Class 11th

61. Let T and C be sets of students taking tea and coffee.

Then, n (T) = 150, number of students taking tea

n (C) = 225, number of students taking coffee

n (T$\cap$C) = 100, number of students taking both tea and coffee.

So, Number of students taking either tea or coffee is.

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

= 150 + 225 100

= 275

Number of students taking neither tea coffee

= Total number of students No of students taking either tea or coffee

= 600 275

= 325.

60. Let A = {x, y}

B = {y, z}

C = {x, z}

So, A$\cap$B = {x, y}$\cap$ {y, z} = {y}≠$?$

B$\cap$C = {y, z} $\cap$ {x, z} = {z}≠$?$

A$\cap$C = {x, y}$\cap$ {x, z} = {x}≠$?$

But A$\cap$B$\cap$C = (A$\cap$B)$\cap$ C

= {y} $\cap$ (x, z}

=$?$

59. Let A, B and x be sets such that,

A$\cap$x = B$\cap$x =$?$ and A$\cup$x = B$\cup$x.

We know that,

A = A$\cap$ (A$\cup$x)

= A$\cap$ (B$\cup$x) [? A$\cup$x = $\cup$Bx]

= (A$\cap$B) (A$\cap$x) [by distributive law]

= (A$\cap$B) ∪? [? A∩x =? ]

=> A = A∩ B [? A ∪? = A]

And B = $\cap$ (B$\cup$x)

= B$\cap$ (A$\cup$x) [? B$\cup$x = A$\cup$x]

= (B$\cap$A)$\cup$ (B$\cap$x) [By distributive law]

= (B$\cap$A) ∪$?$ [? B$\cap$x =? ]

B = B$\cap$A [? A$\cup$ $?$= A]

So, A = B = A$\cap$B.

58. Let A = {a}, B = {a, b}, C = {a, c}

So, A$\cap$B = {a} $\cap$ {a, b} = {a}

A$\cap$C = {a} $\cap$ {a, c} = {a}

i.e., A$\cap$B = A$\cap$C = {a}

But B ≠C. as b$\cancel{\in }$B but $b\cancel{\ne }C$ vice-versa

57. (i) We know that,

A $\subset$A

(A $\cap$B)$\subset$ A

[A$\cup$ (A $\cap$B)] $\subset$ (A)

[A $\cup$ (A $\cap$B)] $\subset$A

and also

A$\subset$ [A $\cup$ (A $\cap$B)]

So, A$\cup$ (A$\cap$ B) = A.

(ii) A$\cap$ (A$\cup$ B) = (A$\cap$ A) $\cup$ (A$\cap$ B) [By distributive law]

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

= A as (A $\cap$B) $\subset$A

56. Here,

(A$\cap$B)$\cup$ (A - B) = (A$\cap$B)$\cup$ (A$\cap$ B') as (A -  B) = A$\cap$ B'

= A $\cap$ (B$\cup$ B') [ by converse of distributive law]

= A $\cap$U [ B $\cup$B' = U, sample space set or universal set]

= A

And (A $\cup$ (B - A) = A $\cup$ (B$\cap$ A') [as B -  A = B$\cap$ A']

= (A$\cup$B) $\cap$ (A$\cup$A')

= (A$\cup$B)$\cap$ U [ A$\cup$A' = U, universal set]

= A$\cup$ B.

55. Let A = {a}, B = {b}.

So, P (A) = $\{?,\{a\left\}\right\}B(A)=\{?,(b\left\}\right\}$

So, P (A) $\cup$P {B} = $\left\{?,\left\{a\right\},\left\{b\right\}\right\}$ ______ (1)

Now, A$\cup$B = {a, b}.

P (A$\cup$B) = $\left\{?,\left\{a\right\},\left\{b\right\},\{a,b\}\right\}$ ____ (2)

So. From (1) and (2) we see that,

P (A)$\cup$ P (B) ≠P (A$\cup$B)

54. Given, P (A) = P (B) where P is power set

Let x$\cancel{\in }$A.

Then, {x} $\subset$P (A)$\subset$ P (B)

i.e., x$\cancel{\in }$ B

A$\subset$ B

Similarly, B $\subset$A

A = B

53. Given, A $\subset$B.

Let x$\cancel{\in }$ C - B then x $\cancel{\in }$C but X$B$ .

However, A $\subset$B, elements of B should have elements of A

i.e., X$B$ 

So, x $\cancel{\in }$C A i.e., x$\cancel{\in }$ C but X$A$

C - B $\subset$C-  A

52. (i) Let A-  B≠$?$ , our assumption.

i.e., x$\in$ A But x≠ B where x is an element.

But as A B, the above condition of assumption is wrong if A $\subset$B then A -  B =$?$

(ii) Let x$\in$ A.

As A - B =$?$ we can say that x$\in$ B because if $x\cancel{\in }B,$ A - B ≠$?$

if A - B =$?$ then A $\subset$B.

(iii) We know that,

B $\subset$A$\cup$ B always true

Let x$\in$A$\cup$B i.e., x$\in$ A or x$\in$ B.

As A $\subset$B,

If x $\in$A then x$\in$ B, all elements of A are among the elements of B

So, (A$\cup$B) = B

(iv) We know that,

(A$\cap$ B) $\subset$A as A$\subset$ B.

Let x $\in$A then x $\in$B

So, x $\in$(A$\cap$ B)

i.e., A$\subset$ (A$\cap$ B)

So, A = (A $\cap$B)

Hence, A$\subset$ B

A - B = $?$.

A$\cup$ B = B

A $\cap$B = A.

i.e., the 4 conditions are equivalent.