1.

Soln:

a. A U B = {1, 2, 3, 4} U {3, 5, 7, 8} = {1, 2, 3, 4, 5, 7, 8}

 

b. A ∩ C = {1, 2, 3, 4} ∩ {1, 2, 7, 8} {1, 2}

 

c. (A – B) ∩ C = ({1, 2, 3, 4} – {3, 5, 7, 8}) ∩ {1, 2, 7, 8}

= {1, 2, 4} ∩ {1, 2, 7, 8} = {1, 2}

 

d. A U C = {1, 2, 3, 4} U {1, 2, 7, 8}

So, A U C = {1, 2, 3, 4, 7, 8}

Now, $\overline {{\rm{AUC}}} $ = U – (AUC)

So, $\overline {{\rm{AUC}}} $ = {5, 6, 9, 10}

 

e. ${\rm{\bar B}}$ = U – B = {1, 2, 3, …9, 10} – {3, 5, 7, 8}

= {1, 2, 4, 6, 9, 10}

Or, ${\rm{\bar C}}$ = U – C = {1, 2, 3, ….9, 10} – {1, 2, 7, 8}

Or, ${\rm{\bar B\: }}$U ${\rm{\bar C}}$ = {1, 2, 4, 6, 9, 10} U {3, 4, 5, 6, 9, 10}

Or, ${\rm{\bar B\: }}$U ${\rm{\bar C}}$ = {1, 2, 3, 4, 5, 6, 9, 10}

 

f. (AUB) – C = {1, 2, 3, 4, 5, 7, 8} – {1, 2, 7, 8}          [from1(a)]

= {3, 4, 5}

 

2.

Soln:

a. AUB = {b, c, d, e} U {d, e, f, g, h, i} = {b, c, d, e, f, g, h, i}

 

b. A∩C = {b, c, d, e} ∩ {a, e, i, o, u} = {e}

 

c. A – B = {b, c, d, e} – {d, e, f, g, h, i} = {b, c}

 

d. A – C = {b, c, d, e} – {a, e, i, o, u} = {b, c, d}

 

e. (A – C) ∩ C = {b, c, d} ∩ {a, e, i, o, u} [from 2(d)] = ɸ

 

f. A $\Delta $ D = (A – D) U (D – A)

=({b, c, d, e} – {b, d, j, k}) U ({b, d, j, k} – {b, c, d, e})

= {c, e} U {j, k} = {c, e, j, k}

 

g. (AUB) – C = {b, c, d, e, f, g, h, i} – {a, e, i, o, u} [from 2(a)]

= {b, c, d, f, g, h}

 

h. ${\rm{\bar B}}$ = U – B = {a, b, c, d, ….i, j, k} – {d, e, f, g, h, i} = {a, b, c}

 

i. A – B = U – (A – B)

= {a, b, c, d, ….i, j, k} – {b, c} = {a, d, e, f, g, h, i, j, k}

 

3.

a.

Soln:

Here, U = {x:x is a positive integer les than 12}

So, U = {1, 2, 3, ….11}

Now, A U B = {3, 5, 7, 9} U {1, 2, 3, 8, 9} = {1, 2, 3, 5, 7, 8, 9}

So, $\overline {{\rm{AUB}}} $ = U – (AUB)

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} – P1, 2, 3, 5, 7, 8, 9}

= {4, 6, 10, 11}

Again, A – B = {3, 5, 7, 9} – {1, 2, 3, 8, 9} = {5, 7}

So, (A – B) U C = {5, 7} U {1, 4, 5, 7, 10} = {1, 4, 5, 7, 10}

And,

A – C = {3, 5, 7, 9} – {1, 4, 7, 10} = {3, 5, 9}

So, (A – C) ∩ B = {3, 5, 9} ∩ {1, 2, 3, 8, 9} {3, 9}

 

b.

Soln:

Here, U = {x:x is a natural number upto 20}

So, U = {1, 2, 3,...20}

A = {x:x≥ 6} = {6, 7, 8, ….20}

B = {x:x ≤ 8} = {1, 2, 3, ….8}

C = {x:10< x < 15} = {11, 12, 13, 14}

Now, B U C = {1, 2, 3, 4, 5, 6, 7, 8} U {11, 12, 13, 14}

So, B U C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

Again, A∩B = {6, 7, 8, …20} ∩ {1, 2, 3, ….8} = {6, 7, 8}

And, A – C = {6, 7, 8….20} – {11, 12, 13, 14}

= {6, 7, 8, 9, 10, 15, 16, 17, 18, 19, 20}

And A U B = {6, 7, 8, …., 20} U {1, 2, 3….8}

= {1, 2, 3, …..20} = U.

So, $\overline {{\rm{AUB}}} $ = U – (AUB) = U – U = ɸ.

 

4.

a.

Soln: Here,

A = {x:x = 2n + 1, n≤5, nԑN}

So, A = {3, 5, 7, 9, 11}

B = {x:x = 3n – 2, n≤4, nԑN}

So, B = {1, 4, 7, 10}

Now, A U B = {3, 5, 7, 9, 11} U {1, 4, 7, 10}

= {1, 3, 4, 5, 7, 9, 10, 11}

A ∩ B = {3, 5, 7, 9, 11} ∩ {1, 4, 7, 10} = {7}

And, B – A = {1, 4, 7, 10} – {3, 5, 7, 9, 11} = {1, 4, 10}

 

b.

Soln:
here, U = {1, 2, 3, ….},

A = {3, 6, 9, 12, 15, 18}

B = {6, 12, 18, 24, 30}

So, A ∩ B = {3, 6, 9, 12, 15, 18} ∩ {6, 12, 18, 24, 30}.

= {6, 12, 18}

And A – B = {3, 6, 9, 12, 15, 18} – {6, 12, 18, 24, 30} = {3, 9, 15}

 

c.

Soln;

Here, U = {x: –1≤x – 2 ≤7} = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A ={x:x is a prime number} = {2, 3, 5, 7}

B = {x:x is a odd number} = {1, 3, 5, 7, 9}

Now,

(i) The sets if elements which are either prime or odd = A U B = {2, 3, 5, 7} U {1, 3, 5, 7, 9} = {1, 2, 3, 5, 7, 9}

 

(ii) The sets if elements which are prime as well as odd = A U B = {2, 3, 5, 7} ∩ {1, 3, 5, 7, 9} = {3, 5, 7}

 

(iii) The sets if elements which are prime but not odd = A – B = {2, 3, 5, 7} – {1, 3, 5, 7, 9} = {2}

 

5.

Soln: Here,

A U B = {a, b, c, d} U {c, d, e, f} = {a, b, c, d, e, f}

A U C = {a, b, c, d} U {d, e, f, g, h} = {a, b, c, d, e, f, g, h}

B ∩ C = {c, d, e, f} U {d, e, f, g, h} = {d, e, f}

a.

L.H.S. = A U (B ∩ C)

= {a, b, c, d, e, f} U {d, e, f} = {a, b, c, d, e, f}

R.H.S. = (A U B) ∩ (A U C)

= {a, b, c, d, e, f} ∩ {a, b, c, d, e, f, g, h} = {a, b, c, d, e, f}

Hence, A U (B ∩ C) = (A U B) ∩ (A U C)

 

b.

A ∩ B = {a, b, c, d} ∩ {c, d, e, f} = {c, d}

A ∩ C = {a, b, c, d} ∩ {d, e, f, g, h} = {d}

B U C = {c, d, e, f} U {d, e, f, g, h} = {c, d, e, f, g, h}

Now, LHS = A ∩ (B U C) = {a, b, c, d} ∩ {c, d, e, f, g, h} = {c, d}

RHS = (A∩B) U (A ∩ C) = {c, d} U {d} = {c, d}

Hence. A ∩ (B U C) = (A∩B) U (A ∩ C)

 

c.

soln: ${\rm{\bar A}}$ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c, d} = {e, f, g, h}

${\rm{\bar B}}$ = U – B = {a, b, c, d, e, f, g, h} – {c, d, e, f} = {a, b, g, h}

Now, LHS = $\overline {{\rm{AUB}}} $

= U – (AUB)

= {a, b, c, d, e, f, g, h} – {a, b, c, d, e, f} = {g, h}

R.H.S. = ${\rm{\bar A}}\mathop\cap\nolimits^ {\rm{\bar B}}$ = {e, f, g, h} ∩ {a, b, g, h} = {g, h}

Hence, $\overline {{\rm{AUB}}}  = {\rm{\bar A}}\mathop\cap\nolimits^ {\rm{\bar B}}$.

 

d.

Here,

A – B = {a, b, c, d} – {c, d, e, f} = {a, b}

A – C = {a, b, c, d} – {d, e, f, g, h} = {a, b, c}

Now, LHS = A – (B U C) = {a, b, c, d} – {c, d, e, f, g, h} = {a, b}

RHS = (A – B) ∩ (A – C) = {a, b} ∩ {a, b, c} = {a, b}

Hence, A – (B U C)= (A – B) ∩ (A – C).

 

6.

a.

Soln:

(i)

We have,

n (A U B) = n(A) + n(B) – n(A∩B)

Or, 80 = 75 + 40 – n (A∩B)

Or, n(A∩B) = 155 – 80 = 35

 

(ii) n(A – B) = n(A) – n (A∩B) = 75 – 35 = 40

 

(iii) $\overline {{\rm{n}}\left( {{\rm{AUB}}} \right)} $ = n(U) – n(A U B) = 100 – 80 = 20

 

b.

Soln:

(i) n(AUB) = n(B)              [A ⊂ B]

So, n (A U B) = 50

 

(ii) n(AnB) = n(A)              [A⊂B]

So, n(AnB) = 37.

 

7.

Soln:

Let B = basketball, V = volleyball

So, n(B) = 32, n(V) = 25, n(B∩V) = 13

Thus, the number of students playing at least one game,

n(BUV) = n(B) + n(V) – n(B∩V) = 32+25 – 13 = 44.

 

8.

Soln:

Here, Let N = Nepali, Ne = newari,

So, n(N) = 60%, n(Ne) = 50%, n(N u Ne) = 100%.

n(N ∩ Ne) = ?,

We have,

n(N U Ne) = n(N) + n(Ne) – n(N ∩ Ne)

or, 100% = 60% + 50% – n(N ∩ Ne)

So, n(N ∩ Ne) = 110% – 100% = 10%.

 

9.

Soln: Here, let M = Mathematics, S = statistics,

So, n(U) = 120, n(M) = 90, n(S) = 72.

n$\overline {{\rm{MUS}}} $ = 10, n(M∩S) = ?

We have,

n(U) = n(M) + n(S) – n(M∩S) + n$\overline {{\rm{MUS}}} $

or, 120 = 90 + 72 – n(M∩S) + 10

so, n(M∩S) = 52.

 

10.

Soln:

(a)

Let T = tea and C = coffee

So, n(U) = 65, n(T) = 38, n (T – C) = 20

n$\overline {{\rm{TUC}}} $ = 15.

To find, (a) n(T∩C) (b) n(C – T)

= 38 – 20

= 18.

 

(b)

n(C – T) = ?         [first we have to find n(C)]

Since, n(U) = n(T) + n(C) – n(T∩C) + n $\overline {{\rm{TUC}}} $

Or, 65 = 38 + n(C) – 18 + 15

So, n(C) = 65 – 35 = 30.

So, n(C – T) = n(C) – n(T ∩ C) = 30 – 18 = 12.

 

11.

Soln:

Let M = mathematics, S = statistics

So, n(U) = 100%, n(M) = 27%, n(S) = 31%.

n(M∩S) = 60%

Failed in examination, n(MUS) = ?

We have, n(MUS) = n(M) + n(S) – n(M∩S)

= 27% + 31% – 6% = 58% – 6% = 52%.

Passed in both subjects, $\overline {{\rm{n}}\left( {{\rm{MUS}}} \right)} $ = 100% – 52% = 48%.

 

12.

Soln:

Let M = Mathematics, P = physics,

S = Statistics,

n(M) = 12,

n(P) = 11

n(S) = 15

n_o(M) = 4

n_o(S) = 7

n_o(S∩P) = 3

n_o(M∩S) = 1

Now,

(a)

Students reading all three subjects,

n(M∩P∩S) = n(S) – n_o(M) – n_o(M∩S) – n_o(S∩P) – n_o(S)

= 15 – 1 – 3 – 7 = 4

 

(b)

Students reading math’s and physics only,

n_o(M∩P) = n(M) – n_o(M) – n_o(M∩S) – n(M∩P∩S)

= 12 – 4 – 1 – 4 = 3.

 

(c)

Students reading physics only, n_o(P)

We have, n_o(P) = n(P) – n_o(M∩P) – n_o(S∩P) – n(M∩P∩S)

= 11 – 3 – 3 – 4 = 1.

 

(d)

Total students, from Venn – diagram,

n(M U S U P) = 4 + 1 + 7 + 3 + 1 + 3 + 4

so, n(M U S U P) = 23.

 

13.

Soln:

Let F = football, H = hockey, B = basketball,
So, n(U) = 500, n(F) = 285, n(H) = 195, n(B) = 115.

N(F∩B) = 45, n(F∩H) = 70, n(H∩B) = 50.

n$\overline {{\rm{FUHUB}}} $ = 50,

 

(a)

How many watch all three games, n(F∩H∩B) = ?

We have,

n(U) = n(F) + n(H) + n(B) – n(F∩H) – n(F∩B) – n(H∩B) + n(F∩H∩B) + n$\overline {{\rm{FUHUB}}} $

or, 500 = 285 + 195 + 115 – 70 – 45 – 50 + n(F∩H∩B) + 50

or, 500 = 595 – 115 + n(F∩H∩B)

or, n(F∩H∩B) = 500 – 480 = 20.

 

(b)

From the given Venn – diagram, the viewers watching exactly one of the tree games,

= n_o(F) + n_o(H) + n_o(B)

= 190 + 95 + 40.

= 325