P^{-1}\right) P^{-1}$<br/><br/>
$\Rightarrow A=(P)^2 C\left(P^{-1}\right)^2 \Rightarrow A=(P)^2 \cdot C \cdot\left(P^2\right)^{-1}$<br/><br/>
$\therefore(A, C) \in R$ for matrix $P^2$.<br/><br/>
$\therefore R$ is transitive relation.<br/><br/>
Hence, $R$ is an equivalence relation. | mcq | jee-main-2021-online-18th-march-evening-shift |
1krxllilh | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let N be the set of natural numbers and a relation R on N be defined by $$R = \{ (x,y) \in N \times N:{x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0\} $$. Then the relation R is : | [{"identifier": "A", "content": "symmetric but neither reflexive nor transitive"}, {"identifier": "B", "content": "reflexive but neither symmetric nor transitive"}, {"identifier": "C", "content": "reflexive and symmetric, but not transitive"}, {"identifier": "D", "content": "an equivalence relation"}] | ["B"] | null | $${x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0$$<br><br>$$ \Rightarrow x({x^2} - {y^2}) - 3y({x^2} - {y^2}) = 0$$<br><br>$$ \Rightarrow (x - 3y)(x - y)(x + y) = 0$$<br><br>Now, x = y $$\forall$$(x, y) $$\in$$N $$\times$$ N so reflexive but not symmetric & transitive.<br><br>See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) & (1, $$-$$1) satisfies but (3, $$-$$1) does not. | mcq | jee-main-2021-online-27th-july-evening-shift |
1ktipm2vd | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Which of the following is not correct for relation R on the set of real numbers ? | [{"identifier": "A", "content": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ 0 < |x| $$-$$ |y| $$\\le$$ 1 is neither transitive nor symmetric."}, {"identifier": "B", "content": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ 0 < |x $$-$$ y| $$\\le$$ 1 is symmetric and transitive."}, {"identifier": "C", "content": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ |x| $$-$$ |y| $$\\le$$ 1 is reflexive but not symmetric."}, {"identifier": "D", "content": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ |x $$-$$ y| $$\\le$$ 1 is reflexive nd symmetric."}] | ["B"] | null | Note that (a, b) and (b, c) satisfy 0 < |x $$-$$ y| $$\le$$ 1 but (a, c) does not satisfy it so 0 $$\le$$ |x $$-$$ y| $$\le$$ 1 is symmetric but not transitive.
<br><br>For example,
<br><br>x = 0.2, y = 0.9, z = 1.5
<br><br>0 ≤ |x – y| = 0.7 ≤ 1
<br><br>0 ≤ |y – z| = 0.6 ≤ 1
<br><br>But |x – z| = 1.3 > 1
<br><br>So, (b) is correct. | mcq | jee-main-2021-online-31st-august-morning-shift |
1l54578k2 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let a set A = A<sub>1</sub> $$\cup$$ A<sub>2</sub> $$\cup$$ ..... $$\cup$$ A<sub>k</sub>, where A<sub>i</sub> $$\cap$$ A<sub>j</sub> = $$\phi$$ for i $$\ne$$ j, 1 $$\le$$ j, j $$\le$$ k. Define the relation R from A to A by R = {(x, y) : y $$\in$$ A<sub>i</sub> if and only if x $$\in$$ A<sub>i</sub>, 1 $$\le$$ i $$\le$$ k}. Then, R is :</p> | [{"identifier": "A", "content": "reflexive, symmetric but not transitive."}, {"identifier": "B", "content": "reflexive, transitive but not symmetric."}, {"identifier": "C", "content": "reflexive but not symmetric and transitive."}, {"identifier": "D", "content": "an equivalence relation."}] | ["D"] | null | <p>$$R = \{ (x,y):y \in {A_i},\,iff\,x \in {A_i}\,1 \le i \ge k\} $$</p>
<p>(1) Reflexive</p>
<p>(a, a) $$\Rightarrow$$ $$a \in {A_i}$$ iff $$a \in {A_i}$$</p>
<p>(2) Symmetric</p>
<p>(a, b) $$\Rightarrow$$ $$a \in {A_i}$$ iff $$b \in {A_i}$$</p>
<p>(b, a) $$\in$$R as $$b \in {A_i}$$ iff $$a \in {A_i}$$</p>
<p>(3) Transitive</p>
<p>(a, b) $$\in$$R & (b, c) $$\in$$R.</p>
<p>$$\Rightarrow$$ $$a \in {A_i}$$ iff $$b \in {A_i}$$ & $$b \in {A_i}$$ iff $$c \in {A_i}$$</p>
<p>$$\Rightarrow$$ $$a \in {A_i}$$ iff $$c \in {A_i}$$</p>
<p>$$\Rightarrow$$ (a, c) $$\in$$ R.</p>
<p>$$\Rightarrow$$ RElation is equivalnece.</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l55h1x3k | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let R<sub>1</sub> = {(a, b) $$\in$$ N $$\times$$ N : |a $$-$$ b| $$\le$$ 13} and</p>
<p>R<sub>2</sub> = {(a, b) $$\in$$ N $$\times$$ N : |a $$-$$ b| $$\ne$$ 13}. Then on N :</p> | [{"identifier": "A", "content": "Both R<sub>1</sub> and R<sub>2</sub> are equivalence relations"}, {"identifier": "B", "content": "Neither R<sub>1</sub> nor R<sub>2</sub> is an equivalence relation"}, {"identifier": "C", "content": "R<sub>1</sub> is an equivalence relation but R<sub>2</sub> is not"}, {"identifier": "D", "content": "R<sub>2</sub> is an equivalence relation but R<sub>1</sub> is not"}] | ["B"] | null | $R_{1}=\{(a, b) \in N \times N:|a-b| \leq 13\}$ and
<br/><br/>
$R_{2}=\{(a, b) \in N \times N:|a-b| \neq 13\}$
<br/><br/>
In $R_{1}: \because|2-11|=9 \leq 13$
<br/><br/>
$\therefore \quad(2,11) \in R_{1}$ and $(11,19) \in R_{1}$ but $(2,19) \notin R_{1}$
<br/><br/>
$\therefore \quad R_{1}$ is not transitive
<br/><br/>
Hence $R_{1}$ is not equivalence
<br/><br/>In $R_{2}:(13,3) \in R_{2}$ and $(3,26) \in R_{2}$ but $(13,26) \notin R_{2}$
$(\because|13-26|=13)$
<br/><br/>
$\therefore R_{2}$ is not transitive
<br/><br/>
Hence $R_{2}$ is not equivalence. | mcq | jee-main-2022-online-28th-june-evening-shift |
1l6jaxbfq | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$R_{1}$$ and $$R_{2}$$ be two relations defined on $$\mathbb{R}$$ by</p>
<p>$$a \,R_{1} \,b \Leftrightarrow a b \geq 0$$ and $$a \,R_{2} \,b \Leftrightarrow a \geq b$$</p>
<p>Then,</p> | [{"identifier": "A", "content": "$$R_{1}$$ is an equivalence relation but not $$R_{2}$$"}, {"identifier": "B", "content": "$$R_{2}$$ is an equivalence relation but not $$R_{1}$$"}, {"identifier": "C", "content": "both $$R_{1}$$ and $$R_{2}$$ are equivalence relations"}, {"identifier": "D", "content": "neither $$R_{1}$$ nor $$R_{2}$$ is an equivalence relation"}] | ["D"] | null | <p>$$a\,{R_1}\,b \Leftrightarrow ab \ge 0$$</p>
<p>So, definitely $$(a,a) \in {R_1}$$ as $${a^2} \ge 0$$</p>
<p>If $$(a,b) \in {R_1} \Rightarrow (b,a) \in {R_1}$$</p>
<p>But if $$(a,b) \in {R_1},(b,c) \in {R_1}$$</p>
<p>$$\Rightarrow$$ Then $$(a,c)$$ may or may not belong to R<sub>1</sub></p>
<p>{Consider $$a = - 5,b = 0,c = 5$$ so $$(a,b)$$ and $$(b,c) \in {R_1}$$ but $$ac < 0$$}</p>
<p>So, R<sub>1</sub> is not equivalence relation</p>
<p>$$a\,{R_2}\,b \Leftrightarrow a \ge b$$</p>
<p>$$(a,a) \in {R_2} \Rightarrow $$ so reflexive relation</p>
<p>If $$(a,b) \in {R_2}$$ then $$(b,a)$$ may or may not belong to R<sub>2</sub></p>
<p>$$\Rightarrow$$ So not symmetric</p>
<p>Hence it is not equivalence relation</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6m5n24e | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>For $$\alpha \in \mathbf{N}$$, consider a relation $$\mathrm{R}$$ on $$\mathbf{N}$$ given by $$\mathrm{R}=\{(x, y): 3 x+\alpha y$$ is a multiple of 7$$\}$$. The relation $$R$$ is an equivalence relation if and only if :</p> | [{"identifier": "A", "content": "$$\\alpha=14$$"}, {"identifier": "B", "content": "$$\\alpha$$ is a multiple of 4"}, {"identifier": "C", "content": "4 is the remainder when $$\\alpha$$ is divided by 10"}, {"identifier": "D", "content": "4 is the remainder when $$\\alpha$$ is divided by 7"}] | ["D"] | null | <p>$$R = \{ (x,y):3x + \alpha y$$ is multiple of 7$$\} $$, now R to be an equivalence relation</p>
<p>(1) R should be reflexive : $$(a,a) \in R\,\forall \,a \in N$$</p>
<p>$$\therefore$$ $$3a + a\alpha = 7k$$</p>
<p>$$\therefore$$ $$(3 + \alpha )a = 7k$$</p>
<p>$$\therefore$$ $$3 + \alpha = 7{k_1} \Rightarrow \alpha = 7{k_1} - 3$$</p>
<p>$$ = 7{k_1} + 4$$</p>
<p>(2) R should be symmetric : $$aRb \Leftrightarrow bRa$$</p>
<p>$$aRb:3a + (7k - 3)b = 7\,m$$</p>
<p>$$ \Rightarrow 3(a - b) + 7kb = 7\,m$$</p>
<p>$$ \Rightarrow 3(b - a) + 7ka = 7\,m$$</p>
<p>So, $$aRb \Rightarrow bRa$$</p>
<p>$$\therefore$$ R will be symmetric for $$a = 7{k_1} - 3$$</p>
<p>(3) Transitive : Let $$(a,b) \in R,\,(b,c) \in R$$</p>
<p>$$ \Rightarrow 3a + (7k - 3)b = 7{k_1}$$ and</p>
<p>$$3b + (7{k_2} - 3)c = 7{k_3}$$</p>
<p>Adding $$3a + 7kb + (7{k_2} - 3)\,c = 7({k_1} + {k_3})$$</p>
<p>$$3a + (7{k_2} - 3)\,c = 7\,m$$</p>
<p>$$\therefore$$ $$(a,c) \in R$$</p>
<p>$$\therefore$$ R is transitive</p>
<p>$$\therefore$$ $$\alpha = 7k - 3 = 7k + 4$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1ldo4r397 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$P(S)$$ denote the power set of $$S=\{1,2,3, \ldots ., 10\}$$. Define the relations $$R_{1}$$ and $$R_{2}$$ on $$P(S)$$ as $$\mathrm{AR}_{1} \mathrm{~B}$$ if $$\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}}\right) \cup\left(\mathrm{B} \cap \mathrm{A}^{\mathrm{c}}\right)=\emptyset$$ and $$\mathrm{AR}_{2} \mathrm{~B}$$ if $$\mathrm{A} \cup \mathrm{B}^{\mathrm{c}}=\mathrm{B} \cup \mathrm{A}^{\mathrm{c}}, \forall \mathrm{A}, \mathrm{B} \in \mathrm{P}(\mathrm{S})$$. Then :</p> | [{"identifier": "A", "content": "only $$R_{2}$$ is an equivalence relation"}, {"identifier": "B", "content": "both $$R_{1}$$ and $$R_{2}$$ are not equivalence relations"}, {"identifier": "C", "content": "both $$R_{1}$$ and $$R_{2}$$ are equivalence relations"}, {"identifier": "D", "content": "only $$R_{1}$$ is an equivalence relation"}] | ["C"] | null | $\begin{aligned} & \mathrm{S}=\{1,2,3, \ldots \ldots 10\} \\\\ & \mathrm{P}(\mathrm{S})=\text { power set of } \mathrm{S} \\\\ & \mathrm{AR}_1 \mathrm{B} \Rightarrow(\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})=\phi \\\\ & \mathrm{R}_1 \text { is reflexive, symmetric } \\\\ & \text { For transitive } \\\\ & (\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})=\phi ;\{\mathrm{a}\}=\phi=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & (\mathrm{B} \cap \overline{\mathrm{C}}) \cup(\overline{\mathrm{B}} \cap \mathrm{C})=\phi \therefore \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \text { equivalence. }\end{aligned}$
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5faj1z/fe5f5c16-ed0d-474b-8146-8a10ec18f85d/5e420070-ad0c-11ed-a86d-8dfe0389db88/file-1le5faj20.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5faj1z/fe5f5c16-ed0d-474b-8146-8a10ec18f85d/5e420070-ad0c-11ed-a86d-8dfe0389db88/file-1le5faj20.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Evening Shift Mathematics - Sets and Relations Question 42 English Explanation 1">
<br><br>$$
\mathrm{R}_2 \equiv \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B}
$$
<br><br>$\mathrm{R}_2 \rightarrow$ Reflexive, symmetric
<br><br><b>For transitive :</b>
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5faj1z/fe5f5c16-ed0d-474b-8146-8a10ec18f85d/5e420070-ad0c-11ed-a86d-8dfe0389db88/file-1le5faj20.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5faj1z/fe5f5c16-ed0d-474b-8146-8a10ec18f85d/5e420070-ad0c-11ed-a86d-8dfe0389db88/file-1le5faj20.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Evening Shift Mathematics - Sets and Relations Question 42 English Explanation 2">
<br><br>$\begin{aligned} & \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B} \Rightarrow\{\mathrm{a}, \mathrm{c}, \mathrm{d}\}=\{\mathrm{b}, \mathrm{c}, \mathrm{d}\} \\\\ & \{\mathrm{a}\}=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & \mathrm{B} \cup \overline{\mathrm{C}}=\overline{\mathrm{B}} \cup \mathrm{C} \Rightarrow \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \quad \therefore \mathrm{A} \cup \overline{\mathrm{C}}=\overline{\mathrm{A}} \cup \mathrm{C} \therefore \text { Equivalence }\end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift |
ldo96adc | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Among the relations
<br/><br/>$\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0\right\}$
<br/><br/> and $\mathrm{T}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}, \mathrm{a}^{2}-\mathrm{b}^{2} \in \mathbb{Z}\right\}$, | [{"identifier": "A", "content": "$\\mathrm{S}$ is transitive but $\\mathrm{T}$ is not\n"}, {"identifier": "B", "content": "both $\\mathrm{S}$ and $\\mathrm{T}$ are symmetric"}, {"identifier": "C", "content": "neither $S$ nor $T$ is transitive"}, {"identifier": "D", "content": "$T$ is symmetric but $S$ is not"}] | ["D"] | null | For relation $\mathrm{T}=\mathrm{a}^{2}-\mathrm{b}^{2}=-\mathrm{I}$
<br/><br/>Then, $(\mathrm{b}, \mathrm{a})$ on relation $\mathrm{R}$
<br/><br/>$\Rightarrow \mathrm{b}^{2}-\mathrm{a}^{2}=-\mathrm{I}$
<br/><br/>$\therefore \mathrm{T}$ is symmetric
<br/><br/>$\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0\right\}$
<br/><br/>$2+\frac{\mathrm{a}}{\mathrm{b}}>0 \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}>-2, \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}<\frac{-1}{2}$
<br/><br/>If $(b, a) \in \mathbf{S}$ then
<br/><br/>$2+\frac{\mathrm{b}}{\mathrm{a}}$ not necessarily positive
<br/><br/>$\therefore \mathrm{S}$ is not symmetric | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldomayox | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$R$$ be a relation on $$\mathbb{R}$$, given by $$R=\{(a, b): 3 a-3 b+\sqrt{7}$$ is an irrational number $$\}$$. Then $$R$$ is</p> | [{"identifier": "A", "content": "an equivalence relation"}, {"identifier": "B", "content": "reflexive and symmetric but not transitive"}, {"identifier": "C", "content": "reflexive and transitive but not symmetric"}, {"identifier": "D", "content": "reflexive but neither symmetric nor transitive"}] | ["D"] | null | <b>For reflexive :</b>
<br/><br/>$3 a-3 a+\sqrt{7}$ is an irrational number $\forall a \in R R$ is reflexive
<br/><br/><b>For symmetric :</b>
<br/><br/>Let $3 a-3 b+\sqrt{7}$ is an irrational number
<br/><br/>$\Rightarrow 3 b-3 a+\sqrt{7}$ is an irrational number
<br/><br/>For example, Let $3 a-3 b=\sqrt{7}$
<br/><br/>$\sqrt{7}+\sqrt{7}$ is irrational but $-\sqrt{7}+\sqrt{7}$ is not.
<br/><br/>$\therefore R$ is not symmetric
<br/><br/><b>For transitive :</b>
<br/><br/>Let $3 a-3 b+\sqrt{7}$ is irrational and $3 b-3 c+\sqrt{7}$ is irrational.
<br/><br/>$\Rightarrow 3 a-3 c+\sqrt{7}$ is irrational.
<br/><br/>For example, take $a=0, b=-\sqrt{7}, c=\frac{\sqrt{7}}{3}$
<br/><br/>$R$ is not transitive. | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldprx4f6 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$\mathrm{R}$$ be a relation on $$\mathrm{N} \times \mathbb{N}$$ defined by $$(a, b) ~\mathrm{R}~(c, d)$$ if and only if $$a d(b-c)=b c(a-d)$$. Then $$\mathrm{R}$$ is</p> | [{"identifier": "A", "content": "symmetric and transitive but not reflexive"}, {"identifier": "B", "content": "reflexive and symmetric but not transitive"}, {"identifier": "C", "content": "transitive but neither reflexive nor symmetric"}, {"identifier": "D", "content": "symmetric but neither reflexive nor transitive"}] | ["D"] | null | Given, $(a, b) R(c, d) \Rightarrow a d(b-c)=b c(a-d)$
<br/><br/><b>Symmetric :</b>
<br/><br/>(c, d) $R(a, b) \Rightarrow \operatorname{cb}(\mathrm{d}-\mathrm{a})=\mathrm{da}(\mathrm{c}-\mathrm{b}) $
<br/><br/>$\Rightarrow$ Symmetric.
<br/><br/><b>Reflexive :</b>
<br/><br/>(a, b) R (a, b) $\Rightarrow a b(b-a) \neq b a(a-b) $
<br/><br/>$\Rightarrow$ Not reflexive.
<br/><br/><b>Transitive : </b>
<br/><br/>$(2,3) \mathrm{R}(3,2)$ and $(3,2) \mathrm{R}(5,30)$ but
<br/><br/>$((2,3),(5,30)) \notin \mathrm{R} $
<br/><br/>$\Rightarrow$ Not transitive. | mcq | jee-main-2023-online-31st-january-morning-shift |
1ldr6q8ea | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>The minimum number of elements that must be added to the relation $$ \mathrm{R}=\{(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c})\}$$ on the set $$\{a, b, c\}$$ so that it becomes symmetric and transitive is :</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["A"] | null | <p>For symmetric $$(b,a),(c,b)\in R$$</p>
<p>For transitive $$(a,c)\in R$$</p>
<p>$$\Rightarrow (c,a)\in R$$</p>
<p>$$\therefore (a,b),(b,a)\in R$$</p>
<p>$$\Rightarrow (a,a)\in R$$</p>
<p>$$(b,c),(c,b)\in R$$</p>
<p>$$\Rightarrow (b,b)\in R,(c,c)\in R$$</p>
<p>7 elements must be added</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldselj5c | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let R be a relation defined on $$\mathbb{N}$$ as $$a\mathrm{R}b$$ if $$2a+3b$$ is a multiple of $$5,a,b\in \mathbb{N}$$. Then R is</p> | [{"identifier": "A", "content": "an equivalence relation"}, {"identifier": "B", "content": "non reflexive"}, {"identifier": "C", "content": "symmetric but not transitive"}, {"identifier": "D", "content": "transitive but not symmetric"}] | ["A"] | null | <p>a R b if 2a + 3b = 5m, m $$\in$$ $$l$$</p>
<p>(1) $$(a,a) \in R$$ as $$2a + 3a = 5a,a \in N$$</p>
<p>Hence, R is reflexive</p>
<p>(2) If $$(a,b) \in R$$ then $$2a + 3 = 5m$$</p>
<p>Now, $$5(a + b) = 5n$$</p>
<p>$$3a + 2b + 2a + 3b = 5n$$</p>
<p>$$\therefore$$ $$3a + 2b = 5(n - m)$$</p>
<p>$$\therefore$$ $$(b,a) \in R$$</p>
<p>$$\therefore$$ R is symmetric</p>
<p>(3) If $$(a,b) \in R$$ and $$(b,c) \in R$$ then</p>
<p>$$2a + 3b = 5m,2b + 3c = 5n$$</p>
<p>$$ \Rightarrow 2a + 5b + 3c = 5(m + n)$$</p>
<p>$$ \Rightarrow 2a + 3c = 5(m = n - b)$$</p>
<p>$$\therefore$$ $$(a,c) \in R$$</p>
<p>$$\therefore$$ R is transitive</p>
<p>Hence, R is equivalence relation.</p>
<p>Option (1) is correct.</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldwxc9aj | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>The minimum number of elements that must be added to the relation R = {(a, b), (b, c), (b, d)} on the set {a, b, c, d} so that it is an equivalence relation, is __________.</p> | [] | null | 13 | $R=\{(a, b)(b, c)(b, d)\}$
<br/><br/>
$S:\{a, b, c, d\}$
<br/><br/>
Adding $(a, a),(b, b),(c, c),(d, d)$ make reflexive.
<br/><br/>
Adding $(b, a),(c, b),(d, b)$ make Symmetric
<br/><br/>
And adding $(a, d),(a, c)$ to make transitive
<br/><br/>
Further $(d, a) \&(c, a)$ to be added to make Symmetricity.
<br/><br/>
Further $(c, d) \&(d, c)$ also be added.
<br/><br/>
So total 13 elements to be added to make equivalence. | integer | jee-main-2023-online-24th-january-evening-shift |
1ldyavo47 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>The relation $$\mathrm{R = \{ (a,b):\gcd (a,b) = 1,2a \ne b,a,b \in \mathbb{Z}\}}$$ is :</p> | [{"identifier": "A", "content": "reflexive but not symmetric"}, {"identifier": "B", "content": "transitive but not reflexive"}, {"identifier": "C", "content": "symmetric but not transitive"}, {"identifier": "D", "content": "neither symmetric nor transitive"}] | ["D"] | null | <p>Given,</p>
<p>(a, b) belongs to relation R if $$\gcd (a,b) = 1, 2a \ne b$$.</p>
<p>Here $$\gcd $$ means greatest common divisor. $$\gcd $$ of two numbers is the largest number that divides both of them.</p>
<p>(1) For Reflexive,</p>
<p>In $$aRa,\,\gcd (a,a) = a$$</p>
<p>$$\therefore$$ This relation is not reflexive.</p>
(2) For Symmetric:<br/><br/>
Take $a=2, b=1 \Rightarrow \operatorname{gcd}(2,1)=1$
Also $2 a=4 \neq b$
<br/><br/>Now $$\gcd (b,a) = 1$$ $ \Rightarrow \operatorname{gcd}(1,2)=1$<br/><br/>
and 2b should not be equal to a<br/><br/>
But here, $2 b=2=a$<br/><br/>
$\Rightarrow \mathrm{R}$ is not Symmetric<br/><br/>
(3) For Transitive:<br/><br/>
Let $\mathrm{a}=14, \mathrm{~b}=19, \mathrm{c}=21$<br/><br/>
$\operatorname{gcd}(\mathrm{a}, \mathrm{b})=1, 2a \ne b$<br/><br/>
$\operatorname{gcd}(\mathrm{b}, \mathrm{c})=1, 2b \ne c$<br/><br/>
$\operatorname{gcd}(\mathrm{a}, \mathrm{c})=7, 2a \ne c$<br/><br/>
Hence not transitive<br/><br/>
$\Rightarrow R$ is neither symmetric nor transitive. | mcq | jee-main-2023-online-24th-january-morning-shift |
1lgoxyjgf | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$\mathrm{A}=\{-4,-3,-2,0,1,3,4\}$$ and $$\mathrm{R}=\left\{(a, b) \in \mathrm{A} \times \mathrm{A}: b=|a|\right.$$ or $$\left.b^{2}=a+1\right\}$$ be a relation on $$\mathrm{A}$$. Then the minimum number of elements, that must be added to the relation $$\mathrm{R}$$ so that it becomes reflexive and symmetric, is __________</p> | [] | null | 7 | $$
\begin{aligned}
A & =\{-4,-3,-2,0,1,3,4\} \\\\
R= & \{(-4,4),(-3,3),(0,0),(1,1) \\
& (3,3),(4,4),(0,1),(3,-2)\}
\end{aligned}
$$
<br/><br/>Relation to be reflexive $(a, a) \in R \forall a \in A$
<br/><br/>$\Rightarrow (-4,-4),(-3,-3),(-2,-2)$ also should be added in $R$.
<br/><br/>Relation to be symmetric if $(a, b) \in R$, then $(b, a) \in R \forall a, b \in A$
<br/><br/>$\Rightarrow (4,-4),(3,-3),(1,0),(-2,3)$ also should be added in $R$
<br/><br/>$\Rightarrow$ Minimum number of elements to be added to $R=3+4=7$ | integer | jee-main-2023-online-13th-april-evening-shift |
1lgrgm8hz | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>The number of relations, on the set $$\{1,2,3\}$$ containing $$(1,2)$$ and $$(2,3)$$, which are reflexive and transitive but not symmetric, is __________.</p> | [] | null | 3 | <p>To find the number of such relations, let's first understand what it means for a relation to be reflexive, transitive, and not symmetric.</p>
<p>A relation $$R$$ on a set $$S$$ is <strong>reflexive</strong> if every element is related to itself. That is, $$(a, a) \in R$$ for all $$a \in S$$.</p>
<p>A relation is <strong>transitive</strong> if whenever $$(a, b) \in R$$ and $$(b, c) \in R$$, then it must also be the case that $$(a, c) \in R$$.</p>
<p>A relation is <strong>symmetric</strong> if whenever $$(a, b) \in R$$, then $$(b, a) \in R$$ also.</p>
<p>Since we are looking for relations that are reflexive and transitive but <em>not</em> symmetric, we will need to include certain elements and exclude others.</p>
<p>First, for the relation to be reflexive on the set $$\{1,2,3\}$$, it must contain $$(1,1), (2,2),$$ and $$(3,3)$$.</p>
<p>Given that the relation must contain $$(1,2)$$ and $$(2,3)$$ and be transitive, it must also contain $$(1,3)$$ because if $$(1,2)$$ and $$(2,3)$$ are included, then to maintain transitivity, $$(1,3)$$ must be included as well.</p>
<p>So far, the must-have elements of the desired relation are:</p>
<ul>
<li>For reflexivity: $$(1,1), (2,2), (3,3)$$</li>
<br/><li>Given in the problem: $$(1,2), (2,3)$$</li>
<br/><li>For transitivity (induced by given elements): $$(1,3)$$</li>
</ul>
<p>"To maintain a relation that is reflexive, transitive, and not symmetric, we must carefully select additional pairs beyond the reflexive minimum $(1,1), (2,2), (3,3)$, and the given $(1,2), (2,3)$, which also necessitates $(1,3)$ due to transitivity. While including pairs such as $(2,1)$, $(3,1)$, or $(3,2)$ could potentially introduce symmetry, we can include some of these pairs as long as the resulting relation does not fulfill the condition for symmetry for all elements. This means we can include one or more of these pairs if doing so does not result in every pair being mirrored (i.e., for every $(a,b) \in R$, $(b,a) \in R$ is not required), thus keeping the relation not symmetric. The key is ensuring that the inclusion of any such pair does not lead to a situation where for every $(a,b)$ in the relation, the reverse $(b,a)$ also exists, which would make the relation symmetric, contradicting our requirement."</p>
<br/>1. R<sub>1</sub> = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}
<br/><br/> Here, none of (2,1), (3,2), (3,1) are in R<sub>1</sub>.
<br/><br/>2. R<sub>2</sub> = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (2,1)}
<br/><br/> Here, only (2,1) is in R<sub>2</sub>, and neither (3,2) nor (3,1) are in R<sub>2</sub>.
<br/><br/>3. R<sub>3</sub> = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (3,2)}
<br/><br/> Here, only (3,2) is in R<sub>3</sub>, and neither (2,1) nor (3,1) are in R<sub>3</sub>. | integer | jee-main-2023-online-12th-april-morning-shift |
1lgyld2f2 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$\mathrm{A}=\{1,2,3,4,5,6,7\}$$. Then the relation $$\mathrm{R}=\{(x, y) \in \mathrm{A} \times \mathrm{A}: x+y=7\}$$ is :</p> | [{"identifier": "A", "content": "reflexive but neither symmetric nor transitive"}, {"identifier": "B", "content": "transitive but neither symmetric nor reflexive"}, {"identifier": "C", "content": "symmetric but neither reflexive nor transitive"}, {"identifier": "D", "content": "an equivalence relation"}] | ["C"] | null | Here, $A=\{1,2,3,4,5,6,7\}$
<br/><br/>Since, $x+y=7 \Rightarrow y=7-x$
<br/><br/>So, $\mathrm{R}=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$
<br/><br/>$\because(a, b) \in \mathrm{R} \Rightarrow(b, a) \in \mathrm{R}$
<br/><br/>$\therefore \mathrm{R}$ is symmetric only. | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh00d5h8 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$A=\{0,3,4,6,7,8,9,10\}$$ and $$R$$ be the relation defined on $$A$$ such that $$R=\{(x, y) \in A \times A: x-y$$ is odd positive integer or $$x-y=2\}$$. The minimum number of elements that must be added to the relation $$R$$, so that it is a symmetric relation, is equal to ____________.</p> | [] | null | 19 | We have, $A=\{0,3,4,6,7,8,9,10\}$
<br/><br/>Case I : $x-y$ is odd, if one is odd and one is even and $x>y$.
<br/><br/>$\therefore$ Possibilites are $\{(3,0),(4,3),(6,3),(7,6),(7,4)$, $(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10$, $7),(10,3)\}$
<br/><br/>No. of cases $=15$
<br/><br/>Case II : $x-y=2$
<br/><br/>$\therefore$ Possibilities are $\{(6,4),(8,6),(9,7),(10,8)\}$
<br/><br/>$\therefore$ No. of cases $=4$
<br/><br/>So, minimum ordered pair to be added $=15+4=19$ | integer | jee-main-2023-online-8th-april-morning-shift |
lsam6kw5 | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Consider the relations $R_1$ and $R_2$ defined as $a R_1 b \Leftrightarrow a^2+b^2=1$ for all $a, b \in \mathbf{R}$ and $(a, b) R_2(c, d) \Leftrightarrow$ $a+d=b+c$ for all $(a, b),(c, d) \in \mathbf{N} \times \mathbf{N}$. Then : | [{"identifier": "A", "content": "$R_1$ and $R_2$ both are equivalence relations"}, {"identifier": "B", "content": "Only $R_1$ is an equivalence relation"}, {"identifier": "C", "content": "Only $R_2$ is an equivalence relation"}, {"identifier": "D", "content": "Neither $R_1$ nor $R_2$ is an equivalence relation"}] | ["C"] | null | <p>To determine if the given relations $R_1$ and $R_2$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity.</p>
<p>Let's start by analysing $R_1$:</p>
<p>Reflexivity: A relation $R$ on a set $S$ is reflexive if every element is related to itself, that is, for every $a \in S$, the pair $(a,a)$ is in $R$. In the case of $R_1$, for an arbitrary $a \in \mathbf{R}$, we need to check if $a R_1 a$ holds, which translates to checking if $a^2 + a^2 = 1$. This would mean $2a^2 = 1$ or $a^2 = \frac{1}{2}$. Since this is not true for every real number $a$, $R_1$ is not reflexive.</p>
<p>Symmetry: A relation $R$ is symmetric if whenever $a R b$, then $b R a$. For $R_1$, if $a^2 + b^2 = 1$, then it is also true that $b^2 + a^2 = 1$. Thus, $R_1$ is symmetric.</p>
<p>Transitivity: A relation $R$ is transitive if whenever $a R b$ and $b R c$, then $a R c$. For $R_1$, suppose $a R_1 b$ and $b R_1 c$, this means $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$. However, if we add these two, we get $a^2 + 2b^2 + c^2 = 2$. There is no guarantee that $a^2 + c^2 = 1$. Therefore, $R_1$ is not transitive.</p>
<p>Conclusion: Relation $R_1$ is not reflexive or transitive, and hence it is not an equivalence relation.</p>
<p>Now let's consider $R_2$:</p>
<p>Reflexivity: For any $(a, b) \in \mathbf{N} \times \mathbf{N}$, it's clear that $a + b = b + a$, which is just a reiteration of the commutative property of addition. Therefore, $(a, b) R_2 (a, b)$, and $R_2$ is reflexive.</p>
<p>Symmetry: If $(a, b) R_2 (c, d)$, meaning $a + d = b + c$, then by reordering the terms we can similarly have $c + b = d + a$, which means $(c, d) R_2 (a, b)$, so $R_2$ is symmetric.</p>
<p>Transitivity: Suppose $(a, b) R_2 (c, d)$ and $(c, d) R_2 (e, f)$, i.e., $a + d = b + c$ and $c + f = d + e$, we want to show that $(a, b) R_2 (e, f)$. Adding the two equations, we get $a + d + c + f = b + c + d + e$. By rearranging and simplifying, we get $a + f = b + e$, thus, $(a, b) R_2 (e, f)$.</p>
<p>So $R_2$ is reflexive, symmetric, and transitive, and therefore it is an equivalence relation.</p>
<p>Conclusion: According to the above examination, only $R_2$ is an equivalence relation. Thus, the correct answer is:</p>
<p>Option C: Only $R_2$ is an equivalence relation.</p> | mcq | jee-main-2024-online-1st-february-evening-shift |
lsbkh6se | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | Let $S=\{1,2,3, \ldots, 10\}$. Suppose $M$ is the set of all the subsets of $S$, then the relation <br/><br/>$\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{A} \cap \mathrm{B} \neq \phi ; \mathrm{A}, \mathrm{B} \in \mathrm{M}\}$ is : | [{"identifier": "A", "content": "symmetric only"}, {"identifier": "B", "content": "reflexive only"}, {"identifier": "C", "content": "symmetric and reflexive only"}, {"identifier": "D", "content": "symmetric and transitive only"}] | ["A"] | null | <p>Let $$S=\{1,2,3, \ldots, 10\}$$</p>
<p>$$R=\{(A, B): A \cap B \neq \phi ; A, B \in M\}$$</p>
<p>For Reflexive,</p>
<p>$$M$$ is subset of '$$S$$'</p>
<p>So $$\phi \in \mathrm{M}$$</p>
<p>for $$\phi \cap \phi=\phi$$</p>
<p>$$\Rightarrow$$ but relation is $$\mathrm{A} \cap \mathrm{B} \neq \phi$$</p>
<p>So it is not reflexive.</p>
<p>For symmetric,</p>
<p>$$\begin{array}{ll}
\text { ARB } & \mathrm{A} \cap \mathrm{B} \neq \phi, \\
\Rightarrow \mathrm{BRA} & \Rightarrow \mathrm{B} \cap \mathrm{A} \neq \phi,
\end{array}$$</p>
<p>So it is symmetric.</p>
<p>For transitive,</p>
<p>$$\begin{aligned}
\text { If } A & =\{(1,2),(2,3)\} \\
B & =\{(2,3),(3,4)\} \\
C & =\{(3,4),(5,6)\}
\end{aligned}$$</p>
<p>$$\mathrm{ARB}$$ & $$\mathrm{BRC}$$ but $$\mathrm{A}$$ does not relate to $$\mathrm{C}$$ So it not transitive</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lseymk2g | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$R$$ be a relation on $$Z \times Z$$ defined by $$(a, b) R(c, d)$$ if and only if $$a d-b c$$ is divisible by 5. Then $$R$$ is</p> | [{"identifier": "A", "content": "Reflexive and transitive but not symmetric\n"}, {"identifier": "B", "content": "Reflexive and symmetric but not transitive\n"}, {"identifier": "C", "content": "Reflexive but neither symmetric nor transitive\n"}, {"identifier": "D", "content": "Reflexive, symmetric and transitive"}] | ["B"] | null | <p>$$(a, b) R(a, b)$$ as $$a b-a b=0$$</p>
<p>Therefore reflexive</p>
<p>Let $$(a, b) R(c, d) \Rightarrow a d-b c$$ is divisible by 5</p>
<p>$$\Rightarrow \mathrm{bc}-\mathrm{ad}$$ is divisible by $$5 \Rightarrow(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})$$</p>
<p>Therefore symmetric</p>
<p>Relation not transitive as $$(3,1) \mathrm{R}(10,5)$$ and $$(10,5) \mathrm{R}(1,1)$$ but $$(3,1)$$ is not related to $$(1,1)$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
lv2er45n | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let a relation $$\mathrm{R}$$ on $$\mathrm{N} \times \mathbb{N}$$ be defined as: $$\left(x_1, y_1\right) \mathrm{R}\left(x_2, y_2\right)$$ if and only if $$x_1 \leq x_2$$ or $$y_1 \leq y_2$$.
Consider the two statements:</p>
<p>(I) $$\mathrm{R}$$ is reflexive but not symmetric.</p>
<p>(II) $$\mathrm{R}$$ is transitive</p>
<p>Then which one of the following is true?</p> | [{"identifier": "A", "content": "Only (II) is correct.\n"}, {"identifier": "B", "content": "Both (I) and (II) are correct.\n"}, {"identifier": "C", "content": "Neither (I) nor (II) is correct.\n"}, {"identifier": "D", "content": "Only (I) is correct."}] | ["D"] | null | <p>$$\begin{aligned}
& \left(x_1, y_1\right) R\left(x_2, y_2\right) \\
& \text { If } x_1 \leq x_2 \text { or } y_1 \leq y_2
\end{aligned}$$</p>
<p>For reflexive;</p>
<p>$$\begin{aligned}
& \left(x_1, y_1\right) R\left(x_1, y_1\right) \\
& \Rightarrow x_1 \leq x_1 \text { or } y_1 \leq y_1
\end{aligned}$$</p>
<p>So, $$R$$ is reflexive</p>
<p>For symmetric</p>
<p>When $$\left(x_1, y_1\right) R\left(x_2, y_2\right)$$</p>
<p>$$\Rightarrow x_1 \leq x_2 \text { or } y_1 \leq y_2$$</p>
<p>For $$\left(x_2, y_2\right) R\left(x_1, y_1\right)$$</p>
<p>$$\Rightarrow x_2 \leq x_1 \text { or } y_2 \leq y_1$$</p>
<p>Not true for $$(1,2)$$ and $$(3,4)$$</p>
<p>For transitive</p>
<p>Take pairs as $$(3,9),(4,6),(2,7)$$</p>
<p>$$(3,9) R(4,6)$$</p>
<p>as $$4 \geq 3$$</p>
<p>$$(4,6) R(2,7)$$</p>
<p>As $$7 \geq 6$$</p>
<p>But $$(3,9) R(2,7)$$</p>
<p>As neither $$2 \geq 3$$ nor $$7 \geq 9$$</p>
<p>So not transitive</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv3vefnm | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$A=\{2,3,6,8,9,11\}$$ and $$B=\{1,4,5,10,15\}$$. Let $$R$$ be a relation on $$A \times B$$ defined by
$$(a, b) R(c, d)$$ if and only if $$3 a d-7 b c$$ is an even integer. Then the relation $$R$$ is</p> | [{"identifier": "A", "content": "reflexive but not symmetric.\n"}, {"identifier": "B", "content": "an equivalence relation.\n"}, {"identifier": "C", "content": "reflexive and symmetric but not transitive.\n"}, {"identifier": "D", "content": "transitive but not symmetric."}] | ["C"] | null | <p>$$(a, b) R(c, d) \Rightarrow 3 a d-7 b c \in$$ even</p>
<p>For reflexive</p>
<p>$$(a, b) R(a, b) \Rightarrow 3 a b-7 b a=-4 a b \in$$ even</p>
<p>For symmetric</p>
<p>$$(a, b) R(c, d)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4jkysz/b1310f77-4268-428b-aeb5-13316c958a9a/413519a0-10ec-11ef-8836-ed8ff380bbe9/file-1lw4jkyt6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4jkysz/b1310f77-4268-428b-aeb5-13316c958a9a/413519a0-10ec-11ef-8836-ed8ff380bbe9/file-1lw4jkyt6.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Evening Shift Mathematics - Sets and Relations Question 4 English Explanation"></p>
<p>then</p>
<p>$$(c, d) R(a, b)=3 b c-7 a d$$</p>
<p>$$\in$$ even</p>
<p>Now check for transitive</p>
<p>$$\begin{aligned}
& \begin{array}{ll}
(a, b) R(c, d) \text { and }(c, d) R(e, f) \text { then }(a, b) R(e, f) \\
3 a d-7 b c=2 m \quad \Rightarrow (2,5) R(6,8) \text { and } \\
3 c f-7 e d=2 n \quad (6,8) R(9,4) \\
\text { then } 3 a f-7 e b \neq \text { even } \notin(2,5) R(9,4) \\
\Rightarrow \text { Not transitive option }(3)
\end{array}
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lvb294aa | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let $$\mathrm{A}=\{1,2,3,4,5\}$$. Let $$\mathrm{R}$$ be a relation on $$\mathrm{A}$$ defined by $$x \mathrm{R} y$$ if and only if $$4 x \leq 5 \mathrm{y}$$. Let $$\mathrm{m}$$ be the number of elements in $$\mathrm{R}$$ and $$\mathrm{n}$$ be the minimum number of elements from $$\mathrm{A} \times \mathrm{A}$$ that are required to be added to R to make it a symmetric relation. Then m + n is equal to :</p> | [{"identifier": "A", "content": "23"}, {"identifier": "B", "content": "26"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "24"}] | ["C"] | null | <p>$$\begin{aligned}
& A=\{1,2,3,4,5\} \\
& x R y \Leftrightarrow 4 x \leq 5 y \\
& 4 x \leq 5 y \quad \Rightarrow \quad \frac{x}{y} \leq \frac{5}{4} \quad \Rightarrow \frac{x}{y} \leq 1.25
\end{aligned}$$</p>
<p>$$\begin{aligned}
& R=\{(1,2),(1,3),(1,4),(1,5),(1,1),(2,2),(2,3),(2,4), \\
& (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\} \\
& \therefore \quad n(R)=m=16
\end{aligned}$$</p>
<p>Elements to be added to $$R$$ to make it symmetric</p>
<p>$$(1,2) \in R \quad \Rightarrow \quad(2,1)$$ should be added</p>
<p>Similarly, $$(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3), (5,3)$$</p>
<p>$$\therefore 9$$ elements should be added</p>
<p>$$\begin{array}{ll}
\therefore & n=9 \\
\therefore & m+n=25
\end{array}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
lvc57b2n | maths | sets-and-relations | symmetric,-transitive-and-reflexive-properties | <p>Let the relations $$R_1$$ and $$R_2$$ on the set $$X=\{1,2,3, \ldots, 20\}$$ be given by $$R_1=\{(x, y): 2 x-3 y=2\}$$ and $$R_2=\{(x, y):-5 x+4 y=0\}$$. If $$M$$ and $$N$$ be the minimum number of elements required to be added in $$R_1$$ and $$R_2$$, respectively, in order to make the relations symmetric, then $$M+N$$ equals</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "10"}] | ["D"] | null | <p>$$\begin{aligned}
& R_1=\{(x, y): 2 x-3 y=2\} \\
& R_2=\{(x, y):-5 x+4 y=0\} \\
& 2 x-3 y=2
\end{aligned}$$</p>
<p>So $$2 x$$ and $$3 y$$ both has to be even or odd simultaneously and $$2 x$$ can't be odd so $$2 x$$ and $$3 y$$ both will be even</p>
<p>$$R_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\}$$</p>
<p>For symmetric we need to add 6 elements as</p>
<p>$$\begin{aligned}
& (2,4),(4,7),(6,10),(8,13),(10,16),(12,19) \\
& M=6
\end{aligned}$$</p>
<p>For $$R_2-5 x+4 y=0$$</p>
<p>$$5 x$$ and $$4 y$$ has to be equal $$4 y$$ is always even so $$5 x$$ will also be even</p>
<p>$$R_2=\{(4,5),(8,10),(12,15),(16,20)\}$$</p>
<p>For symmetric we need to add 4 element as</p>
<p>$$\begin{aligned}
& (5,4)(10,8)(15,12)(20,16) \\
& N=4 \\
& M+N=6+4=10
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
ljalzt2e | maths | sets-and-relations | venn-diagram | If $A, B$ and $C$ are three sets such that $A \cap B=A \cap C$ and $A \cup B=A \cup C$, then : | [{"identifier": "A", "content": "$A=C$"}, {"identifier": "B", "content": "$B=C$"}, {"identifier": "C", "content": "$A \\cap B=\\phi$"}, {"identifier": "D", "content": "$A=B$"}] | ["B"] | null | <p>From the given conditions, we have :</p>
<ol>
<li><p>A ∩ B = A ∩ C : The intersection of set A with set B is the same as the intersection of set A with set C. This indicates that all elements common to A and B are also common to A and C, and vice versa.</p>
</li>
<br/><li><p>A ∪ B = A ∪ C : The union of set A with set B is the same as the union of set A with set C. This indicates that all elements in A, B, and C are the same.</p>
</li>
</ol>
<p>From these two conditions, we can infer that set B is equal to set C because every element of B is also an element of C and vice versa. Hence, <b>Option B :</b> B = C is the correct answer.</p> | mcq | aieee-2009 |
l7sKsDq2kn1zBOJ3jUvQF | maths | sets-and-relations | venn-diagram | In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "102"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "38"}] | ["D"] | null | We're given that there are 140 students numbered from 1 to 140.
<br/><br/>1. Define the set $A$ to be the set of even numbered students. The cardinality of $A$ (the number of elements in $A$), denoted as $n(A)$, can be computed as the greatest integer less than or equal to $140/2$. Hence, $n(A) = \left[\frac{140}{2}\right] = 70$. ([.] denotes greatest integer function)
<br/><br/>2. Similarly, let $B$ be the set of students whose number is divisible by 3. Hence, $n(B) = \left[\frac{140}{3}\right] = 46$. ([.] denotes greatest integer function)
<br/><br/>3. Let $C$ be the set of students whose number is divisible by 5. Hence, $n(C) = \left[\frac{140}{5}\right] = 28$.
<br/><br/>So far, we've found the number of students who opted for Mathematics ($n(A)$), Physics ($n(B)$), and Chemistry ($n(C)$).
<br/><br/>We also need to consider the students who have opted for multiple subjects :
<br/><br/>1. $n(A \cap B)$ represents the count of numbers that are divisible by both 2 and 3 (i.e., divisible by 6).
<br/><br/>So, $n(A \cap B) = \left[\frac{140}{6}\right] = 23$.
<br/><br/>2. $n(B \cap C)$ represents the count of numbers that are divisible by both 3 and 5 (i.e., divisible by 15).
<br/><br/>So, $n(B \cap C) = \left[\frac{140}{15}\right] = 9$.
<br/><br/>3. $n(C \cap A)$ represents the count of numbers that are divisible by both 2 and 5 (i.e., divisible by 10).
<br/><br/>So, $n(C \cap A) = \left[\frac{140}{10}\right] = 14$.
<br/><br/>Finally, $n(A \cap B \cap C)$ represents the count of numbers that are divisible by 2, 3, and 5 (i.e., divisible by 30).
<br/><br/>So, $n(A \cap B \cap C) = \left[\frac{140}{30}\right] = 4$.
<br/><br/>Now we use the principle of inclusion and exclusion to compute the number of students who have opted for at least one subject. The principle states :
<br/><br/>$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$$
<br/><br/>Substituting the values we calculated above :
<br/><br/>$$n(A \cup B \cup C) = (70+46+28) - (23+9+14) + 4 = 102$$
<br/><br/>Hence, the number of students who opted for at least one subject is 102. Therefore, the number of students who did not opt for any of the subjects is <br><br>Total $$-$$ n(A $$ \cup $$ B $$ \cup $$ C)
<br><br>= 140 $$-$$ 102 = 38 | mcq | jee-main-2019-online-10th-january-morning-slot |
bm95iHHu7AMPxaoEUQ18hoxe66ijvwvu5os | maths | sets-and-relations | venn-diagram | Two newspapers A and B are published in a city.
It is known that 25% of the city populations reads
A and 20% reads B while 8% reads both A and
B. Further, 30% of those who read A but not B
look into advertisements and 40% of those who
read B but not A also look into advertisements,
while 50% of those who read both A and B look
into advertisements. Then the percentage of the
population who look into advertisement is :- | [{"identifier": "A", "content": "13.5"}, {"identifier": "B", "content": "13"}, {"identifier": "C", "content": "12.8"}, {"identifier": "D", "content": "13.9"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265966/exam_images/s3wpew3c8wnvsbkrzigh.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263345/exam_images/cerp3qr3wsysa2fxukmv.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264435/exam_images/to8fimmktaygs2umiznl.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Sets and Relations Question 78 English Explanation"></picture>
<p>The total population to be 100 (for simplicity's sake) and the percentages can be treated as actual numbers of people in this context.</p>
<p>The percentage of people who read newspaper A is given as 25. However, among these, there are people who read both newspapers A and B, given as 8. To find the number of people who read only newspaper A, we subtract the number of people who read both from the total number of people who read A. That is,</p>
<p>n(A only) = 25 – 8 = 17</p>
<p>Similarly, the number of people who read only newspaper B is calculated as :</p>
<p>n(B only) = 20 – 8 = 12</p>
<p>Now, we are given the percentage of each of these groups that look into the advertisements:</p>
<ul>
<li>30% of those who read A but not B,</li>
<li>40% of those who read B but not A,</li>
<li>50% of those who read both A and B.</li>
</ul>
<p>To find the total percentage of the population that looks into advertisements, we add up the contributions from each of these groups. We calculate each group's contribution by multiplying the size of the group by the percentage of that group that looks at advertisements :</p>
<p>= $${{30} \over {100}} \times 17$$ (from A only) + $${{40} \over {100}} \times 12 $$(from B only) + $${{50} \over {100}} \times 8 $$(from both A and B)</p>
<p>= 5.1 (from A only) + 4.8 (from B only) + 4 (from both A and B)</p>
<p>Adding these up, we get</p>
<p>= 13.9</p>
<p>This means that 13.9% of the total population looks into the advertisements.</p>
<p>So, the correct answer is :</p>
<p>Option D : 13.9.</p>
| mcq | jee-main-2019-online-9th-april-evening-slot |
O7qOj5q59aigrfqtMA3rsa0w2w9jxae4toq | maths | sets-and-relations | venn-diagram | Let A, B and C be sets such that $$\phi $$ $$ \ne $$ A $$ \cap $$ B $$ \subseteq $$ C. Then which of the following statements is not true ? | [{"identifier": "A", "content": "If (A \u2013 B) $$ \\subseteq $$ C, then A $$ \\subseteq $$ C"}, {"identifier": "B", "content": "B $$ \\cap $$ C $$ \\ne $$ $$\\phi $$"}, {"identifier": "C", "content": "(C $$ \\cup $$ A) $$ \\cap $$ (C $$ \\cup $$ B) = C"}, {"identifier": "D", "content": "If (A \u2013 C) $$ \\subseteq $$ B, then A $$ \\subseteq $$ B "}] | ["D"] | null | According to the question, we have the following Venn diagram.
<br><br>Here, $A \cap B \subseteq C$ and $A \cap B \neq \phi$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lj7pt113/8d56bc6b-c920-4fcb-9e3c-2bea678573a5/22404c90-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pt13t.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lj7pt113/8d56bc6b-c920-4fcb-9e3c-2bea678573a5/22404c90-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pt13t.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Sets and Relations Question 77 English Explanation 1">
<br>Now, from the Venn diagram, it is clear that $B \cap C \neq \phi$, is true
<br><br>Also, $(C \cup A) \cap(C \cup B)=C \cup(A \cap B)=C$ is true.
<br><br>If $(A-B) \subseteq C$, for this statement the Venn diagram is
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lj7pw2t5/50f84c21-de84-4885-acda-aeb2d0014e55/76fea290-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pw2t6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lj7pw2t5/50f84c21-de84-4885-acda-aeb2d0014e55/76fea290-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pw2t6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Sets and Relations Question 77 English Explanation 2">
<br>From the Venn diagram, it is clear that if $A-B \subseteq C$, then $A \subseteq C$.
<br><br>Now, if $(A-C) \subseteq B$, for this statement the Venn diagram.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lj7pvyrh/71535ada-ad4c-4284-8e05-1212a5834ebf/73de76d0-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pvyri.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lj7pvyrh/71535ada-ad4c-4284-8e05-1212a5834ebf/73de76d0-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pvyri.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Sets and Relations Question 77 English Explanation 3">
<br>From the Venn diagram, it is clear that
<br><br>$A \cap B \neq \phi, A \cap B \subseteq C$ and $A-C=\phi \subseteq B$ but $A \subseteq B$ | mcq | jee-main-2019-online-12th-april-evening-slot |
0lH3KaKfqfK9WMpOSp7k9k2k5kheoi8 | maths | sets-and-relations | venn-diagram | If A = {x $$ \in $$ R : |x| < 2} and B = {x $$ \in $$ R : |x – 2| $$ \ge $$ 3};
then : | [{"identifier": "A", "content": "A \u2013 B = [\u20131, 2)"}, {"identifier": "B", "content": "A $$ \\cup $$ B = R \u2013 (2, 5)"}, {"identifier": "C", "content": "A $$ \\cap $$ B = (\u20132, \u20131)"}, {"identifier": "D", "content": "B \u2013 A = R \u2013 (\u20132, 5)"}] | ["D"] | null | A : x $$ \in $$ (–2, 2);
<br><br>B : x $$ \in $$ (–$$\infty $$, –1] $$ \cup $$ [5, $$\infty $$)
<br><br>$$ \Rightarrow $$ B – A = R – (–2, 5)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267710/exam_images/ndmd3v807ggdl7pdczek.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Evening Slot Mathematics - Sets and Relations Question 75 English Explanation"> | mcq | jee-main-2020-online-9th-january-evening-slot |
D2PrxSbbi6MwpADcmujgy2xukf8zgpnm | maths | sets-and-relations | venn-diagram | A survey shows that 63% of the people in a city read newspaper A whereas 76% read
newspaper B. If x% of the people read both the newspapers, then a possible value of x can be: | [{"identifier": "A", "content": "37"}, {"identifier": "B", "content": "65"}, {"identifier": "C", "content": "29"}, {"identifier": "D", "content": "55"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263695/exam_images/le1gtktn6hvqfqyl9vnt.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267027/exam_images/vaqvolhmyxkkppgwspnb.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266258/exam_images/s2o6kxbybhgir34w3g6x.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 4th September Morning Slot Mathematics - Sets and Relations Question 70 English Explanation"></picture>
<br><br>A $$ \cup $$ B = 63 - x + x + 76 - x = 139 - x
<br><br>As 139 - x $$ \le $$ 100
<br><br>$$ \Rightarrow $$ x $$ \ge $$ 39
<br><br>From venn diagram, you can see x should be less than 76 and 63 otherwise only A newspaper or only B newspaper reader will be negative number.
<br><br>Intersection of x $$ \le $$ 63 and x $$ \le $$76 is = x $$ \le $$ 63.
<br><br>$$ \therefore $$ 39 $$ \le $$ x $$ \le $$ 63
<br><br>From options possible value of x = 55. | mcq | jee-main-2020-online-4th-september-morning-slot |
A6axaqgESreYfYvDBFjgy2xukfg6ejul | maths | sets-and-relations | venn-diagram | A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be : | [{"identifier": "A", "content": "63"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "54"}, {"identifier": "D", "content": "38"}] | ["B"] | null | C $$ \to $$ person like coffee
<br><br>T $$ \to $$ person like Tea
<br><br>n(C) = 73
<br><br>n(T) = 65
<br><br>n(C $$ \cup $$ T) $$ \le $$ 100
<br><br>n(C) + n(T) – n (C $$ \cap $$ T) $$ \le $$ 100
<br><br>73 + 65 – x $$ \le $$ 100
<br><br>x $$ \ge $$ 38
<br><br>73 – x $$ \ge $$ 0 $$ \Rightarrow $$ x $$ \le $$ 73
<br><br>65 – x $$ \ge $$ 0 $$ \Rightarrow $$ x $$ \le $$ 65
<br><br>$$ \therefore $$ 38 $$ \le $$ x $$ \le $$ 65 | mcq | jee-main-2020-online-5th-september-morning-slot |
XyHNmjyUgZXbsPIFQb1kmjb97ia | maths | sets-and-relations | venn-diagram | In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?<br/><br/><img src="data:image/png;base64,UklGRoQbAABXRUJQVlA4IHgbAABwggCdASoTArYAPm00lkgkIqIhI5KbEIANiWlu/HyXDcW+TUvlCOTKOR8J/0n8iPel3/fc/7f/cv9V/cfIn8//f/yZ/eD3nf7Ppkf8f+G9458b/Gv7H/v/7z85P0T/CfxP+n/8/+7+j/vO/hv5n/Tv1a+QL1t/iv4z/Xf9b/YPT7/zP4B/Ov8z4bdev95/M/YF9qvon+F/j39W/0/9m9Cj+l/m/9O9kfrd/uv5t8AP6vf6n+K/1L/o/3z5o/x3/F/g/9k83bxb9Mv2A+gH+o/2D/Tfy3+zf7f+y/TZ+qf6L+H/17/1f4D////f4m/nv9R/3n8Y/uv/k/u3///BH+h/zr/OfxH+qf+H+7f/////bv/1f4B79P189l39V/+V/Ef3/KXzJ9qpqgWllolTkiLu0juRF4vRp/utllndVqblVZvsyGcLyohR5AuEpDVR/MJ/EkTlVWBr0Tm2cDgaqJzpisUBjZV38DHiA6NurHWtG9Ko+6nWK4YkPw2UrPfdcpiNUFe+852MKHspvALsG1SSkHPzrSORlaIk/G80ZiNmoYghsJFM7aBdd7eXLKzGfC8ExyCu+ykAqszubczPL8IJYxXWljb7gh80+Czv4hvHKAbDTImYSFa/QNPd9HCxh6CaiN+EVPt+c4dm2d89XyAbWz1y5FaxvOFXxrDcmQOWSY0FiAoeb/5U/fJAa/uv+EkPbB099i2C3uVYIqTqafl7/iK2AM5HTnXsUmTo6gnhveW50CskWpN0hnshBburvyrA7N9jccyYq/Pt3YPedEMX4PKtmUO/GP/BGX8IgUxY4WpNvfse/3/zKqlSD+t1VamK5PEE3KBv7gOlBnhA50CjBooptTPQ9KH5zJQp5xNIAwLkJE/szDdB9Inh1Yowrh2wz6uqIZ8G+dcggJD3S/bnIg7Hd6YXY8cwGj/Ssj+GmxT3ehOaY5ovYqZYcy36UUc8XJCZVRmwao7TJLO0yl55Y6Oa0crbANqfV7Qp0GWUcBtitVQ6vQB+IzwJxPpUv9u0MkIvYoG5PF02n27aOhMIYcVjVN/MCBleU/ZdEU0Zlpf0pMZ5GcdSZayc2NLt/Vc7Fp5lqfzza8ne7Cmiz3SoMQikFdg//CAvaX7kaC+f3PaA3tOiPh8tjwNnUKwvFG8R8cEIZl2S+8OvjPwsf9MIXmTYNf24Uej1FCCizFMvkLw4IQf/48Ih0Z+7v1xs6GfspBeIxCbW+wWWUp4QO/EDjiEf/7cROdXbCrW/ogRZkki8nn5eKz5Qx1A8iKqRt8MfGKUF/YeUtVg7/Lf/+fWvljQhaUzLXcIcGw5Xg4NiVEKPI56kCxkE6tYFXcIcGwgYrkhS/nhrD9nHkc9SEFFfYOObMaNCiFHkjTaKHBsSnNmeEF3CHBsUVa3aQAD+6/vaV4Ab4vVWdL0otwCTwzgw1GHHG773FycJ3egPoioU70Dsme9u9fLKOBNBdBZeoE8qRS5L+D+7+rjf1fRnp4tpDaifzTHR5QUvDRydH1oNCyNlIMPOpJnR6GFZh1FLZwMJ2m6giznhUQfivmBqb44UcVwzBLkH1JgYrgE71XT9yn2nlnLk2fbj2atTsFrRug1jnX4Bn41+zKetCNkPatg1Hvtx3gjHQrpDC8XSwftVVSW13LIVSM1Iz459WnkyP19mimahPwTYHHL4ClIK/pijmQ8IVQ5w+3HW4yqNFrLkBWW7UNCPrLl74Awd449ZhRhgMoJRNwG5GF8qqml+hoKceTL6jWJXqKioq8TPJphmbTwwiGB5zXMB2t9KYPemhOhNphW/iMpuDw8XqBTXaL20XrC9VeXTXJuaGKg3sUrwAy4Crm0SFuY+t2fbxXj8F3039fpDrP/KgR/2KL8qTVYxyXLVn97ZzfV/Y+CWTjdq1cFPmE3itnKogsHhxvKRLV+qspAIA+lZFXHFJwKSELacqWtvubdiQ8RJYiUs2UrpyFpUJ2XupAQWmje++CVqOqHhcW1lw2onfHDPXVbfXtBz5EOZofAfeO7FtEU6D01WyKcLW1MFhzEDHcRwG6ft141C1d9nQPzkAgeKgUjTQjzCgCSx4WMhMaH70A1ulZcjdvnGqEMfD6qTQfNiTuJEh19sarJNUD06A9RILWh2dpnufMYKbZG2GMRl05duibu4fY9ozhPKA/opPillzguI8WakvvviCDvNAW+19JVBZGHqyUfCw/imI12rEcuGgiD+KkCHgAV4Fst5yuiyLccjb//15S6FQYoCAiQc64w4jeUCogATJ9lkmUN2mXO4zW8iVEO53jVPMx/HvGMoEkY6cDgypPeuLNOWvAAkrJym3sSF4GhSQDhnWtEG1V2b8vMN1LgKZj+vWmmIR8Y54EXrhyT3iF8ER8tHj81YVbFT1/pJFE2ZotLEKRwUOqq5SIqze/k2KWw+SMqwLER1JIRBK9v9g+yOkxo37MYAxl140GuJAfkMT1/h9qxL3B9xVZI5pC/K+vmDpepPRiWB9eQBWV8qVORbs+72yxRi9lS3dxN+jPlhCpuNYhBuT5gBomVZMDNo3bMwgdIuGU3ikrgXUDu9hGDp/z12Eu1Gmri6lrDmV7jKwWOdEqDYoTg8pCmFYNdodPqSTlyTuP4959aMZbRKkJy9LEszJQeLoCeuplEhfWGCNBmxCjbtLWlV2nlOuuClEUrwvnbGZTCgpzuHtnBf8gWm/zjWgJIKz5LBcmUV9aADi6wuHjBFMGrsYBJDnXxHrT2nhuJPZ8V9OIivvYESIzOk5i/GAHLLrppBL1mxFoc17T8RqhIyY9rUTaSNNjNijL12mAi0/K3Gw98BkqKjwlO80ZBcYp/V94aFVDSg3IxV8w83NTXfEH/uQFBIR5zk6xlebplfsx204tj3r7/45Uywz7OjrJKVmWjbYwQMDU0QNcY0pjbaCqz3wWtY3svj52GKtITNYXow2eWxorj67XF6wLDDvt43+Z3JxgFulQIFvmOBOSYamfP/BtPKbQXy6HqQv6rEK5HqU/7q0SxXr090LQaXzSpugym43nUN71VtaqFq2HbssCMrHMrfw7MjcBD40ioCZ2RcFdCmGWkUR0gC8DCc0BnL11pslqW63mLHPvH2Q6W0vWLE3m+J9ege1bXS8aD6weZGok7kaNPFW4dkKLspyJvtx6BSvTRG2PTp9RtTI7E9zvbxOex6+MVA65uEO2yZ7UX4w8y19tPw/q7935p43HdqYqgQjC5nHnpkX/CDBQKbvc4xX4Dv9+gKfwqZdlvEPXAFNeINLfQ+Cvx/cNFS81bVETkMsuPjkh/uIDBObML3XsYbqiLNJnQDY1tTSHC+kCuABZO7l/iC85Wfwnw3pnYvv8dToMIXgNlC3XuQDmParhRI/ZDxpIWgyP3sr+KebGbH45cJamZgSYLCdmOXkb0chUSD3d1uG2LBF+oxwQHdNRWQJjErWYraBGiLkUpgeoPTebGOSiVt2Ao33yWQvidhittvxVkPpRMakH6nK7wvHoGI6yRx8uCXFTNBDFoOWz41tp8RIbh+T++4V7GlBoYHeihZfyaFyy3a/0w0lcu3Nxi3r1Itb3pIcoJjEhcD7REfToZywj5ObuNS/lxCwXNEfkgahoHNIRgPutElA3xQF1dto5Hls3eIAMYhUqqljQxW5mfxr04A/GOrWedGBYiN/tUiT7FxAMnXdJYkZRpyCt29RBs+UtqbBVjmA148xbHhg8LekWj/+tT+Qx8Q1f5px/pbE5AFOYpChwu3RfoXs37L/C1g0UTIRhwjqhQTDBUFU7rrlaKwDytYUKGSbxnBK2Y/P/Ou8m7yDVvb8MM9AOc6tNNtqZf/4ep9BcPQM8qYKft8FBWb0d8198rGP7ajKq+F9zY/YjETk4EwV9a0ABfH+p0q5VM4Lls8Bs8DJYJD1ZPaVVKTwjxDhV9qUHTZ54GLxWsSQdlIt9MlSp6vekrFFzScpPSahQf1pfYvDmD+RNQup7LVZItkZvh0WppYJy736puuOSFCUO/G1CBdfBDVjl3W5MJPtpapNn9CfewRWHvlUXFd7KFCc09iy4qvWhd+UAkIlOzNNqkobLvPh8MTPiklmbY9SF4IA/rUWXjwH+E8Jr7XohM3/Rpda3QsadWxQ4gifbwkN2kQGikBEANIymXW+So5Sk5lsljIiIDBTj56hs2Jq/PGG7S5W0qeL6kE3mDfEJooqAaicl2akz7k4CLcPkqOd6H48baiAaF0xZZikJLeQVqYgjF9GGt9/rHjVAZSPhBxJXNWbxpHbEaCPjo3Ozh/dAO7hORhx5KQuW20vY65yJA5EiWB3tZaIR4haflhoNT6KXAY570hxmnfr5NOWRzt+Kxj/IycfClVC49MPnOm/LnPX+EB8gY/dIkR2gB/ZC5IqrzRb67j/Kx60HcBPEoVqEA1ZcQwnlQBCfyot1EWeiTPb794c4ozvdF6ZkDIiHYDGgKL/GbG41Y0Pp8n+hbkkIBboFqGoAgj4wVZ7cNpHSfeO6uc30YgXPbKZsYIFQAejp41dAHkT8GanoulkcCA9I4fWAuHmHaXc93GWVLyqj1CYKLv7cuVxXCGRnYQKN9vQzpCWmNPy06/EsjJtycj12tOXuvaJdVBjLa7u6gKvtmCPLl/FMLNWgJp9WWlp0VCqeWK5RoP5j0h1MTLCcZbxA6cauVoA9xuy4OsVqh0dwR91lAtiFTMquRSkttcJMMZHCFGdSXXh/BeW8wF3MnUfk0Jj3/hGphKgnM+7DtD9L6I0f/S960mYuWW1hli2nOLS/6mfUuoCqHuY7Dq49uFUtNZYfhXpvoJpY7hsha9vUoOo2RD4WeEEDT749iFI93js8dwGqMPxxBYF0yL5hPZSbUZSFN3yCPGzZ8fIS4rRzIqSGtE8KU0+aEU1795++tDlyTElaNQWjdeEYkM+Ssj+EtlhZUUC2PW14vCC+c8ZwTJl3Zyq3unBpsYrunCqtc2tMBpYpa8Mm77r6t/s2Iv+WD77LBogGLOstf382jsi2ApgSXMQdn+pxHichiTJTO3/RFaNa8DmnKdu1ZXaXBb8SG3e7bcFCQUzM8elWRIimNLVhvl/SvmTkB+PNVpeIXj3bTEYmgixEp+Xus8zvgxu5umjcTfXZNTURDiBCe55tLxEGWgHTDPWKQZsRFH83T8fC/9EF0RJPXESc1QVejS/vHJx8Qb4TqgvnVqOrk2713TzOF1OuaIlJiN1Z+WL/nKVQPONTmTD/zLBmLDdINgNdAk0zHZhHKE0NIWsAqDU6HVW9loYRoczEkymLhr3NfuMoOomT7iRKeY4ZXbKGDYq6kiqTPYATx+iJ4r0amGV/x8lEik5O5eugpFdMbhZ+deQGyfkaSKtbshtTvLeIuvUVwHW6O7/p4DAUf2gntDMd/CncueVXJGHdZylRZ9xcapRDBo7OZLLx1Qya860FcZIRBLs8NWRzs5Xcyv2Fr20FQfxcGKzy1Y4k33iiSUmqabJNQOtER8NLEWVqk34rXuQ+BgOFdNpqV5jhnDLsI/YWby5RUqC7R7tfe9R5ropQESB6QAVgrSCyyx7bNo0RfcIECRhFg00N5eieQSfiBL8FDVR6hMFF3+MoiBhYOw16L/qCSzAjJdxOChRxxnw5Ni6LS96HSX6YXjpqs3QH4bD6z4EVjtnTMr9YBBjlDYkaH+QNhyQS3mG6OrKzgb1gC3V8qsjcs/oe0ij7ZxLmHjjZ0snifVcvGqoaG/yNOHTfiQ5N43uKEDjsjxWH6kO30Ff5iPGdY6+U784cM57MBGPAq0SSFmSzu/WWu6Gcj7+B20FNAMr2Yqe2twePY/SCCYgfsHaHI94Ha/wWQ2ZzH9StBRKCBz0G2du+Ia33LAUVXeOI4pvU6271Y7ZE9jExn+lNYoJIVx2wJFIFoDU8+G8wGTuKTW/OroQsRb/JcBb/kG9tgaqA224buYLrDFTcxMPw2oaiAZ7EWur2XKlFtnYHOrG3gIlRf4QNz+SFLwcY4mEc9hvFAzJEpHNlvw7Cx61uN4+yHG3cfDRhvN4Z3wJWc2DknBAYTaBPqyqJEyyOM2a8t38hyfvrqQjD75Coz/9qi7RcEgLziY0bm17TXSaD7UsrYyN3fY/WdXJl11T4P0dqI0V/q6biYZHDJe9BvqWzDIqCTgb6Z5WVsjlkm9bCCJ/djyPEVNooTp8sUUVCgkse5md0UsFkQeM4FsA3x4YR/d33jlCf0noQF6tms/67/MkMBLiLEF9Omar+jvib3V940oER2v47IjBm1Ip43M5NeS2j9sWF1mUkbRa4DGDfBQt/fH8iUkDgHUscXkjyIXmqlmMr3d7InVLYgu2MHXUwG1qoY+Tz17IAE1Iw70UFlAcIN7n0Iq+yj4RojTOVuwMDRIy8PENiBcLViZeAls0I0sTa5l7A/8qB43DDBuwwAQ2ENqL0hZEiGTaFqCDEDHXriR1SJ4ECyZA/+shnjPLPH72KOQpm+rE1SrDMrOJBei+nchEQtrhVe4D0xD0tuHGiGadivRnvgsfXEgc9lQxrdZgTIogqWvaQk8Z0qJedeURpXxJCKLhcV+i/JuTj2RjujOEKSxRQIUuoeXAF9oOccxw8OzmMwjShzHJZzSzjLx/5qxRJ4EANu1DC8LSNYRomsdUBZ1/X96hZSFL0GwdQIEEgSN3gkzvev0UbalyyzdUdRj46Z6zpoo7RBmQ1xhJNxcPy9u270bSaBG7X22OSKYp8ORIvTlPrA53cSZyKiPV8OvPHzpRCsT405EFbG+Wlzru0qlPWP9XcCwbI4gIlTpmCZvfvvBw813ViJYw4gtcZX7s7RUJZiQ+CmIFUvoabMYZkjgBzUZ576RGLiRDQYL9Ug6l25FtjjoXgiHLK9m2JxUMPS2x5aVkK8cgLZ4X3OaSbhsEpcWd9jNhcVrHgQFre+s5pxjMwNXj02qITXu919gxttLhRVqUOqA2wzPy/KDvr8hyLKtMppKyq+u/XRb5L0WpsxmMObqUWFWGh5Oa/S61yfS5U7GGU0buxTBBZRSXL7CE7n0DMGEcpHndLB8I3uvlfvif7ACcyfeD0mluz2nly2a58f2XslL2H3Qg/TN8ILR23+3Xrhi0JhL06C7wnhAwZHxw/++SeKfcvwG/4/QTeVwfzZE5lIrpfIEfNub1kWQ0yrBNGpsHqKvUnutrk6DU4EMA3h6btjj4xp5CpZ6AYb07ru04hRSV5rwfy/kb01cC4+N8gYks7OYlUpiqFeDNevvKCaoJfiNIPAjwuO6HonwU948J2odlmdp2HODSkG+pc5zRCQJzuJzL1doHU5476hK6vTR2vCEToSi7Mu2UoxYp74VWQ6utqNcqaxG79SuX2Zj9TPB47StTDbbfkPycj9uhlDIxaM0ytXlzIAnib3mTP1queKiFHDMCMt0D+Nii5Gg6pLq1IaXrpn4OJvoXus4PR5n4CJ+f1t9scPB2dRVHFMFkBXPMYrCQhOGkG4McNpACpBzXAgjXRa97bzETUUvCZMjNPJW0FdLWUHyq8XFU3e5Gl8zGES6J4X3p3GzWoRRmoPsVZvw16lje9j4qkSoaatCsT1bn4ePcfvXGkcAFmM7wLynOJPdt3BAVpRvbEQugXK5iNe7tVLmJPPW3hkuyrxPA4/ah7u2VOmjlXJ76ZDbPMTIYpwcCuRb7+p67XNWxk46wAYCxTs2mlIbSI41ypKSwqh6pdfl6y2hcHWtplHMxa+WyQnmSPgQx9cSGEEUuV3c2vCdFf/kAzy3uHEqvxRJeA1lTPhGTRPaJwuEjuGfq83Y1KBUbCBBy7iA3KSntTrqZ8bmtYkmqJyNgww/nl8pKPz3MxNOkGvpxLwCKK42CqXUxn5sa2X3kjIuF/Zvwg4WuGJACiBMN99lr+9RCOQ6p6yqV6FamohJ+mH7H7DkGKWLKVFnD5fK1OIaJ3gFFB8HB7NXVqQcM1OEBNPte3dUU+uXoT0EP90tmO/GOCWi4lf0JIdxt3x6sqDmuC3H/3sBW/aVUWOu25S/aM2cjDnlR7zfTbnWgyZn/IWMqSCtlD3OSA2Jw0jLbHTQ2V3wj3qCkiM70ioQIZaiy7pHAkijH4L93fjMcAqkLkV68JFLeOufwDp6V7p/iW86Aja2Ci9pBYCAXQUnvjM0BfJ/Kaie5F7tE7WpfiQeBW6WonNLRviAc0qHbzqvIC+WtAziX8Z9u5pqS3lOSSfTQK4l3iVLYb/+QurQSwMTpOEaiG7ltDZ8a3M/Izb8yyblwdla4wwFrU8FHOl+m0rwG8A8fQ2gGCIjiWPNI2YgZgwZ1Rpv3gnyiBjT1EsrQEshvwL3W8RQ/RePqBUlBG2gLhZLW664DTbf8AwoR6iqD9shlBX1o1IgpZd6YHVW9aetXZ7Rr9K461Z2MEFHy5SrGPkc641NGhKgwsercTkPcqPqA0ccVNvaxGGQ2J1EroIbwzZNO6M7Qtmam0DTBujZ47Q/Pd185wHeIb4ijyOEbXOzJKghxNS47X0DmkIwH3h3ebp2a6xQXNDwrRanR7GIhmXd72wh5fxQxEi75uP4XpeeR/DL1EclrEowvmKFq5Zqu3rosEfzf99Oc1fEPBQ9nM6pPfINb85Slztjgj1ZMnEkxRuX1ZV5mbI7ad1ErebR78AljXeNlJ1o4a62s8CQAzimHskabK04/4jBbwUpBoDxjZh/vUIek2a/sKQ2iTJ/an+YqkjWueGRe4nxSYPUvsDcrW2JqN03JJUb3gjJfwXDFNT/6+8McDG+JNqoQ1cJo+Y9RicBZ8uhQTi/syDra64rjJNV6OCqURUDnTBhY2z9qosHONr72H5mC5c8rHhwQikIAqGXlA8PcCHj3pVfI0qUV649hKQilLCbC2AFQYwPcIrc4xJqiqPvAtx8sR8KHsQHV/ehM0DK0PIFO1Mn6goKjkD0YaIViLr4cGDxjTQr6amPi99InqVVWZMcANfJ+A3Whdgw1JJayAb8D3jwyeQqpLLcibkbqpIJJ7X6v8lDcUskB5MO/dHSCP6GCtWamWZ+2OzSUPSry+1qaTpFgxe48OkGbBSGYBVHBIJgCBKgJzI0vTsUgprDkGCUgkYSg05RjLIrrNnpxDx1w2zPoTubP1x+hodYO1Fdp3BZM01DAvtQrp9DdGN/MFqixtTooAsc/4TvS0EzrYU50J7ugQjgyXsNac7UquV+wfpo2THkZFWjvkQfb1lnGKPHSNZexh31dhQk8oDgUQrc09R5vRnCrgopNleXQv5N71bCUlLovTktcIN5P9+C0GY54W1B7dGSStTzmIDj2IGR5UKxuPDM84zuwQWH7tuV7w36cdC757gLB6DfkyExjgXxxBwF9kv9zZN/6o4YWPU7abfFiJddQD5GD+CDva26mwACy8+7eHvK9AGXaW6dhZAAAAA="/> | [{"identifier": "A", "content": "Q and R"}, {"identifier": "B", "content": "None of these"}, {"identifier": "C", "content": "P and R"}, {"identifier": "D", "content": "P and Q"}] | ["B"] | null | As none play all three games the intersection of all
three circles must be zero.
<br><br>Hence none of P, Q, R justify the given statement | mcq | jee-main-2021-online-17th-march-morning-shift |
1ktbcom3g | maths | sets-and-relations | venn-diagram | Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If K% of them are suffering from both ailments, then K can not belong to the set : | [{"identifier": "A", "content": "{80, 83, 86, 89}"}, {"identifier": "B", "content": "{84, 86, 88, 90}"}, {"identifier": "C", "content": "{79, 81, 83, 85}"}, {"identifier": "D", "content": "{84, 87, 90, 93}"}] | ["C"] | null | <p>This solution begins by applying the principle of inclusion and exclusion, which in the context of this problem, is represented by the formula : </p>
<p>n(A ∪ B) ≥ n(A) + n(B) - n(A ∩ B)</p>
<p>Here, n(A ∪ B) represents the total number of patients in the hospital, which is 100%. n(A) represents the proportion of patients with a heart ailment (89%), and n(B) represents the proportion of patients with a lung infection (98%). </p>
<p>By rearranging this formula, the solution establishes an inequality for n(A ∩ B), the proportion of patients suffering from both ailments :</p>
<p>100% ≥ 89% + 98% - n(A ∩ B)</p>
<p>Therefore, </p>
<p>n(A ∩ B) ≥ 87%</p>
<p>Next, the solution notes that n(A ∩ B) cannot be greater than the smaller of n(A) and n(B), since it cannot be larger than the smallest group. Thus, we have another inequality :</p>
<p>n(A ∩ B) ≤ 89%</p>
<p>Combining these two inequalities gives :</p>
<p>87% ≤ n(A ∩ B) ≤ 89%</p>
<p>Hence, the proportion of patients suffering from both ailments must be a value between 87% and 89% inclusive. So, the set of values {79,81,83,85} which are all less than 87% are values that n(A ∩ B) cannot belong to. Therefore, option C is the correct answer.</p>
| mcq | jee-main-2021-online-26th-august-morning-shift |
1lguwarzq | maths | sets-and-relations | venn-diagram | <p>An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "9"}] | ["C"] | null | <ol>
<li><p>We are given the number of medals for events A, B, and C which are 48, 25, and 18 respectively. We are also given that the total number of unique medal recipients across all events is 60 and that 5 people received a medal in all three events.</p>
</li>
<br><li><p>Using the Principle of Inclusion and Exclusion (PIE), we know that the total number of unique medal recipients can be calculated by adding the number of medal recipients in each event, subtracting the number of people who received a medal in any two events (to correct for double counting), and then adding back the number of people who received a medal in all three events (since we subtracted these people too much). </p>
<br><p>Mathematically, this can be represented as :</p>
<br><p>|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| + |A ∩ B ∩ C|</p>
<br><p>However, we want to find the total number of people who received a medal in any two events (which is represented by |A ∩ B| + |B ∩ C| + |C ∩ A| in the equation).</p>
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljaginws/88d6b048-49b9-4f92-be73-c0b995c4c9c3/29c085b0-12ce-11ee-9fec-35dc3a4c73b2/file-6y3zli1ljaginwt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ljaginws/88d6b048-49b9-4f92-be73-c0b995c4c9c3/29c085b0-12ce-11ee-9fec-35dc3a4c73b2/file-6y3zli1ljaginwt.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Sets and Relations Question 28 English Explanation">
</li>
<br><li><p>To find this, we rearrange the PIE formula to solve for |A ∩ B| + |B ∩ C| + |C ∩ A| :</p>
<br><p>|A ∩ B| + |B ∩ C| + |C ∩ A| = |A| + |B| + |C| + |A ∩ B ∩ C| - |A ∪ B ∪ C|</p>
<br><p>Substituting the given values, we find that the total number of people who received a medal in any two events is 48 + 25 + 18 + 5 - 60 = 36.</p>
</li>
<br><li><p>However, this includes people who received a medal in all three events, and we want to find the number of people who received a medal in exactly two events. Therefore, we need to subtract the people who received a medal in all three events from our calculated value. </p>
<br><p>Since each person who received a medal in all three events is counted three times in |A ∩ B| + |B ∩ C| + |C ∩ A| (once for each pair of events), we subtract three times the number of people who received a medal in all three events from our calculated value:</p>
<br><p>Number of people who received a medal in exactly two events = |A ∩ B| + |B ∩ C| + |C ∩ A| - 3 $$ \times $$ |A ∩ B ∩ C|</p>
<br><p>Substituting the values we know, we find that the number of people who received a medal in exactly two events is 36 - 3 $$ \times $$ 5 = 21.</p>
</li>
</ol>
<p>Therefore, 21 people received a medal in exactly two of the three events.</p>
| mcq | jee-main-2023-online-11th-april-morning-shift |
lv0vxdqq | maths | sets-and-relations | venn-diagram | <p>In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let $$m$$ and $$n$$ respectively be the least and the most number of students who studied all the three subjects. Then $$\mathrm{m}+\mathrm{n}$$ is equal to ___________.</p> | [] | null | 45 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lx36wai5/74d61a18-6cda-4479-99f2-da165f8c36c1/14310dd0-23fa-11ef-9cd5-572f1863d2cb/file-6y3zli1lx36wai6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lx36wai5/74d61a18-6cda-4479-99f2-da165f8c36c1/14310dd0-23fa-11ef-9cd5-572f1863d2cb/file-6y3zli1lx36wai6.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Morning Shift Mathematics - Sets and Relations Question 6 English Explanation">
<p>$\begin{aligned} & 125 \leq \mathrm{m}+90-\mathrm{x} \leq 130 \\\\ & 85 \leq \mathrm{P}+70-\mathrm{x} \leq 95 \\\\ & 75 \leq \mathrm{C}+80-\mathrm{x} \leq 90 \\\\ & \mathrm{~m}+\mathrm{P}+\mathrm{C}+120-2 \mathrm{x}=210 \\\\ & \Rightarrow 15 \leq \mathrm{x} \leq 45 \& 30-\mathrm{x} \geq 0 \\\\ & \Rightarrow 15 \leq \mathrm{x} \leq 30 \\\\ & 30+15=45\end{aligned}$</p> | integer | jee-main-2024-online-4th-april-morning-shift |
lvc57bbi | maths | sets-and-relations | venn-diagram | <p>Let $$A=\{n \in[100,700] \cap \mathrm{N}: n$$ is neither a multiple of 3 nor a multiple of 4$$\}$$. Then the number of elements in $$A$$ is</p> | [{"identifier": "A", "content": "300"}, {"identifier": "B", "content": "310"}, {"identifier": "C", "content": "290"}, {"identifier": "D", "content": "280"}] | ["A"] | null | <p>$$n \in[100,700]$$</p>
<p>$$n(A)=$$ Total $$-$$ (multiple of $$3$$ + multiple of 4) + (multiple of 12)</p>
<p>Total $$=601$$</p>
<p>Multiple of $$3=102,105, \ldots, 699$$</p>
<p>$$\begin{aligned}
& n=699=102+(n-1) 3 \\
& \Rightarrow n=200
\end{aligned}$$</p>
<p>Multiple of $$4=100,104 \ldots ., 700$$</p>
<p>$$\begin{aligned}
& n=700=100+(n-1) 4 \\
& \frac{600}{4}+1=n \\
& \Rightarrow n=151
\end{aligned}$$</p>
<p>Multiple of $$12=108,120 \ldots .696$$</p>
<p>$$\begin{aligned}
& n=696=108+(n-1) 12 \\
& n=50 \\
& \therefore n(A)=601-(200+151)+50 \\
& =300
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
Y7H3QCu3IBSaOqbi | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average marks of the girls? | [{"identifier": "A", "content": "73"}, {"identifier": "B", "content": "65"}, {"identifier": "C", "content": "68"}, {"identifier": "D", "content": "74"}] | ["B"] | null | Given that, total students = 100
<br><br>and number of boys = 70
<br><br>$$\therefore$$ No. of girls = 100 - 70 = 30
<br><br>Average of 100 students = 72
<br><br>$$\therefore$$ Total marks of 100 students = 100 $$ \times $$ 72 = 7200
<br><br>Average of 70 boys = 75
<br><br>$$\therefore$$ Total marks of 70 boys = 70 $$ \times $$ 75 = 5250
<br><br>$$\therefore$$ Total marks of 30 girls = 7200 $$ - $$ 5250 = 1950
<br><br>$$\therefore$$ Average marks of 30 girls = $${{1950} \over {30}}$$ = 65 | mcq | aieee-2002 |
S2CpfA5xzaUED7bD | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of
the set is increased by 2, then the median of the new set : | [{"identifier": "A", "content": "is increased by 2"}, {"identifier": "B", "content": "is decreased by 2"}, {"identifier": "C", "content": "is two times the original median"}, {"identifier": "D", "content": "remains the same as that of the original set "}] | ["D"] | null | Here total no of observation is 9 which is a odd number. As we know for odd number 9 the median will be the 5<sup>th</sup> term.
<br><br>Now question says, you increase largest 4 number by 2 which does not affect the 5<sup>th</sup> term so the new median will be the same. | mcq | aieee-2003 |
IEaBRIby7TZbChYp | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | If in a frequency distribution, the mean and median are 21 and 22 respectively, then
its mode is approximately : | [{"identifier": "A", "content": "20.5"}, {"identifier": "B", "content": "22.0"}, {"identifier": "C", "content": "24.0"}, {"identifier": "D", "content": "25.5"}] | ["C"] | null | Given that,
<br><br>Mean = 21 and median = 22
<br><br>We know,
<br><br>Mode + 2 Mean = 3 Median
<br><br>$$\therefore$$ Mode = 3 $$ \times $$ 22 $$-$$ 2 $$ \times $$ 21
<br><br>= 66 $$-$$ 42
<br><br>= 24 | mcq | aieee-2005 |
voIKBn3jwUXSNIhC | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys
and girls combined is 50. The percentage of boys in the class is | [{"identifier": "A", "content": "80"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "40"}, {"identifier": "D", "content": "20"}] | ["A"] | null | Let x and y are number of boys and girls in a class respectively.
<br><br>$$\therefore$$ 52x + 42y = 50 (x + y)
<br><br>$$ \Rightarrow$$ 52x + 42y = 50x + 50y
<br><br>$$ \Rightarrow$$ 2x = 8y
<br><br>$$ \Rightarrow$$ x = 4y
<br><br>$$\therefore$$ Total no. of students = x + y = 4y + y = 5y
<br><br>$$\therefore$$ Percentage of boys = $${{4y} \over {5y}} \times 100\,\% $$ = 80% | mcq | aieee-2007 |
lB1b64UMGJQfxOW0 | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted
and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is : | [{"identifier": "A", "content": "15.8"}, {"identifier": "B", "content": "14.0"}, {"identifier": "C", "content": "16.8"}, {"identifier": "D", "content": "16.0"}] | ["B"] | null | Initially we have $$16$$ observations and among them one is $$16.$$
<br><br>So, we have $$15$$ unknowns. Let those are $${a_1},a{}_2,{a_3}.....{a_{15}}$$
<br><br>$$\therefore\,\,\,$$ Mean of $$16$$ datal set
<br><br>$$ = {{{a_1} + {a_2} + .....{a_{15}} + 16} \over {16}}$$
<br><br>According to the question,
<br><br>$${{{a_1} + {a_2} + .....{a_{15}} + 16} \over {16}} = 16$$
<br><br>$$ \Rightarrow {a_1} + {a_2} + ...... + {a_{15}} = 256 - 16 = 240$$
<br><br>Now we deleted $$16$$ and replaced by there new numbers $$3,4,$$ and $$5.$$
<br><br>So, new mean
<br><br>$$ = {{{a_1} + {a_2} + .......{a_{15}} + \left( {3 + 4 + 5} \right)} \over {18}}$$
<br><br>$$ = {{240 + \left( {3 + 4 + 5} \right)} \over {18}}$$
<br><br>$$ = {{240 + 12} \over {18}}$$
<br><br>$$=14$$
| mcq | jee-main-2015-offline |
yW2sKypZRX7KsEL99JymR | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is : | [{"identifier": "A", "content": "25"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "35"}, {"identifier": "D", "content": "40"}] | ["C"] | null | Mean $$\left( {\overline x } \right)$$ = $${{{x_1} + {x_2}..... + {x_n}} \over n}$$ = $${{\sum x } \over n}$$
<br><br>Here, Mean = 40 of 25 teachers
<br><br>$$\therefore$$ 40 = $${{\sum x } \over {25}}$$
<br><br>$$ \Rightarrow $$ $$\sum x $$ = 40 $$ \times $$ 25 = 1000
<br><br>After retireing of a 60 year old teacher, total age of 24 teachers,
<br><br>x<sub>1</sub> + x<sub>2</sub> + . . . . . .x<sub>24</sub> = 1000 $$-$$ 60 = 940
<br><br>Now a new teacher of age A year is appointed.
<br><br>$$\therefore$$ Now total age of this 25 teachers
<br><br>x<sub>1</sub> + x<sub>2</sub> + x<sub>3</sub> + . . . . . + x<sub>25</sub> = 940 + A
<br><br>$$\therefore$$ Mean age = $${{940 + A} \over {25}}$$
<br><br>According to question,
<br><br>$${{940 + A} \over {25}}$$ = 39
<br><br>$$ \Rightarrow $$ A = 35 | mcq | jee-main-2017-online-8th-april-morning-slot |
neBwTeBhdFgcTfYjwLPWU | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | The mean of set of 30 observations is 75. If each observation is multiplied by a non-zero number $$\lambda $$ and then each of them is decreased by 25, their mean remains the same. Then $$\lambda $$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${10 \\over 3}$$"}] | ["C"] | null | As mean is a linear operation, so if each observation is multiplied by $$\lambda $$ and decreased by 25 then the mean becomes 75$$\lambda $$$$-$$25.
<br><br>According to the question,
<br><br>75$$\lambda $$ $$-$$ 25 = 75 $$ \Rightarrow $$ $$\lambda $$ = $${4 \over 3}$$. | mcq | jee-main-2018-online-15th-april-morning-slot |
GqaYQGm3JSi2zukbEp18hoxe66ijvww34j1 | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | The mean and the median of the following ten
numbers in increasing order 10, 22, 26, 29, 34, x,
42, 67, 70, y are 42 and 35 respectively, then $${y \over x}$$ is equal to | [{"identifier": "A", "content": "$${7 \\over 2}$$"}, {"identifier": "B", "content": "$${8 \\over 3}$$"}, {"identifier": "C", "content": "$${9 \\over 4}$$"}, {"identifier": "D", "content": "$${7 \\over 3}$$"}] | ["D"] | null | Given ten numbers are 10, 22, 26, 29, 34, x,
42, 67, 70, y.
<br><br>As the numbers are in increasing order so
<br><br>Mediun = $${{34 + x} \over 2}$$ = 35
<br><br>$$ \Rightarrow $$ x = 36
<br><br>Also given mean = 42
<br><br>$$ \Rightarrow $$ $${{10 + 22 + 26 + 29 + 34 + x + 42 + 67 + 70 + y} \over {10}}$$ = 42
<br><br>$$ \Rightarrow $$ $${{300 + x + y} \over {10}}$$ = 42
<br><br>$$ \Rightarrow $$ 300 + x + y = 420
<br><br>$$ \Rightarrow $$ 336 + y = 420
<br><br>$$ \Rightarrow $$ y = 84
<br><br>$$ \therefore $$ $${y \over x} = {{84} \over {36}}$$ = $${7 \over 3}$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
h2VdOffcBfbNtOpUQ73rsa0w2w9jwy0yzk9 | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | If for some x $$ \in $$ R, the frequency distribution of the marks obtained by 20 students in a test is :<br/><br/>
<style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg th{font-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg .tg-s6z2{text-align:center}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-hgcj{font-weight:bold;text-align:center}
</style>
<table class="tg">
<tbody><tr>
<th class="tg-hgcj">Marks</th>
<th class="tg-s6z2">2</th>
<th class="tg-baqh">3</th>
<th class="tg-baqh">5</th>
<th class="tg-baqh">7</th>
</tr>
<tr>
<td class="tg-hgcj">Frequency</td>
<td class="tg-s6z2">(x + 1)<sup>2</sup></td>
<td class="tg-baqh">2x - 5</td>
<td class="tg-baqh">x<sup>2</sup> - 3x</td>
<td class="tg-baqh">x</td>
</tr>
</tbody></table>
<br/>then the mean of the marks is | [{"identifier": "A", "content": "3.0"}, {"identifier": "B", "content": "2.8"}, {"identifier": "C", "content": "2.5"}, {"identifier": "D", "content": "3.2"}] | ["B"] | null | Number of students<br><br>
$$ \Rightarrow {\left( {x + 1} \right)^2} + (2x - 5) + \left( {{x^2} - 3x} \right) + x = 20$$<br><br>
$$ \Rightarrow 2{x^2} + 2x - 4 = 20$$<br><br>
$$ \Rightarrow {x^2} + x - 12 = 0$$<br><br>
$$ \Rightarrow (x + 4)(x - 3) = 0$$<br><br>
$$x = 3$$<br><br>
<style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;width:100%}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg th{font-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-yla0{font-weight:bold;text-align:left;vertical-align:middle}
.tg .tg-nrix{text-align:center;vertical-align:middle}
</style>
<table class="tg">
<tbody><tr>
<th class="tg-yla0">Marks</th>
<th class="tg-nrix">2</th>
<th class="tg-nrix">3</th>
<th class="tg-nrix">5</th>
<th class="tg-baqh">7</th>
</tr>
<tr>
<td class="tg-yla0">No. of students</td>
<td class="tg-nrix">16</td>
<td class="tg-nrix">1</td>
<td class="tg-nrix">0</td>
<td class="tg-baqh">3</td>
</tr>
</tbody></table><br>
Average marks = $${{32 + 3 + 21} \over {20}} = {{56} \over {20}} = 2.8$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
KqISKJ8PMSjrFVZan4jgy2xukg4n89i7 | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | Consider the data on x taking the values<br/> 0, 2, 4,
8,....., 2<sup>n</sup> with frequencies<br/> <sup>n</sup>C<sub>0</sub>
,
<sup>n</sup>C<sub>1</sub>
,
<sup>n</sup>C<sub>2</sub>
,....,
<sup>n</sup>C<sub>n</sub>
respectively. If the<br/> mean of this data is $${{728} \over {{2^n}}}$$, then n is equal to _________ .
| [] | null | 6 | Mean = $${{\sum {{x_1}.{f_1}} } \over {\sum {{f_1}} }}$$
<br><br>= $${{0.{}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}} \over {{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}}$$
<br><br>We know,
<br><br>(1 + x)<sup>n</sup> = $${{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}}$$ ...(1)
<br><br>Put x = 2, at (1) we get
<br>$$ \Rightarrow $$ 3<sup>n</sup> - 1 = $${2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}}$$
<br><br>And Putting x = 1 in (1), we get
<br><br>2<sup>n</sup> = $${{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}$$
<br><br>$$ \therefore $$ Mean = $${{{3^n} - 1} \over {{2^n}}}$$
<br><br>According to question,
<br><br>$${{{3^n} - 1} \over {{2^n}}}$$ = $${{728} \over {{2^n}}}$$
<br><br>$$ \Rightarrow $$ 3<sup>n</sup> = 729
<br><br>$$ \Rightarrow $$ n = 6 | integer | jee-main-2020-online-6th-september-evening-slot |
pFSajzAuBaaQVZQ3Qj1kmlm1y0e | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is _________. | [] | null | 35 | Mean $$\left( {\overline x } \right)$$ = $${{{x_1} + {x_2}..... + {x_n}} \over n}$$ = $${{\sum x } \over n}$$
<br><br>Here, Mean = 40 of 25 teachers
<br><br>$$\therefore$$ 40 = $${{\sum x } \over {25}}$$
<br><br>$$ \Rightarrow $$ $$\sum x $$ = 40 $$ \times $$ 25 = 1000
<br><br>After retireing of a 60 year old teacher, total age of 24 teachers,
<br><br>x<sub>1</sub> + x<sub>2</sub> + . . . . . .x<sub>24</sub> = 1000 $$-$$ 60 = 940
<br><br>Now a new teacher of age A year is appointed.
<br><br>$$\therefore$$ Now total age of this 25 teachers
<br><br>x<sub>1</sub> + x<sub>2</sub> + x<sub>3</sub> + . . . . . + x<sub>25</sub> = 940 + A
<br><br>$$\therefore$$ Mean age = $${{940 + A} \over {25}}$$
<br><br>According to question,
<br><br>$${{940 + A} \over {25}}$$ = 39
<br><br>$$ \Rightarrow $$ A = 35 | integer | jee-main-2021-online-18th-march-morning-shift |
1krub3sh2 | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | Consider the following frequency distribution :<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-0lax{text-align:left;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-0lax">Class :</th>
<th class="tg-baqh">0-6</th>
<th class="tg-baqh">6-12</th>
<th class="tg-baqh">12-18</th>
<th class="tg-baqh">18-24</th>
<th class="tg-baqh">24-30</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">Frequency :</td>
<td class="tg-baqh">$$a $$</td>
<td class="tg-baqh">$$b$$</td>
<td class="tg-baqh">12</td>
<td class="tg-baqh">9</td>
<td class="tg-baqh">5</td>
</tr>
</tbody>
</table><br/>If mean = $${{309} \over {22}}$$ and median = 14, then the value (a $$-$$ b)<sup>2</sup> is equal to _____________. | [] | null | 4 | <table class="tg">
<thead>
<tr>
<th class="tg-0lax">Class</th>
<th class="tg-0lax">Frequency</th>
<th class="tg-0lax">$${x_i}$$</th>
<th class="tg-0lax">$${f_i}{x_i}$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">0-6</td>
<td class="tg-0lax">a</td>
<td class="tg-0lax">3</td>
<td class="tg-0lax">3a</td>
</tr>
<tr>
<td class="tg-0lax">6-12</td>
<td class="tg-0lax">b</td>
<td class="tg-0lax">9</td>
<td class="tg-0lax">9b</td>
</tr>
<tr>
<td class="tg-0lax">12-18</td>
<td class="tg-0lax">12</td>
<td class="tg-0lax">15</td>
<td class="tg-0lax">180</td>
</tr>
<tr>
<td class="tg-0lax">18-24</td>
<td class="tg-0lax">9</td>
<td class="tg-0lax">21</td>
<td class="tg-0lax">189</td>
</tr>
<tr>
<td class="tg-0lax">24-30</td>
<td class="tg-0lax">5</td>
<td class="tg-0lax">27</td>
<td class="tg-0lax">135</td>
</tr>
<tr>
<td class="tg-0lax"></td>
<td class="tg-0lax">$$N = (26 + a + b)$$</td>
<td class="tg-0lax"></td>
<td class="tg-0lax">$$(504 + 3a + 9b)$$</td>
</tr>
</tbody>
</table><br><br> Mean = $${{3a + 9b + 180 + 189 + 135} \over {a + b + 26}} = {{309} \over {22}}$$<br><br>$$ \Rightarrow 66a + 198b + 11088 = 309a + 309b + 8034$$<br><br>$$ \Rightarrow 243a + 111b = 3054$$<br><br>$$ \Rightarrow 81a + 37b = 1018$$ $$\to$$ (1)<br><br>Now, Median $$ = 12 + {{{{a + b + c} \over 2} - (a + b)} \over {12}} \times 6 = 14$$<br><br>$$ \Rightarrow {{13} \over 2} - \left( {{{a + b} \over 4}} \right) = 2$$<br><br>$$ \Rightarrow {{a + b} \over 4} = {9 \over 2}$$<br><br>$$ \Rightarrow a + b = 18$$ $$\to$$ (2)<br><br>From equation (1) $ (2)<br><br>a = 8, b = 10<br><br>$$\therefore$$ $${(a - b)^2} = {(8 - 10)^2}$ = 4 | integer | jee-main-2021-online-22th-july-evening-shift |
1krw28thi | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | Consider the following frequency distribution :<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-0lax{text-align:left;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-0lax">Class :</th>
<th class="tg-baqh">10-20</th>
<th class="tg-baqh">20-30</th>
<th class="tg-baqh">30-40</th>
<th class="tg-baqh">40-50</th>
<th class="tg-baqh">50-60</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">Frequency :</td>
<td class="tg-baqh">$$\alpha $$</td>
<td class="tg-baqh">110</td>
<td class="tg-baqh">54</td>
<td class="tg-baqh">30</td>
<td class="tg-baqh">$$\beta $$</td>
</tr>
</tbody>
</table><br/><br/>If the sum of all frequencies is 584 and median is 45, then | $$\alpha$$ $$-$$ $$\beta$$ | is equal to _______________. | [] | null | 164 | $$\because$$ Sum of frequencies = 584<br><br>$$\Rightarrow$$ $$\alpha$$ + $$\beta$$ = 390<br><br>Now, median is at $${{584} \over 2}$$ = 292<sup>th</sup><br><br>$$\because$$ Median = 45 (lies in class 40 - 50)<br><br>$$\Rightarrow$$ $$\alpha$$ + 110 + 54 + 15 = 292<br><br>$$\Rightarrow$$ $$\alpha$$ = 113, $$\beta$$ = 277<br><br>$$\Rightarrow$$ | $$\alpha$$ $$-$$ $$\beta$$ | = 164 | integer | jee-main-2021-online-25th-july-morning-shift |
1ktise4jh | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | The mean of 10 numbers 7 $$\times$$ 8, 10 $$\times$$ 10, 13 $$\times$$ 12, 16 $$\times$$ 14, ....... is ____________. | [] | null | 398 | 7 $$\times$$ 8, 10 $$\times$$ 10, 13 $$\times$$ 12, 16 $$\times$$ 14 ........<br><br>T<sub>n</sub> = (3n + 4) (2n + 6) = 2(3n + 4) (n + 3)<br><br>= 2(3n<sup>2</sup> + 13n + 12) = 6n<sup>2</sup> + 26n + 24<br><br>S<sub>10</sub> = $$\sum\limits_{n = 1}^{10} {{T_n}} = 6\sum\limits_{n = 1}^{10} {{n^2}} + 26\sum\limits_{n = 1}^{10} n + 24\sum\limits_{n = 1}^{10} 1 $$<br><br>$$ = {{6(10 \times 11 \times 21)} \over 6} + 26 \times {{10 \times 11} \over 2} + 24 \times 10$$<br><br>$$ = 10 \times 11(21 + 13) + 240$$<br><br>= 3980<br><br>Mean $$ = {{{S_{10}}} \over {10}} = {{3980} \over {10}} = 398$$ | integer | jee-main-2021-online-31st-august-morning-shift |
1l6m6xl9a | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | <p>Let $$x_{1}, x_{2}, x_{3}, \ldots, x_{20}$$ be in geometric progression with $$x_{1}=3$$ and the common ratio $$\frac{1}{2}$$. A new data is constructed replacing each $$x_{i}$$ by $$\left(x_{i}-i\right)^{2}$$. If $$\bar{x}$$ is the mean of new data, then the greatest integer less than or equal to $$\bar{x}$$ is ____________.</p> | [] | null | 142 | <p>$${x_1},{x_2},{x_3},\,.....,\,{x_{20}}$$ are in G.P.</p>
<p>$${x_1} = 3,\,r = {1 \over 2}$$</p>
<p>$$\overline x = {{\sum {x_i^2 - 2{x_i}i + {i^2}} } \over {20}}$$</p>
<p>$$ = {1 \over {20}}\left[ {12\left( {1 - {1 \over {{2^{40}}}}} \right) - 6\left( {4 - {{11} \over {{2^{18}}}}} \right) + 70 \times 41} \right]$$</p>
<p>$$\left\{ {\matrix{
{S = 1 + 2\,.\,{1 \over 2} + 3\,.\,{1 \over {{2^2}}}\, + \,....} \cr
{{S \over 2} = {1 \over 2} + {2 \over {{2^2}}}\, + \,....} \cr
} } \right.$$</p>
<p>$$\left. {{S \over 2} = 2\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}} = 4 - {{11} \over {{2^{18}}}}} \right\}$$</p>
<p>$$\therefore$$ $$[\overline x ] = \left[ {{{2858} \over {20}} - \left( {{{12} \over {240}} - {{66} \over {{2^{18}}}}} \right)\,.\,{1 \over {20}}} \right]$$</p>
<p>$$ = 142$$</p> | integer | jee-main-2022-online-28th-july-morning-shift |
1lsgack8c | maths | statistics | calculation-of-mean,-median-and-mode-of-grouped-and-ungrouped-data | <p>Let M denote the median of the following frequency distribution</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-0lax{text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 458px">
<colgroup>
<col style="width: 81px"/>
<col style="width: 79px"/>
<col style="width: 74px"/>
<col style="width: 76px"/>
<col style="width: 74px"/>
<col style="width: 74px"/>
</colgroup>
<thead>
<tr>
<th class="tg-0lax">Class</th>
<th class="tg-baqh">0 - 4</th>
<th class="tg-baqh">4 - 8</th>
<th class="tg-baqh">8 - 12</th>
<th class="tg-baqh">12 - 16</th>
<th class="tg-baqh">16 - 20</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">Frequency</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">9</td>
<td class="tg-baqh">10</td>
<td class="tg-baqh">8</td>
<td class="tg-baqh">6</td>
</tr>
</tbody>
</table></p>
<p>Then 20M is equal to :</p> | [{"identifier": "A", "content": "104"}, {"identifier": "B", "content": "52"}, {"identifier": "C", "content": "208"}, {"identifier": "D", "content": "416"}] | ["C"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 456px">
<colgroup>
<col style="width: 141px">
<col style="width: 153px">
<col style="width: 162px">
</colgroup>
<thead>
<tr>
<th class="tg-amwm">Class</th>
<th class="tg-amwm">Frequency</th>
<th class="tg-amwm">Cumulative<br>frequency</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">0-4</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">3</td>
</tr>
<tr>
<td class="tg-baqh">4-8</td>
<td class="tg-baqh">9</td>
<td class="tg-baqh">12</td>
</tr>
<tr>
<td class="tg-baqh">8-12</td>
<td class="tg-baqh">10</td>
<td class="tg-baqh">22</td>
</tr>
<tr>
<td class="tg-baqh">12-16</td>
<td class="tg-baqh">8</td>
<td class="tg-baqh">30</td>
</tr>
<tr>
<td class="tg-baqh">16-20</td>
<td class="tg-baqh">6</td>
<td class="tg-baqh">36</td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
& \mathrm{M}=1+\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\right) \mathrm{h} \\
& \mathrm{M}=8+\frac{18-12}{10} \times 4 \\
& \mathrm{M}=10.4 \\
& 20 \mathrm{M}=208
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
q3HGzZnYw0060luY | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | In an experiment with 15 observations on $$x$$, then following results were available:
<br/>$$\sum {{x^2}} = 2830$$, $$\sum x = 170$$
<br/>One observation that was 20 was found to be wrong and was replaced by the correct value
30. Then the corrected variance is : | [{"identifier": "A", "content": "188.66"}, {"identifier": "B", "content": "177.33"}, {"identifier": "C", "content": "8.33"}, {"identifier": "D", "content": "78.00"}] | ["D"] | null | Given that,
<br><br>N = 15, $$\sum {x{}^2} = 2830,\,\sum x = 170$$
<br><br>As, 20 was replaced by 30 then,
<br><br>$$\sum x = 170 - 20 + 30 = 180$$
<br><br>and $$\sum {{x^2}} = 2830 - 400 + 900 = 3330$$
<br><br>So, the corrected variance
<br><br>$$ = {{\sum {{x^2}} } \over N} - {\left( {{{\sum x } \over N}} \right)^2}$$
<br><br>$$ = {{3330} \over {15}} - {\left( {{{180} \over {15}}} \right)^2}$$
<br><br>$$ = 222 - {12^2}$$
<br><br>$$ = 222 - 144$$
<br><br>$$ = 78$$ | mcq | aieee-2003 |
KcC0DnzzKdYqXQua | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Consider the following statements:
<br/>(a) Mode can be computed from histogram
<br/>(b) Median is not independent of change of scale
<br/>(c) Variance is independent of change of origin and scale.
<br/> Which of these is/are correct? | [{"identifier": "A", "content": "only (a)"}, {"identifier": "B", "content": "only (b)"}, {"identifier": "C", "content": "only (a) and (b)"}, {"identifier": "D", "content": "(a), (b) and (c) "}] | ["C"] | null | <p>The statements are analyzed as follows :</p>
<p>(a) Mode can be computed from histogram : This is correct. The mode is the value that appears most frequently in a data set. A histogram provides a graphical representation of the frequency of each data value. The data value corresponding to the highest bar in a histogram is the mode.</p>
<p>(b) Median is not independent of change of scale : This is correct. The median is the middle value in a data set when the values are arranged in ascending or descending order. When you change the scale (e.g., by multiplying all data points by a constant), the median changes as well.</p>
<p>(c) Variance is independent of change of origin and scale : This is not correct. Variance, which measures the dispersion of a set of data points, is affected by changes in scale (it is not independent of scale). If you multiply all the values in a dataset by a constant, the variance gets multiplied by the square of that constant. However, it is true that variance is independent of the change of origin (it remains the same if you add or subtract a constant from all data points in the set).</p>
<p>Therefore, the correct answer is <b>Option C :</b> only (a) and (b) are correct.</p> | mcq | aieee-2004 |
Vtsx1SkaZm1Q91Mf | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | In a series of 2n observations, half of them equal $$a$$ and remaining half equal $$–a$$. If the
standard deviation of the observations is 2, then $$|a|$$ equals | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\sqrt 2 $$"}, {"identifier": "C", "content": "$${1 \\over n}$$"}, {"identifier": "D", "content": "$${{\\sqrt 2 } \\over n}$$"}] | ["A"] | null | Mean $$\left( A \right) = {{a - a} \over {2n}} = 0$$
<br><br>Given standard deviation (S.D) = 2
<br><br>$$\therefore\,\,\,$$ $$\sqrt {{{\sum {{{\left( {x - A} \right)}^2}} } \over {2n}}} = 2$$
<br><br>$$ \Rightarrow \,\,\,\sqrt {{{{{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2} + ..... + {{\left( {0 - a} \right)}^2}} \over {2n}}} = 2$$
<br><br>$$ \Rightarrow \,\,\,\sqrt {{{{a^2} + {a^2}........2n\,times} \over {2n}}} = 2$$
<br><br>$$ \Rightarrow \,\,\,\sqrt {{{2n\,.\,{a^2}} \over {2n}}} = 2$$
<br><br>$$ \Rightarrow \,\,\,\sqrt {{a^2}} = 2$$
<br><br>$$ \Rightarrow \,\,\,\left| a \right| = 2$$ | mcq | aieee-2004 |
KXk4NEHZZHI9u4CB | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let x<sub>1</sub>, x<sub>2</sub>,...........,x<sub>n</sub> be n observations such that
<br/><br/>$$\sum {x_i^2} = 400$$ and $$\sum {{x_i}} = 80$$. Then a
possible value of n among the following is | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "9"}] | ["A"] | null | As we know,
<br><br>$${\sigma ^2} \ge 0$$
<br><br>$$\therefore\,\,\,$$ $${{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} \ge 0$$
<br><br>$$ \Rightarrow \,\,\,{{400} \over n} - {{6400} \over {{n^2}}} \ge 0$$
<br><br>$$ \Rightarrow \,\,\,n \ge 16$$
<br><br>$$\therefore\,\,\,$$ Possible value of n according to the option is = 18 | mcq | aieee-2005 |
jAJXr12O9W6IUO6E | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Suppose a population A has 100 observations 101, 102,........, 200, and another
population B has 100 observations 151, 152,......., 250. If V<sub>A</sub> and V<sub>B</sub> represent the
variances of the two populations, respectively, then $${{{V_A}} \over {{V_B}}}$$ is | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${9 \\over 4}$$"}, {"identifier": "C", "content": "$${4 \\over 9}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["A"] | null | Series A = 101, 102 ............ 200
<br><br>Series B = 151, 152 ............ 250
<br><br>Here series B can be obtained if we change the origin of A by 50 units.
<br><br>And we know the variance does not change by changing the origin.
<br><br>So, $$\,\,\,\,$$ $${V_A} = {V_B}$$
<br><br>$$ \Rightarrow \,\,\,\,\,{{{V_A}} \over {{V_B}}} = 1$$ | mcq | aieee-2006 |
Swo3ZuYKZZWDz5zc | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following
gives possible values of a and b? | [{"identifier": "A", "content": "a = 0, b = 7"}, {"identifier": "B", "content": "a = 5, b = 2"}, {"identifier": "C", "content": "a = 1, b = 6"}, {"identifier": "D", "content": "a = 3, b = 4"}] | ["D"] | null | Given that,
<br><br>Mean of a, b, 8, 5, 10 = 6
<br><br>$$\therefore\,\,\,$$ $${{a + b + 8 + 5 + 10} \over 5} = 6$$
<br><br>$$ \Rightarrow \,\,\,$$ a + b + 23 = 30
<br><br>$$ \Rightarrow \,\,\,$$ a + b = 7 ....... (1)
<br><br>Variance $$ = {{\sum {{{\left( {{x_i} - A} \right)}^2}} } \over n} = 6.8$$
<br><br>$$ \Rightarrow \,\,\,{{{{\left( {6 - a} \right)}^2} + {{\left( {6 - b} \right)}^2} + {{\left( {6 - 8} \right)}^2} + {{\left( {6 - 5} \right)}^2} + {{\left( {6 - 10} \right)}^2}} \over 5} = 6.5$$
<br><br>$$ \Rightarrow \,\,\,{\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} + 4 + 1 + 16 = 34$$
<br><br>$$ \Rightarrow {\left( {6 - a} \right)^2} + {\left( {6 - b} \right)^2} = 13$$
<br><br>$$ \Rightarrow \,\,\,{a^2} + {b^2} = 25$$
<br><br>By using (1) we get,
<br><br>$${a^2} + \left( {7 - a} \right){}^2 = 25$$
<br><br>$$ \Rightarrow \,\,\,a{}^2 - 7a + 12 = 0$$
<br><br>$$\therefore\,\,\,$$ a = 3, 4
<br><br>then b = 4, 3. | mcq | aieee-2008 |
O8gldTfNXQMBhdGy | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <b>Statement - 1 :</b> The variance of first n even natural numbers is $${{{n^2} - 1} \over 4}$$
<br/><br/><b>Statement - 2 :</b> The sum of first n natural numbers is $${{n\left( {n + 1} \right)} \over 2}$$ and the sum of squares of first n natural numbers is $${{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}$$ | [{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1"}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1"}, {"identifier": "C", "content": "Statement-1 is true, Statement-2 is false"}, {"identifier": "D", "content": "Statement-1 is false, Statement-2 is true"}] | ["D"] | null | Let first n even natural numbers = 2,4, 6, 8 ...... 2n
<br><br>$$\therefore$$ Sum of those num = 2 + 4 + 6 + ..... 2n
<br><br>= 2 (1 + 2 + ..... n)
<br><br>= $$2.{{n\left( {n + 1} \right)} \over 2}$$
<br><br>= n (n + 1)
<br><br>$$\therefore\,\,\,$$ Mean $$\left( {\overline x } \right) = {{n\left( {n + 1} \right)} \over n}$$ $$ = n + 1$$
<br><br>$$\therefore\,\,\,$$ Variance $$ = {1 \over n}\sum {x_i^2} - {\left( {\overline x } \right)^2}$$
<br><br>$$ = {1 \over n}\left[ {{2^2} + {4^2} + ...... + {{\left( {2n} \right)}^2}} \right] - {\left( {n + 1} \right)^2}$$
<br><br>$$ = {1 \over n}{2^2}\left[ {{1^2} + {2^2} + .......n{}^2} \right] - \left( {n + 1} \right){}^2$$
<br><br>$$ = {4 \over n}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}} \right] - {\left( {n + 1} \right)^2}$$
<br><br>$$ = {{\left( {n + 1} \right)\left[ {2\left( {2n + 1} \right) - 3\left( {n + 1} \right)} \right]} \over 3}$$
<br><br>$$ = {{\left( {n + 1} \right)\left( {n - 1} \right)} \over 3}$$
<br><br>$$ = {{{n^2} - 1} \over 3}$$
<br><br>$$\therefore$$ Statement 1 is false.
<br><br>Statement 2 is true as those are standard formula.
| mcq | aieee-2009 |
gghxq0uoNUICOkl2 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean deviation of number 1, 1 + d, 1 + 2d,........, 1 + 100d from their mean is 255, then the d is
equal to | [{"identifier": "A", "content": "20.0"}, {"identifier": "B", "content": "10.1"}, {"identifier": "C", "content": "20.2"}, {"identifier": "D", "content": "10.0"}] | ["B"] | null | Mean $$\,\,\,\,\left( {\overline x } \right) = {{Sum\,\,of\,\,numbers} \over n}$$
<br><br>$$ = {{{n \over 2}\left( {a + l} \right)} \over n}$$
<br><br>$$ = {1 \over 2}\left( {1 + l + 100d} \right)$$
<br><br>$$ = 1 + 50\,d.$$
<br><br>Mean deviation (M.D) $$ = {1 \over n}\sum\limits_{i = 1}^{101} {\left| {{x_i} - \overline x } \right|} $$
<br><br>$$ = {1 \over {101}}$$[ 50d + 49d + 48d + .......d + 0 + ..... + 50d]
<br><br>$$ = {1 \over {101}}.2d\left( {1 + 2 + .... + 50} \right)$$
<br><br>$$ = {1 \over {101}}.2d.{{50 \times 51} \over 2}$$
<br><br>$$ = {{50 \times 51\,d} \over {101}}$$
<br><br>Given that M.D = 255
<br><br>$$\therefore\,\,\,$$ $${{50 \times 51\,d} \over {101}} = 255$$
<br><br>$$ \Rightarrow \,\,\,d = {{101 \times 255} \over {51 \times 50}}$$
<br><br>$$ = 10.1$$ | mcq | aieee-2009 |
ZY7XixgriqKYkqZL | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding
means are given to be 2 and 4, respectively. The variance of the combined data set is | [{"identifier": "A", "content": "$${5 \\over 2}$$"}, {"identifier": "B", "content": "$${11 \\over 2}$$"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "$${13 \\over 2}$$"}] | ["B"] | null | Given that,
<br><br>$${\sigma _1}^2 = 4$$
<br><br>and $${\sigma _2}^2 = 5$$
<br><br>And also given,
<br><br>$$\overline x = 2\,\,$$ and $$\overline y = 4\,\,$$
<br><br>So, $$\,\,\,$$ $${{\sum {{x_i}} } \over 5} = 2$$
<br><br>$$ \Rightarrow \sum {{x_i}} = 10$$ v
<br><br>and $${{\sum {{y_i}} } \over 5} = 4$$
<br><br>$$ \Rightarrow \,\,\,\sum {{y_i}} = 20$$
<br><br>$$\therefore\,\,\,$$ $${\sigma _1}^2 = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}$$
<br><br>$$ \Rightarrow \,\,\,4 = {{\sum {x_i^2} } \over 5} - 4$$
<br><br>$$ \Rightarrow \,\,\,\sum {x_i^2} = 40$$
<br><br>and $${\sigma _2}^2 = {{\sum {y_i^2} } \over 5} - {\left( {\overline y } \right)^2}$$
<br><br>$$ \Rightarrow \,\,\,\,5 = {{\sum {y_i^2} } \over 5} - 16$$
<br><br>$$ \Rightarrow \,\,\,\sum {y_i^2} = 105$$
<br><br>Variance of combined data set
<br><br>$${\sigma ^2} = {1 \over {10}}\left( {\sum {x_i^2 + \sum {y_i^2} } } \right) - {\left( {{{\overline x + \overline y } \over 2}} \right)^2}$$
<br><br>$$ = {1 \over {10}}\left( {40 + 105} \right) - 9$$
<br><br>$$ = {{145 - 90} \over {10}}$$
<br><br>$$ = {{55} \over {10}}$$
<br><br>$$ = {{11} \over {2}}$$ | mcq | aieee-2010 |
3mIl7ePHUNR0dez9 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean deviation about the median of the numbers a, 2a,........., 50a is 50, then |a| equals | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["A"] | null | <b>NOTE :</b>
<br><br>If total no of terms are even then median
<br><br>$$ = {1 \over 2}$$ [ $${n \over 2}$$th term $$ + \left( {{n \over 2} + 1} \right)$$ th term]
<br><br>Here total terms $$ = 50,$$ which is even
<br><br>$$\therefore$$ $$\,\,\,$$ Median $$ = {1 \over 2}$$ [ $${{50} \over 2}$$ th term $$ + \left( {{{50} \over 2} + 1} \right)$$ th term]
<br><br>$$ = {1 \over 2}$$ [ $$25$$ th term $$+$$ $$26$$ th term ]
<br><br>$$ = {1 \over 2}$$ [ $$25a$$ $$+$$ $$26a$$ ]
<br><br>$$=25.5a$$
<br><br>Mean deviation (M.D.) about the median
<br><br>$$ = {{\sum\limits_{i = 1}^{50} {\left| {{x_i} - Median} \right|} } \over N} = 50$$ (given)
<br><br>$$\therefore$$ $${1 \over {50}}\left[ {2 \times \left| a \right| \times \left( {0.5 + 1.5 + 2.5 + ....24.5} \right)} \right] = 50$$
<br><br>$$ \Rightarrow 2\left| a \right| \times {{25} \over 2} \times 25 = 2500$$
<br><br>$$ \Rightarrow \left| a \right| = {{2500} \over {25 \times 25}} = 4$$ | mcq | aieee-2011 |
xBO8kMOnfG3Qz0d1 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let x<sub>1</sub>, x<sub>2</sub>,........., x<sub>n</sub> be n observations, and let $$\overline x $$ be their arithematic mean and $${\sigma ^2}$$ be their variance.
<br/><br/><b>Statement 1 :</b> Variance of 2x<sub>1</sub>, 2x<sub>2</sub>,......., 2x<sub>n</sub> is 4$${\sigma ^2}$$.
<br/><b>Statement 2 :</b> : Arithmetic mean of 2x<sub>1</sub>, 2x<sub>2</sub>,......, 2x<sub>n</sub> is 4$$\overline x $$. | [{"identifier": "A", "content": "Statement 1 is false, statement 2 is true"}, {"identifier": "B", "content": "Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1"}, {"identifier": "C", "content": "Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1"}, {"identifier": "D", "content": "Statement 1 is true, statement 2 is false"}] | ["D"] | null | Given that,
<br><br>for $${x_1},{x_2},....{x_n},$$ $$A.M = \overline x $$
<br><br>and variance $$ = {\sigma ^2}$$
<br><br>Now A.M of
<br><br>$$2{x_1},2x{}_2.....2{x_n} = {{2\left( {{x_1} + {x_2} + ....{x_n}} \right)} \over n} = 2\overline x $$
<br><br>But given $$A.M = 4\overline x $$
<br><br>$$\therefore\,\,\,$$ Statement $${\rm I}{\rm I}$$ is false.
<br><br>Variance of $$2{x_1},2{x_2}......2{x_n}$$
<br><br>$$=$$ Variance of $$\left\{ {2{x_i}} \right\}$$
<br><br>$$ = {2^2}$$ Variance of $$\left\{ {{x_i}} \right\} = 4{\sigma ^2}$$
<br><br>So, statement $${\rm I}$$ is correct. | mcq | aieee-2012 |
yLM7mSC8LrFHWos6 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10
to each of the students. Which of the following statistical measures will not change even after the grace
marks were given? | [{"identifier": "A", "content": "median"}, {"identifier": "B", "content": "mode"}, {"identifier": "C", "content": "variance"}, {"identifier": "D", "content": "mean "}] | ["C"] | null | As we know variance does not change with the change of origin. So, here even after adding grace marks $$10$$, the variance will be same.
<br><br>Let's see with an example,
<br><br>Assume initial variance $$ = {{\sum {{{\left( {{x_i} - \overline x } \right)}^2}} } \over N}$$
<br><br>After adding grace marks $$10$$ with each student,
<br><br>the final variance $$ = {{{\sum {\left[ {\left( {{x_i} + 10} \right) - \left( {\overline x + 10} \right)} \right]} } \over N}^2}$$
<br><br>$$ = {{\sum {{{\left( {{x_i} - \overline x } \right)}^2}} } \over N}$$
<br><br>$$ = $$ Initial variance. | mcq | jee-main-2013-offline |
weiErB44wHyXOgoj | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The variance of first 50 even natural numbers is | [{"identifier": "A", "content": "833"}, {"identifier": "B", "content": "437"}, {"identifier": "C", "content": "$${{437} \\over 4}$$"}, {"identifier": "D", "content": "$${{833} \\over 4}$$"}] | ["A"] | null | Here is total $$50$$ numbers, so $$N=50$$
<br><br>Variance $$ = $$ $${{\sum {{x^2}} } \over {50}} - {\left( {{{\sum x } \over {50}}} \right)^2}$$
<br><br>Here $$\sum {{x^2}} = $$ sum of square of first $$50$$ even natural number.
<br><br>$$ = {2^2} + {4^2} + ..... + {100^2}$$
<br><br>$$ = {2^2}\left[ {{1^2} + {2^2} + ....... + {{50}^2}} \right]$$
<br><br>$$ = 4\left[ {{{50 \times 51 \times 101} \over 6}} \right]$$
<br><br>So, $${{\sum {{x^2}} } \over {50}} = {{4 \times 51 \times 101} \over 6} = 3434$$
<br><br>$$\sum {x = } $$ sum of first $$50$$ even natural numbers
<br><br>$$ = 2 + 4 + ...... + 100$$
<br><br>$$ = 2\left[ {1 + 2 + .... + 50} \right]$$
<br><br>$$ = 2\left[ {{{50 \times 51} \over 2}} \right]$$
<br><br>$$ = 50 \times 51$$
<br><br>$$\therefore\,\,\,$$ $${\left( {{{\sum x } \over {50}}} \right)^2} = {\left( {{{50 \times 51} \over {50}}} \right)^2} = 2601$$
<br><br>$$\therefore\,\,\,$$ Variance $$ = 3434 - 2601$$ $$=833$$ | mcq | jee-main-2014-offline |
TvmSJULEl6IoTAWP | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true? | [{"identifier": "A", "content": "3$$a$$<sup>2</sup> - 26$$a$$ + 55 = 0"}, {"identifier": "B", "content": "3$$a$$<sup>2</sup> - 32$$a$$ + 84 = 0"}, {"identifier": "C", "content": "3$$a$$<sup>2</sup> - 34$$a$$ + 91 = 0"}, {"identifier": "D", "content": "3$$a$$<sup>2</sup> - 23$$a$$ + 44 = 0"}] | ["B"] | null | The formula for standard deviation (S.D)
<br><br>$$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {{{\sum {{x_i}} } \over n}} \right)}^2}} $$
<br><br>Where $$\sum {x_i^2 = } $$ Sum of square of the numbers
<br><br>$$ = {2^2} + {3^2} + {a^2} + {11^2}$$
<br><br>$$ = 4 + 9 + {a^2} + 121$$
<br><br>$$ = 134 + {a^2}$$
<br><br>$$\sum {{x_i}} = $$ Sum of numbers
<br><br>$$ = 2 + 3 + a + 11$$
<br><br>$$ = 16 + a$$
<br><br>$$\therefore\,\,\,$$ $$SD = \sqrt {{{134 + a{}^2} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} $$
<br><br>$$ \Rightarrow \sqrt {{{134 + {a^2}} \over 4} - {{\left( {{{16 + a} \over 4}} \right)}^2}} = 3.5 = {7 \over 2}$$ (given)
<br><br>$$ \Rightarrow {{134 + {a^2}} \over 4} - {\left( {{{16 + a} \over 4}} \right)^2} = {{49} \over 4}$$
<br><br>$$ \Rightarrow 4\left( {134 + {a^2}} \right) - \left( {256 + 32a + {a^2}} \right) = 4 \times 49$$
<br><br>$$ \Rightarrow 3{a^2} - 32a + 84 = 0$$ | mcq | jee-main-2016-offline |
o2CYmg1ei6ZV5R3QOmfZp | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean of 5 observations is 5 and their variance is 124. If three of the observations
are 1, 2 and 6 ; then the mean deviation from the mean of the data is : | [{"identifier": "A", "content": "2.4"}, {"identifier": "B", "content": "2.8"}, {"identifier": "C", "content": "2.5"}, {"identifier": "D", "content": "2.6"}] | ["B"] | null | Let 5 observations are x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, x<sub>5</sub>
<br><br>given, x<sub>1</sub> = 1, x<sub>2</sub> = 2, x<sub>3</sub> = 6
<br><br>Mean = 5
<br><br>$$ \therefore $$ Mean$$\left( {\overline x } \right)$$ = $${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$$ = 5
<br><br>$$ \Rightarrow $$ 1 + 2 + 6 + x<sub>4</sub> + x<sub>5</sub> = 25
<br><br>$$ \therefore $$ x<sub>4</sub> + x<sub>5</sub> = 16
<br><br>$$ \Rightarrow $$ (x<sub>4</sub> $$-$$ 5) + (x<sub>5</sub> $$-$$ 5) + 10 = 16
<br><br>$$ \Rightarrow $$ (x<sub>4</sub> $$-$$ 5) + (x<sub>5</sub> $$-$$ 5) = 6
<br><br>$$ \therefore $$ Mean deviation about mean,
<br><br>= $${{\sum {\left| {{x_i} - \overline x } \right|} } \over n}$$
<br><br>= $${{\left| {1 - 5} \right| + \left| {2 - 5} \right| + \left| {6 - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|} \over 5}$$
<br><br>= $${{4 + 3 + 1 + 6} \over 5}$$
<br><br>= $${{14} \over 5}$$
<br><br>= 2.8 | mcq | jee-main-2016-online-10th-april-morning-slot |
fDAndzgC3lC0xV4A5gqpU | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean deviation of the numbers 1, 1 + d, ..., 1 +100d from their mean is 255, then a value of d is : | [{"identifier": "A", "content": "10.1"}, {"identifier": "B", "content": "20.2"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "5.05"}] | ["A"] | null | Given numbers are,
<br><br>1, 1 + d, 1 + 2d . . . . . 1 + 100d
<br><br>$$ \therefore $$ Total 101 number are present.
<br><br>$$ \therefore $$ n = 101
<br><br>$$ \therefore $$ mean $$\left( {\overline x } \right)$$ = $${{1 + \left( {1 + d} \right) + ......\left( {1 + 100d} \right)} \over {101}}$$
<br><br>= $${1 \over {101}} \times {{101} \over 2}$$ [1 + (1 + 100d)]
<br><br>= 1 + 50d
<br><br>$$ \therefore $$ Mean deviation from mean
<br><br>= $${1 \over {101}}$$ $$\left[ {\left| {1 - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + d} \right) - \left( {1 + 50d} \right)} \right|} \right.$$
<br><br> $$\left. { + ...... + \left| {\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|} \right]$$
<br><br>= $${{2\left| d \right|} \over {101}}$$ ( 1 + 2 + 3 + . . . . . + 50)
<br><br>= $${{2\left| d \right|} \over {101}} \times {{50 \times 51} \over 2}$$
<br><br>= $${{2550} \over {101}}\left| d \right|$$
<br><br>From question,
<br><br>$${{2550} \over {101}}\left| d \right|$$ = 255
<br><br>$$ \Rightarrow $$ $$\left| d \right|$$ = 10.1 | mcq | jee-main-2016-online-9th-april-morning-slot |
Zd4npDUYRa223mH398w05 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The sum of 100 observations and the sum of their squares are 400 and 2475,
respectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If
the incorrect observations are omitted, then the variance of the remaining observations
is : | [{"identifier": "A", "content": "8.25 "}, {"identifier": "B", "content": "8.50"}, {"identifier": "C", "content": "8.00"}, {"identifier": "D", "content": "9.00"}] | ["D"] | null | <p>We have</p>
<p>$$\sum\limits_{i = 1}^{100} {{x_i} = 400} $$</p>
<p>$$\sum\limits_{i = 1}^{100} {x_i^2 = 2425} $$</p>
<p>The variance of the remaining observations is</p>
<p>$${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\left( {{{\sum {{x_i}} } \over N}} \right)^2}$$</p>
<p>$$ \Rightarrow {{2425} \over {97}} - {\left( {{{388} \over {97}}} \right)^2}$$</p>
<p>$$ \Rightarrow {{2425} \over {97}} - 16$$</p>
<p>$$ \Rightarrow {{2425 - 1552} \over {97}} = {{873} \over {97}} = 9$$</p> | mcq | jee-main-2017-online-9th-april-morning-slot |
vx5i8PURp9Ov44S5 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If $$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9$$ and
<br/><br/>$$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$$, then the standard deviation of the 9 items
<br/>$${x_1},{x_2},.......,{x_9}$$ is | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}] | ["D"] | null | <u>IMPORTANT POINT :-</u>
<br><br>When every number is added or subtracted by a fixed number then the standard Deviation remain unchanged.
<br><br>so let $${x_i} - 5 = {y_i}$$
<br><br>So, new equation is $$\sum\limits_{i = 1}^9 {{y_i}} = 9$$
<br><br>and $$\sum\limits_{i = 1}^9 {y_i^2} = 45$$
<br><br>As, we know. Standard Deviation (S.D)
<br><br>$$ = \sqrt {{{\sum\limits_{i = 1}^9 {y_i^2} } \over 9} - {{\left( {{{\sum\limits_{i = 1}^9 {yi} } \over 9}} \right)}^2}} $$
<br><br>$$ = \sqrt {{{45} \over 9} - {{\left( {{9 \over 9}} \right)}^2}} $$
<br><br>$$ = \sqrt {5 - 1} $$
<br><br>$$=$$ $$\sqrt 4 $$
<br><br>$$=2$$ | mcq | jee-main-2018-offline |
mplrVzKRZMtSkE8492dXI | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean of the data : 7, 8, 9, 7, 8, 7, $$\lambda $$, 8 is 8, then the variance of this data is : | [{"identifier": "A", "content": "$${7 \\over 8}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$${9 \\over 8}$$"}, {"identifier": "D", "content": "2"}] | ["B"] | null | $$\overline x $$ = $${{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8} \over 8}$$ = 8
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{54 + \lambda } \over 8}$$ = 8 $$ \Rightarrow $$ $$\lambda $$ = 10
<br><br>Now variance = $$\sigma $$<sup>2</sup>
<br><br>= $${{{{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {9 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {10 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2}} \over 8}$$
<br><br>$$ \Rightarrow $$ $$\sigma $$<sup>2</sup> = $${{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0} \over 8}$$ = $${8 \over 8}$$ = 1
<br><br>Hence, the variance is 1. | mcq | jee-main-2018-online-15th-april-evening-slot |
RGk9zKugGjSBF8zj5MoXV | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and the standard deviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["C"] | null | Here mean = $$\overline x $$ = 9
<br><br>$$ \Rightarrow $$ $$\overline x $$ = $${{\sum {{x_i}} } \over n}$$ = 9
<br><br>$$ \Rightarrow $$ $${\sum {{x_i}} }$$ = 9 $$ \times $$ 5 = 45
<br><br>Now, standard deviation = 0
<br><br>$$\therefore\,\,\,$$ all the five terms are same i.e.; 9
<br><br>Now for changed observation
<br><br>$${\overline x _{new}}$$ = $${{36 + {x_5}} \over 5} = 10$$
<br><br>$$ \Rightarrow $$ x<sub>5</sub> = 14
<br><br>$$\therefore\,\,\,$$ $$\sigma $$<sub>new</sub> = $$\sqrt {{{\sum {{{\left( {{x_i} - {{\overline x }_{new}}} \right)}^2}} } \over n}} $$
<br><br>= $$\sqrt {{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}} \over 5}} $$ = 2 | mcq | jee-main-2018-online-16th-april-morning-slot |
AjX0JjPR9fp5X27YCGJ9H | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | A student scores the following marks in five tests
:
<br/><br/>45, 54, 41, 57, 43.
<br/><br/>His score is not known for the
sixth test. If the mean score is 48 in the six tests,
then the standard deviation of the marks in six tests
is | [{"identifier": "A", "content": "$$100 \\over {\\sqrt 3}$$"}, {"identifier": "B", "content": "$$10 \\over {\\sqrt 3}$$"}, {"identifier": "C", "content": "$$10 \\over3$$"}, {"identifier": "D", "content": "$$100 \\over3$$"}] | ["B"] | null | Let the score in the sixth test = x
<br><br>Given, Mean ($$\overline x $$) = 48
<br><br>$$ \Rightarrow $$ $${{45 + 54 + 41 + 57 + 43 + x} \over 6}$$ = 48
<br><br>$$ \Rightarrow $$ x = 48
<br><br>Standard deviation (SD)
<br><br>= $$\sqrt {{{\sum\limits_{i = 1}^N {{{\left( {{x_i} - \overline x } \right)}^2}} } \over N}} $$
<br><br>= $$\sqrt {{\matrix{
{\left( {45 - 48} \right)^2} + {\left( {54 - 48} \right)^2} \hfill \cr
+ {\left( {41 - 48} \right)^2} + {\left( {57 - 48} \right)^2} \hfill \cr
+ {\left( {43 - 48} \right)^2} + {\left( {48 - 48} \right)^2} \hfill \cr} \over 6}} $$
<br><br>= $$\sqrt {{{9 + 36 + 49 + 81 + 25} \over 6}} $$
<br><br>= $$\sqrt {{{200} \over 6}} $$
<br><br>= $$\sqrt {{{100} \over 3}} $$
<br><br>= $${{10} \over {\sqrt 3 }}$$ | mcq | jee-main-2019-online-8th-april-evening-slot |
ia0Ybgi0hXqiDtEOXS3rsa0w2w9jx620ebc | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the data x<sub>1</sub>, x<sub>2</sub>,......., x<sub>10</sub> is such that the mean of first four of these is 11, the mean of the remaining six is
16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is : | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "2$$\\sqrt 2 $$"}, {"identifier": "D", "content": "4"}] | ["B"] | null | $${\sigma ^2} = {{\sum {x_i^2} } \over {10}} - {\left( {{{\sum {{x_i}} } \over {10}}} \right)^2} \to (i)$$<br><br>
Now x<sub>1</sub> + x<sub>2</sub> + x<sub>3</sub> + x<sub>4</sub> = 44 & x<sub>5</sub> + x<sub>6</sub> + ......... + x<sub>10</sub> = 96<br><br>
Hence $${\sigma ^2}$$ = $${{2000} \over {10}} - {\left( {{{140} \over {10}}} \right)^2}$$ = 200 - 196 = 4<br><br>
Hence $$\sigma $$ = 2 | mcq | jee-main-2019-online-12th-april-morning-slot |
8FSQCWDSs03z11aXVQ3rsa0w2w9jx2fqton | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If both the mean and the standard deviation of 50 observations x<sub>1</sub>, x<sub>2</sub>,..., x<sub>50</sub> are equal to 16, then the mean of (x<sub>1</sub> – 4)<sup>2</sup>
, (x<sub>2 </sub>– 4)<sup>2</sup>
,....., (x<sub>50</sub> – 4)<sup>2</sup>
is : | [{"identifier": "A", "content": "400"}, {"identifier": "B", "content": "480"}, {"identifier": "C", "content": "380"}, {"identifier": "D", "content": "525"}] | ["A"] | null | $$Mean(\mu ) = {{\sum {{x_i}} } \over {50}} = 16$$<br><br>
$$ \therefore $$ $$\sum {{x_i}} = 16 \times 50$$<br><br>
$$S.D.\left( \sigma \right) = \sqrt {{{\sum {{x_i}^2} } \over {50}} - {{\left( \mu \right)}^2}} = 16$$<br><br>
$$ \Rightarrow {{\sum {{x_i}^2} } \over {50}} = 256 \times 2$$<br><br>
Required mean = $${{\sum {{{\left( {{x_i} - 4} \right)}^2}} } \over {50}}$$<br><br>
$$ \Rightarrow {{\sum {{x_i}^2} + 16 \times 50 - 8\sum {{x_i}} } \over {50}}$$<br><br>
$$ \Rightarrow $$ 256 $$ \times $$ 2 + 16 - 8 $$ \times $$ 16<br><br>
$$ \Rightarrow $$ 400 | mcq | jee-main-2019-online-10th-april-evening-slot |
Gf3bGbvx41RN1hxaDf3iQ | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the standard deviation of the numbers
–1, 0, 1, k is $$\sqrt 5$$ where k > 0, then k is equal to | [{"identifier": "A", "content": "2$$\\sqrt 6 $$"}, {"identifier": "B", "content": "$$\\sqrt 6 $$"}, {"identifier": "C", "content": "$$2\\sqrt {{{5} \\over 6}} $$"}, {"identifier": "D", "content": "$$2\\sqrt {{{10} \\over 3}} $$"}] | ["A"] | null | standard deviation = $$\sqrt 5$$
<br><br>$$ \therefore $$ Variance = $${\left( {\sqrt 5 } \right)^2}$$ = 5
<br><br>Also variance = $${{\sum {x_i^2} } \over N} - {\mu ^2}$$
<br><br>Where $$\mu $$ = Mean = $${{ - 1 + 0 + 1 + k} \over 4}$$ = $${k \over 4}$$
<br><br>$$ \therefore $$ Variance = $${{{{\left( { - 1} \right)}^2} + 0 + {1^2} + {k^2}} \over 4}$$ - $${{{k^2}} \over {16}}$$
<br><br>$$ \Rightarrow $$ 5 = $${{2 + {k^2}} \over 4}$$ - $${{{k^2}} \over {16}}$$
<br><br>$$ \Rightarrow $$ 8 + 3k<sup>2</sup> = 80
<br><br>$$ \Rightarrow $$ k<sup>2</sup> = 24 = 2$$\sqrt 6 $$ | mcq | jee-main-2019-online-9th-april-morning-slot |
fQlKI7V234cc0jyKXcWQh | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and variance of seven observations are
8 and 16, respectively. If 5 of the observations are
2, 4, 10, 12, 14, then the product of the remaining
two observations is :
| [{"identifier": "A", "content": "40"}, {"identifier": "B", "content": "48"}, {"identifier": "C", "content": "49"}, {"identifier": "D", "content": "45"}] | ["B"] | null | Given mean ($$\mu $$) = 8
<br><br>variance ($${\sigma ^2}$$) = 16
<br><br>No of observations (N) = 7
<br><br>Let the two unknown observation = x and y
<br><br>We know,
<br><br>$${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$$ = 16
<br><br>$$ \Rightarrow $$ $${{{2^2} + {4^2} + {{10}^2} + {{12}^2} + {{14}^2} + {x^2} + {y^2}} \over 7} - {\left( 8 \right)^2}$$ = 16
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup> = 100 ........(1)
<br><br>We know,
<br><br>$$\mu $$ = $${{\sum {{x_i}} } \over N}$$ = 8
<br><br>$$ \Rightarrow $$ $${{2 + 4 + 10 + 12 + 14 + x + y} \over 7}$$ = 8
<br><br>$$ \Rightarrow $$ x + y = 14 ...........(2)
<br><br>As (x + y)<sup>2</sup> = x<sup>2</sup> + y<sup>2</sup> + 2xy
<br><br>$$ \Rightarrow $$ (14)<sup>2</sup> = 100 + 2xy
<br><br>$$ \Rightarrow $$ 196 = 100 + 2xy
<br><br>$$ \Rightarrow $$ xy = 48
| mcq | jee-main-2019-online-8th-april-morning-slot |
u2v4ONloLZkkoEVwLPxuI | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are
3, 4 and 4 ; then the absolute value of the difference of the other two observations, is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["B"] | null | mean $$\overline x $$ = 4, $$\sigma $$<sup>2</sup> = 5.2, n = 5, . x<sub>1</sub> = 3 x<sub>2</sub> = 4 = x<sub>3</sub>
<br><br>$$\sum {{x_i}} = 20$$
<br><br>x<sub>4</sub> + x<sub>5</sub> = 9 . . . . . . (i)
<br><br>$${{\sum {x_i^2} } \over x} - {\left( {\overline x } \right)^2} = \sigma \Rightarrow \sum {x_i^2} = 106$$
<br><br>$$x_4^2 + x_5^2 = 65$$ . . . . . .(ii)
<br><br>Using (i) and (ii) (x<sub>4</sub> $$-$$ x<sub>5</sub>)<sup>2</sup> = 49
<br><br>$$\left| {{x_4} - {x_5}} \right| = 7$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
6iCyo9dsvN73Uh53IzStp | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is : | [{"identifier": "A", "content": "31"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "30"}] | ["A"] | null | $$\sum\limits_{i = 1}^{50} {\left( {{x_i} - 30} \right) = 50} $$
<br><br>$$\sum {{x_i}} = 50 \times 30 = 50$$
<br><br>$$\sum {{x_i}} = 50 + 50 + 30$$
<br><br>Mean $$ = \overline x = {{\sum {{x_i}} } \over n} = {{50 \times 30 + 50} \over {50}}$$
<br><br>$$ = 30 + 1 = 31$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
Bjkqh6oBuyTv8c8pUXMQ4 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The outcome of each of 30 items was observed; 10 items gave an outcome $${1 \over 2}$$ – d each, 10 items gave outcome $${1 \over 2}$$ each and the remaining 10 items gave outcome $${1 \over 2}$$+ d each. If the variance of this outcome data is $${4 \over 3}$$ then |d| equals :
| [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "C", "content": "$${\\sqrt 2 }$$"}, {"identifier": "D", "content": "2"}] | ["C"] | null | Variance is independent of region. So we shift the given data by $${1 \over 2}$$.
<br><br>so, $${{10{d^2} + 10 \times {0^2} + 10{d^2}} \over {30}} - {\left( 0 \right)^2} = {4 \over 3}$$
<br><br>$$ \Rightarrow $$ d<sup>2</sup> $$=$$ 2 $$ \Rightarrow $$ $$\left| d \right| = \sqrt 2 $$ | mcq | jee-main-2019-online-11th-january-morning-slot |
BdCTyhbNUBEqTym6WfHMu | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If mean and standard deviation of 5 observations x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, x<sub>5</sub> are 10 and 3, respectively, then the variance of 6 observations x<sub>1</sub>, x<sub>2</sub>, ….., x<sub>5</sub> and –50 is equal to | [{"identifier": "A", "content": "582.5 "}, {"identifier": "B", "content": "507.5"}, {"identifier": "C", "content": "586.5"}, {"identifier": "D", "content": "509.5"}] | ["B"] | null | $$\overline x = 10 \Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 50} $$
<br><br>S.D. $$ = \sqrt {{{\sum\limits_{i = 1}^5 {x_i^2} } \over 5} - {{\left( {\overline x } \right)}^2}} = 8$$
<br><br>$$ \Rightarrow \,\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} = 109$$
<br><br>variance $$ = \,\,{{\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} + {{\left( { - 50} \right)}^2}} \over 6} - \left( {\sum\limits_{i = 1}^5 {{{{x_i} - 50} \over 6}} } \right)$$
<br><br>$$ = \,\,507.5$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
Sfh5V4TIKbn0jrKQt2JZt | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is - | [{"identifier": "A", "content": "6 : 7"}, {"identifier": "B", "content": "10 : 3"}, {"identifier": "C", "content": "4 : 9"}, {"identifier": "D", "content": "5 : 8"}] | ["C"] | null | Let two observations are x<sub>1</sub> & x<sub>2</sub>
<br><br>mean = $${{\sum {{x_i}} } \over 5} = 5 $$<br><br>$$\Rightarrow 1 + 3 + 8 + {x_1} + {x_2} = 25$$
<br><br>$$ \Rightarrow {x_1} + {x_2} = 13$$ . . . . (1)
<br><br>variance $$\left( {{\sigma ^2}} \right)$$ = $${{\sum {x_i^2} } \over 5} - 25 = 9.20$$
<br><br>$$ \Rightarrow $$ $${\sum {x_i^2 = 171} }$$
<br><br>$$ \Rightarrow $$ $$x_1^2 + x_2^2 = 97$$ . . . . . (2)
<br><br>$$ \Rightarrow $$(x<sub>1</sub> + x<sub>2</sub>)<sup>2</sup> $$-$$ 2x<sub>1</sub>x<sub>2</sub> = 97
<br><br>$$ \Rightarrow $$ 169 - 2x<sub>1</sub>x<sub>2</sub> = 97
<br><br>or x<sub>1</sub>x<sub>2</sub> = 36
<br><br>$$ \therefore $$ x<sub>1</sub> : x<sub>2</sub> = 4 : 9 | mcq | jee-main-2019-online-10th-january-morning-slot |
aGocSe3Maheyl9wAc4gct | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | A data consists of n observations : x<sub>1</sub>, x<sub>2</sub>, . . . . . . ., x<sub>n</sub>.
<br/><br/>If $$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$$ and
<br/><br/>$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$$
<br/><br/>then the standard deviation of this data is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\sqrt 5 $$"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$\\sqrt 7 $$"}] | ["B"] | null | $$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n $$
<br><br>$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} + 2\sum\limits_{i = 1}^n {{x_i}} + n = 9n\,\,\,\,\,...\,(1)$$
<br><br>$$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n $$<br><br>$$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} - 2\sum\limits_{i = 1}^n {{x_i}} + n = 5n\,\,\,\,\,...\,(2)$$
<br><br>Performing (1) + (2), we get
<br><br>$$2\sum\limits_{i = 1}^n {x_i^2} + 2n = 14n$$
<br><br>$$\sum\limits_{i = 1}^n {x_i^2} = 6n$$
<br><br>Performing (1) $$-$$ (2), we get
<br><br>$$ \Rightarrow 4\sum\limits_{i = 1}^n {{x_i}} = 4n$$
<br><br>$$ \Rightarrow $$$$ \Rightarrow \sum\limits_{i = 1}^n {{x_i}} = n$$
<br><br>S.D($$\sigma $$)$$ = \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}} $$
<br><br>$$\sigma $$ $$ = \sqrt {{{6n} \over n} - \left( 1 \right)} $$
<br><br>$$\sigma $$ $$ = \sqrt 5 $$ | mcq | jee-main-2019-online-9th-january-evening-slot |
IME5czDs8NRxifwSDCVpU | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | 5 students of a class have an average height 150 cm and variance 18 cm<sup>2</sup>. A new student, whose height is 156 cm, joined them. The variance (in cm<sup>2</sup>) of the height of these six students is : | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "18"}] | ["C"] | null | Average height of 5 students,
<br><br>$$\overline x = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = 150$$
<br><br>$$ \Rightarrow \,\,\,\sum\limits_{i = 1}^5 {{x_i}} = 750$$
<br><br>We know,
<br><br>Variance $$\left( \sigma \right) = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}$$
<br><br>given that,
<br><br>$${{\sum {x_i^2} } \over 5} - {\left( {150} \right)^2} = 18$$
<br><br>$$ \Rightarrow \,\,\,\sum {x_i^2} = 112590$$
<br><br>Height of new student, x<sub>6</sub> $$=$$ 156 cm
<br><br>New average height $$\left( {{{\overline x }_{new}}} \right) = {{750 + 156} \over 6} = 151$$
<br><br>New variance $$ = {{\,\sum\limits_{i = 1}^6 {x_i^2} } \over 6} - {\left( {{{\overline x }_{new}}} \right)^2}$$
<br><br>$$ = {{112590 + {{\left( {156} \right)}^2}} \over 6} - {\left( {151} \right)^2}$$
<br><br>$$ = 22821 - 22801$$
<br><br>$$ = 20$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
CxpoxX2uEkwDPImF9Vjgy2xukewquink | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let X = {x
$$ \in $$ N : 1
$$ \le $$ x
$$ \le $$ 17} and
<br/>Y = {ax + b: x
$$ \in $$ X and a, b $$ \in $$ R, a > 0}. If mean
<br/>and variance of elements of Y are 17 and 216
<br/>respectively then a + b is equal to : | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "-7"}, {"identifier": "D", "content": "-27"}] | ["C"] | null | Mean of X = $${{\sum\limits_{x = 1}^{17} x } \over {17}}$$ = $${{17 \times 18} \over {17 \times 2}}$$ = 9
<br><br>Mean of Y = $${{\sum\limits_{x = 1}^{17} {\left( {ax + b} \right)} } \over {17}}$$ = 17
<br><br>$$ \Rightarrow $$ $$a{{\sum\limits_{x = 1}^{17} x } \over {17}} + b$$ = 17
<br><br>$$ \Rightarrow $$ 9a + b = 17 ....(1)
<br><br>Given Var(Y) = 216
<br><br>$$ \Rightarrow $$ $${{\sum\limits_{x = 1}^{17} {{{\left( {ax + b} \right)}^2}} } \over {17}} - {\left( {17} \right)^2}$$ = 216
<br><br>$$ \Rightarrow $$ $${\sum\limits_{x = 1}^{17} {{{\left( {ax + b} \right)}^2}} }$$ = 8585
<br><br>$$ \Rightarrow $$ (a + b)<sup>2</sup> + (2a + b)<sup>2</sup> +....+ (17a + b)<sup>2</sup> = 8585
<br><br>$$ \Rightarrow $$ 105a<sup>2</sup> + b<sup>2</sup> + 18ab = 505 ....(2)
<br><br>From equation (1) & (2)
<br><br>a = 3 & b = -10
<br><br>$$ \therefore $$ a + b = –7 | mcq | jee-main-2020-online-2nd-september-morning-slot |
oPM41URzGqus9Ki3Ccjgy2xukfw0pvwk | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If $$\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n$$ and $$\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na$$
<br/>(n, a > 1) then the standard deviation of n
<br/>observations x<sub>1</sub>
, x<sub>2</sub>
, ..., x<sub>n</sub>
is :
| [{"identifier": "A", "content": "$$a$$ \u2013 1"}, {"identifier": "B", "content": "$$n\\sqrt {a - 1} $$"}, {"identifier": "C", "content": "$$\\sqrt {n\\left( {a - 1} \\right)} $$"}, {"identifier": "D", "content": "$$\\sqrt {a - 1} $$"}] | ["D"] | null | S.D = $$\sqrt {{{\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} } \over n} - {{\left( {{{\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} } \over n}} \right)}^2}} $$
<br><br>= $$\sqrt {{{na} \over n} - {{\left( {{n \over n}} \right)}^2}} $$
<br><br>= $$\sqrt {a - 1} $$ | mcq | jee-main-2020-online-6th-september-morning-slot |
0lLhsD5yfJG9GnWwsSjgy2xukfqbxcbz | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean and the standard deviation of the
<br/>data 3, 5, 7, a, b are 5 and 2 respectively, then
a and b are the roots of the equation : | [{"identifier": "A", "content": "x<sup>2</sup> \u2013 20x + 18 = 0"}, {"identifier": "B", "content": "2x<sup>2</sup> \u2013 20x + 19 = 0\n"}, {"identifier": "C", "content": "x<sup>2</sup> \u2013 10x + 18 = 0"}, {"identifier": "D", "content": "x<sup>2</sup> \u2013 10x + 19 = 0\n"}] | ["D"] | null | Mean = $${{3 + 5 + 7 + a + b} \over 5}$$ = 5
<br><br>$$ \Rightarrow $$ $$a$$ + b = 10
<br><br>Variance = $${{{3^2} + {5^2} + {7^2} + {a^2} + {b^2}} \over 5}$$ - (5)<sup>2</sup> = 4
<br><br>$$ \Rightarrow $$ $${{a^2} + {b^2}}$$ = 62
<br><br>$$ \Rightarrow $$ $${\left( {a + b} \right)^2} - 2ab$$ = 62
<br><br>$$ \Rightarrow $$ $$ab$$ = 19
<br><br>So $$a$$ and b are the roots of the equation
<br><br>x<sup>2</sup> – 10x + 19 = 0 | mcq | jee-main-2020-online-5th-september-evening-slot |
gm77u9cthH2QxdEMuHjgy2xukfg6f9cx | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["A"] | null | $$\overline x = {{2 + 4 + + 10 + 12 + 14 + x + y} \over 7} = 8$$<br><br>x + y = 14 ....(i)<br><br>$${(\sigma )^2} = {{\sum {{{({x_i})}^2}} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2}$$<br><br>$$ \Rightarrow $$ $$16 = {{4 + 16 + 100 + 144 + 196 + {x^2} + {y^2}} \over 2} - {8^2}$$<br><br>$$ \Rightarrow $$ $$16 + 64 = {{460 + {x^2} + {y^2}} \over 7}$$<br><br>$$ \Rightarrow $$ 560 = 460 + x<sup>2</sup> + y<sup>2</sup><br><br>$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup> = 100 ......(ii)<br><br>Clearly by (i) and (ii),
<br><br>(x + y)<sup>2</sup> - 2xy = 100
<br><br>$$ \Rightarrow $$ (14)<sup>2</sup> - 2xy = 100
<br><br>$$ \Rightarrow $$ 2xy = 96
<br><br>$$ \Rightarrow $$ xy = 48
<br><br>Now, |x - y| = $$\sqrt {{{\left( {x + y} \right)}^2} - 4xy} $$
<br><br>= $$\sqrt {196 - 192} $$
<br><br>= 2 | mcq | jee-main-2020-online-5th-september-morning-slot |
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