NCERT Exemplar Solutions Class 11 Maths Sets — Long Answer and Objective Type Q23–43

NCERT Exemplar Problems, Class 11 Mathematics — Sets, Long Answer and Objective Type questions 23 to 43. All 21 questions solved.

Question 23 is the distributive law; Questions 24 to 28 are the survey problems solved by the inclusion–exclusion formula $$n(A \cup B \cup C) = n(A) + n(B) + n(C) – n(A \cap B) – n(B \cap C) – n(A \cap C) + n(A \cap B \cap C).$$ The fifteen MCQs (29 to 43) mix counting subsets ($2^n$), simplifying a set expression with De Morgan’s laws, and recognising a set from its description.

Two answers in the printed key are wrong: Question 37 and Question 42. Question 42 is also misprinted, because the correct value is not among the four options. Both are explained where they occur.

Key insight. Never count a survey by adding the groups directly. Find the number in the union with inclusion–exclusion, then subtract it from the total for “neither”, and build the single-subject counts from the centre outwards: start with $A \cap B \cap C$, then each pairwise overlap minus the centre.

Question 23

Let $A$, $B$ and $C$ be sets. Show that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Solution. Prove the two inclusions.

Let $x \in A \cap (B \cup C)$. Then $x \in A$ and ($x \in B$ or $x \in C$). If $x \in B$, then $x \in A \cap B$; if $x \in C$, then $x \in A \cap C$. Either way $x \in (A \cap B) \cup (A \cap C)$.

Conversely, let $x \in (A \cap B) \cup (A \cap C)$. Then $x \in A \cap B$ or $x \in A \cap C$. In both cases $x \in A$, and $x$ lies in $B$ or in $C$, so $x \in B \cup C$. Hence $x \in A \cap (B \cup C)$.

The inclusions in both directions give equality.

Question 24

Out of $100$ students, $15$ passed in English, $12$ passed in Mathematics, $8$ in Science, $6$ in English and Mathematics, $7$ in Mathematics and Science, $4$ in English and Science, $4$ in all the three. Find how many passed (i) in English and Mathematics but not in Science (ii) in Mathematics and Science but not in English (iii) in Mathematics only (iv) in more than one subject only.

Solution. Let $E, M, S$ be the sets of students who passed in English, Mathematics and Science, with $n(E \cap M \cap S) = 4$. A pairwise overlap includes the centre, so “only that pair” is the overlap minus $4$.

(i) English and Mathematics but not Science: $6 – 4 = 2$.

(ii) Mathematics and Science but not English: $7 – 4 = 3$.

(iii) Mathematics only: $12 – 2 – 3 – 4 = 3$. (Subtract the two pair-only regions and the centre from $n(M)$.)

(iv) The third pair, English and Science but not Mathematics, is $4 – 4 = 0$. Students who passed in more than one subject, meaning two or three subjects, number $2 + 3 + 0 + 4 = 9$. This matches the printed key. If “only” were read as “exactly two subjects”, the count would be $2 + 3 + 0 = 5$; the answer below follows the key’s reading, at least two subjects.

Question 25

In a class of $60$ students, $25$ students play cricket and $20$ students play tennis, and $10$ students play both the games. Find the number of students who play neither.

Solution. Students playing at least one game: $n(C \cup T) = 25 + 20 – 10 = 35$. The rest play neither: $60 – 35 = 25$.

Question 26

In a survey of $200$ students of a school, it was found that $120$ study Mathematics, $90$ study Physics and $70$ study Chemistry, $40$ study Mathematics and Physics, $30$ study Physics and Chemistry, $50$ study Chemistry and Mathematics and $20$ none of these subjects. Find the number of students who study all the three subjects.

Solution. Students studying at least one subject: $200 – 20 = 180$. By inclusion–exclusion, with $x$ the number studying all three, $$180 = 120 + 90 + 70 – 40 – 30 – 50 + x = 160 + x,$$ so $x = 20$. A check on the Venn regions: Mathematics only $= 120 – 40 – 50 + 20 = 50$, Physics only $= 90 – 40 – 30 + 20 = 40$, Chemistry only $= 70 – 30 – 50 + 20 = 10$, pair-only regions $20, 10, 30$, centre $20$, giving $50 + 40 + 10 + 20 + 10 + 30 + 20 = 180$.

Question 27

In a town of $10{,}000$ families it was found that $40\%$ families buy newspaper $A$, $20\%$ families buy newspaper $B$, $10\%$ families buy newspaper $C$, $5\%$ families buy $A$ and $B$, $3\%$ buy $B$ and $C$ and $4\%$ buy $A$ and $C$. If $2\%$ families buy all the three newspapers, find (a) the number of families which buy newspaper $A$ only, (b) the number of families which buy none of $A$, $B$ and $C$.

Solution. Convert the percentages of $10{,}000$ families into numbers: $n(A) = 4000$, $n(B) = 2000$, $n(C) = 1000$, $n(A \cap B) = 500$, $n(B \cap C) = 300$, $n(A \cap C) = 400$, $n(A \cap B \cap C) = 200$.

(a) Newspaper $A$ only is $A$ with both overlaps removed, adding back the centre because it was removed twice: $4000 – 500 – 400 + 200 = 3300$.

(b) The union is $4000 + 2000 + 1000 – 500 – 300 – 400 + 200 = 6000$, so $10{,}000 – 6000 = 4000$ families buy none.

Question 28

In a group of $50$ students, the number of students studying French, English, Sanskrit were found to be as follows: French $= 17$, English $= 13$, Sanskrit $= 15$, French and English $= 09$, English and Sanskrit $= 4$, French and Sanskrit $= 5$, English, French and Sanskrit $= 3$. Find the number of students who study (i) French only (ii) English only (iii) Sanskrit only (iv) English and Sanskrit but not French (v) French and Sanskrit but not English (vi) French and English but not Sanskrit (vii) at least one of the three languages (viii) none of the three languages.

Solution. The centre is $3$. Each pair-only region is the pair overlap minus $3$: (iv) $4 – 3 = 1$, (v) $5 – 3 = 2$, (vi) $9 – 3 = 6$. Each single-language region is the total minus the two pair-only regions that touch it minus the centre:

(i) French only $= 17 – 6 – 2 – 3 = 6$. (ii) English only $= 13 – 6 – 1 – 3 = 3$. (iii) Sanskrit only $= 15 – 2 – 1 – 3 = 9$.

(vii) At least one language: $6 + 3 + 9 + 1 + 2 + 6 + 3 = 30$, which agrees with $17 + 13 + 15 – 9 – 4 – 5 + 3 = 30$.

(viii) None: $50 – 30 = 20$.

Choose the correct answers from the given four options in each of Exercises 29 to 43 (M.C.Q.).

Question 29

Suppose $A_1, A_2, \dots, A_{30}$ are thirty sets each having $5$ elements and $B_1, B_2, \dots, B_n$ are $n$ sets each with $3$ elements, let $\displaystyle\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^{n} B_j = S$ and each element of $S$ belongs to exactly $10$ of the $A_i$’s and exactly $9$ of the $B_j$’s. Then $n$ is equal to

Solution. Count the pairs (element of $S$, set containing it) in two ways. Using the $A_i$: there are $30 \times 5 = 150$ such pairs. Each element of $S$ is in exactly $10$ of the sets, so $10\,|S| = 150$ and $|S| = 15$. Using the $B_j$: there are $3n$ pairs, and each element lies in exactly $9$ of them, so $3n = 9 \times 15 = 135$ and $n = 45$.

Question 30

Two finite sets have $m$ and $n$ elements. The number of subsets of the first set is $112$ more than that of the second set. The values of $m$ and $n$ are, respectively,

Solution. A set with $k$ elements has $2^k$ subsets, so $2^m – 2^n = 112$. Factor out the smaller power: $2^n(2^{m-n} – 1) = 112 = 16 \times 7$. The factor $2^{m-n} – 1$ is odd, so it must be the odd part $7$, giving $m – n = 3$, and then $2^n = 16$ gives $n = 4$ and $m = 7$. Check: $128 – 16 = 112$.

Question 31

The set $(A \cap B’)’ \cup (B \cap C)$ is equal to

Solution. By De Morgan, $(A \cap B’)’ = A’ \cup B$. The expression becomes $A’ \cup B \cup (B \cap C)$. Since $B \cap C \subset B$, it is absorbed by $B$, leaving $A’ \cup B$.

Question 32

Let $F_1$ be the set of parallelograms, $F_2$ the set of rectangles, $F_3$ the set of rhombuses, $F_4$ the set of squares and $F_5$ the set of trapeziums in a plane. Then $F_1$ may be equal to

Solution. Rectangles, rhombuses and squares are all parallelograms, so $F_2, F_3, F_4 \subset F_1$. Then $F_2 \cup F_3 \cup F_4 \cup F_1 = F_1$ exactly, because the three smaller sets are absorbed. The other options cannot equal $F_1$: $F_2 \cap F_3$ and $F_3 \cap F_4$ are both the set of squares (a square is a rectangle and a rhombus), far smaller than all parallelograms. For $F_2 \cup F_5$, if a trapezium means exactly one pair of parallel sides then $F_5$ contains no parallelogram, so a rhombus that is not a square is missed; if a trapezium means at least one pair, then $F_5$ also contains non-parallelograms. Either way $F_2 \cup F_5 \ne F_1$.

Question 33

Let $S$ be the set of points inside the square, $T$ the set of points inside the triangle and $C$ the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square, then

Solution. The triangle and the circle both lie inside the square, so $T \subset S$ and $C \subset S$. Hence $S \cup T \cup C = S$, which is (C). The others fail: since the triangle and circle overlap, $S \cap T \cap C = T \cap C \ne \phi$ (A); the union is $S$, the largest set, not $C$ (B); and $S \cup T = S$ while $S \cap C = C$, which is smaller than $S$ (D).

Question 34

Let $R$ be the set of points inside a rectangle of sides $a$ and $b$ ($a, b > 1$) with two sides along the positive direction of the $x$-axis and $y$-axis. Then

Solution. The rectangle has corners $(0, 0)$, $(a, 0)$, $(a, b)$, $(0, b)$. Points strictly inside it do not lie on any of the four sides, so both coordinates satisfy strict inequalities, $0 < x < a$ and $0 < y < b$. Options (A), (B) and (C) each include some boundary points.

Question 35

In a class of $60$ students, $25$ students play cricket and $20$ students play tennis, and $10$ students play both the games. Then, the number of students who play neither is

Solution. At least one game: $25 + 20 – 10 = 35$. Neither: $60 – 35 = 25$. The tempting wrong choice is (C) $35$, which is the number who play at least one game.

Question 36

In a town of $840$ persons, $450$ persons read Hindi, $300$ read English and $200$ read both. Then the number of persons who read neither is

Solution. Persons reading at least one language: $450 + 300 – 200 = 550$. Neither: $840 – 550 = 290$.

Question 37

If $X = \{8^n – 7n – 1 \mid n \in \mathbf{N}\}$ and $Y = \{49n – 49 \mid n \in \mathbf{N}\}$. Then

Solution. Expand $8^n = (1 + 7)^n = 1 + 7n + \binom n2 7^2 + \binom n3 7^3 + \dots$, so $$8^n – 7n – 1 = 49\left[\binom n2 + \binom n3 \cdot 7 + \dots\right],$$ which is a multiple of $49$. Every element of $X$ is therefore a multiple of $49$, and since $Y = \{0, 49, 98, 147, \dots\}$ contains every non-negative multiple of $49$, we get $X \subset Y$. The sets are not equal: $98 \in Y$ (take $n = 3$ in $49n – 49$), but $X = \{0, 49, 490, \dots\}$ jumps from $49$ (for $n = 2$) to $490$ (for $n = 3$) and never takes the value $98$. So $Y \not\subset X$, $X \ne Y$, and they certainly have common elements ($0$ and $49$), ruling out (D).

The printed key says (C) $X = Y$. That is wrong: $X$ misses $98$, which is in $Y$. The correct answer is (A).

Question 38

A survey shows that $63\%$ of the people watch a News Channel whereas $76\%$ watch another channel. If $x\%$ of the people watch both channels, then

Solution. The overlap cannot exceed the smaller group, so $x \le 63$. The union cannot exceed $100\%$: $63 + 76 – x \le 100$, so $x \ge 39$. Hence $39 \le x \le 63$, and only that range is forced; $x$ is not a single value.

Question 39

If sets $A$ and $B$ are defined as $A = \left\{(x, y) \mid y = \dfrac1x,\ 0 \ne x \in \mathbf{R}\right\}$ and $B = \{(x, y) \mid y = -x,\ x \in \mathbf{R}\}$, then

Solution. A common point would need $\frac1x = -x$, that is $x^2 = -1$, which has no real solution. So the hyperbola and the line do not meet, and $A \cap B = \phi$.

Question 40

If $A$ and $B$ are two sets, then $A \cap (A \cup B)$ equals

Solution. $A \subset A \cup B$, so the intersection of $A$ with the larger set $A \cup B$ is $A$ itself (the absorption law).

Question 41

If $A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\}$, $B = \{2, 4, \dots, 18\}$ and $\mathbf{N}$, the set of natural numbers, is the universal set, then $A’ \cup ((A \cup B) \cap B’)$ is

Solution. Simplify $(A \cup B) \cap B’ = (A \cap B’) \cup (B \cap B’) = A \cap B’$. The odd numbers in $A$ are not in $B$, so $A \cap B’ = A$. Then the whole set is $A’ \cup A = \mathbf{N}$.

Question 42

Let $S = \{x \mid x$ is a positive multiple of $3$ less than $100\}$ and $P = \{x \mid x$ is a prime number less than $20\}$. Then $n(S) + n(P)$ is

(A) $34$ (B) $31$ (C) $33$ (D) $30$

Solution. None of the four options is correct, and the key’s (B) is wrong. The positive multiples of $3$ below $100$ are $3, 6, \dots, 99$, that is $3 \times 1, \dots, 3 \times 33$, so $n(S) = 33$. The primes below $20$ are $2, 3, 5, 7, 11, 13, 17, 19$, so $n(P) = 8$. Then

$$n(S) + n(P) = 33 + 8 = 41.$$

The key’s (B) $31$ would need $n(S) = 23$, which is the count of multiples of $3$ below $70$, not below $100$; with the sets as printed it cannot be reached. The value $41$ is not among the options, so none is marked above. Forgetting the primes gives $33$, which is why (C) looks tempting.

Question 43

If $X$ and $Y$ are two sets and $X’$ denotes the complement of $X$, then $X \cap (X \cup Y)’$ is equal to

Solution. By De Morgan, $(X \cup Y)’ = X’ \cap Y’$. Then $X \cap X’ \cap Y’ = \phi \cap Y’ = \phi$, since $X \cap X’ = \phi$.

Common mistakes

• Question 24(iii) and 28(i) to (iii). Subtracting only the pairwise overlaps from a total, instead of the pair-only regions plus the centre. Pairwise overlaps already contain the centre, so always peel off “pair minus centre” first.
• Question 24(iv). The phrase “more than one subject only” is ambiguous; the key uses at least two subjects ($9$). Answering $5$ is correct only if “only” means exactly two.
• Question 26. Putting the $20$ students who study none into the union, or forgetting to remove them from $200$ first.
• Question 30. Setting $m – n = 112$ instead of $2^m – 2^n = 112$ (confusing the number of elements with the number of subsets).
• Question 37. Believing $X = Y$ because both sets are multiples of $49$. The set $Y$ contains every multiple of $49$; $X$ only some of them, so the inclusion is one way.
• Question 42. Counting only the multiples of $3$ ($33$, option (C)) and forgetting that $n(P) = 8$ must be added, or dropping the prime $2$; then picking the nearest printed option instead of noticing that $41$ is not offered.

Practise next

• The other pages of this unit: Short Answer Type, Q1–22 and Fill in the Blanks and True or False, Q44–58.
• NCERT textbook practice on the same ideas: Exercise 1.5 (complements and the survey-style problems), Exercise 1.4 and the Miscellaneous Exercise.