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NCERT Exemplar · Class 11 Mathematics Sets

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EXERCISE 1.3 21–30 (part 3 of 6)

  1. Using properties of sets prove the statements given in Exercises $\displaystyle 18$ to $\displaystyle 22$

    Exercise 21

    For all sets A and B,(A∪B)−B=A−B\displaystyle \mathrm{B},(\mathrm{A} \cup \mathrm{B})-\mathrm{B}=\mathrm{A}-\mathrm{B}

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    \[(A\cup B)-B=(A\cup B)\cap B' \quad \text{(} X-Y=X\cap Y' \text{)} \] \[=(A\cap B')\cup(B\cap B') \quad \text{(distributive law)} \] \[=(A\cap B')\cup\varnothing \quad \text{(} B\cap B'=\varnothing \text{)} \] \[=A\cap B'=A-B \] Answer: \(\displaystyle (A\cup B)-B=A-B\).

  2. Exercise 22

    Let T={x∣ x+5x−7−5=4x−4013−x}\displaystyle \mathrm{T}=\left\{x \left\lvert\, \frac{x+5}{x-7}-5=\frac{4 x-40}{13-x}\right.\right\}. Is T an empty set? Justify your answer.

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    NCERT’s answer
    \(\displaystyle \mathrm{T}=\{10\}\)
    Domain: \(\displaystyle x\neq 7,\ x\neq 13\). \[\frac{x+5}{x-7}-5=\frac{x+5-5x+35}{x-7}=\frac{40-4x}{x-7} \] \[\frac{40-4x}{x-7}=\frac{4x-40}{13-x}=-\frac{40-4x}{13-x} \] \[(40-4x)\left[\frac{1}{x-7}+\frac{1}{13-x}\right]=0 \] \[(40-4x)\cdot\frac{6}{(x-7)(13-x)}=0 \] The fraction never vanishes, so \[40-4x=0 \Rightarrow x=10 \] Check ($\displaystyle 10$ is in the domain): \[\frac{15}{3}-5=0=\frac{0}{3} \] Answer: No. \(\displaystyle T=\{10\}\), so \(\displaystyle T\) is not empty.

  3. Exercise 23

    Let A, B and C be sets. Then show that A∩(B∪C)=(A∩B)∪(A∩C)\mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})=(\mathrm{A} \cap \mathrm{B}) \cup(\mathrm{A} \cap \mathrm{C})

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    Show each side lies inside the other. \[x\in A\cap(B\cup C) \Rightarrow x\in A \ \text{and}\ (x\in B \ \text{or}\ x\in C) \] \[\Rightarrow (x\in A\ \text{and}\ x\in B)\ \text{or}\ (x\in A\ \text{and}\ x\in C) \] \[\Rightarrow x\in(A\cap B)\cup(A\cap C) \] \[A\cap(B\cup C)\subset(A\cap B)\cup(A\cap C) \quad (1) \] Conversely, \[x\in(A\cap B)\cup(A\cap C) \Rightarrow (x\in A\ \text{and}\ x\in B)\ \text{or}\ (x\in A\ \text{and}\ x\in C) \] \[\Rightarrow x\in A \ \text{and}\ (x\in B \ \text{or}\ x\in C) \] \[\Rightarrow x\in A\cap(B\cup C) \] \[(A\cap B)\cup(A\cap C)\subset A\cap(B\cup C) \quad (2) \] From ($\displaystyle 1$) and ($\displaystyle 2$): \[A\cap(B\cup C)=(A\cap B)\cup(A\cap C) \]

  4. Exercise 24

    Out of 100\displaystyle 100 students; 15\displaystyle 15 passed in English, 12\displaystyle 12 passed in Mathematics, 8\displaystyle 8 in Science, 6\displaystyle 6 in English and Mathematics, 7\displaystyle 7 in Mathematics and Science; 4\displaystyle 4 in English and Science; 4\displaystyle 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

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    NCERT’s answer
    (i)
    $\displaystyle 2$ (ii) $\displaystyle 3$
    (iii)
    $\displaystyle 3$
    (iv)
    $\displaystyle 9$
    Let \(\displaystyle E, M, S\) be the students passing English, Mathematics, Science. Start from the students who passed all three, then work outwards, subtracting what is already counted. \[n(E\cap M\cap S)=4 \] \[n(E\cap M\cap S')=n(E\cap M)-n(E\cap M\cap S)=6-4=2 \] \[n(M\cap S\cap E')=n(M\cap S)-n(E\cap M\cap S)=7-4=3 \] \[n(E\cap S\cap M')=n(E\cap S)-n(E\cap M\cap S)=4-4=0 \] \[n(M\ \text{only})=12-(2+3+4)=3 \] \[n(E\ \text{only})=15-(2+0+4)=9,\qquad n(S\ \text{only})=8-(3+0+4)=1 \] More than one subject: the three exactly-two-subject counts plus all three. \[2+3+0+4=9 \] Answer: (i) $\displaystyle 2$ (ii) $\displaystyle 3$ (iii) $\displaystyle 3$ (iv) $\displaystyle 9$

  5. Exercise 25

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

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    NCERT’s answer
    $\displaystyle 25$
    \[n(C \cup T) = n(C) + n(T) - n(C \cap T) = 25 + 20 - 10 = 35 \] \[n\big((C \cup T)'\big) = 60 - 35 = 25 \] Answer: $\displaystyle 25$ students play neither game.

  6. Exercise 26

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

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    NCERT’s answer
    $\displaystyle 20$
    \[n(M \cup P \cup C) = 200 - 20 = 180 \] \[n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(C \cap M) + n(M \cap P \cap C) \] \[180 = 120 + 90 + 70 - 40 - 30 - 50 + n(M \cap P \cap C) \] \[180 = 160 + n(M \cap P \cap C) \] \[n(M \cap P \cap C) = 20 \] Answer: $\displaystyle 20$ students study all three subjects.

  7. Exercise 27

    In a town of 10,000\displaystyle 10,000 families it was found that 40\displaystyle 40% families buy newspaper A, 20\displaystyle 20% families buy newspaper B, 10\displaystyle 10% families buy newspaper C, 5\displaystyle 5% families buy A and B, 3\displaystyle 3% buy B and C and 4\displaystyle 4% buy A and C. If 2\displaystyle 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

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    NCERT’s answer
    (a)
    $\displaystyle 3300$
    (b)
    $\displaystyle 4000$
    \[n(A) = \tfrac{40}{100} \times 10000 = 4000, \quad n(B) = 2000, \quad n(C) = 1000 \]
    \[n(A \cap B) = 500, \quad n(B \cap C) = 300, \quad n(A \cap C) = 400, \quad n(A \cap B \cap C) = 200 \]
    (a)
    \[n(A \text{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C) \]
    \[= 4000 - 500 - 400 + 200 = 3300 \]
    (b)
    \[n(A \cup B \cup C) = 4000 + 2000 + 1000 - 500 - 300 - 400 + 200 = 6000 \]
    \[n(\text{none}) = 10000 - 6000 = 4000 \]
    Answer: (a) $\displaystyle 3300$ families; (b) $\displaystyle 4000$ families.

  8. Exercise 28

    In a group of 50\displaystyle 50 students, the number of students studying French, English, Sanskrit were found to be as follows:
    French = 17\displaystyle 17, English = 13\displaystyle 13, Sanskrit = 15\displaystyle 15
    French and English = 09\displaystyle 09, English and Sanskrit = 4\displaystyle 4
    French and Sanskrit =5\displaystyle =5, English, French and Sanskrit =3\displaystyle =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

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    NCERT’s answer
    (i)
    $\displaystyle 6$, (ii) $\displaystyle 3$,
    (iii)
    $\displaystyle 9$,
    (iv)
    $\displaystyle 1$,
    (v)
    $\displaystyle 2$, (vi) $\displaystyle 6$,
    (vii)
    $\displaystyle 30$,
    (viii)
    $\displaystyle 20$
    Let \(\displaystyle F, E, S\) be the sets of students studying French, English and Sanskrit, and \(\displaystyle U\) the $\displaystyle 50$ students.One language only: subtract both of its pairs; the $\displaystyle 3$ who study all three are subtracted twice, so add them back once. \[\text{(i)}\quad n(F \cap E' \cap S') = n(F) - n(F \cap E) - n(F \cap S) + n(F \cap E \cap S) = 17 - 9 - 5 + 3 = 6 \] \[\text{(ii)}\quad n(E \cap F' \cap S') = 13 - 9 - 4 + 3 = 3 \] \[\text{(iii)}\quad n(S \cap F' \cap E') = 15 - 4 - 5 + 3 = 9 \] Exactly two languages: remove the $\displaystyle 3$ from the pair. \[\text{(iv)}\quad n(E \cap S \cap F') = n(E \cap S) - n(F \cap E \cap S) = 4 - 3 = 1 \] \[\text{(v)}\quad n(F \cap S \cap E') = n(F \cap S) - n(F \cap E \cap S) = 5 - 3 = 2 \] \[\text{(vi)}\quad n(F \cap E \cap S') = n(F \cap E) - n(F \cap E \cap S) = 9 - 3 = 6 \] \[\text{(vii)}\quad n(F \cup E \cup S) = 17 + 13 + 15 - 9 - 4 - 5 + 3 = 30 \] \[6 + 3 + 9 + 1 + 2 + 6 + 3 = 30 \quad \text{(the seven parts agree)} \] \[\text{(viii)}\quad n(F' \cap E' \cap S') = n(U) - n(F \cup E \cup S) = 50 - 30 = 20 \] Answer: (i) $\displaystyle 6$, (ii) $\displaystyle 3$, (iii) $\displaystyle 9$, (iv) $\displaystyle 1$, (v) $\displaystyle 2$, (vi) $\displaystyle 6$, (vii) $\displaystyle 30$, (viii) 20.
  9. Choose the correct answers from the given four options in each Exercises $\displaystyle 29$ to $\displaystyle 43$ (M.C.Q.).

    Exercise 29

    Suppose A1, A2,…, A30\displaystyle \mathrm{A}_1, \mathrm{~A}_2, \ldots, \mathrm{~A}_{30} are thirty sets each having 5\displaystyle 5 elements and B1, B2,…, Bn\displaystyle \mathrm{B}_1, \mathrm{~B}_2, \ldots, \mathrm{~B}_n are n\displaystyle n sets each with 3\displaystyle 3 elements, let ⋃i=130 Ai=⋃j=1n Bj=S\displaystyle \bigcup_{i=1}^{30} \mathrm{~A}_i=\bigcup_{j=1}^n \mathrm{~B}_j=\mathrm{S} and each element of S belongs to exactly 10\displaystyle 10 of the Ai\displaystyle \mathrm{A}_i 's and exactly 9\displaystyle 9 of the B,'S. then n\displaystyle n is equal to
    (A)
    15\displaystyle 15
    (B)
    3\displaystyle 3 (C) 45\displaystyle 45
    (D)
    35\displaystyle 35

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    NCERT’s answer
    C
    (C) \(\displaystyle 45\). Count pairs (element, set containing it) in two ways. \[30 \times 5 = 10 \, n(S) \] \[n(S) = 15 \] \[n \times 3 = 9 \, n(S) = 135 \] \[n = 45 \]

  10. Exercise 30

    Two finite sets have m\displaystyle m and n\displaystyle n elements. The number of subsets of the first set is 112\displaystyle 112 more than that of the second set. The values of m\displaystyle m and n\displaystyle n are, respectively,
    (A)
    4\displaystyle 4, 7\displaystyle 7
    (B)
    7\displaystyle 7, 4\displaystyle 4
    (C)
    4\displaystyle 4, 4\displaystyle 4
    (D)
    7\displaystyle 7, 7\displaystyle 7

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    NCERT’s answer
    B
    (B) \(\displaystyle 7, 4\). The odd factor of $\displaystyle 112$ is $\displaystyle 7$, so the power of $\displaystyle 2$ is fixed. \[2^m - 2^n = 112 \] \[2^n \left( 2^{m-n} - 1 \right) = 2^4 \times 7 \] \[2^n = 2^4, \quad 2^{m-n} - 1 = 7 \] \[n = 4, \quad m - n = 3 \Rightarrow m = 7 \]