### important Questions class 11 chapter 1 sets

• 1. created by Akstudy1024 1 /akstudy1024 Maths Class 11 Chapter 1 SETS Most Important Questions for Exams SOLUTION OF ONE MARK QUESTIONS Q1. Write the set A = {x : x is a letter of the word MATHEMATICS} in the roster form. Sol 1. Distinct letters in the word ‘MATHEMATICS’ are : M, A, T, H, E, I, C, S ⇒ In roster form, set A = {M, A, T, H, E, I, C, S} Q2. Write the set {1, 4, 9, …, 100} in the set-builder form. Sol 2. We observe that the elements of set A are the squares of all natural numbers less than or equal to 10. ⇒ In set-builder form, set A = {x : x = n2 , where n ∈ N and n ≤ 10} Q3. Find the value of X – Y if X = {a, b, c, d} and Y = {b, d, f, g}. Sol 3. X = {a, b, c, d}, Y = {b, d, f, g}. ⇒ X – Y = {a, c} Q4. If A = {-1, 0, 1}, then write the power set of A. Imp Note : (i) The order in which the elements are written in a set makes no difference. For example - A = {M, A, T, H, E, I, C, S} = {T, A, M, H, E, I, C, S} (ii) The repetition of element has no effect. For example - A = {M, A, T, H, E, M, A, T, I, C, S} = {M, A, T, H, E, I, C, S}
• 2. created by Akstudy1024 2 /akstudy1024 Sol 4. Subsets of A are : 𝜙, {-1}, {0}, {1}, {-1, 0}, {0, 1}, {-1, 1}, {-1, 0, 1} ⇒ P(A) = { 𝝓, {-1}, {0}, {1}, {-1, 0}, {0, 1}, {-1, 1}, {-1, 0, 1}} Q5. If A = {1, 3, 5, 7, 9}, how many elements has P(A)? Sol 5. No. of elements in A, n(A) = 5 ⇒ Total no. of subsets of set A = 2(No. of elements in A) = 25 = 32 ∴ Total no. of elements in P(A) = Total no. of subsets of set A = 32 SOLUTION OF TWO MARK QUESTIONS Q6. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 3, 5, 6}, B = {1, 4, 5, 6, 7} and C = {3, 4, 8, 9}. Find (A - B) ∪ C’. Sol 6. U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 3, 5, 6}, B = {1, 4, 5, 6, 7} and C = {3, 4, 8, 9} ⇒ A – B = {2, 3} and C’ = {1, 2, 5, 6, 7} ⇒ (A - B) ∪ C’ = {1, 2, 3, 5, 6, 7} Q7. Let A and B be two sets having 4 and 7 elements respectively. Then write the (i) minimum number of elements that A ∪ B can have (ii) maximum number of elements that A ∩ B can have Sol 7. (i) Minimum number of elements that A ∪ B can have is 7 (greater no. b/w 4 and 7) (ii) Maximum number of elements that A ∩ B can have is 4 (smaller no. b/w 4 and 7)
• 3. created by Akstudy1024 3 /akstudy1024 Q8. Using properties of sets, prove that A ∪ (A ∩ B) = A. Sol 8. To prove, A ∪ (A ∩ B) = A L.H.S. A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) [∵ ∪ is distributive over ∩] = A ∩ (A ∪ B) [∵ A ∪ A = A] ⇒ A ∪ (A ∩ B) = A ∩ (A ∪ B) …(i) Let a ∈ A, ⇒ a ∈ A ∪ B ⇒ a ∈ A ∩ (A ∪ B) [∵ a ∈ A and a ∈ A ∪ B] ⇒ a ∈ A ∪ (A ∩ B) [∵ from (i)] ∴ a ∈ A ⇒ a ∈ A ∪ (A ∩ B) ⇒ A ⊂ A ∪ (A ∩ B) …(ii) Let b ∈ A ∪ (A ∩ B), ⇒ b ∈ A ∩ (A ∪ B) [∵ from (i)] ⇒ b ∈ A and b ∈ A ∪ B ⇒ b ∈ A and (b ∈ A or b ∈ B) ⇒ b ∈ A ∴ b ∈ A ∪ (A ∩ B) ⇒ b ∈ A ⇒ A ∪ (A ∩ B) ⊂ A …(iii) From (ii) and (iii), A ∪ (A ∩ B) = A Hence proved.
• 4. created by Akstudy1024 4 /akstudy1024 SOLUTION OF FOUR MARK QUESTIONS Q9. From the adjoining venn diagram, write the value of the following : (i) A’ (ii) B’ (iii) (A ∩ B)’ Sol 9. U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (i) A = {2, 4, 3, 0, 6} ⇒ A’ = {1, 5, 7, 8, 9} (ii) B = {0, 3, 5, 6, 7} ⇒ B’ = {1, 2, 4, 8, 9} (iii) A ∩ B = {3, 0, 6} (A ∩ B)’ = {1, 2, 4, 5, 7, 8, 9} Q10. If U = {x : x ∈ N and x ≤ 10}, A = {x : x is prime and x ≤ 10} and B = {x : x is a factor of 24} then verify the following result (i) A – B = A ∩ B’ (ii) (A ∪ B)’ = A’ ∩ B’ (iii) (A ∩ B)’ = A’ ∪ B’ (iv) A ∪ (B - A) = (A ∪ B) Sol 10. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 3, 5, 7} A B 1 8 2 3 4 5 6 7 9 0 U
• 5. created by Akstudy1024 5 /akstudy1024 and B = {1, 2, 3} (i) B’ = {4, 5, 6, 7, 8, 9, 10} A – B = {5, 7} …(i) A ∩ B’ = {5, 7} …(ii) From (i) & (ii), A – B = A ∩ B’ Hence verified. (ii) A’ = {1, 4, 6, 8, 9, 10} A ∪ B = {1, 2, 3, 5, 7} …(iii) (A ∪ B)’ = {4, 6, 8, 9, 10} …(iv) A’ ∩ B’ = {4, 6, 8, 9, 10} …(v) From (iv) & (v), (A ∪ B)’ = A’ ∩ B’ Hence verified. (iii) A ∩ B = {2, 3} (A ∩ B)’ = {1, 4, 5, 6, 7, 8, 9, 10} …(vi) A’ ∪ B’ = {1, 4, 5, 6, 7, 8, 9, 10} …(vii) From (vi) & (vii), (A ∩ B)’ = A’ ∪ B’ Hence verified. (iv) B - A = {1} A ∪ (B - A) = {1, 2, 3, 5, 7} …(viii) From (iii) & (viii), A ∪ (B - A) = (A ∪ B) Hence verified.
• 6. created by Akstudy1024 6 /akstudy1024 Q11. On the real axis, if A = [0, 3) and B = [2, 6], then find the following (i) A’ (ii) A ∪ B (iii) A ∩ B (iv) A - B Sol 11. A = [0, 3) B = [2, 6] (i) A’ = (-∞, 0) ∪ [3, ∞) (ii) A ∪ B = [0, 6] (iii) A ∩ B = [2, 3) (iv) A – B = [0, 2) Q12. In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice. Sol 12. Let U = Set of all surveyed students, A = Set of students drinking Apple Juice and O = Set of students drinking Orange Juice then, n(U) = 400, n(A) = 100, n(O) = 150 and n(A ∩ O) = 75 To find : n(A’ ∩ O’) +∞−∞ 0 3 +∞−∞ 0 2 6 +∞−∞ 0 3 +∞−∞ 0 6 +∞−∞ 0 2 3 +∞−∞ 0 2
• 7. created by Akstudy1024 7 /akstudy1024 now, n(A ∪ O) = n(A) + n(O) – n(A ∩ O) = 100 + 150 – 75 = 250 – 75 = 175 ⇒ n(A ∪ O) = 175 ⇒ n(A’ ∪ O’) = n(U) - n(A ∪ O) = 400 – 175 = 225 SOLUTION OF SIX MARK QUESTIONS Q13. There are 200 individuals with a skin disorder, 120 has been exposed to the chemical C1, 50 to chemical C2, and 30 to both the chemicals C1 and C2. Find the number of individuals exposed to (i) Chemical C1 but not Chemical C2 (ii) Chemical C2 but not Chemical C1 (iii) Chemical C1 or Chemical C2 Sol 13. Let U = Set of individuals suffering from the skin disorder, A = Set of individuals exposed to the chemical C1 and B = Set of individuals exposed to the chemical C2 then, n(U) = 200, n(A) = 120, n(B) = 50 and n(A ∩ B) = 30 To find : (i) n(A − B) (ii) n(B − A) (iii) n(A ∪ B) (i) From the Venn diagram, we have A = (A – B) ∪ (A ∩ B) A BU A - B B - A (A ∩ B)
• 8. created by Akstudy1024 8 /akstudy1024 ⇒ n (A) = n(A – B) + n(A ∩ B) [Since A – B and A ∩ B are disjoint sets] ⇒ n (A – B) = n (A) – n (A ∩ B) = 120 – 30 = 90 Hence, the number of individuals exposed to chemical C1 but not to chemical C2 is 90. (ii) From the Venn diagram, we have B = (B – A) ∪ (A ∩ B) ⇒ n (B) = n (B – A) + n (A ∩ B) [Since B – A and A ∩ B are disjoint sets] ⇒ n(B – A) = n (B) – n (A ∩ B) = 50 – 30 = 20 Hence, the number of individuals exposed to chemical C2 but not to chemical C1 is 20. (iii) n (A ∪ B) = n (A) + n (B) – n (A ∩ B) = 120 + 50 – 30 = 170 – 30 = 140 The number of individuals exposed either to chemical C1 or to chemical C2 is 140. Q14. A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all the three sports. How many received medals in exactly two of the three sports? Sol 14. Let F = Set of men who received medals in football, B = Set of men who received medals in basketball and C = Set of men who received medals in cricket then, n(F) = 38, n(B) = 15, n(C) = 20 n (F ∪ B ∪ C) = 58 and n(F ∩ B ∩ C) = 3 ⇒ n(F ∪ B ∪ C) = n(F) + n(B) + n(C) - n(F ∩ B) - n(B ∩ C) - n(F ∩ C) + n(F ∩ B ∩ C) Putting values, 58 = 38 + 15 + 20 - n(F ∩ B) - n(B ∩ C) - n(F ∩ C) + 3
• 9. created by Akstudy1024 9 /akstudy1024 F B U C a b c d ⇒ 58 = 76 - n(F ∩ B) - n(B ∩ C) - n(F ∩ C) ⇒ n(F ∩ B) + n(B ∩ C) + n(F ∩ C) = 76 – 58 = 18 ⇒ n(F ∩ B) + n(B ∩ C) + n(F ∩ C) = 18 …(i) Consider the Venn diagram, Here, a = n(F ∩ B) = number of men who got medals in football and basketball only, b = n(F ∩ C) = number of men who got medals in football and cricket only, c = n(B ∩ C) = number of men who got medals in basket ball and cricket only and d = n(F ∩ B ∩ C) = number of men who got medal in all the three Putting values in (i), we get a + d + b + d + c + d = 18 ⇒ a + b + c + 3d = 18 ⇒ a + b + c + 3(3) = 18 [∵ n(F ∩ B ∩ C) = d = 3] ⇒ a + b + c + 9 = 18 ⇒ a + b + c = 18 – 9 ⇒ a + b + c = 9 ∴ The number of people who got medals in exactly two of the three sports is 9. Q15. In a town of 10,000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B and 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 the no. of families which buy
• 10. created by Akstudy1024 10 /akstudy1024 A U Bd f g e C a b c (i) A only (ii) B only (iii) exactly two newspapers (iv) A and C but not B (v) exactly one newspaper (vi) atleast one of A, B and C (vii) none of A, B and C. Sol 14. Let U = Set of families in a town, A = Set of families buying newspaper A, B = Set of families buying newspaper B and C = Set of families buying newspaper C then, n(U) = 10,000 n(A) = 40% of 10,000 = 40 100 × 10,000 = 4,000 n(B) = 20% of 10,000 = 20 100 × 10,000 = 2,000 n(C) = 10% of 10,000 = 10 100 × 10,000 = 1,000 n(A ∩ B) = 5% of 10,000 = 5 100 × 10,000 = 500 n(B ∩ C) = 3% of 10,000 = 3 100 × 10,000 = 300 n(A ∩ C) = 4% of 10,000 = 4 100 × 10,000 = 400 n(A ∩ B ∩ C) = 2% of 10,000 = 2 100 × 10,000 = 200
• 11. created by Akstudy1024 11 /akstudy1024 Consider the Venn diagram, Here, a = no. of families buying newspaper A only, b = no. of families buying newspaper B only, c = no. of families buying newspaper C only, d = no. of families buying newspaper A and B but not C, e = no. of families buying newspaper B and C but not A, f = no. of families buying newspaper A and C but not B and g = no. of families buying all the three newspapers ⇒ g = n(A ∩ B ∩ C) = 200 now, n(A ∩ B) = d + g ⇒ 500 = d + 200 [Putting values of n(A ∩ B) and g] ⇒ d = 500 – 200 = 300 n(B ∩ C) = e + g ⇒ 300 = e + 200 [Putting values of n(B ∩ C) and g] ⇒ e = 300 – 200 = 100 n(A ∩ C) = f + g ⇒ 400 = f + 200 [Putting values of n(A ∩ C) and g] ⇒ f = 400 – 200 = 200 n(A) = a + d + f + g ⇒ 4,000 = a + 300 + 200 + 200 [Putting values of n(A), d, f and g] ⇒ a = 4,000 - 300 - 200 - 200 ⇒ a = 4,000 – 700 = 3,300 n(B) = b + d + e + g ⇒ 2,000 = b + 300 + 100 + 200 [Putting values of n(B), d, e and g] ⇒ b = 2,000 - 300 - 100 - 200
• 12. created by Akstudy1024 12 /akstudy1024 ⇒ b = 2,000 – 600 = 1,400 n(C) = c + e + f + g ⇒ 1,000 = c + 100 + 200 + 200 [Putting values of n(C), e, f and g] ⇒ c = 1,000 - 100 - 200 - 200 ⇒ c = 1,000 – 500 = 500 (i) A only = a = 3,300 (ii) B only = a = 1,400 (iii) Exactly two newspapers = d + e + f = 300 + 100 + 200 = 600 (iv) A and C but not B = f = 200 (v) Exactly one newspaper = a + b + c = 3,300 + 1,400 + 500 = 5,200 (vi) Atleast one of A, B and C, n(A ∪ B) = a + b + c + d + e + f + g = 3,300 + 1,400 + 500 + 300 + 100 + 200 + 200 = 6,000 (vii) None of A, B and C = n(U) – n(A ∪ B) = 10,000 – 6,000 = 4,000