VIEWS: 1 PAGES: 7 CATEGORY: Algebra POSTED ON: 9/14/2012 Public Domain
SET THEORY & ELE. NUMBER SYSTEM( AIEEE level) 9-5-2010 1 Section - I (Single Answer Type) 1. If A = {a, b, {c}, φ} ; then which one is true. 4 × 2220 ⇒ y = = 60 148 (a) c∈A (b) {c} ⊂ A 6. If a natural number X is divided by 7, then the set (c*) {c} ∈ A (d) {φ} ∈ A of all possible remainders is. Ans. c (a) {2, 3, 5} (b) {1, 3, 5, 7} Sol. Q {c} is an element of A, (c) {1, 2, 3, 5, 7} (*d) {0, 1, 2, 3, 4, 5, 6} Q {c} ∈ A, Ans. d 2. Which one of the given sets is empty. Sol. On dividing any natural no. by 7, the possible remain- (a) {x : x is an even prime number} der can be 0, 1,2,3,4,5,6 (b*) {x : x2 + 1 = 0 and x ∈ N} Q Set of possible remainders is {0,1,2,3,4,5,6} (c) {x : x2 = x and x ∈ N} 7. If A = {3x : x = N}, B = {5x : x = 5N} then de- (d) {x : G.C.D. (x, 1) = 1 and x ∈ N} scribe set A ∩ B. Ans. b (a) {3x : x ∈ N} (b) {5x : x ∈ N} Sol. x2 + 1 =0 ⇒ x2 = -1 ⇒ x = ± −1 ∉ N (*c) {15x : x ∈ N} (d) none of these ⇒ Set in opt(b) is empty Ans c Sol. A = {3, 6 9, 12, 15 ......} 3. .in If N = set of natural numbers, W = set of whole numbers, Z = set of integers, then which set can serve as the universal set for this problem. B = {5, 10, 15, 20, .....} (A ∩ B) = {15, 30, 45 ........} .co (a) N (b) W = {15 x : x ∈ N} (c) Z = set of positive integer + 8. The product of two +ve integer is 24 & their (d*) Z greatest common divisor is 2. Their least com- Ans. d mon divisor is du Sol : (a) 12 (b) 24 Q N, W, Z are the subsets of set Z (set of integers) (*c) 1 (d) 2 Q Z can serve as a universal set for this problem. Ans. c ite 4. If X and Y are two sets such that n(X) = 17, n(Y) = Sol. Least Common divisor of any two natural no. is One 28 and n(X ∪ Y) = 40, then n(X ∩ Y) is equal to. (1) Note : L.C.Divisor means Least +ve common Di- (a) 15 (b*) 5 visor. mi (c) 25 (d) 10 9. The values of x, y, z from the venn- A x 4 y B Ans. b diagram shown, if is known that 6 5 8 Sol. n(x ∩ y) = n(x) + n(y) - n(x ∪ y) x, y, z are three consecutive z C = 17 + 28 - 40 = 5 prime numbers and n(A ∪ B ∪ C) 5. If the H.C.F. and L.C.M. of two numbers are re- = 33 are. spectively 4 and 2220 respectively. If one of the (a) 2, 3, 7 (b*) 2, 3, 5 numbers is 148, then what is the other number. (c) 5, 7, 11 (d) No solution (a*) 60 (b) 90 Ans. b (c) 30 (d) 40 Sol. n(A ∪ B ∪ C) = 33 Ans. a ⇒ x + y + z + 4 + 5 + 6 + 8 = 33 Sol. x . y = H.C.F. x L.C.M. ⇒ 148 y = 4 x 2220 ⇒ x + y + z = 10 S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706 SET THEORY & ELE. NUMBER SYSTEM( AIEEE level) 9-5-2010 2 Q x, y, z are numbers having sum equal to 10. Ans. a ⇒ x, y , z should be 2, 3, 5 respectively. Sol. Total no of divsors = 15 10. If U = { 2, 3, 5, 7, 11, 13, 17 } ; A = { 2, 3, 5, 13}; ⇒ (a + 1)(b + 1) = 15 B = {3, 5, 7, 17}. The set representing A ∆ B will ⇒ a + 1 = 3, b + 1 = 5 or a + 1 = 5, b + 1 = 3 or be : a + 1 = 1 , b + 1 = 15 or a + 1 = 15, b + 1 = 1 (a) {2, 7, 13, 17} (b) {2, 7, 13} ⇒ a = 2 , b = 4 or a = 4, b = 2, or a = 0, b = 14 or (c) {2, 13, 17} (d) none of these a = 14, b = 0 Ans. a ⇒ (a, b) can be (2, 4); (4, 2); (0, 4); (4, 0) Sol. A = U − A = {7,11,17} and B = U - B = {2, 11, 13} But in last two pairs N = paqb becomes q14 and p14 A∆B = (A − B) ∪ (B − A) = {7, 17} ∪ {2, 13} respectively i.e. they have only one prime and not = {2, 7, 13, 17} two distinct prime. 11. If A & B are two disjoint sets then which one true ∴ Last two pairs are rejected (a) A ⊂ BC and B ⊄ AC ∴ Only two pairs of a and b are possible. (b) A ⊄ BC and B ⊂ AC 14. The number of the proper common divisors of x = (c) A ⊂ BC and B ⊂ AC (2)4 (3)5 and y = (2)3 . (3)6 is : (d) None of these (a) 23 (b) 24 Ans. c Sol. A and B are disjoint sets, ⇒ .in A ∩ B = {} Ans. a (c) 25 (d) none x ∈ A ⇒ x ∉ B [ Q A and B have no element in common] Sol. Sol. x = (2)4 (3)5 and y = (2)3 . (3)6 H.C.F. of divisors .co of x and y = (2)3 (3)5 ⇒ x ∈ Bc ⇒ A ⊂ Bc Common divisors of x & y are the divisors of ((2)3 (3)5) similarly B ⊂ Ac = 4 . 6 = 24 du 12. The number of proper even factors of 684 is : ∴ Number of proper common divisors = 24 – 1 = 23. (a) 12 (b) 16 15. Number of non-empty subsets of set containing (c) 4 (d*) 11 10 elements in all are (a) 1023 (b) 1024 ite Ans. d Sol. 684 = (2)2 × (3)2 × (19)1 (c) 210 – 2 (d) 29 No. of even factors of 684 Ans a ∴ Sol. Number of subsets of set containing 10 elements mi = (3)(3)(2) = 12 = 210 . 1024. ∴ No. of proper even factors = 12 - 1 = 11 ∴ Number of subsets containing 1 empty subset φ. 13. A natural number N is primerly factorised as paqb, where p and q are two distinct primes. Such that Number non empty containing 1 empty subset total number of divisors of N be 15 then numbers = 1240 – 1 = 1023. of possible values of (a, b) will be : ∴ option (a) (a) 2 (b) 4 (c) 1 (d) None of these S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706 SET THEORY & ELE. NUMBER SYSTEM( AIEEE level) 9-5-2010 3 (Assertion & Reason Type) 17. A: A quadrilateral having equal sides can be 16. A: LCM of √2 & 3 is 3√2 rhombus or square. R: If x & y are two coprime numbers then their R: A compound statement with an exclusive OR LCM is x. y. is true when only one component statement Ans. D is true not both. Sol. Multiples of √2 = √2, 2√2, 3√2 ....... Ans. a Multiples of 3 = 3, 6, 6, 9 ..... Sol. A is true and R is true ∴ LCM does not exist 18. A: x is divisible by 15, this implies x is divisible ∴ A is false but R is true. by 3 and x is divisible by 5 ∴ (d) option R: x is not divisible by 3 or 5, implies x is not divisible by 15. Ans. A Comprehensive Type) .in Paragraph for question nos. 19 - 22 If A and B are two sets, then the set of all elements 21. Let C = set of players of cricket B = set of players of basketball which belong to A but not B i.e. n (A – B) = { x : x ∈ A H = set of players of Hockey, and n (C) = 40, n (B) .co but x ∉ B}, & the set of all elements which belong to = 39, n(H) = 36, n( C ∩ B) = 9, n(C ∩ H) = 15, n(B ∩ only one of the sets A & B is known as Symmetric H) = 11 n(C ∩ B ∩ H) = 5, then the number of players Difference of sets A and B, denoted by A ∆ B i.e. A ∆ B = who play only cricket is : du (A - B) ∪ (B - A) = {x :x ∈ A or x ∈ B but x ∉ ( A ∩ B) }. (*a) 21 (b) 22 Now answer the following questions. (c) 23 (d) 24 Ans. a 19. If A = {a, e, i, o, u} and B = {e, i, u, m, n} then (A ite ~ B) is : Sol. n(only cricket) (a) {a, e} (*b) {a, o} = n(c) - n(C ∩ B)-n(C ∩ H)+n(C ∩ B ∩ H) (c) {o, u} (d) {u} = 40 - 9 - 15 + 5 = 21 mi Ans. b 22. If AC = {1, 3, 5, 6, 9, 10} & BC = {1, 2, 3, 5, 6, 8} Sol. A ~ B = {x : x ∈ A and x ∉ B} = {a, 0} then (A ∪ B)C is : 20. If A = {a, e, i, o, u} and B = {e, i, u, m, n}, then A (a) {1, 3, 5, 9} ∆ B is : (b) {1, 3, 5, 6, 7, 8, 9, 10} (a) {a, e, m, n} (b) {o, m, n} (c) {3, 5, 6, 9} (*d) {1, 3, 5, 6} (*c) {a, o, m, n} (d) {e, o, m, n} Ans. d Ans. c Sol. By Demorgan’s theorem Sol. A ∆ B = (A - B) ∪ (B - A) (A ∪ B)c = Ac ∩ Bc = {a, o} ∪ {m, n} ={1,3,5,6,9,10} ∩ {1,2,3,5,6,8} = {a, o, m, n} ={1,3,5,6} S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706 SET THEORY & ELE. NUMBER SYSTEM( AIEEE level) 9-5-2010 4 Paragraph for Question nos. 23 - 26 Given A = {x : x = 3m, m ∈ N & x ≤ 17} , B = n(A ∪ (B ∩ C)) = 6 {y : y = 2k , k ∈ W & k ≤ 6} & C = {x : x is prime number 25. Number of elements in ( A − C ) ∪ (C − B) is equal & x ≤ 7}. Where W is the set of whole numbers and N is to the set of natural numbers. Based on this information (a) 6 (b*) 7 (c) 8 now answer the following questions (d) 10 Sol. A = {3, 6, 9, 12, 15} Sol. (A – C) ∪ (C – B) = {6, 9, 12, 15} ∪ {3, 5, 7} B = {0, 2, 4, 6, 8, 10, 12} = {3, 5, 6, 7, 9, 12, 15} C = {2, 3, 5, 7} number of element = 7. 23. Number of elements in ( A ∪ B) will be 26. Number of elements in A ∆ B (The symmetric differ- (a) 9 (b) 6 (c) 8 ence of A & B) is (d*) None of these (a) 12 (b) 10 (c*) 8 Sol. (A ∪ B) = {0, 2, 3, 4, 6, 8, 9, 10, 12, 15} (d) None of these n(A ∪ B) = 10 Sol. (A ∆ B) = (A – B) ∪ (B – A) 24. (a) 12 (b) .in Number of elements in A ∪ (B ∩ C) will be 10 (c) 8 = {3, 9, 15} ∪ {0, 2, 4, 8, 10} n(A ∆ B) = {0, 2, 3, 4, 8, 9, 10, 15} (*d) None of these .co number of elements = 8. Sol. (A ∪ (B ∩ C)) = {2, 3, 6, 9, 12, 15} Section II (More than one Type) du 27. Which of following sets is/are empty sets Sol. (a) A = {x : x ∈ R and x – 4 = 0} 2 (a) A = {5} (b) B = {x : x ∈ R and x + 4 = 0} 4 (b) B = {6, –6} (c) C = {x : x ∈ N and x = 1} 3 (c) C = {6} ite (d) D = {x : x ∈ N and x + 2x + 1 = 0} 2 (d) x2 = – 1 ∴ x is not real Ans. b, d ∴ D = {} ∴ a, c are singleton sets Sol. A = {2, –2} ∴ non empty 29. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {x : x ∈ N, 30 < mi (a) x =–4 4 ∴ not reals x2 < 70} B = {x : x is a prime number less than ∴ B=φ 10} then which of following is false ? (c) C = {1} ∴ non empty (a) A' ∩ B' = {1, 4, 9} (b) A–B={2,3,5} (d) D = {–1} ∴ empty (c) B – A = {6, 8} (d) A∆B = {2,3,5,6,8}. but x ∈ N ∴ (b), (d) Ans. b, c 28. Which of the following is a singleton set ? Sol. A = {6, 7, 8} B = {2, 3, 5, 7} (a) A = {x : |x| = 5, x ∈ N} (a) A' ∩ B' (b) A = {x : |x| = 6, x ∈ Z} {1, 2, 3, 4, 5, 9} ∩ {1, 4, 6, 8, 9} ∩ {1, 4, 9} is true (c) A = {x : 5 < x ≤ 6, x ∈ Z} (b) A – B = {6, 8} is true (d) A = {x : x ∈ R and x + 1 = 0 } 2 (c) B – A = {2, 3, 5} is false Ans. a, c (d) A ∆ B = (A – B) ∪ (B – A) = {2, 3, 5, 6, 8} is true S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706 SET THEORY & ELE. NUMBER SYSTEM( AIEEE level) 9-5-2010 5 ∴ (b), (c) are false. 30. Consider the following relations : A = (A ∩ B) ∪ (A – B) is correct (3) is false (1) A-B = A – (A ∩ B) (2) A =(A ∩ B) ∪ (A – B) ∴ 1 and 2 are true that is (a) and (b) is the correct (3) A – (B ∪ C) = (A-B) ∪ (A – C) answer. which of these is correct 31. If A = {x : x is a factor of 800} then which of the *(a) 1 (*b) 2 following is true ? (c) 3 (d) None of these (a) n(A) = 18 (b) n[p(A)] = 2 18 Ans. a, b (c) n(A) = 10 (d) n[(p(A)] = 2 10 Sol. A – B = A – (A ∩ B) is correct. Ans. a, b A B Sol. 800 = 25 . 52 ∴ n(A) = no. of factor of A & n(P (A)) = 2 18 = 6×3 = 18. ∴ So (a), (b) is correct.. A-B A-(A ∩ B) Section III (Column Matching Type) 32. Match the shaded portions in Venn diagram in column A with their rotations in column B. Column A Column B A .in B .co (i) (a) A – (B ∪ C) C A du (ii) (b) AC ∩ (B ∪ C) C B ite A B mi (iii) (c) A ∪ (B ∩ C) ~ (A ∩ B) C B A (iv) (d) CC ∩ (A – B) C (e) {(A – B) ∪ (B – C) ∪ (C – A)} Ans. (i) c (ii) a, d (iii) b (iv) e S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706 SET THEORY & ELE. NUMBER SYSTEM( AIEEE level) 9-5-2010 6 33. Column A Column B (i) {x : x2 = 25 and x is positive even integer} (a) Singleton set (ii) {x : x ∈ R and 1 < x < 1.01} (b) Infinite set (iii) Set of all concentric circle with centre (0, 0) (c) Uncountable set (iv) A = {1, 2, 3}, B = {3, 4}, C = {3, 4, 5} (d) null set then A ∩ (B ∩ C) (e) countable set Ans. (i) d, e (ii) c, b (iii) b, c (iv) a, e Solution. ber and are uncountable. (i) {+5, –5} there is no positive even integer (iv) (B ∩ C) = {3, 4} ∴ It is null set A ∩ (B ∩ C) = {1, 2, 3} ∩ {3, 4} = {3} (ii) There are infinite real numbers between two num- ∴ Singeton set. Section IV (Subjective Type) 34. In a city, three daily newspapers A, B, C are and 1546 = xq2 + r published 42% of the people in that city read A, on Subtraction. 248 = x (q2 − q1 ) 51% read B and 68% read C, 30% read A and B; ⇒ x is a divisor of 248 .in 28% read B and C 36% read A and C 8% do not read any of the three newspapers. The percentage of persons who read all the three papers is Now, two digit factors of 248 are 31 and 62 ∴ If x = 31, then 1298 = (31) (41) +27 ⇒ λ = 27 .co Ans. 25% If x = 62, then, 1298 = (62) (20) + 58 Sol. Let the no. of persons in the city by 100. Then we have ⇒ λ = 58 n(A) = 42, n(B) = 51, b(C) = 68; λ → 27, 58 du n(A ∩ B) = 30, n(B ∩ C) = 28, n(A ∩ C) = 36 ∴ sum = 27 + 58 = 85 n(A ∪ B ∪ C) = 100 – 8 = 92. 36. If x = 560, and Using ite (i) L = Number of proper divisors of x n(A ∪ B ∪ C) = n(A) + n(B) + n(C)– n(A ∩ B) – n(B ∩ C) (ii) M = Number of proper odd divisors of x – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) (iii) N = Number of proper even divisors of x Substituting the above values, we have mi Find L – (M + N) 92 = 42 + 51 + 68 – 30 – 28 – 36 + n(A ∩ B ∩ C) Ans. 0 ⇒ n(A ∩ B ∩ C) = 92 – 161 + 94 Sol. 560 = (2)4 . (5)1 . (7)1 ⇒ n(A ∩ B ∩ C) = 92 – 67 = 25. (i) Total No. of divisors of Hence, 25% of the people read all the three papers. x = (5) × (2) × (2) = 20 35. When 1298 and 1546 are divided by a certain two digit number, the remainders are equal. Find sum L = Number of proper divisors of x = 20 - 2 = 18 of all possible remainders. (ii) No. of odd divisors of x = (b + 1) (c + 1) Ans. 85 = ( 2) (2) = 4 Sol. Let x be the two digit numbers ∴ M = no. of proper odd divisors = 4 – 1 = 3 ∴ 1298 = xq1 + r (iii) No. of even divisors of x = (a) ( b +1)( c + 1) S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706 SET THEORY & ELE. NUMBER SYSTEM( AIEEE level) 9-5-2010 7 = ( 4) (2) (2) = 16 passed in chemistry. Maximum 19 students passed ∴ N = No of proper even divisors of x = 16 - 1 = 15 in Maths and Physics, Maximum 29 students passed in Maths and Chemistry and Maximum 20 ∴ L – (M + N) = 0 Ans. students in Physics and Chemistry. Find the number 37. A = {x : x = n2 and n = 4, 5, 6} B = {y : y = n ∈ N of maximum students passed in three subjects. and 15 ≤ y ≤ 30, n ∈ N} and C = {p ∈ A and p ∉ B} Ans. 14 Find number of proper odd factors of p. C = {p ∈ A}. Sol. Q n(M ∪ P ∪ C) = 50, n(M) = 37, n(P) = 24 Ans. 2 n(C)=43, n(M ∩ P) ≤ 19, n(M ∩ C) ≤ 29, Sol. A = {16, 25, 36} B = {16, 25} n(P ∩ C) ≤ 20 (A – B) = {36} = p Now, n(M ∪ P ∪ C) = n (M)+n(P) + n(C) ∴ p = 36 = 22 . 32. – n (M ∩ P) -n (M ∩ C)- n(P ∩ C)+n (M ∩ P ∩C) number of odd factors = 1.(3) = 3 ⇒ 50 = 37 +24 +43 – n (M ∩ P) – n (M ∩ C) – n(P ∩ C) + n(M ∩ P ∩ C) number of odd proper factors = 3 – 1 = 2 ⇒ n(M ∩ P ∩ C) =n (M ∩ P) +n(M ∩ C) + N(P ∩ C) – 54 38. Out of 50 students, in examinations of Maths, Physics and Chemistry, 37 students passed in ⇒ n (M ∩ P ∩ C) ≤ 19 + 29 +20 – 54 .in Maths, 24 students in Physics and 43 students ⇒ n (M ∩ P ∩ C) ≤ 14 .co du ite mi S.C.O 73, Sec - 20 C, Chandigarh, 2775646, 9417317019, S.C.F 73, Sector 10, & S.C.O. 214, Sec. 14, Panchkula, 2583315, 9463087706