MCQ 1011 Mark
Choose the correct answers from the given four option: Suppose $A_1, A_2, ..., A_{30}$ are thirty sets each having $5$ elements and $B_1, B_2, ..., Bn$ are $n$ sets each with $3$ elements, let $\bigcup\limits_{\text{i}=1}^{30}\text{A}_\text{i}=\bigcup\limits_{\text{j}=1}^\text{n}\text{B}_\text{j}=\text{S}$ and each element of $S$ belongs to exactly $10$ of the $A_i$ ’s and exactly $9$ of the $B, 'S.$ then $n$ is equal to.
AnswerNumber of elements in $\text{A}_1\cup\text{A}_2\cup\text{A}_3\ ..... \cup \text{A}_{30}=30\times5=150$ $($When repetition is not allowed$)$
But each element is repeated $10$ times
$\therefore \text{n(S)}=\frac{30\times5}{10}=\frac{150}{10}=15\ .....\text{(i)}$
Number of elements in $\text{B}_1\cup\text{B}_2\cup\text{B}_3\ ...... \text{B}_\text{n}=3\text{n} ($when repetitiom is not allowed$)$
But each element is repeated $09$ times
$\therefore \text{n(S)}=\frac{3\text{n}}{9}=\frac{\text{n}}{3}\ .....\text{(ii)}$
From $(i)$ and $(ii)$ we get
$\frac{\text{n}}{3}=15$
$\Rightarrow \text{n}=15\times3=45$
Hence, the corrrect option is $(c).$
View full question & answer→MCQ 1021 Mark
Which of the following sets are null sets.
- A
$(\text{x:}\mid\text{x }\mid<-4,\text{x}?\text{ N})$
- ✓
$2$ and $3$
- C
Set of all prime numbers between $15$ and $19$
- D
$(\text{x : }\text{x}<5,\text{x}>6)$
AnswerCorrect option: B. $2$ and $3$
$2$ and $3$ is the null set.
View full question & answer→MCQ 1031 Mark
Which of the following is correct for $A - B?$
- A
$\text{A } \cap \text{ B}$
- B
$\text{A }' \cap \text{ B}$
- ✓
$\text{A } \cap \text{ B}'$
- D
$\text{A }' \cap \text{ B}'$
AnswerCorrect option: C. $\text{A } \cap \text{ B}'$
View full question & answer→MCQ 1041 Mark
The set of intelligent students in a class is:
- A
- B
- C
- ✓
Not a well defined collection
AnswerCorrect option: D. Not a well defined collection
View full question & answer→MCQ 1051 Mark
Choose the correct answers from the given four option:
If $A$ and $B$ are two sets, then $\text{A} \cap (\text{A} \cup \text{B})$ equals.
- ✓
$\text{A}$
- B
$\text{B}$
- C
$\phi$
- D
$\text{A}\cap\text{B}$
AnswerCorrect option: A. $\text{A}$
Given that: $\text{A}\cap(\text{A}\cup\text{B})$
Let $\text{x}\in\text{A}\cap(\text{A}\cup\text{B})$
$\Rightarrow \text{x}\in\text{A}$ and $\text{x}\in(\text{A}\cup\text{B})$
$\Rightarrow \text{x}\in\text{A}$ and $(\text{x}\in\text{A}$ or $\text{x}\in\text{B})$
$\Rightarrow (\text{x}\in\text{A}$ and $\text{x}\in\text{A})$ or $(\text{x}\in\text{A and x}\in\text{B})$
$\Rightarrow \text{x}\in\text{A}$ or $\text{x}\in(\text{A}\cap\text{B})$
$\Rightarrow \text{x}\in\text{A}$
Hence, the correct option is $(a).$
View full question & answer→MCQ 1061 Mark
If $A = (6, 7, 8, 9), B = (4, 6, 8, 10)$ and $C = \{x : x \in N : 2 < x ≤ 7\}$ ; find: $A - B$
- A
$\{6, 8\}$
- ✓
$\{7, 9\}$
- C
$\{6, 9\}$
- D
$\{6, 7, 9, 10\}$
AnswerCorrect option: B. $\{7, 9\}$
$A − B = A − A ∩ B = \{7, 9\}$
View full question & answer→MCQ 1071 Mark
Let $A$ and $B$ have $3$ and $6$ elements respectively. What can be the minimum number of elements in $\ce{A ∪ B}?$
Answer$\ce{n(A ∪ B) = n(A) + n(B) − n(A ∩ B)}$
Now $A$ has $3$ elements and $B$ has $6 $elements.
If they are disjoint, then $\ce{n(A ∩ B)} = 0.$
$\therefore \ce{n(A ∪ B)} = 6 + 3 = 9$
If $\ce{A ⊂ B}$ then $\ce{A ∪ B = B}$
$\therefore (A ∪ B) = n(B) = 6$
$B$ cannot be a subset of $A$ and hence the other possibility of $\ce{A ∪ B = A}$ is ruled out.
View full question & answer→MCQ 1081 Mark
Let $A = \{1, 2, 3, 4, 5, 6\}$.How many subsets of A can be formed with just two elements, one even and one odd?
View full question & answer→MCQ 1091 Mark
If $A$ and $B$ are two non$-$empty sets, then $\text{B}\cap({\text{A }} \cup \text { B})\text{c},$ where $Xc$ denotes the complement of $X$, is equal to:
AnswerCorrect option: D. $\text{f}$
View full question & answer→MCQ 1101 Mark
If $A = \{2, 3, 4, 5, 7\}, B = \{7, 8, 9\}$, then find $\ce{n(A ∪ B).}$
Answer$A = \{2, 3, 4, 5, 7\}$
$n(A) = 5$
$B =\{7, 8, 9\}$
$n(B) = 3$
$n(A ∩ B) = 1$
$\therefore \ce{(A ∪ B) = n(A) + n(B) − n(A ∩ B)}$
$\ce{(A ∪ B)} = 5 + 3 − 1 = 7$
View full question & answer→MCQ 1111 Mark
If $A = \{1, 2, 3, 4\},$ then the number of subsets of $A$ that contain the element $2$ but not $3$, is:
AnswerThe subsets are be $\{1, 2, 4\}, \{1, 2\}, \{2, 4\}, \{2\}$
Number of subsets of $A$ that contain the element $2$ but not $3$ is $4.$
View full question & answer→MCQ 1121 Mark
Let $P$ be a set of squares, $Q$ be set of parallelograms, $R$ be a set of quadrilaterals and $S$ be a set of rectangles. Consider the following:
$1. \ce{P Ì Q}$
$2. \ce{R Ì P}$
$3. \ce{P Ì S}$
$4. \ce{S Ì R}$
Which of the above are correct?
- A
$1, 2$ and $3$
- ✓
$1, 3$ and $4$
- C
$1, 2$ and $4$
- D
$3$ and $4$
AnswerCorrect option: B. $1, 3$ and $4$
View full question & answer→MCQ 1131 Mark
In a science talent examination, $50\%$ of the candidates fail in Mathematics and $50\%$ fail in Physics. If $20\%$ fail in both these subjects, then the percentage who pass in both Mathematics and Physics is:
- A
$0\%$
- ✓
$20\%$
- C
$25\%$
- D
$50\%$
AnswerCorrect option: B. $20\%$
By set theory
$\ce{n(M ∪ P) = n(M) + n(P) − n(M ∩ P)}$ where $M$ and $P$ are sets of students failing in respective subjects.
$= 0.5 + 0.5 − 0.2$
$= 0.8$
This indicates $80\%$ of the class fails in at least one of the given subjects while $20\%$ pass in both.
View full question & answer→MCQ 1141 Mark
How many rational and irrational numbers are possible between $0$ and $1$?
AnswerThere are infinite many rational and irrational numbers are possible between $0$ and $1$
This is because between any two numbers, there are infinite numbers.
View full question & answer→MCQ 1151 Mark
In an examination $80\%$ passed in English, $85\%$ in Maths, $75\%$ in both and $40$ students failed in both subjects. Then the number of students appeared are
Answer$\ce{n(E)} = 80$
$\ce{n(M)} = 85$
$\ce{n(E ∩ M)} = 75$
$\ce{n(E ∪ M) = n(E) + n(M) − n(E ∩ M)} = 80 + 85 − 75 = 90$
$\ce{n(E∪M)′} = 10$
Let $n$ be the total number of students appeared
$\frac{100}{100}\times\text{n}=40$
$\therefore n = 400$
View full question & answer→MCQ 1161 Mark
Two finite sets have $N$ and $M$ elements. The number of elements in the power set of first set is $48$ more than the total number of elements in power set of the second test. Then the value of $M$ and $N$ are.
- A
$7, 6$
- ✓
$6, 4$
- C
$7, 4$
- D
$6, 3$
AnswerCorrect option: B. $6, 4$
Let $A$ and $B$ be two sets having $m$ and $n$ numbers of elements respectively
Number of subsets of $A = 2m$
Number of subsets of $B = 2n$
Now, according to question
$2m - 2n = 48$
$\Rightarrow 2n(2m - n - 1) = 24(22 - 1)$
So, $n = 4$ and $m - n = 2$
$\Rightarrow m – 4 = 2$
$\Rightarrow m = 2 + 4$
$\Rightarrow m = 6$
View full question & answer→MCQ 1171 Mark
Let $A$ and $B$ be two sets in the same universal set. Then, $A - B =$
- A
$\text{A}\cap\text{B}$
- B
$\text{A}'\cap\text{B}$
- ✓
$\text{A}\cap\text{B}'$
- D
AnswerCorrect option: C. $\text{A}\cap\text{B}'$
$A - B$ belongs to those elements of $A$ that do not belong to $B.$
$\therefore\text{A} - \text{B = A}\cap\text{B}'.$
View full question & answer→MCQ 1181 Mark
If $A$ and $B$ are non empty sets and $A'$ and $B'$ represents their compliments respectively then:
- A
$\ce{A − B = A′ − B′}$
- B
$\ce{A − A ′ = B − B′}$
- ✓
$\ce{A − B = B′ − A′}$
- D
$\ce{A − B′ = A′ − B}$
AnswerCorrect option: C. $\ce{A − B = B′ − A′}$
Let $U \rightarrow$ Universal set
$X \rightarrow \ce{U - (A + B)}$
$\ce{B′ = X + A}$
$\ce{A′ = X + B}$
$\ce{B′ - A′ = X + A - (X + B)}$
$= \ce{X + A - X - B}$
$\ce{B′ - A′ = A - B}$
View full question & answer→MCQ 1191 Mark
If $\text{A}\cap\text{B}=\text{B},$ then:
AnswerCorrect option: B. $\text{B}\subset\text{A}$
View full question & answer→MCQ 1201 Mark
In a group of $15,7$ have studied German, $8$ have studied French, and $3$ have not studied either. How many of these have studied both German and French?
AnswerSince $3$ have neither studied German nor French
$\ce{n(G ∪ F)} = 15 − 3 = 12$
By set theory
$\ce{n(G ∩ F) = n(G) + n(F) − n(G ∪ F)}$
$= 7 + 8 − 12$
$= 3$
View full question & answer→MCQ 1211 Mark
In an examination, $34\%$ of the candidates fail in Arithmetic and $42\%$ in English. If $20\%$ fail in Arithmetic and English, the percentage of those passing in both subjects is:
AnswerLet $A$ denote students fail in Arithmetic, $B$ denote students fail in English
$\ce{n(A)} = 34$
$\ce{n(B)} = 42$
$\ce{n(A ∩ B)} = 20$
$\ce{n(A ∪ B) = n(A) + n(B) − n(A ∩ B)} = 34 + 42 − 20 = 56$
$\ce{n(A ∪ B)′ = 100 − n(A ∪ B)} = 100 − 56 = 44$
View full question & answer→MCQ 1221 Mark
If $X$ and $Y$ are any two non empty sets then what is $(X−Y)′$ equal to?
- A
$X′−Y′$
- B
$X′∩Y$
- ✓
$X′∪Y$
- D
$X−Y′$
AnswerCorrect option: C. $X′∪Y$
$\ce{X - Y = \{x : x \in X, x \in Y\}}$
$=\ce{ \{x : x \in X, x \in Y′\}}$
$\Rightarrow \ce{\{x : x \in X ∩ Y′\}}$
$\Rightarrow \ce{(X − Y)′ = (X ∩ Y)′}$
$= \ce{X′ ∪(Y′) = X′ ∪ Y}$
View full question & answer→MCQ 1231 Mark
Let $V = \{a, e, i, o, u\}$ and $B = \{a, i, k, u\}.$ Value of $\ce{V - B}$ and $\ce{B - V}$ are respectively.
AnswerCorrect option: A. $\{e, 0\}$ and $\{k\}$
View full question & answer→MCQ 1241 Mark
AnswerIn mathematics, and more specifically set theory, the empty set is the unique set having no elements and its size or cardinality $($count of elements in a set$)$ is zero.
So, an empty set is a finite set.
View full question & answer→MCQ 1251 Mark
In a school with an envolment of $950$ students, each student must join either the lions club or the country club or both. Given that $646$ students are members of the lions club and $532$ are members of the country club, calculate the number of students who are members of both clubs:
AnswerTotal number of students $= 950$
Number of students who are members of lion club $= 646$
Number of students who are members of country club $= 532$
Number of students who are members of both club $= (646 + 532) - 950 = 1178 - 950 = 228$
$\therefore$ There are $228$ students who joined both clubs.
View full question & answer→MCQ 1261 Mark
If $I$ is the set of isosceles triangle and $E$ is the equilateral triangles then $............$
AnswerCorrect option: B. $\text{E}\subset\text{I}$
Given, $I$ is the set of isosceles triangle and $E$ is the equilateral triangles.
We know that every equilateral triangle is an isosceles triangle but the converse is not true.
View full question & answer→MCQ 1271 Mark
In a group, if $60\%$ of people drink tea and $70\%$ drink coffee.What is the maximum possible percentage of people drinking either tea or coffee but not both?
- A
$100\%$
- ✓
$70\%$
- C
$30\%$
- D
$10\%$
AnswerCorrect option: B. $70\%$
To find maximum possible percentage of people drinking either coffee or tea, we can assume everyone drinks at least either of the options.
Hence
$a + b = 100$
$a + 2b = 60 + 70 = 130$
$b = 30$
View full question & answer→MCQ 1281 Mark
If $X$ and $Y$ are two sets such that $\ce{n(X) = 45, n(X ∪ Y) = 76, n(X ∩ Y)} = 12,$ find $\ce{n(Y):}$
Answer$\ce{n(X ∪ Y) = n(X) + n(Y) -n(X ∩ Y)}$
$76 = 45 + \ce{n(Y)} - 12$
$\ce{n(Y)} = 43$
View full question & answer→MCQ 1291 Mark
Which of the following statements is false:
- A
$\text{A} - \text{B = A}\cap\text{B}'$
- B
$\text{A} - \text{B = A} - \text{(A}\cap\text{B)}$
- ✓
$\text{A} - \text{B = A}-\text{B}'$
- D
$\text{A} - \text{B = (A}\cup\text{B)}-\text{B.}$
AnswerCorrect option: C. $\text{A} - \text{B = A}-\text{B}'$
It includes all those elements of $A$ which do not belongs to complement of $B$ which is equal to $\text{A}\cap\text{B}$ but not equal to $A - B.$
$\therefore (c)$ is false.
View full question & answer→MCQ 1301 Mark
If $A = \{2, 4, 6 ,8\}$ and $B = \{1, 4, 7, 8\}$ then $\ce{A - B}$ and $\ce{B - A}$ will be respectively:
- ✓
$\{2, 6\} ; \{1, 7\}$
- B
$\{1, 7\} ; \{4, 8\}$
- C
$\{1, 7\} ; \{2, 6\}$
- D
$\{4, 8\} ; \{1, 7\}$
AnswerCorrect option: A. $\{2, 6\} ; \{1, 7\}$
$A = \{2, 4, 6, 8\}$ and $B = \{1, 4, 7, 8\}$
$A − B = \{2, 6\}$ and $B − A = \{1, 7\}$
View full question & answer→MCQ 1311 Mark
Let $A$ and $B$ be two sets such that $A ∩ B = ϕ.$ Find the value of $(A ∪ B′) =$
AnswerGiven, $\ce{A ∩ B} = ϕ.$
Now,
$\ce{(A ∪ B′)}$
$= \ce{(A′ ∩ B)′} [$Using De Morgan's law$]$
$= \ce{B′}.[$ As $\ce{A′ ∩ B = B − (A ∩ B) = B}$ since $\ce{A ∩ B} = ϕ]$
View full question & answer→MCQ 1321 Mark
The number of elements of the set $\{x : x \in Z,x^2= 1\}$ is:
Answer$x^2 = 1 \Rightarrow x = 1, -1$
Since both solutions are integers the set has $2$ elements.
View full question & answer→MCQ 1331 Mark
If $A, B$ and $C$ are any three sets, then $\text{A}-(\text{B }\cup\text{ C})$ is equal to.
- A
$(\text{A - B ) }\cup\ (\text{A - C})$
- B
$(\text{A - B ) }\cup\ \text{C}$
- C
$(\text{A - B ) }\cap\ \text{C}$
- ✓
$(\text{A - B ) }\cap\ (\text{A - C ) }$
AnswerCorrect option: D. $(\text{A - B ) }\cap\ (\text{A - C ) }$
Given $A, B$ and $C$ are any three sets.
Now $\text{A }-(\text{B }\cup\text{ C})=(\text{A - B ) }\cap\ (\text{A - C ) }$
View full question & answer→MCQ 1341 Mark
In a flight $50$ people speak Hindi, $20$ speak English and $10$ speak both English and Hindi. The number of people who speak atleast one of the two languages is:
AnswerLet $H =$ People who speak Hindi
$E =$ People who speak English
According to the questions,
$\ce{n(H) = 50, n(E) = 20, n(H ∩ E) = 10}$
Number of people who speak at least two language $= \ce{n(H ∪ E)}$
$= \ce{n(H) + n(E) − n(H ∩ E)}$
$= 50 + 20 − 10 = 60$
View full question & answer→MCQ 1351 Mark
Let $A$ and $B$ be two sets that $\text{n(A)} = 16, \text{ n(B)} = 14,\text{ n(A}\cup\text{B)}=25.$ Then, $\text{n(A}\cap\text{B)}$ is equal to:
AnswerWe know:
$\text{n(A}\cup\text{B) = n(A) + n(B)} - \text{n(A}\cap\text{B)}$
Now,
$\text{n(A}\cap\text{B) = n(A) + n(B)} -\text{n(A}\cup\text{B)}$
$=16+14-25$
$=5.$
View full question & answer→MCQ 1361 Mark
$A$ is set haveing $6$ distinct elements. The number of distinct functions from $A$ to $A$ which are not objections is:
- ✓
$6! − 6$
- B
$6^6 − 6$
- C
$6^6 − 6!$
- D
$6!$
AnswerCorrect option: A. $6! − 6$
Since set $A$ has $6$ distinct elements.
Total number of function from $A$ to $A = 6!$
Number of objective function from $A$ to $A$ is $6$
Therefore the number of function ehich are not objective $= 6! − 6.$
View full question & answer→MCQ 1371 Mark
If $A = [5, 6, 7]$ and $B = [7, 8, 9]$ then $\text{A } \cup \text{ B}$ is equal to.
- A
$[5, 6, 7]$
- ✓
$[5, 6, 7, 8, 9]$
- C
$[7, 8, 9]$
- D
AnswerCorrect option: B. $[5, 6, 7, 8, 9]$
Given $A = [5, 6, 7]$ and $B = [7, 8, 9]$
then $\text{A } \cup \text{ B}=[5, 6, 7, 8, 9]$
View full question & answer→MCQ 1381 Mark
Out of $500$ first year students, $260$ passed in the first semester and $210$ passed in the second semester.If $170$ did not pass in either semester, how many passed in both semesters?
AnswerLet $A$ be the set of students who passed first semester so $n(A) = 260$
and $B$ be the set of students who passed second semester so $n(B) = 210.$
Now $170$ did not passed any semester.
So, $(500 − 170 = 330)$ students passed atleast one of the semesters.
$\therefore n(A∪B) = 330$
Now $\ce{n(A ∪ B) = n(A) + n(B) − n(A ∩ B)}$
$330 = 260 + 210 − \ce{n(A ∩ B)}$
$\ce{n(A ∩ B)} = 140$
View full question & answer→MCQ 1391 Mark
$A = \{1, 3, 5\}, B = \{2, 4, 6\}$ and $C = \{0, 2, 4, 6, 8\}.$ Which of the following may be considered as universal set for all the three sets $A, B$ and $C?$
- A
$\{0, 1, 2, 3, 4, 5, 6\}$
- B
$f$
- ✓
$\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
- D
$\{1, 2, 3, 4, 5, 6, 7, 8\}$
AnswerCorrect option: C. $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
View full question & answer→MCQ 1401 Mark
In a group of $15$ women, $7$ have nose studs, $8$ have ear rings and $3$ have neither. How many of these have both nose studs and ear rings?
AnswerSince $3$ women have neither nose studs nor earrings
$\ce{n(N∪E)} = 15 − 3 = 12$
By set theory
$\ce{n(N∩E) = n(N) + n(E) − n(N∪E)}$
$= 7 + 8 − 12$
$= 3$
View full question & answer→MCQ 1411 Mark
The number of binary operations on the set $\{1, 2, 3\}$ is $..........?$
AnswerLet us denote this set by $S$, then $\ce{∣S∣} = 3.$
A binary relation defined on the elements of $S$ maps all elements in $S \times S$ to elements in $S$ by definition.
In this case any binary relation will thus have $32 = 9$ inputs each of which is an ordered pair of elements from $S$ and only $3$ number of possible outputs.
If all possible binary operations are considered then it is possible to assign any of the $3$ outputs to any of the $9$ inputs.
So the number of all binary operations would exactly be $3^9$.
View full question & answer→MCQ 1421 Mark
Choose the correct answers from the given four option:
Let $S = \{x | x$ is a positive multiple of $3$ less than $100\} P = \{x | x$ is a prime number less than $20\}.$ Then $\ce{n(S) + n(P)}$ is.
AnswerGiven that: $S = \{x | x$ is a positive multiple of $3 < 100\}$
$\therefore S = \{3, 6, 9, 12, 15 18, ....., 99\}$
$n(S) = 33$
$T = (x | x$ is a prime number $< 20)$
$\therefore T = \{2, 3, 5, 7, 11, 13, 17, 19\}$
$n(T) = 8$
So, $n(S) + n(T) = 33 + 8 = 41$
Hence, the correct option is $(b).$
View full question & answer→MCQ 1431 Mark
The number of subsets of the set $\{10, 11, 12\}$ is:
AnswerNo. of subsets $= 2^3 = 8.$
We have $2$ choices with each of the elements: either put them in a subset or to not put them in a subset. Hence there are $8$ subsets.
View full question & answer→MCQ 1441 Mark
Sets $A$ and $B$ have $3$ and $6$ elements respectively. What can be the minimum number of elements in $\ce{A ∪ B} ?$
AnswerLet $A$ be the left circle and $B$ be the right circle.There are $3$ elements in $A$ and $6$ elements in $B.$
The union of $A$ and $B$ contains elements that are in either circle.
Thus,the union of $A$ and $B$ will be all of the elements in $A$ along with all of the element $B.$
However, you have to subtract the elements that are in the overlapping area because you are counting twice.
If $A$ and $B$ don't overlap at all,then the union will ontain $9$ elements.If $A$ is completely inside $B$ then the union will only contain $6$ elements,which is the minimum no. of elements in the union of $A$ and $B.$
let $A = 1, 2, B = 2, 3$
$\therefore A ∪ B = 1, 2, 3$ which is $3$ elements.
$\therefore A$ has $2$ elements, $B$ has $2$ elements, and there is $1$ element overlapping.
$\therefore 2 + 2 − 1 = 3! = 3 \times 2 \times 1 = 6$
View full question & answer→MCQ 1451 Mark
From among the given alternatives select the one in which the set of numbers is most like the set of numbers given in the question.
Given set : $(7, 15, 31):$
- A
$7, 13, 28$
- B
$5, 13, 28$
- C
$9, 13, 26$
- ✓
$5, 13, 29$
AnswerCorrect option: D. $5, 13, 29$
Let us find the Relation between the numbers of the set $(7, 15, 31).$
$(7) \times 2 + 1 = (15)$
$(15) \times 2 + 1 = (31)$
Options $A,B,C$ are not of similar type of above set
This similar type of relation is shown by option $D (7, 15, 31).$
$(5) \times 2 + 3 = (13)$
$(13) \times 2 + 3 = (29)$
View full question & answer→MCQ 1461 Mark
If $S$ and $T$ are two sets such that $S$ has $21$ elements, $T$ has $32$ elements and $\ce{S ∩ T}$ has $11$ elements, then find the number of elements in $\ce{S ∪ T.}$
AnswerGiven $\ce{n(S) = 21, n(T) = 32, n(S ∩ T)} = 11$
Now $\ce{n(S) + n(T) = n(S ∩ T) + n(S ∪ T)}$
$\Rightarrow \ce{n(S ∪ T)} = 21 + 32 - 11 = 42.$
View full question & answer→MCQ 1471 Mark
If $A = \{1, 2, 3\}, B = \{4, 5\},$ then find $A - B.$
- A
$\{1, 4, 5\}$
- B
$\{1, 4, 3\}$
- ✓
$\{1, 2, 3\}$
- D
$\{4, 5\}$
AnswerCorrect option: C. $\{1, 2, 3\}$
Given, $A = {1, 2, 3}$ and $B = {4, 5}$.
Since $A$ and $B$ are two disjoint sets
i.e. $\ce{A ∩ B} = ϕ$ then we've,
$\ce{A − B}$
$= \{1, 2, 3\}.$
View full question & answer→MCQ 1481 Mark
If $A$ and $B$ are two sets, then $\text{A} \cap (\text{A} \cup \text{B})'$ equals.
View full question & answer→MCQ 1491 Mark
The set $\{x : x$ is an even prime number$\}$ can be written as.
- ✓
$\{2\}$
- B
$\{2, 4\}$
- C
$\{2, 14\}$
- D
$\{2, 4, 14\}$
AnswerCorrect option: A. $\{2\}$
View full question & answer→MCQ 1501 Mark
In a class of $120$ students numbered $1$ to $120$, all even numbered students opt for Physics, whose numbers are divisible by $5$ opt for Chemistry and those whose numbers are divisible by $7$ opt for Math. How many opt for none of the three subjects?
AnswerThe number of students who took at least one of the three subjects can be found by finding out $\text{A }\cup \text{ B}\cup \text{ C},$
where $A$ is the set of those who took Physics, $B$ the set of those who took Chemistry and $C$ the set of those who opted for Math.
Now $\text{A }\cup \text{ B}\cup \text{ C}=\text{A + B + C}-(\text{A }\cap \text{ B}+\text{B } \cap \text{ C}+\text{C } \cap \text{ A})(\text{A } \cap\text{ B } \cap \text{ C})$
$A$ is the set of those who opted for Physics $=\frac{120}{2}=60\text{ Students}$
$B$ is the set of those who opted for Chemistry $=\frac{120}{5}=24$
$C$ is the set of those who opted for Math $=\frac{120}{7}=17$
The $10^{th}, 20^{th}, 30^{th}…..$ numbered students would have opted for both Physics and Chemistry.
Therefore $\text{A }\cap \text{ B}=\frac{120}{10}=12$
The $14^{th}, 28^{th}, 42^{nd}……$ Numbered students would have opted for Physics and Math.
Therefore, $\text{C }\cap \text{ A}=\frac{120}{14}=8$
The $35^{th}, 70^{th}….$ numbered students would have opted for Chemistry and Math.
Therefore, $\text{B }\cap \text{ C}=\frac{120}{35}=3$
And the $70^{th}$ numbered student would have opted for all three subjects.
Therefore, $\text{A }\cup \text{ B }\cup \text{ C}= 60 + 24 + 17 - (12 + 8 + 3) + 1 = 79$
Number of students who opted for none of the three subjects $= 120 – 79 = 41$
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