Maths

50. (i) False Let A = {a}, a $\in$A then B = {a}, b} I e,  $a\cancel{\in B.}$

(ii) False. Let A. = {a}, if A $\subset$B, B = {a, b} and B $\in$C I e, C = {a, b}, c} I e, A = {a}. 

(iii) True. Let x$\in$A, if A$\subset$ B then x$\in$B and if B$\subset$ C, x$\in$ C I e, elements of A are also elements of C. A$\subset$ C.

(iv) False. Let A = {a} and B = {b} then A$\not\subset$$B$ . Let C = {a, c} then B$\not\subset$$\subset$ but a $\in$A and a c, i.e., A$\subset$ C.

(v) False. Let A = {a} and B = {b} so, A$\not\subset$$B$ i.e., A$B$ .

(vi) True. Let A$\subset$ B such that y$\in$ B i.e., y A But x$\in$$B$ and suppose x$\in$A.

Then by above definition,  A $\subset$B i.e., x$\in$ B and x$\in$ A which is not the case

49. A = {x: xR and x satisfy x² - 8x + 12 = 0}

So, x² - 8x + 12 = 0

x² - 6x 2x + 12 = 0

x (x- 6) 2 (x -6) = 0

(x -6) (x- 2) = 0

x = 6, 2.

So, A = {2, 6) B = {2, 4, 6} C = {2, 4, 6, 8, ….}

D = {6}

D⊂ A ⊂B ⊂C

i e, D ⊂A, D⊂ B, D⊂ C, A⊂ B, A ⊂C and B⊂ C.

48. Let A and B be set of people who speaks Frenchand Spanish respectively. Then,

n (A) = 50, people speak French

n (B) = 20, people speak Spanish.

n (A $\cap$B) = 10, speaks both French& Spanish.

So, number of people who speaks at least one of these two languages

= n (A $\cup$B)

= n (A) + n (B) n (A $\cap$B)

= 50 + 20 10

= 60

47. Let C and T be the set of people who likes cricket and tennis respectively. Then,

n (C) = 40, people who likes cricket

n (C $\cup$T) = 65, people who likes either cricket or tennis

and n (C $\cap$T) = 10, people who likes both

So, n (C $\cup$T) = n (C) + n (T) n (C $\cap$T)

65 = 40 + n (T) 10

n (T) = 35

So, 35 people likes tennis

And number of people who likes tennis only and not cricket

(TC) = n (T) n (C $\cap$T) = 35 10 = 25.

46. Let A and B but the set of people who likes coffee and tea.

Then, n (A) = 37. no. of people who like coffee

n (B) = 52, no. of people who like tea.

As each person likes at least one of the two drink,

n (A $\cup$B) = 70.

So using, n (A $\cup$B) = n (A) + n (B) n (A $\cap$B)

n (A $\cap$B) = n (A) + n (B) n (A $\cup$B)

= 37 + 52 70

= 89 70

= 19

So, 19 people likes both coffee and tea.

45. Given, n (X) = 40

n (Y $\cup$Y) = 60.

n (X $\cap$Y) = 10.

n (Y) =?

Using, n (X $\cup$Y) = n (X) + n (Y) n (X $\cap$Y)

60 = 40 + n (y) 10.

n (Y) = 60 - 40 + 10

n (Y) = 30.

44. Given, n (S) = 21

n (T) = 32

n (S $\cap$T) = 11.

Using, n (S $\cup$T) = n (S) + n (T) n (S $\cap$T)

= 21 + 32 11

= 42.

43. Let H and E be set if people who speak Hindi and English respectively. there,

n (H) = 250, people speak Hindi n (E) = 200, people speak English

n (H $\cup$E) = 400, people speaks either Hindi or English

So, n (H $\cap$E) = n (H) + n (E) n (H $\cap$E)

= 250 + 200 400

= 50

So, 50 people can speak both Hindi and English.

42. Given, n (X $\cup$Y) = 18.

n (X) = 8.n (Y) = 15.n (X $\cap$Y) =?

Using n (X $\cap$Y) = n (X) + n (Y) n (X $\cup$Y)

= 8 + 15 18.

= 5.

41. Given, n (X)= 17

n (Y) = 23

n (X $\cup$Y)= 38.

So, n (X $\cup$ Y) = n (X) + n (Y) n (X $\cap$Y)

n (X $\cap$Y) = n (X) + n (Y) n (X $\cup$Y)

= 17 + 23 38.

= 2