Download QUESTION PAPERS SOLUTION UNIT 1 Set Theory Define power

QUESTION PAPERS SOLUTION
UNIT 1
Set Theory
1. Define power set of a set. Determine power sets of the following sets. (4m) Jun
2015 / Jan 2014
Sol: Power Set: The collection of all subsets of a set A is called the power set of A,
and is represented P(A).
For instance, if A = {1, 2, 3,4}, then
P(A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3},{1,2,3},{2,3,4},
{1,3,4},{1,2,4}. A}.
2. Using laws of set theory show that A – (A – B) = A – (A ∩ B) (5m) Jun 2015/Jan
2014
Sol: A – (A – B) = A – (A ∩ B1) [i.e. A – B = A ∩ B1]
= A ∩ (A ∩ B1)1 [i.e. A – B = A ∩ B1]
= A ∩ (A1 U B) [DeMorgan’s Law]
= (A ∩ A1) U (A ∩ B) [Distributive Law]
= ø U (A ∩ B) [i.e. A ∩ A1 = ø]
= A ∩B
3. In a survey of 260 college students, the following data were obtained: 64 had
taken a mathematics course, 94 had taken a computer science course, 58 had
taken a business course, 28 had taken both a mathematics and a business course,
26 had taken both a mathematics and a computer science course, 22 had taken
both a computer science and a business course, and 14 had taken all three types
i.
How many of these students had taken none of the three courses?
ii.
How many had taken only a computer science courses ? (11m) Jun 2015 / Jan
2014
Solution:
Given: U=260 |A|=64 |B|=94 |C|=58 |A ∩ C| = 28 |A ∩ B| = 26 |B ∩ C| = 22 |A ∩ B ∩ C| =
14
|A U B U C| =? |B1| =?
|A U B U C| = A + B + C – |A ∩ B| – |B ∩ C| – |A ∩ C| + |A ∩ B ∩ C|
= 64 + 94 + 58 – 26 – 22 – 28 + 14
= 154
i. |A U B U C| = U – |A U B U C|
= 260 – 154 = 106 ii.|B1| = |B| - |B ∩ C| - |B ∩ A| + |A ∩ B ∩ C|
= 94 – 22 – 26 + 14
= 60
4. For any two sets A and B, prove the following (4m)
jun 2013
A – (A – B) = A ∩ B
Solution: A – (A – B) = A – (A ∩ B1) [i.e A – B = A ∩ B1]
= A ∩ (A ∩ B1)1 [i.e A – B = A ∩ B1]
= A ∩ (A1 U B) [Demorgan’s Law]
= (A ∩ A1) U (A ∩ B) =
ø U (A ∩ B)
= A ∩B
5.
State and prove De – Morgan’s law of set theory.
(6m)
jun 2013
Solution: The complement of union of 2 sets is equal to the intersection of complement of
the sets.
AUB=A∩B
Consider RHS, A B = { x / x ε A and x ε B }
= { x / x ¢ A and x ¢ B }
={x/x¢
(A U B) }
=AUB
• The union of compliment of 2 sets is equal to the compliment of intersection of the
2 sets.
AUB=A∩B
Consider LHS, A U B = { x / { x / x ε A or x ε B }
= { x / x ¢ A or x ¢ B }
(A ∩ B) }
={x/x¢
=A∩B
6. In a survey of 260 college students, the following data were obtained: 64 had taken
a mathematics course, 94 had taken a computer science course, 58 had taken a
business course, 28 had taken both a mathematics and a business course, 26 had
taken both a mathematics and a computer science course, 22 had taken both a
computer science and a business course, and 14 had taken all three types of courses.
(8m)
i.
How many of these students had taken none of the three courses?
ii.
How many had taken only a computer science courses jun 2013
Solution:
Given: U=260 |A|=64 |B|=94 |C|=58 |A ∩ C| = 28 |A ∩ B| = 26 |B ∩ C| = 22 |A ∩ B ∩ C| =
14
|A U B U C| =? |B1| =?
|A U B U C| = A + B + C – |A ∩ B| – |B ∩ C| – |A ∩ C| + |A ∩ B ∩ C|
= 64 + 94 + 58 – 26 – 22 – 28 + 14
= 154
iii. |A U B U C| = U – AUBU
C|
= 260 – 154 = 106 iv.
∩ A| + |A ∩ B ∩ C|
|B1| = |B| - |B ∩ C| - B
= 94 – 22 – 26 + 14
= 60
7. For any two sets A and B, prove the following (2m)
jun 2013
A – (A – B) = A ∩ B
Solution:
A – (A – B) = A – (A ∩ B1) [i.e A – B = A ∩ B1]
= A ∩ (A ∩ B1)1 [i.e A – B = A ∩ B1]
= A ∩ (A1 U B) [Demorgan’s Law]
= (A ∩ A1) U (A ∩ B) [Distributive Law] =øU
(A ∩ B) [i.e. A ∩ A1 = ø]
= A ∩B
8. Determine the sets A and B given that A – B = {1, 2, 4}, B – A = {7, 8} and
AUB = {1, 2, 4, 5, 7, 8, 9}
(4m)
jan 2015
Solution:
Set A
A = (A U B) – (B – A)
= [{1,2,4,5,7,8,9}] –[{7,8}]
= {1, 2, 4, 5, 9}
Set B
B = (A U B) – (A – B)
= [{1,2,4,5,7,8,9}] –[{1,2,4}]
= {5, 7, 8, 9}
9. Let M, P and C be the sets of students taking Mathematics courses, Physics courses
and Computer Science courses respectively in a university. Assume |M| = 300, |P| =
350, |C| = 450, |M \ P| = 100, |M \ C| = 150, |P \ C| = 75, |M \ P \ C| = 10.
How many students are taking exactly one of those courses?
(7m)
Jan 2015
Solution:
We see that |(M\P)-(M\P\C)| = 100-10 = 90, |(M\C)-(M\ P \C)| = 150-10 = 140 and |(P \C)(M \P \C)| = 75-10 = 65.
Then the region corresponding to students taking Mathematics courses only has cardinality
300-(90+10+140) = 60. Analogously we compute the number of students taking Physics
courses only (185) and taking Computer Science courses only (235). The sum 60 + 185 +
235 = 480 is the number of students taking exactly one of those courses.
10. For any three sets A,B and C prove that (A-B)-C =A – (BUC) = (A-C) – (B-C)
jan 2015
Solution: (A-B) – C = (A∩B) ∩C
= A∩ (B∩C)
= A∩ (B∩C)
= A – (B∩C)
(A – C) – (B – C) = (A –C) ∩ (B – C)
= (A ∩ C) ∩ (B – C1)
= (A ∩ C) ∩ (BU C)
= [(A ∩ C) ∩ B)] U [(A ∩ C) ∩C]
= [(A ∩ (C∩B)] U [A ∩ (C∩C)]
= [A ∩ (CUB) ] U (A∩Ø)
= [A ∩ (BUC) ] UØ
= [A ∩ (BUC) ]
= [A - (BUC)]
(8m)
11. Explain the laws of set theory:
jun 2014
(6m)
12. Determine the sets A and B given that A – B = {1, 3, 7, 11}, B – A = {2, 6,
8} and A∩B = {4, 9}
Solution:
(5m)
jun 2014
set A
A = (A – B) + (A∩B)
= [{1, 3, 7, 11}] + [{4, 9}]
= {1, 3, 4, 7, 9, 11}
Set B
B = (B-A) + (A∩B)
= [{2, 6, 8}] – [{4, 9}]
= {2, 4, 6, 8, 9}
13. Prove that: A▲B
. (4m) jun 2014
Solution: A▲B
Let A=P1 U P2
-A) U (A-B).
B=P2 U P3
A▲B = (AUB)-(A∩B) = (P1 U P2 UP3) – (P2) = P1 U P3 ………… (1)
B – A = P3, A – B = P1
Therefore (B – A) U (A – B) = P3 U P1……………………… (2)
A=U - A = P3 U P4, AND B=U – B = P1U P4
B ∩ A= P3, A ∩ B = P1 Therefore (B ∩ A1) U (A ∩ B1) = P3 U P1………(3)
From (1), (2) and (3), A▲B= (B∩A) U (A∩B) = (B-A) U (A-B).
14. Using Venn diagram, prove the following property of the symmetric difference:
A ▲ (B▲C) = (A▲B) ▲ C (4m)
jan 2015
Solution: By using Venn diagram,
Where, A = P1 U P2 U P3 U P4 B = P2 U P3 U P5 U P6 C = P3 U P4 U P6 U P7
L.H.S (B▲C) = (B U C) – (B∩C) = P2 U P4 U P5 U P7
A U (B▲C) = P1 U P3 U P2 U P4 U P5 U P7 A ∩ (B▲C) = P2 U P4 A
▲ (B▲C) = A U (B▲C) - A ∩ (B▲C) = P1 U P3 U P5 U P7.
R.H.S (A▲B) = (A U B) – (B ∩ A) = P1 U P4 U P5 U P7
(A ▲ B) U C) = P1 U P4 U P5 U P6 U P3 U P7
(A ▲ B) ∩ C) = P4 U P6
(A ▲ B)▲C) = (A ▲ B) U C) - (A ▲ B) ∩ C) = P1 U P5 U P3 U P7 So
that L.H.S = R.H.S Hence proved
15. Thirty cars are assembled in a factory. The options available are a transistor, an air
conditioner and power windows. It is known that 15 of the cars have transistor, 8
of them have conditioners and 6 of them have power windows. Moreover, 3 of them
have all three options. Determine at least how many cars do not have any options
at all. (5m) Jan 2014
Solution: Given data:
|U| = 30 Total cars |A| = 15 Transistors |B| =8 Air conditioners
|C|= 6 Power windows |A ∩ B ∩ C| = 3 All options
i) |A U B U C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| +
|A ∩ B ∩ C|
Here A ∩ B, B ∩ C, C ∩ A is a subset of A ∩ B ∩ C So
|A U B U C|
(A U B U C) is the set of cars that do not have any option.
|A U B U C|
- |A U B U C|
≥ 30 – 23 = 7
Therefore a minimum of 7 cars has none of the options.
16. A survey on a sample of 25 new cars showed that the cars had the following
15 cars had air conditioners
12 cars had radios
11 cars had power windows
5 cars had air conditioners and power windows
9 cars had air conditioners and radios
4 cars had radios and power windows
3 cars had all the three options
Find the number of cars that had
i) only power windows
ii) at least one option
(7m)
Jan 2014
Solution: Given data:
|U| = 25 Total cars
|A| = 15 Air conditioners |R|
= 12 Radios
|W|= 11 Power windows
|A ∩ R| = 9 Air conditioners and Radios
|A ∩ W| = 5 Air conditioners and Power windows
|R ∩ W| = 4 Radios and Power windows
|A ∩ R ∩ W| = 3 All options i)
only power windows:
|W| = |W – A – R |
= |W| - |A ∩ W| - |R ∩ W| + |A ∩ R ∩ W|
= 11 – 5 – 4 + 3
=5
5 CARS HAVE ONLY POWER WINDOWS ii)
At least one option:
|W U A U R| = |W| + |A| + |R| - |A ∩ R| - |A ∩ W| - |R ∩ W|
+ |A ∩ R ∩ W|
= 11 + 15 + 12 – 5 – 9 – 4 + 3
= 23
17. A survey of 500 television viewers of sports channel produced the following
information: 285 watch cricket, 195 watch hockey, 115 watch foot ball, 45 watch
cricket and foot ball, 70 watch cricket and hockey, 50 watch hockey and foot ball
and 50 do not watch any of the three kinds of games
i) How many viewers in survey watch all three kinds of games?
ii) How many viewers watch exactly one sport?
Solution: At least one:
(8m)
Jan 2014
|CU H U F| = |U| - |C U F U H|1
=500 – 50 = 450 do watch any of the games
|C U H U F| = |C| + |F| +|F| -|C∩F|-|C∩H|-|H∩F|+|C∩F∩H|
But |C∩F∩H| means those viewers who watch all the 3 games.
So, |C∩F∩H|= |C U H U F| -|C| - |H| -|F| +|C∩F|+|C∩H|+|H∩F|
= 450 – 285 – 195 – 115 +45 +70 +50
= 20
Therefore 20 viewers watch all the 3 games.
2) Only cricket viewers:
|C1| = |C – H - F|
= |C|-|C∩F|-|C∩H|+|C∩F∩H|
= 285 -70 -45 +20
=190
3) only hockey viewers:
|H1| = | H - C - F|
= |H|-|H∩F|-|C∩H|+|C∩F∩H|
= 195 -50 -70 +20
= 95
4) only football viewers:
|F1| = | F- C - H|
= |F|-|H∩F|-|C∩F|+|C∩F∩H|
=115 – 45 -50 +20
= 40
Therefore number of viewers those who watch exactly 1 game is
|C1| + |H1| + |F1|
= 190 + 95 + 40
= 325.
18. The freshman class of a private engineering college has 300 students. It is known
that 180 can program in PASCAL, 120 in FORTRAN, 30 in c++, 12 in PASCAL
and c++, 18 in FORTRAN and c++, 12 in PASCAL and FORTRAN, and 6 in all
three languages If two students are selected at random, what is the probability that
they can
i) Both program in PASCAL?
ii) Both program only in PASCAL?
(6m)
Jan 2014
300
19. In a survey of 120 passengers, an airline found that 48 enjoyed wine with their
meals, 78 enjoyed mixed drinks, 66 enjoyed iced tea. In addition, 36 enjoyed any given
pair of these beverages and 24 enjoyed them all. If two passengers are selected at
random from thee survey sample of 120, what is the probability that they both want
only iced tea with their meals? (7m) Jan 2014 Solution: from the information provided,
we construct the Venn diagram .The sample space constants of the pairs of passengers we
can select from the sample of 120.
120
So |S| =
c =7140
2
A) Only iced tea
|T–W–C|
= |T| -|T∩W| - |T∩C| - |T∩W∩C|
= 66 – 36 – 36 + 24 = 18
The Venn diagram indicates that there are 18 passengers who drink only ice tea.
So |A|
153
Therefore p (A) = |A| / |S|
= 153 / 7140
= 51 / 2380.
B) Exactly 2 of the 3 type
Only T and C = |T∩C| - |T∩W∩C| ………………….i
36 – 24 = 12
Only C and W = |W∩C| - |T∩W∩C| ………………..ii
36 – 24 = 12
Only W and T =|W∩T| - |T∩W∩C| ………………iii
36 – 24 = 12
Adding i, ii, iii we get = 12 + 12 + 12
= 36 = |B|
20. Find the probability of getting a sum different from 10 or 12 after rolling two dice.
(5m) Jan 2015
Solution : We can get 10 in 3 different ways:
4 + 6, 5 + 5, 6 + 4, so P (10) — 3/36. Similarly we get that P (12) — 1/36. Since they are
mutually exclusive events, the probability of getting 10 or 12 is P (10) + P (12) — 3/36 +
1/36 — 4/36 — 1/9. So the probability of not getting 10 or 12 is 1 — 1/9 — 8/9.
21. Explain set operations: (6m)
Jan 2015
Solution:
1. Intersection : The common elements of two sets:
A ∩ B = {x | (x e A) ∧ (x e B)} .
If A ∩ B = ∅, the sets are said to be disjoint.
2. Union : The set of elements that belong to either of two sets:
A ∪ B = {x | (x e A) ∨ (x e B)} .
3. Complement : The set of elements (in the universal set) that do not belong to a given set:
A = {x e U | x /e A} .
4. Difference or Relative Complement : The set of elements that belong to a set but not to
another:
A - B = {x | (x e A) ∧ (x /e B)} = A ∩ B .
5. Symmetric Difference : Given two sets, their symmetric differ- ence is the set of elements
that belong to either one or the other set but not both.
A ⊕ B = {x | (x e A) ⊕ (x e B)} .
22. A compuer services company has 300 programmers. It is known that 180 of these
can program pascal, 120 in FORTRAN, 30 in c++, 12 in pascal and c++, 18 in
FORTRAN and c++, 12 in pascal and FORTRAN and 6 in all the three.
a) If a programmer is selected at random what is the probability that she
can program in exactly two languages?
Page
b) If two programmers are selected at random what is the probability that
they can both program in pascal? (10m) Jan 2014
Solution:
a) We note that the number of programmers who can program in
1) pascal and FORTRAN only is 6
2) pascal and C++ only is 6
3) FORTRAN and c++ only is 12
Thus 6+6+12=24 can program in exactly two languages.
Pr(A)=24/300=8%
b) the number of ways of selecting 2 programmers from 300 is
300 c 2. Same way selecting 2 in pascal is 180 c 2.
Same way selecting 2 programmers who can program in pascal is 162 c 2.
The probability of selecting two who can program in pascal is
Pr(B)=180 c 2/300 c 2= 36%
The probability of selecting 2 who can program in both is
Pr©=162 c 2/300 c 2=29%
23. Define power set of a set. Obtain all the power sets of A={1,2,3,4} (3m)Jan 2014
Sol: Power Set: The collection of all subsets of a set A is called the power set of A,
and is represented P(A). For instance, if A = {1, 2, 3,4}, then
P(A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3},{1,2,3},{2,3,4},
{1,3,4},{1,2,4}. A} .
24. Simplify the following expression:
(4m)
Jan 2014
Sol:
UNIT 2
Fundamentals of Logic
1. Let p, q be primitives statements for which implication p → q is false. Determine
the truth values of the following. Jun 2015 / Jan 2014
Sol:
5 marks
pvq
qvp
(p v q) ↔ (q v p)
p
q
T
T
T
T
T
T
F
T
T
T
F
T
T
T
T
F
F
F
F
T
2. By constructing the truth table. Show that the compound propositions pAND(~q V
r) Jun 2015 / Jan 2014
4. Let p, q be primitives statements for which implication .June 2015 / Jan 2014
Sol:
5 marks
5. Discuss the basic connectives that are used in logic. (6m)
Jun 2013
Solution: new proposition are often formed by starting with given propositions with the
aid of words or phrases like “not”, “and”, “if…then”, and “if and only if” such words or
phrases w=are called connectives
a)
Negation: a proposition obtained by inserting the word “not” at an appropriate places
in a given propositions called negation of given proposition and is denoted by ~p ( p is any
proposition).
b)
Conjunction: a compound proposition obtained by combining two given
propositions by inserting the word “and” in between them is called conjunction (denoted by
p ^ q)
c)
Disjunction: a compound proposition obtained by inserting the word “or” in between
them is called disjunction (denoted by p v q)
d)
Conditional: a compound proposition obtained by combining two given proposition
by using the word “if and then” at appropriate place is called a conditional proposition and
is denoted by “p →q”
6. Given p and q statements, explain the following terms
(7m) Jun 2013
a) Conjunction b) disjunction c) logically Equivalence d) tautology Solution:
a) conjunction: a compound proposition obtained by combining two given propositions
by inserting the word “and” in between them is called conjunction (denoted by p ^ q)
b) Disjunction: a compound proposition obtained by inserting the word “or” in between them
is called disjunction (denoted by p v q)
c) Logical equivalence : Two propositions p and q are said to be logical equivalent where p
and q have the same truth value or equivalently the biconditional p ↔ q is tautology.
Then we denote p ≡ q.
d) Tautology: a compound proposition which is always true regardless of the truth values of
its components is called a Tautology.
7. show that (p v q) ↔ (q v p) is a tautology.
(3m)
Jun 2013
Solution: we have to show that (p v q) ↔ (q v p) is Tautology we can write the truth
table as follows.
p
q
pvq
qvp
(p v q) ↔ (q v p)
T
T
T
T
T
T
F
T
T
T
F
T
T
T
T
F
F
F
F
T
8. Define converse, inverse and contra positive of a statement:
Solution:
(4m) Jun 2013
Consider a conditional (p→q) , Then :
1) q→p is called the converse of p→q
2) �p→�q is called the inverse of p→q
3) �q→�p is called the contrapositive of p→q
4) The converse of a conditional proposition p → q is the proposition q → p. As we
have seen, the bi- conditional proposition is equivalent to the conjunction of a
conditional proposition an its converse.
p ↔ q ≡ (p → q) ∧ (q → p)
So, for instance, saying that “John is married if and only if he has a spouse”
is the same as saying “if John is married then he has a spouse” and “if he has
a spouse then he is married”.
Note that the converse is not equivalent to the given conditional proposition,
for instance “if John is from Chicago then John is from Illinois” is true, but
the converse “if John is from Illinois then John is from Chicago” may be
false.
9. Find the truth value of p,q,r for the following using truth tables:
(5m)
Sol:
10. Prove the following tautologies: (5m) Jan 2015
Sol:
11. Prove the following:
(6m)
Jan 2015
Jun 2013
Sol:
12. Find the truth values for the following logical expressions: (4m) Jan 2015 Sol:
13. Write the truth table for the following:
(8m)
jun 2014
Sol:
14. Simplify the following compound statements:
(6m)
jun 2014
Sol:
15. Verify whether the following logical expressions are tautology or contradiction
using truth tables: (6m) jun 2014
Sol:
16. Prove the following logical statement is a tautology: (5m) jun 2014
Sol:
17. Prove the following logical statement is a tautology:
(5m)
Sol:
18. Prove the following logical statement is a tautology: (5m) jan 2014
Sol:
19. Prove the following logical statement is a tautology. (5m) jan 2014
Sol:
jun 2014
UNIT 3
Fundamentals of Logic contd
1) Find inverse, converse and contra positive of the following (6m) Jan 2014
Sol:
2) Find inverse, converse and contra positive of the following:
S o l:
(6m)
Ja n 20 14
3) Simplify the following with reasons:
(4m)
Jan 2014
Sol:
4. P rrove the fo llow ing pr im itive statem ents:
(7 m )
Jun 201 3
5. Prove below open statements:
(4m)
6. Prove below quantifiers:
Jun 2013
(6m)
Jun 2013 sol:
Sol:
7. Write inverse, converse and contra-positive:
Sol:
8. Prove the below open statements: (7m) jan 2015
(5m)
Jun 2013
Sol:
9. Check the validity of the following arguments:
Sol:
(3m)
jan 2015
11. Prove the following rules of inferences
(6m)
jun 2014
(4m)
jun 2014
Sol:
12. Prove the following rules of inferences
Sol:
13. Verify the rules of inference from the following truth tables: (6m) jun 2015
Sol:
14. Pro ve the fo llow ing open state ments:
(4 m )
jun 201 5
Sol:
15. Find converse inverse and contra positive of the logical expressions given
below:
Sol:
(4m)
jun 2015