Sets (Package Solutions)

8/10/2019 Sets (Package Solutions)

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Solutions of Assignment (Set-2) Sets (Solutions)

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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f

Section-A

Q.No. Solution

1. Answer (3)

There is no well defined criteria for determining good people as it can vary from person to person.

2. Answer (3)

All metropolitan cities can be clearly identified.

3. Answer (1)

2 B, 2 A.

4. Answer (3)

5. Answer (3)

X = {x : x is a letter in the word “SET THEORY”}

= {S, E, T, H, O, R, Y}

(Order of elements is not significant).

6. Answer (2)

Elements of B can be found by putting n = 0, 1, 2, 3 in .2 1

2

n

n

7. Answer (3)

8. Answer (3)

9. Answer (3)

10. Answer (4)

A = {x : x is a positive multiple of 2, 0 < x < 20, x N}

= {2, 4, 6, 8, 10, 12, 14, 16, 18}

So n( A) = 9

1Chapter

Sets



Sets (Solutions) Solutions of Assignment (Set-2)

Aakash Educational Services Pvt . Ltd . - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Q.No. Solution

11. Answer (2)

P = {2, 3, 5, 7, 11, 13, 17, 19} n(P) = 8

M = {6, 12, 18, 24} n(M) = 4

n(P) – n(M) = 8 – 4 = 4

12. Answer (2)

n( A) = 3

n(P( A)) = 23 = 8.

13. Answer (4)

Power set is collection of all the subsets of given set.

14. Answer (3)

A = {1, 4, 9, 16}

and B = {x : x = n2 and 0 < n < 5}

= {1, 4, 9, 16}

For all x A, x B A = B.

15. Answer (3)

1 A but 1 B. So A B.

16. Answer (1)

x A and A B then x B and x B and B C then x C A C

17. Answer (4)

If x A {x} A {x} P( A)

{x} P(B) ( P( A) = P(B))

{x} B

x B

So, A B

A = B

18. Answer (4)

0 is an element of {0, 1, 2}.

19. Answer (3)

The given set has open - closed interval (–3, 15].





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Q.No. Solution

20. Answer (3)

{x : x R, 2 < x < 7} has open interval (2, 7).

21. Answer (3)

Power set of a set is the collection of all the subsets of given set.

22. Answer (3)

M N = {3, 6, 9, 18, 27} {1, 3, 9, 27}

= {1, 3, 6, 9, 18, 27}

23. Answer (2)

The pair of disjoint sets have A B = .

24. Answer (3)

A = {R, A, I, G, H}

B = {P, R, A, T, G, H}

A B = {R, A, G, H} n( A B) = 4.

25. Answer (1)

D = {6, 12, 18}

B = {6, 12, 18, 24}

Then D – B =

26. Answer (4)

B – A = B – ( A B).

27. Answer (2)

( A – B) = A – ( A B)

28. Answer (3)

A – ( A B) = A – B.

29. Answer (4)

A = {2, 3, 5, 7}

B = {3, 4, 5, 6}

Then ( A – B) = A – ( A B) = {2, 3, 5, 7} – {3, 5} = {2, 7}.

30. Answer (1)

X (X Y) = X.





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Q.No. Solution

31. Answer (4)

X (X Y) = X (X Y) (De morgan’s law)

= (X X) Y (Associative law)

= U Y ( X X = U)

= U

32. Answer (3)

M = {7, 8, 9, 10, ……}

then M = U – M

= N – M = {x : x < 7, x N}

= {1, 2, 3, 4, 5, 6}

33. Answer (2)

A – B = (U – A) – (U – B) ( X = U – X)

= U – A – U + B

= B – A

34. Answer (2)

( A – B) B = ( A B) B

= ( A B) (B B)

= ( A B) (U)

= A B

35. Answer (2)

n( A B) = n( A) + n(B) – n( A B)

= 20 + 18 – 5

= 20 + 13

= 33

36. Answer (3)

n( A – B) = n( A B) – n(B) = n( A) – n( A B)

= 20 – 5= 15

37. Answer (2)

n( A) = n(U) – n( A)

= 50 – 20

= 30





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Q.No. Solution

38. Answer (2)

n(( A B)) = 18

n(U) – n( A B) = 18

n( A B) = 32

n(B – A) = n( A B) – n( A)

= 32 – 18

= 14

39. Answer (4)

n(T C) = 280, n(T) = 100

n(T – C) = 75

n(T C) – n(C) = 75

280 – n(C) = 75

n(C) = 280 – 75

n(C) = 205

40. Answer (2)

n(U) = 60

m(M) = 30, n(S) = 25, n(M S) = 15

n(M S) = n(M) + n(S) – n(M S)

= 30 + 25 – 15

= 40

41. Answer (2)

n((M S)) = (U) – n(M S)

= 60 – 40

= 20

42. Answer (1)

ax = 1 x = 0, (a 1)

But if a = 1, then x

R, Hence the set can never be null

43. Answer (4)

Number of subsets = 22 = 4

Number of subsets of power = 24 = 16





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Q.No. Solution

44. Answer (3)

n( A B) = n( A) + n(B) – n( A B) n( A) = 30

n( A B) = 30 + 25 – 10 n(B) = 25

n( A B) = 45 n( A B) = 10

n( A B) = 50 – 45 = 5

45. Answer (2)

Since ( A BC C

C) CC = ( A B

C) ( CC C

C)

= A BC C

C

Since A BC C

C A B C

Therefore ( A B C) ( A BC C

C) = A BC C

C

( A B C) ( A BC C

C) CC = A B

C C

C

But this is not given therefore if BC C

C A then

A BC C

C = BC C

C

46. Answer (2)

AB = ( A – B) (B – A) = {1, 2, 3, 4} {5, 6}

= {1, 2, 3, 4, 5, 6}

47. Answer (2)

n( A) = 3, n(B) = 6

n( A B) = n( A) + n(B) – n( A B)

If n( A B) = 0 then

n( A B) = n( A) + n(B) = 3 + 6 = 9

If n( A B) = 3 because it may be that AB

then n( A B) = 3 + 6 – 3 = 6

therefore 6 n( A B) 9

and 0 n( A B) 3

48. Answer (3)

( A B) (B C) (C A) = ( A A) (B B) (C C)

= A B C

49. Answer (4)

y x=

y

x

(0, 1)

3x

By the graph of y = 3x and y = x it is clear that A B =





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Q.No. Solution

50. Answer (3)

A = {, {}}

Let {} = a

A = {, a}

P( A) = {, {}, {a}, {, a}}

51. Answer (3)

X A and X B

X P( A) and ( )X P B

Thus X = {a}, {b}, {c} 5C1 – 2

{a, b}, {a, c} ….. = 5C2 – 1

{a, b, c}, {a, c, d} ….. = 5C3

{a, b, c, d}, {b, c, d, e} ….. = 5C4

{a, b, c, d, e } = 5C5

Total X = 28

52. Answer (4)

Let

The number of elements in A : 0 Choices for B : 24



Total number of possible subsets is 41

Section-B

Q.No. Solution

1. Answer (3, 4)

a

b c d

A B

a + b + c + d = 14 …(i)

a + c + d = 12 …(ii)

b + c + d = a …(iii)

b + d = 7

b = 2, a = 5, d = 5, c = 2

| A| = b + c = 4

|B| = d + c = 7





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Q.No. Solution

2. Answer (3, 4)

a

b c d

T.V. Radio

a + b + c + d = 14 …(i)

a + c + d = 12 …(ii)

b + c + d = 9 …(iii)

b + d = 7 …(iv)

a = 5, b = 2, c = 2, d = 5

A = b + c = 4

B = c + d = 7

3. Answer (1, 2, 3)X = {1, 2, 3, 4}, Y = {2, 3, 5, 7}, Z = {3, 6, 8, 9}, W = {2, 4, 8, 10}

(1) (X Y) (Z W)

Let K1 = (X Y) = (X – Y) (Y – X) = {1, 4, 5, 7}

K2 = (Z – W) (W – Z) = {3, 6, 9, 2, 4, 10}

Now, K1 K2 = (K1 – K2) (K2 – K1)

{1, 5, 7, 3, 6, 9, 2, 10} = {1, 2, 3, 5, 7, 9, 10}

Similarly by applying

( A B) = ( A – B) (B – A)

Option (1), (2), and (3) are correct but not (4).

4. Answer (1, 4)

P, Q, R

P Q

R

P Q

(P Q) R

( )P Q R





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Q.No. Solution

5. Answer (1, 3, 4)

By using properties of sets and subset only option (2) is correct.

Section-C

Q.No. Solution

Comprehension-I

Ti m es Mi rr or

cbd

a

e

Sun

Now, we know from first two statements times and mirror are subsets of Sun.

Now, According to statement-3, a + b = 11 …(i)

According to statement-4, b + c = 8 …(ii)

According to statement-5, a + c + d = 10 …(iii)

According to statement-6, a + b + d = 14 …(iv)

According to statement-7, a + e = 9 …(v)

Solving these

a = 5, b = 6, c = 2, d = 3, e = 4

Sol-6 a = 5

Sol-7 d = 3

Sol-8 e = 4

1. Answer (2)

2. Answer (1)

3. Answer (3)

Comprehension-II

1. Answer (3)

8000 1000 7000n A B

4000 1000 3000n A B

1000n A B

Required number of persons

35 25 507000 3000 1000 1495

100 100 100





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Q.No. Solution

2. Answer (2)

15000 – [8000 + 4000 – 1000] = 4000

3. Answer (4)

A A B n A B n A B

4000 + 7000 + 3000 = 14000

Section-D

Q.No. Solution

1. Answer (1)

AB =( A B) – ( A B)

Hence in this case AB = R – [2, 4)

2. Answer (4)

If A and B has a elements in common. The number of common elements in ( A × B) and (B × A) are 2so answer is (99)2 so statement 1 is false.

Section-E

Q.No. Solution

1. Answer A(q), B(s), C(p), D(r)

A = {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35}

B = {1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39}

x( A B) = 9

x( A B) = 10

C = {x : x A B, x is prime}

= {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}

x(C) = 11

x(( A – B) × (B – A)) = x( A – B)·x(B – A)

= 3 × 7 = 21

Section-F

Q.No. Solution

1. Answer (1)

For P Q = {3}

Element 3 should be either in P or in Q.

As it is given in P must not in Q and all remaining elements should be same so only 1 way is possible.





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Q.No. Solution

2. Answer (3)

2m – 2n = 56 8 × 7 = 23 × 7

2n(2m – n – 1) = 23 × 7 n = 3 2m – n = 8 = 23 = 3m n

3. Answer (0)

A = {1, 2}, B = {3, 4, 5}, Y = {1, 2, 3, 4, 5}

(Y × A) = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5)}

(Y × B) = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), ..... (5, 4), (5, 5)}

(Y × A) (Y × B) = so no element into this set.

4. Answer (2)

x3 – 11x

2 + 39x – 45 = 0

(x – 3)(x2 – 8x + 15) = (x – 3)(x – 3)(x – 5) = 0

x = 3, 5

5x – 6 + 3x + 1 > x – 1

47 4

7x x

5x – 6 + x – 1 > 3x + 1

83 8

3x x

3x + 1 + x – 1 > 5x – 6

x < 6

8

63

x

5. Answer (7)

4

2 3

2 164

4 82 2

2 22 16

2 22 2K

Section-G

Q.No. Solution

1. Answer (2)Clearly n( A C) = 3

n(B D) = 2

n(P( A C) = 23; n(P(B D) = 22

2. Answer (1)

All are true. Standard Results.





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Section-H

Q.No. Solution

1. Answer (3)

1

( )m

i

i

o A mp

, in which each element of S is added number of times, hence

( )mp

o S

Similarly ( )nq

o S

mp nq

mp = nq

2. Answer (4)

n(P( A)) = 2° = 1

n(P(P( A))) = 21 = 2

n(P(P(P( A)))) = 22 = 4

n(P(P(P(P( A))))) = 24 = 16

n(P(P(P(P(P( A)))))) = 216

3. Answer (2)

n( A B) = n( A B) – n( A B)

for maximum n( A B), n( A B) should be maximum and n( A B) is minimum

For n( A B) to be minimum, A B =

n( A B) = 25 + 15 = 40

n( A B) = 40

For n( A B) to be maximum, A B

n( A B) = 15

n( A B) = 25

n( A B) = 25 – 15 = 10

Range of n( A B) = {10, 11, 12, ….., 40}

4. Answer (1)

(8, 0) (10, 0)

(0, 8)

(0, 10)





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Q.No. Solution

x y n

0

1

2

8

0, 1, 2,....., 8

0, 1,....., 7

0, 1, 2,....., 6

0

8

7

1

9 10Total 45

2

5. Answer (1)

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