In this page, you can find the complete solutions of the second exercise of Logic, Sets, and Real Number System chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, logic, sets, and real number system is the 1st chapter and has four exercises only. Out of which, this is the solution of the second exercise.

Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here. Few questions have been typed while most of them have been updated as pictures.

Sets Exercises



1. If U = {1,2,3,4,...,9,10}, A={1,2,3,4}, B = {3,5,7,8} and C={1,2,7,8}, find
$a) A \cup B$
$b) A \cap C$
$c) (A-B) \cap C$
$d) \overline{A-C}$
$e) \overline{B} \cup \overline{C}$
$f) (A \cup B) - C$

Solution:

Given,
$\text{U} = {1,2,3,4,5,6,7,8,9,10}$

$\text{A} = {1,2,3,4}$
$\text{B} = {3,5,7,8}$
$\text{C} = {1,2,7,8}$

Now,
$a) A \cup B$
= ${1,2,3,4} \cup {3,5,7,8}$
$= {1,2,3,4,5,7,8}$
$b) A \cap C$ = ${1,2,3,4} \cap {1,2,7,8}$

$= {1,2}$


$c) (A-B) \cap C$
= $[{1,2,3,4} - {3,5,7,8}] \cap {1,2,7,8}$

$= {1,2,4} \cap {1,2,7,8}$
$= {1,2}$

$d) \overline{A \cup C}$ 

= $\overline{ {1,2,3,4} \cup {1,2,7,8} }$

$= \overline{ {1,2,3,4,7,8} }$

$= U - {1,2,3,4,7,8}$
$= {5,6,9,10}$

$e) \overline{B} \cup \overline{C}$

$= [ U - B ] \cup [U - C]$
$= [ {1,2,3,4,5,6,7,8,9,10} - {3,5,7,8} ] \cup [{1,2,3,4,5,6,7,8,9,10} - {1,2,7,8}]$

$= {1,2,4,6,9,10} \cup {3,4,5,6,9,10}$

$= {1,2,3,4,5,6,9,10}$


$f) (A \cup B) - C$
$= [{1,2,3,4} \cup {3,5,7,8} ] - {1,2,7,8}$

$= {1,2,3,4,5,7,8} - {1,2,7,8}$

$= {3,4,5}$



2. Let U = {a,b,c,d,e,f,g,h,i,j,k}, A = {b,c,d,e}, B = {d,e,f,g,h,i}, C = {a,e,i,o,u}, D = {b,d,j,k}

Solution:
Here,

$\text{U} = {a,b,c,d,e,f,g,h,i,j,k}$

$\text{A} = {b,c,d,e}$

$\text{B} = {d,e,f,g,h,i}$

$\text{C} = {a,e,i,o,u}$

$\text{D} = {b,d,j,k}$

Perform the following indicated operations

a) $A \cup B$

$\implies {b,c,d,e} \cup {d,e,f,g,h,i}$

$\implies {b,c,d,e,f,g,h,i}$


b) $A \cap C$

$\implies {b,c,d,e} \cap {a,e,i,o,u}$

$\implies {e}$


c) $A - B$

$\implies {b,c,d,e} - {d,e,f,g,h,i}$

$\implies {b,c}$


d) $A - C$

$\implies {b,c,d,e} - {a,e,i,o,u}$

$\implies {b,c,d}


e) $(A-C) \cap C$

$\implies [ {b,c,d,e} - {a,e,i,o,u} ] \cap C$

$\implies {b,c,d} \cap {a,e,i,o,u}$

$\phi$


f) $A \triangle D$

$\implies (A-D) \cup (D-A)$

$\implies [ {b,c,d,e} - {b,d,j,k} ] \cup [ {b,d,j,k} - {b,c,d,e}]$

$\implies {c,e} \cup {j,k}$

$\implies {c,e,j,k}$


g) $(A \cup B) - C$

$\implies [ {b,c,d,e} \cup { d,e,f,g,h,i} ] - C$

$\implies {b,c,d,e,f,g,h,i} - {a,e,i,o,u}$

$\implies {b,c,d,f,g,h}$


h) $A \cap \overline{B}$

$\implies A \cap [ U - B]$

$\implies A \cap [ {a,b,c,d,e,f,g,h,i,j,k} - {d,e,f,g,h,i}]$

$\implies {b,c,d,e} \cap {a,b,c,j,k}$

$\implies {b,c}$


i) $\overline{A-B}$

$\implies U - (A-B)$

$\implies U - [ {b,c,d,e} - {d,e,f,g,h,i} ]$

$\implies {a,b,c,d,e,f,g,h,i,j,k} - {b,c}$

$\implies {a,d,e,f,g,h,i,j,k}$


3 a) Given the sets,
$\text{U} = {x:x is a positive integer less than 12}, \text{A} = {3,5,7,9}$, $B= {1,2,3,8,9}$, $C = {1,4,7,10}$, find $(\overline{A \cup B} ), (A-B) \cup C$, (A-C)\cap B.$

Solution:
Here,

$\text{U} = {x:x is a positive integer less than 12}$

$\implies \text{U} = {1,2,3,4,5,6,7,8,9,10,11}$

$\text{A} = {3,5,7,9}$

$\text{B} = {1,2,3,8,9}$

$\text{C} = {1,4,7,10}$

Now,

$\text{(i)} (\overline{A \cup B})$

$= U - (A \cup B)$

$= U - [{3,5,7,9} \cup {1,2,3,8,9}]$

$= {1,2,3,4,5,6,7,8,9,10,11} - {1,2,3,5,7,8,9}$

$= {4,6,10,11}$


$\text{(ii)} (A - B) \cup C$

$= [{3,5,7,9} - {1,2,3,8,9} ] \cup C$

$= {5,7} \cup {1,4,7,10}$

$= {1,4,5,7,10}$


$\text{(iii)} (A - C) \cap B$

$= [ {3,5,7,9} - {1,4,7,10} ] \cap B$

$= {3,5,9} \cap {1,2,3,8,9}$

$= {3,9}$


3 b) Given U = {x: x is a natural number upto 20}, A = {x:x $\geq}$ 6}, B = {x: x $\leq$ 8} and C = {x:10 < x <15}, find $B \cup C$, $A \cap B}$ , $A -C$, $\overline{A \cup B}$.

Solution:
Here,

$\text{U} = {x: x is a natural number upto 20}$

$implies \text{U} = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}$

$\text{A} = {x:x \geq 6$

$\implies \text{A} = {6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}$

$\text{B} = {x:x \leq 8}$

$\implies \text{B} = {1,2,3,4,5,6,7,8}$

$\text{C} = {x:10 < x< 15}$

$\implies \text{C} = {11,12,13,14}$


Now,

$\text{(i)} B \cup C$

$\implies { 1,2,3,4,5,6,7,8} \cup {11,12,13,14}$

$\implies {1,2,3,4,5,6,7,8,11,12,13,14}$


$\text{(ii)} A \cap B$

$\implies {6,7,8,9,10,11,12,13,14,15,16,17,18,19,20} \cap {1,2,3,4,5,6,7,8}$

$\implies {6,7,8}$


$\text{(iii)} A - C$

$\implies {6,7,8,9,10,11,12,13,14,15,16,17,18,19,20} - {11,12,13,14}$

$\implies {6,7,8,9,10,15,16,17,18,19,20}$
4 a) If A = {x: x = 2n + 1, n $leq$ 5, n $\epsilon$ N} and B = {x: x = 3n - 2, n $\leq$ 4, n $\epsilon$ N}, find A $\cup$ B, A $\cap$ B, and B $-$ A.

Solution:

Here,
For $n \leq 5 \text{and} n \epsilon N$, set A = ${3,5,7,9,11}$

For $n \leq 4 \text{and} n \epsilon N$, set B = ${1, 4,7,10}$

Now,

$(i) A \cup B$

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

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

And,

$(ii) A \cap B$

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

$= {7}$

Also,

$(iii) B - A$

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

$= {1,4,10}$


4 b) If A is the set of all multiples of 3 less than 20, B is the set of multiples of 6 less or equal to 30 and U is the set of all natural numbers, find A $\cap$ B and A $-$ B.

Solution:
According to the question,

$\text{A} = {3,6,9,12,15,18}$

$\text{B} = {6,12,18,24,30}$

$\text{U} = {1,2,3,4,5,6,.....}$

Now,

$(i) A \cap B$

$= {3,6,9,12,15,18} \cap {6,12,18,24,30}$

$= {6,12,18}$

And,

$(ii) A - B$

$= {3,6,9,12,15,18} - {6,12,18,24,30}$

$= {3,9,15}$


4 c) If $\text{U} = {x: -1 \leq x -2 \leq 7}, \text{A} = {x: x is a prime number} and \text{B} = {x: x is an odd number}, \text{find}$
  1. the set of elements which are either prime or odd.
  2. the set of elements which are prime as well as odd.
  3. the set of elements which are prime but not odd.
Solution:

According to the question,

$\text{U} = {1,2,3,4,5,6,7,8,9}$

So the sets A and B must be a subset of set U.

$\text{A} = {2,3,5,7}$ $\implies \text{Set A represents prime numbers}$

$\text{B} = {1,3,5,7,9}$ $\implies \text{Set B represents odd numbers}$

Now,

i) the set of elements which are either prime or odd $\implies A \cup B$

$= {2,3,5,7} \cup {1,3,5,7,9}$

$= {1,2,3,5,7,9}$

And,

ii) the set of elements which are prime as well as odd $\implies A \cap B$

$= {2,3,5,7} \cap {1,3,5,7,9}$

$= {3,5,7}$

Also,

iii) the set of elements which are prime but not odd $\implies A - B$

$= {2,3,5,7} - {1,3,5,7,9}$

$= {2}$


5. Let U = {a,b,c,d,e,f,g,h}, A= {a,b,c,d}, B= {c,d,e,f} and C= {d,e,f,g,h}. Verify the following relations

$\text{a)} A \cup ( B\cap C) = (A \cup B) \cap (A \cup C)$

$\text{b)} A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

$\text{c)} \overline{A \cup B} = \overline{A} \cap \overline{B}$

$\text{d)} A - (B \cup C) = (A - B) \cap (A - C)$

Solution:
Here,
$U = {a,b,c,d,e,f,g,h}$
$A = {a,b,c,d}$
$B= {c,d,e,f}$
$C= {d,e,f,g,h}$

Now,

$\text{a)} A \cup ( B\cap C) = (A \cup B) \cap (A \cup C)$

LHS = $A \cup (B \cap C)$
$= A \cup [ {c,d,e,f} \cap {d,e,f,g,h} ]$

$= {a,b,c,d} \cup {d,e,f}$
$= {a,b,c,d,e,f}$
RHS = (A \cup B ) \cap (A \cup C)$

$= [ {a,b,c,d} \cup {c,d,e,f} ] \cap [ {a,b,c,d} \cup {d,e,f,g,h} ]$

$= {a,b,c,d,e,f} \cap {a,b,c,d,e,f,g,h}$

$= {a,b,c,d,e,f}$
Since LHS = ${a,b,c,d,e,f}$ = RHS, given relation is verified.


And, 

$\text{b)} A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

LHS = $A \cap (B \cup C)$
$= A \cap [ {c,d,e,f} \cup {d,e,f,g,h} ]$

$= {a,b,c,d} \cap {c,d,e,f,g,h}$

$= {c,d}$

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

$= [{a,b,c,d} \cap {c,d,e,f}] \cup {a,b,c,d} \cap {d,e,f,g,h}]$

$= {c,d,} \cup {d}$
$= {c,d}$
Since LHS = ${c,d}$ = RHS, given relation is verified.


Also,
$\text{c)} \overline{A \cup B} = \overline{A} \cap \overline{B}$

LHS = $\overline{A \cup B}$
$= \overline{ {a,b,c,d} \cup {c,d,e,f} }$

$= \overline{ {a,b,c,d,e,f} }$

$= U - {a,b,c,d,e,f}$
$= {a,b,c,d,e,f,g,h} - {a,b,c,d,e,f}$

$= {g,h}$
RHS = \overline{A} \cap \overline{B}$

$= (U -A ) \cap (U - B)$
$= [{a,b,c,d,e,f,g,h} - {a,b,c,d} ] \cap [ {a,b,c,d,e,f,g,h} - {c,d,e,f}]$

$= {e,f,g,h} \cap {a,b,g,h}$
$= {g,h}$
Since LHS = ${g,h}$ = RHS, given relation is verified.


Again,

$\text{d)} A - (B \cup C) = (A - B) \cap (A - C)$

LHS = $A - (B \cup C)$
$= A - [{c,d,e,f} \cup {d,e,f,g,h} ]$

$= {a,b,c,d} - {c,d,e,f,g,h}$
$= {a,b}$
RHS = $(A - B) \cap (A -C)$
$= [{a,b,c,d} - {c,d,e,f} ] \cap [ {a,b,c,d} - {d,e,f,g,h} ]$

$= {a,b} \cap {a,b,c}$
$= {a,b}$
Since LHS = ${a,b}$ = RHS, given relation is verified.

About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)