Sets and Relations
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?
Let $\mathrm{A}=\{2,3,4\}$ and $\mathrm{B}=\{8,9,12\}$. Then the number of elements in the relation $\mathrm{R}=\left\{\left(\left(a_{1}, \mathrm{~b}_{1}\right),\left(a_{2}, \mathrm{~b}_{2}\right)\right) \in(A \times B, A \times B): a_{1}\right.$ divides $\mathrm{b}_{2}$ and $\mathrm{a}_{2}$ divides $\left.\mathrm{b}_{1}\right\}$ is :
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 :
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 :
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
$\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0\right\}$
and $\mathrm{T}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}, \mathrm{a}^{2}-\mathrm{b}^{2} \in \mathbb{Z}\right\}$,
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
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 :
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
The relation $\mathrm{R = \{ (a,b):\gcd (a,b) = 1,2a \ne b,a,b \in \mathbb{Z}\}}$ is :
$\left\{n \in \mathbb{N}: 10 \leq n \leq 100\right.$ and $3^{n}-3$ is a multiple of 7$\}$ is ___________.
Explanation:
Recall that for any integers $a$ and $b$, $a$ is a multiple of $b$ if there exists an integer $k$ such that $a = bk$. So in our case, we need to find how many $n$ satisfy the equation $3^n - 3 = 7k$ for some integer $k$.
Notice that $3^n - 3 = 3(3^{n-1} - 1)$. We want this expression to be a multiple of 7. Let's explore a few powers of 3 modulo 7:
$3^1 \equiv 3 \pmod{7}$
$3^2 \equiv 9 \equiv 2 \pmod{7}$
$3^3 \equiv 27 \equiv 6 \pmod{7}$
$3^4 \equiv 81 \equiv 4 \pmod{7}$
$3^5 \equiv 243 \equiv 5 \pmod{7}$
$3^6 \equiv 729 \equiv 1 \pmod{7}$
We observe that $3^n \pmod{7}$ follows a cycle of length 6. So, $3^{n-1} \pmod{7}$ also follows the same cycle, but shifted:
$3^0 \equiv 1 \pmod{7}$
$3^1 \equiv 3 \pmod{7}$
$3^2 \equiv 2 \pmod{7}$
$3^3 \equiv 6 \pmod{7}$
$3^4 \equiv 4 \pmod{7}$
$3^5 \equiv 5 \pmod{7}$
We want $3(3^{n-1} - 1) \equiv 0 \pmod{7}$, which means that $3^{n-1} - 1 \equiv 0 \pmod{7}$. From the cycle above, we see that this is true when $n-1$ is a multiple of 6, or equivalently, when $n$ is one more than a multiple of 6.
Now let's find the multiples of 6 between 10 and 100:
$12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96$
Adding 1 to each of these values, we get the set of natural numbers $n$ that satisfy the given condition:
$13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97$
There are 15 elements in this set. Therefore, the number of elements in the given set is $\boxed{15}$.
$R=\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\}$. Then the number of elements in $\mathrm{R}$ is ____________.
Explanation:
Given A = {1, 2, 3, 4}, the maximum value of $2a + 3b$ is 20, when (a, b) = (4, 4), and the minimum value of $4c + 5d$ is 9, when (c, d) = (1, 1). Therefore, the possible values for $2a + 3b = 4c + 5d$ are 9, 13, 14, 17, 18, and 19.
Now, let's find the combinations of (a, b), (c, d) that satisfy the given equation:
1. $2a + 3b = 9 \Rightarrow (a, b) = (3, 1) \Rightarrow (c, d) = (1, 1)$
2. $2a + 3b = 13 \Rightarrow (a, b) = (2, 3) \Rightarrow (c, d) = (2, 1)$
3. $2a + 3b = 14 \Rightarrow (a, b) = (4, 2) \Rightarrow (c, d) = (1, 2)$
4. $2a + 3b = 14 \Rightarrow (a, b) = (1, 4) \Rightarrow (c, d) = (1, 2)$
5. $2a + 3b = 17 \Rightarrow (a, b) = (4, 3) \Rightarrow (c, d) = (3, 1)$
6. $2a + 3b = 18 \Rightarrow (a, b) = (3, 4) \Rightarrow (c, d) = (2, 2)$
There are a total of 6 elements in the relation R for the given equation with the specified values of a, b, c, and d.
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 __________
Explanation:
Relation to be reflexive $(a, a) \in R \forall a \in A$
$\Rightarrow (-4,-4),(-3,-3),(-2,-2)$ also should be added in $R$.
Relation to be symmetric if $(a, b) \in R$, then $(b, a) \in R \forall a, b \in A$
$\Rightarrow (4,-4),(3,-3),(1,0),(-2,3)$ also should be added in $R$
$\Rightarrow$ Minimum number of elements to be added to $R=3+4=7$
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 __________.
Explanation:
To find the number of such relations, let's first understand what it means for a relation to be reflexive, transitive, and not symmetric.
A relation $R$ on a set $S$ is reflexive if every element is related to itself. That is, $(a, a) \in R$ for all $a \in S$.
A relation is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then it must also be the case that $(a, c) \in R$.
A relation is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$ also.
Since we are looking for relations that are reflexive and transitive but not symmetric, we will need to include certain elements and exclude others.
First, for the relation to be reflexive on the set $\{1,2,3\}$, it must contain $(1,1), (2,2),$ and $(3,3)$.
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.
So far, the must-have elements of the desired relation are:
- For reflexivity: $(1,1), (2,2), (3,3)$
- Given in the problem: $(1,2), (2,3)$
- For transitivity (induced by given elements): $(1,3)$
"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."
1. R1 = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}
Here, none of (2,1), (3,2), (3,1) are in R1.
2. R2 = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (2,1)}
Here, only (2,1) is in R2, and neither (3,2) nor (3,1) are in R2.
3. R3 = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (3,2)}
Here, only (3,2) is in R3, and neither (2,1) nor (3,1) are in R3.
The number of elements in the set $\{ n \in Z:|{n^2} - 10n + 19| < 6\} $ is _________.
Explanation:
$\Rightarrow-6 < n^2-10 n+19 < 6$
Take, $-6 < n^2-10 n+19$ and $n^2-10 n+19 < 6$
$ \begin{array}{ll} \Rightarrow n^2-10 n+25 > 0 & \text { and }\quad n^2-10 n+13 < 0 \\\\ \Rightarrow(n-5)^2 > 0 & \text { and } n=\frac{10 \pm \sqrt{100-52}}{2}<0 \end{array} $
$\Rightarrow n \in \mathbb{Z}-\{5\}$
$ \begin{array}{lr} & \therefore n \in[5-2 \sqrt{3}, 5+2 \sqrt{3}] \\\\ & \therefore n \in[13,8.3] \\\\ & \therefore n=2,3,4,5,6,7,8 \end{array} $
Thus, number of element in the set is ' 6 '
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 ____________.
Explanation:
Case I : $x-y$ is odd, if one is odd and one is even and $x>y$.
$\therefore$ Possibilites are $\{(3,0),(4,3),(6,3),(7,6),(7,4)$, $(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10$, $7),(10,3)\}$
No. of cases $=15$
Case II : $x-y=2$
$\therefore$ Possibilities are $\{(6,4),(8,6),(9,7),(10,8)\}$
$\therefore$ No. of cases $=4$
So, minimum ordered pair to be added $=15+4=19$
Let $\mathrm{A}=\{1,2,3,4, \ldots ., 10\}$ and $\mathrm{B}=\{0,1,2,3,4\}$. The number of elements in the relation $R=\left\{(a, b) \in A \times A: 2(a-b)^{2}+3(a-b) \in B\right\}$ is ___________.
Explanation:
Given sets :
A={1,2,3,4, ............,10}
B={0,1,2,3,4}
We are looking for pairs $(a,b) \in A \times A$ such that :
$ 2(a-b)^2 + 3(a-b) \in B $
Let's break down the relation :
Case 1 : $ a-b = 0 $
$ 2(a-b)^2 + 3(a-b) = 0 $
Pairs : $(1,1), (2,2), (3,3), \ldots, (10,10)$ which gives 10 pairs.
Case 2 : $ a-b = 1 $
$ 2(a-b)^2 + 3(a-b) = 2(1) + 3(1) = 5 $
But 5 is not in B, so no pairs for this case.
Case 3 : $ a-b = -1 $
$ 2(a-b)^2 + 3(a-b) = 2(1) - 3(1) = -1 $
This value is not in B, so no pairs for this case.
Case 4 : $ a-b = 2 $
$ 2(a-b)^2 + 3(a-b) = 2(4) + 3(2) = 8+6 = 14 $
Again, 14 is not in B, so no pairs for this case.
Case 5 : $ a-b = -2 $
$ 2(a-b)^2 + 3(a-b) = 2(4) - 3(2) = 8 - 6 = 2 $
Pairs : $(1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8,10)$ which gives 8 pairs.
For any other $ a-b $ value, the quadratic will grow larger than the maximum value in B, so we don't need to consider them.
In total, we have $ 10 + 8 = 18 $ pairs in the relation $ R $.
Therefore, the number of elements in the relation $ R $ is 18.
Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _____________.
Explanation:
Elements of the type $3 \mathrm{k}+1=1,7,9$
Elements of the type $3 \mathrm{k}+2=2,5,11$
Subsets containing one element $S_1=1$
Subsets containing two elements
$ S_2={ }^3 C_1 \times{ }^3 C_1=9 $
Subsets containing three elements
$ \mathrm{S}_3={ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1+1+1=11 $
Subsets containing four elements
$
\mathrm{S}_4={ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2=11 $
Subsets containing five elements
$ \mathrm{S}_5={ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2 \times 1=9 $
Subsets containing six elements $\mathrm{S}_6=1$
Subsets containing seven elements $\mathrm{S}_7=1$
$ \Rightarrow \text { sum }=43 $
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 __________.
Explanation:
$S:\{a, b, c, d\}$
Adding $(a, a),(b, b),(c, c),(d, d)$ make reflexive.
Adding $(b, a),(c, b),(d, b)$ make Symmetric
And adding $(a, d),(a, c)$ to make transitive
Further $(d, a) \&(c, a)$ to be added to make Symmetricity.
Further $(c, d) \&(d, c)$ also be added.
So total 13 elements to be added to make equivalence.
Let R be a relation from the set $\{1,2,3, \ldots, 60\}$ to itself such that $R=\{(a, b): b=p q$, where $p, q \geqslant 3$ are prime numbers}. Then, the number of elements in R is :
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 :
Let $R_{1}$ and $R_{2}$ be two relations defined on $\mathbb{R}$ by
$a \,R_{1} \,b \Leftrightarrow a b \geq 0$ and $a \,R_{2} \,b \Leftrightarrow a \geq b$
Then,
Let a set A = A1 $\cup$ A2 $\cup$ ..... $\cup$ Ak, where Ai $\cap$ Aj = $\phi$ for i $\ne$ j, 1 $\le$ j, j $\le$ k. Define the relation R from A to A by R = {(x, y) : y $\in$ Ai if and only if x $\in$ Ai, 1 $\le$ i $\le$ k}. Then, R is :
Let R1 = {(a, b) $\in$ N $\times$ N : |a $-$ b| $\le$ 13} and
R2 = {(a, b) $\in$ N $\times$ N : |a $-$ b| $\ne$ 13}. Then on N :
Let $S=\{4,6,9\}$ and $T=\{9,10,11, \ldots, 1000\}$. If $A=\left\{a_{1}+a_{2}+\ldots+a_{k}: k \in \mathbf{N}, a_{1}, a_{2}, a_{3}, \ldots, a_{k}\right.$ $\epsilon S\}$, then the sum of all the elements in the set $T-A$ is equal to __________.
Explanation:
Here $S = \{ 4,6,9\} $
And $T = \{ 9,10,11,\,\,......,\,\,1000\} $.
We have to find all numbers in the form of $4x + 6y + 9z$, where $x,y,z \in \{ 0,1,2,\,......\} $.
If a and b are coprime number then the least number from which all the number more than or equal to it can be express as $ax + by$ where $x,y \in \{ 0,1,2,\,......\} $ is $(a - 1)\,.\,(b - 1)$.
Then for $6y + 9z = 3(2y + 3z)$
All the number from $(2 - 1)\,.\,(3 - 1) = 2$ and above can be express as $2x + 3z$ (say t).
Now $4x + 6y + 9z = 4x + 3(t + 2)$
$ = 4x + 3t + 6$
again by same rule $4x + 3t$, all the number from $(4 - 1)\,(3 - 1) = 6$ and above can be express from $4x + 3t$.
Then $4x + 6y + 9z$ express all the numbers from 12 and above.
again 9 and 10 can be express in form $4x + 6y + 9z$.
Then set $A = \{ 9,10,12,13,\,....,\,1000\} .$
Then $T - A = \{ 11\} $
Only one element 11 is there.
Sum of elements of $T - A = 11$
Let $A=\{1,2,3,4,5,6,7\}$ and $B=\{3,6,7,9\}$. Then the number of elements in the set $\{C \subseteq A: C \cap B \neq \phi\}$ is ___________.
Explanation:
As C $\cap$ B $\ne$ $\phi$, c must be not be formed by {1, 2, 4, 5}
$\therefore$ Number of subsets of A = 27 = 128
and number of subsets formed by {1, 2, 4, 5} = 16
$\therefore$ Required no. of subsets = 27 $-$ 24 = 128 $-$ 16 = 112
Let $A=\{1,2,3,4,5,6,7\}$. Define $B=\{T \subseteq A$ : either $1 \notin T$ or $2 \in T\}$ and $C=\{T \subseteq A: T$ the sum of all the elements of $T$ is a prime number $\}$. Then the number of elements in the set $B \cup C$ is ________________.
Explanation:
$\because$ $(B \cup C)' = B'\, \cap C'$
B' is a set containing sub sets of A containing element 1 and not containing 2.
And C' is a set containing subsets of A whose sum of elements is not prime.
So, we need to calculate number of subsets of {3, 4, 5, 6, 7} whose sum of elements plus 1 is composite.
Number of such 5 elements subset = 1
Number of such 4 elements subset = 3 (except selecting 3 or 7)
Number of such 3 elements subset = 6 (except selecting {3, 4, 5}, {3, 6, 7}, {4, 5, 7} or {5, 6, 7})
Number of such 2 elements subset = 7 (except selecting {3, 7}, {4, 6}, {5, 7})
Number of such 1 elements subset = 3 (except selecting {4} or {6})
Number of such 0 elements subset = 1
$n(B'\, \cap C') = 21 \Rightarrow n(B \cup C) = {2^7} - 21 = 107$
Let R1 and R2 be relations on the set {1, 2, ......., 50} such that
R1 = {(p, pn) : p is a prime and n $\ge$ 0 is an integer} and
R2 = {(p, pn) : p is a prime and n = 0 or 1}.
Then, the number of elements in R1 $-$ R2 is _______________.
Explanation:
and, set $A=\{1,2,3 \ldots \ldots .50\}$
$p$ is a Prime number which can take 15 values $2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and 47
$\therefore$ We can calculate no. of elements in $\mathrm{R}_1$
$ \mathrm{R}_1=\left(2,2^0\right),\left(2,2^1\right),\left(2,2^2\right)\left(2,2^3\right) \ldots \ldots\left(2,2^5\right)=6$ number of ordered pairs
$\left(3,3^0\right),\left(3,3^1\right),\left(3,3^2\right) \ldots \ldots . .\left(3,3^3\right)=4$ number of order paris
$\left(5,5^0\right),\left(5,5^1\right),\left(5,5^2\right) \ldots \ldots \ldots . .=3$ number of order paris
$\left(7,7^0\right) \ldots \ldots .\left(7,7^2\right) \ldots \ldots \ldots=3$ number of order paris
$\left(11,11^0\right)$ and $\left(11,11^1\right)=2$ number of order paris
$\left(13,13^0\right)$ and $\left(13,13^1\right)=2$ number of order paris
$ \therefore $ For the 11 prime numbers ($11,13,17,19,23,29,31,37,41,43$ and 47), $n$ can only be 0, 1 (two pairs each).
$ \therefore n\left(\mathrm{R}_1\right)=6+4+3+3+(2 \times 11)=38 $
$\mathrm{R}_2=\left(p, p^n\right) $, where n = 0 or 1
$ \left(2,2^0\right),\left(2,2^1\right)\left(3,3^0\right)\left(3,3^1\right) \ldots . .\left(47,47^0\right)\left(47,47^1\right) $
Two ordered pairs of each element $n\left({R}_2\right)=2 \times 15=30$ elements
Hence $ R_1-R_2=38-30=8$
Let A = {n $\in$ N : H.C.F. (n, 45) = 1} and
Let B = {2k : k $\in$ {1, 2, ......., 100}}. Then the sum of all the elements of A $\cap$ B is ____________.
Explanation:
Sum of all elements of A $\cap$ B = 2 [Sum of natural numbers upto 100 which are neither divisible by 3 nor by 5]
$ = 2\left[ {{{100 \times 101} \over 2} - 3\left( {{{33 \times 34} \over 2}} \right) - 5\left( {{{20 \times 21} \over 2}} \right) + 15\left( {{{6 \times 7} \over 2}} \right)} \right]$
$ = 10100 - 3366 - 2100 + 630$
$ = 5264$
Let $A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\min \,\{ i,j\} } } $ and $B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \,\{ i,j\} } } $. Then A + B is equal to _____________.
Explanation:
$\sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\{ i,\,j\} } } $
= {1, 1} {1, 2} {1, 3} ..... {1, 10}
{2, 1} {2, 2} {2, 3} ..... {2, 10}
{3, 1} {3, 2} {3, 3} ..... {3, 10}
$ \vdots $
{10, 1} {10, 2} {10, 3} ..... {10, 10}
Now, $A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {min\{ i,\,j\} } } $
= minimum between i and j in all sets and summation of all those values.
and $B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \{ i,\,j\} } } $
= maximum between i and j in all sets and summation of all those values.
For 1 :
1 is minimum in sets =
{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {1, 9}, {1, 10}, {2, 1}, {3, 1}, {4, 1}, {5, 1}, {6, 1}, {7, 1}, {8, 1}, {9, 1}, {10, 1}
$\therefore$ 1 is minimum in 19 sets
1 is maximum in {1, 1} sets.
$\therefore$ 1 is maximum and minimum in total 20 sets.
$\therefore$ Sum of 1 in all those sets = 1 $\times$ 20 = 20
For 2 :
2 is minimum in sets =
{2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {2, 9}, {2, 10}, {3, 2}, {4, 2}, {5, 2}, {6, 2}, {7, 2}, {8, 2}, {9, 2}, {10, 2}
$\therefore$ 2 is minimum in 17 sets
2 is maximum in sets = {1, 2}, {2, 1}, {2, 2}
$\therefore$ 2 is maximum and minimum in 20 sets.
$\therefore$ Sum of 2 in all those sets = 2 $\times$ 20 = 40
Similarly 3 is maximum and minimum in 20 sets.
$\therefore$ Sum of 3 in all those sets = 20 $\times$ 3 = 60
$ \vdots $
Similarly, 10 is maximum and minimum in 20 sets.
$\therefore$ Sum of 10 in all those sets = 20 $\times$ 10 = 200
$\therefore$ A + B = 20 + 20 $\times$ 2 + 20 $\times$ 3 + ....... + 20 $\times$ 10
= 20(1 + 2 + 3 + ...... + 10)
= 20 $\times$ ${{10 \times 11} \over 2}$
= 1100
The sum of all the elements of the set $\{ \alpha \in \{ 1,2,.....,100\} :HCF(\alpha ,24) = 1\} $ is __________.
Explanation:
The numbers upto 24 which gives g.c.d. with 24 equals to 1 are 1, 5, 7, 11, 13, 17, 19 and 23.
Sum of these numbers = 96
There are four such blocks and a number 97 is there upto 100.
$\therefore$ Complete sum
= 96 + (24 $\times$ 8 + 96) + (48 $\times$ 8 + 96) + (72 $\times$ 8 + 96) + 97
= 1633
If $A=\left\{x \in R / \sqrt{x^2-8 x+15} \in R\right\}$ and $B=\left\{x \in R / \frac{x-3}{2 x-5}<\frac{x-6}{2 x-11}\right\}$, then $A \cap B=$
$\phi$
$\left(\frac{5}{2}, 3\right] \cup\left[5, \frac{11}{2}\right)$
$\left(\frac{5}{2}, \frac{21}{4}\right)$
$\left(\frac{5}{2}, \frac{11}{2}\right)$
205 students take an examination of whom 105 pass in English, 70 students pass in Mathematics and 30 students pass in both. How many students fail in both subjects?
"ARB iff there exists a non-singular matrix P such that PAP$-$1 = B".
Then which of the following is true?
B = {x $\in$ R : $\sqrt {{x^2} - 3} $ > 1},
C = {x $\in$ R : |x $-$ 4| $\ge$ 2} and Z is the set of all integers, then the number of subsets of the
set (A $\cap$ B $\cap$ C)c $\cap$ Z is ________________.
Explanation:
B = ($-$$\infty$, $-$2) $\cup$ (2, $\infty$)
C = ($-$$\infty$, 2] $\cup$ [6, $\infty$)
So, A $\cap$ B $\cap$ C = ($-$$\infty$, $-$2) $\cup$ [6, $\infty$)
z $\cap$ (A $\cap$ B $\cap$ C)' = {$-$2, $-$1, 0, $-$1, 2, 3, 4, 5}
Hence, no. of its subsets = 28 = 256.
Explanation:
Now, n2 $-$ n $\le$ 100 $\times$ 100
$\Rightarrow$ n(n $-$ 1) $\le$ 100 $\times$ 100
$\Rightarrow$ A = {1, 2, ......., 100}.
So, A$\cap$(B $-$ C) = {7, 13, 19, ......., 97}
Hence, sum = ${{16} \over 2}(7 + 97) = 832$
B = {9k + 2: k $ \in $ N}
and C = {9k + $l$: k $ \in $ N} for some $l ( 0 < l < 9)$
If the sum of all the elements of the set A $ \cap $ (B $ \cup $ C) is 274 $ \times $ 400, then $l$ is equal to ________.
Explanation:
1. First, we consider the numbers of the form $9k + 2$ (Set $B$) within the 3-digit range, which starts at 101 and ends at 992.
2. We calculate the sum of these numbers, denoted as $s_1$. To calculate $s_1$, you use the formula for the sum of an arithmetic series :
$(n/2) \times (\text{{first term}} + \text{{last term}})$
Here, $n$ is the total count of such numbers. These are 3-digit numbers of the form $9k + 2$, and we can find the total count by subtracting the smallest such number (101) from the largest (992), dividing the result by 9 (because we're considering numbers with a difference of 9), and then adding 1.
The sum $s_1$ is calculated as follows :
$(100/2) \times (101 + 992) = 54650$
3. According to the problem, the sum of all elements of the set $A \cap (B \cup C)$ is $274 \times 400 = 109600$.
Since the set $A \cap (B \cup C)$ is the union of two disjoint sets (the set of all three-digit numbers of form $9k + 2$ and the set of all three-digit numbers of form $9k + l$), we can write this sum as :
$s_1 $(for numbers of the form 9k + 2) + $s_2$ (for numbers of the form 9k + l) = 109600
4. Solving this equation for $s_2$ (the sum of numbers of the form $9k + l$), we get :
$s_2 = 109600 - s_1 = 109600 - 54650 = 54950$
5. The sum $s_2$ can be expressed as $(n/2) \times (\text{{first term}} + \text{{last term}})$, where $n$ is the count of numbers of the form $9k + l$. The first term here is the smallest 3-digit number of this form, which is $99 + l$, and the last term is the largest such number, which is $990 + l$.
We equate this to $s_2$ to solve for $l$ :
$54950 = (100/2)[(99 + l) + (990 + l)]$
6. Simplifying this equation, we get :
$2l + 1089 = 1099$
Solving for $l$, we find :
$l = 5$
So, the correct answer is 5.
R1 = {(a, b) $ \in $ R2 : a2 + b2 $ \in $ Q} and
R2 = {(a, b) $ \in $ R2 : a2 + b2 $ \notin $ Q},
where Q is the set of all rational numbers. Then :
A = {m $ \in $ R : both the roots of
x2 – (m + 1)x + m + 4 = 0 are real}
and B = [–3, 5).
Which of the following is not true?