Revised tutorial4-is1-14
Tutorial
Exercise 1 The demand and supply functions for a good are given by
Pd  9  0.6Qd -----demand function
Ps  2  0.4Qs -----supply function
Calculate the equilibrium price and quantity graphically. Confirm your answer
algebraically.
Solution
20
16
9 0.6Q
12
2 0.4Q
8
4
0
0
4
8
12
16
20
Q
Graphical solution is approximately Q=7 and P=5.
Algebraically Pd  Ps  9  0.6Q  2  0.4Q  Q  7 and P  9  0.6(7)  9  4.2  4.8
P  0.6Q  9
OR
Subtracting Q  7
P  0.4Q  2
P  2  0.4Q  2  0.4(7)  4.8 .
1
Exercise 2 The demand and supply functions for a good(jeans) are given by
Pd  50  3Qd -----demand function
Ps  14  1.5Qs -----supply function
Calculate the equilibrium price and quantity graphically. Confirm your answer
algebraically.
Solution
50
43.75
37.5
31.25
50 3Q
14 1.5Q
25
18.75
12.5
6.25
0
0
6.25
12.5
18.75
25
31.25
37.5
43.75
50
Q
Graphical solution is approximately Q=7.5 and P=25.5.
Algebraically Pd  Ps  50  3Q  14  1.5Q  4.5Q  36  8 and P  50  3(8)  26
P  3Q  50
OR
Subtracting 4.5Q  36  Q  8
P  1.5Q  14
P  14  1.5Q  14  1.5(8)  26 .
2
Tutorial on Logic
Exercise 1
Define a proposition or statement. Which of the following are statements or
propositions. Explain?
(a)The moon is made of green cheese.
(b) He is certainly a tall man.
(c) Two is a prime number.
(d)Will the game be soon over?
(e) Next year interest rates will rise.
(f) Next year interest rates will fall.
(g) Give me your book.
(h)My laptop computer works.
(i)Set up my computer.
(j)All my computer files are binary files.
Solution
A statement or proposition in logic is a statement that is either true(T) or false(F).
(a) (c) (e) (f) (h) and (j) are propositions because they have truth values that are
either true(T) or false(F).
(b) is not a proposition because it is not well defined—he is unknown.
(d) is not a proposition because it is a question.
(g) is not a proposition because it is a command.
(i) is not a proposition because it is a command.
Exercise 2 Determine whether each of the following sentences is a proposition.
(i)In 1998 William Clinton was president of the United States.
(ii)Compile my computer program.
(iii)What time is it?
(iv)Fifteen is an even number.
Explain.
Solution
(i)In 1998 William Clinton was president of the United States.
Yes this is a proposition as it is either true or false.
(ii)Compile my computer program.
No this is not a proposition as it is command.
(iii)What time is it?
No this is not a proposition as it is question.
(iv)Fifteen is an even number.
Yes this is a proposition as it is either true or false.
3
Exercise 3 If p stands for the proposition “The specification is suitable” and
q stands for “The programming team is happy” write out in English
the meaning of the following symbolic statements.
(i)  p  q , (ii)  p  q    p  q  .
Note not A  A or A
Solution
(a)Let p : The specification is suitable. Let q : The programming team is happy.
(i) p  q : The specification is not suitable and the programming team is happy.
(ii)  p  q    p  q  : If the specification is suitable then the programming
team is happy or the specification is suitable and the
programming team is not happy.
Exercise 4 Using letters for component statements, translate the following compound
statements into symbolic logic notation.
(i)
If Brian wins the election, then tax rates will be reduced.
(ii)
Tax rates will be reduced only if Brian wins the election and the economy
remains strong.
(iii)
Tax rates will be reduced if the economy remains strong.
(iv)
A strong economy will follow from Brian winning the election.
(v)
The economy will remain strong iff Brian wins the election or tax rates are
reduced.
Solution Let A : Brian wins the election.
Let B : Tax rates will be reduced.
Let C : Economy remains strong.
(i) A  B , (ii) B   A  C  , (iii) C  B , (iv) A  C and (v) C   A  C  .
4
Exercise 5 Construct truth tables for the following wffs

(i) A   A  B  ,
(ii)  A  B   A  C   B  C 
Solution
(i) There are 2 variables or letters so we need 2 2  4 rows in truth table
A
B
A
B
A  B 
 A  B  A   A  B 
T
T
F
F
F
T
T
T
F
F
T
T
F
F
F
T
T
F
T
F
F
F
F
T
T
T
F
F
(ii) There are 3 variables or letters so we need 2 3  8 rows in truth table.
A B C  A  B   A  C  B  C   A  C   B  C   A  B   A  C   B  C 
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
F
F
T
T
T
T
T
T
T
T
T
F
T
F
T
T
T
F
T
T
T
F
T
T
T
F
T
T
T
T
T
T
T
T
T
T
T
T
This is known as a tautology, that is that it is always true for every combination of truth
values of its constituent propositions.
5

Exercise 6 Show that  A  B  and A  B are logically equivalent.
Solution
A
T
T
F
F
B
A
B
A  B
A  B
 A  B 
T
F
T
F
F
F
T
T
F
T
F
T
F
F
T
F
T
T
F
T
F
F
T
F

The truth values of the column for  A  B  are the same truth values as the column


for A  B .Hence  A  B   A  B that is  A  B  is equivalent to A  B .
Exercise 7

Prove De Morgan’s law of logic namely (ii)  A  B   A  B  --(6b).
Solution
A
B
A B
 A  B 
A
B
A  B 
T
T
F
F
T
F
T
F
T
F
F
F
F
T
T
T
F
F
T
T
F
T
F
T
F
T
T
T

The truth values of the column for  A  B  are the same truth values as the column


for A  B  .Hence  A  B   A  B  that is  A  B  is equivalent to A  B  .
6
Exercise 10
An algorithm contains the following line
If not ( x  3 and x  6) then…
How could this be written more simply? Explain
Solution Using De Morgan’s Law we can simplify above expression by first
expressing it in symbolic logic as follows.
Let A : x  3 and
Let B : x  6
then If not ( x  3 and x  6) then … expressed in symbolic logic is

if  A  B  then
This is equivalent to
if  A  B then –by applying De Morgan’s law 6b.
But A = not x  3 = x  3 and B  = not x  6 = x  6 .
Hence we can write above line of code more simply as
If ( x  3 or x  6) then…
7
Exercise 11 Verify by constructing truth tables that the following wffs are tautologies.
(i) ( A)  A (ii) A  B  B (iii) A  ( A  B) and (iv) ( A  B)  A  B 
Solution
(i) To prove that ( A)  A is a tautology.
A
A
 A
 A  A
T
F
F
T
T
F
T
T
Hint: Remember the truth table for 
Since in the last column all T 'S ( A)  A is a tautology.
(ii)
To prove that A  B  B is a tautology.
A
T
T
F
F
A B
T
F
F
F
B
T
F
T
F
Since in the last column all T 'S A  B  B
8
is a tautology.
 A  B  B
T
T
T
T
(iii)
A
T
T
F
F
To prove that A  ( A  B) s a tautology.
B
A B
T
T
F
T
T
T
F
F
A   A  B
T
T
T
T
Since in the last column all T 'S A  ( A  B)
is a tautology.
(iv)
To prove that ( A  B)  ( A  B ) is a tautology i.e. De Morgan’s Law 6a
A
B
 A  B
 A  B 
A
B
A  B 
 A  B   ( A  B )
T
T
F
F
T
F
T
F
T
T
T
F
F
F
F
T
F
F
T
T
F
T
F
T
F
F
F
T
T
T
T
T
Hint: Remember the truth table for 
Since in the last column all T 'S ( A  B)  ( A  B )
9
is a tautology.
Exercise 12 Construct truth tables for the following well formed formulae(wffs).Identify
any tautologies or contradictions or contingencies.
(i) ( A  B)  ( A  B) (ii) A  ( A  B )
(iii) ( A  B)  A (iv) ( A  B)  [( A  C )  ( B  C )]
(v) ( A  B)  ( B   A) (vi) ( A  B )  ( A  B) (vii) [( A  B)  C ]  ( A  C )
Solution
(i)
A
T
T
F
F
AB
T
F
T
T
B
T
F
T
F
A
F
F
T
T
B
T
F
T
F
A  B
T
F
T
T
A  B   A  B
T
T
T
T
This  A  B   A  B is a tautology since the logical expression is always true no
matter what the truth values of the individual propositions are i.e. always has the truth
value T in the last column.
Note You must remember the truth table for  i.e.
A
T
T
F
F
B
T
F
T
F
.
10
AB
T
F
F
T
(ii)
A
B
A
B
A  B 
 A  B 
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
F
T
T
T
T
F
F
F

A   A  B  
T
F
F
F

Thus A   A  B  is neither a tautology or a contradiction because of the mixture
of T’ and F’’s in last column that is a contingency.
(iii)
A
T
T
F
F
 A  B
B
T
F
T
F
T
F
F
F
 A  B  A
A
F
F
T
T
F
T
T
T
Thus  A  B  A is neither a tautology or a contradiction because of the mixture
of T’ and F’’s in last column that is a contingency.
(iv)
A
B
C
A B A C
B  C  [A  C   B  C ] A  B  [A  C   B  C ]
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
F
F
T
T
T
T
T
T
T
F
T
T
T
F
T
T
T
T
T
F
T
F
T
T
T
F
T
T
T
T
T
T
T
T
T
T
T
T
Thus  A  B  [ A  C   B  C ] is a tautology because of T’s in the last column.
11
(v)
 A  B
B
F
T
F
T
A
T
T
F
F
B
T
F
T
F
Thus
A  B  B  A is
T
F
F
F
A
F
F
T
T
B  A
A  B  B  A
F
T
T
T
F
F
F
F
a contradiction because of all F’s in the last column.
(vi)
A
B
B
 A  B 
A B
 A  B 
 A  B    A  B 
T
T
F
F
T
F
T
F
F
T
F
T
T
T
F
T
T
F
F
F
F
T
T
T
F
T
F
T

Thus  A  B    A  B  is neither a tautology or a contradiction because of the mixture
of T’ and F’’s in last column that is it is a contingency.
12
Exercise 13 Using a truth table prove the following distributive law in logic namely
A  ( B  C )  ( A  B)  ( A  C ) .
Solution There are 3 statements A, B and C so there will be 8 rows in the truth table.
A
T
T
T
T
F
F
F
F
B
T
T
F
F
T
T
F
F
C
T
F
T
F
T
F
T
F
BC
T
F
F
F
F
F
F
F
A  (B  C)
T
T
T
T
T
F
F
F
( A  B)
T
T
T
T
T
T
F
F
( A  C)
T
T
T
T
T
F
T
F
( A  B)  ( A  C )
T
T
T
T
T
F
F
F
Column for A  ( B  C ) has the same truth values as the column for ( A  B)  ( A  C )
for every combination of truth values of their constituent statements A, B and C .
13
(vii)
A
T
T
T
T
F
F
F
F
B
T
T
F
F
T
T
F
F
C
T
F
T
F
T
F
T
F
A B
T
T
T
T
T
T
F
F
C
F
T
F
T
F
T
F
T
 A  B  C 
F
T
F
T
F
T
F
F
A
F
F
F
F
T
T
T
T
A  C
T
F
T
F
T
T
T
T
[ A  B  C ]   A  C 
T
F
T
F
T
T
T
T
Thus [ A  B  C ]   A  C is neither a tautology or a contradiction because of the
mixture of T’ and F’’s in last column that is, it is a contingency.
14
Tutorial on Simple Interest
Exercise 1 A person borrows € 1000.00 at 9.5% simple interest per year. Find the
interest owed in (a) 30 days (b) 10 months (c) 36 months and (d) 4.5 years.
Solution PV  1000 , r  0.095 .
(a) Since the money is borrowed, bank charges ordinary interest (over 360 days)
30
1
Hence t 
 .
360 12
1
FV  PV (1  r  t )  1000(1  0.095  )
12
 1000(1.007916)
 1007.92
Interest charged  1007.92  1000.00  7.92 .
(b) FV  PV (1  r  t )  1000(1  0.095 
 1000(1.0791667)
10
)
12
 1079.17
Interest charged  1079.17  1000.00  79.17 .
(c) FV  PV (1  r  t )  1000(1  0.095 
 1000(1.285)
36
)
12
 1285
Interest charged  1285  1000.00  285.00 .
(d) FV  PV (1  r  t )  1000(1  0.095  4.5)
 1000(1.4275)
 1427.5
Interest charged  1427.5  1000.00  427.5 .
15
Exercise 2 A person invests € 1500.00 at 7.00% simple interest per year. Find the interest
owed in (a) 60 days (b) 6 months (c) 6 years.
Solution Since it is an investment and not a loan we use the exact interest year 365 days
PV  1500 , r  0.07 .
(a) Since the money is borrowed, bank charges ordinary interest (over 360 days)
60 1
Hence t 
 .
360 6
1
FV  PV (1  r  t )  1500(1  0.07  )
6
 1000(1.0115068)
 1517.26
6 1
 .
(b) t 
12 2
1
FV  PV (1  r  t )  1500(1  0.07  )
2
 1500(1.035)
 1552.50
(c) t  6 .
FV  PV (1  r  t )  1500(1  0.07  6)
 1500(1.42)
 2130
Exercise 3 It is desired that an investment amount to € 5000.00 at a simple interest rate r
in t years. Find the principal or PV that must be invested now if
(a) r  6% , t  3 years .(b) r  8% , t  5 years .
FV
Solution FV  5000.00 . PV 
.
1 r  t
5000
 4237.29 .
(a) r  0.06 . t  3 .Hence PV 
1  0.06  3
5000
 3571.43
(b) r  0.08 . t  5 .Hence PV 
1  0.08  5
16
Tutorial on Compound Interest
Exercise 4 A person invests € 10000.00 for 40 months and interest is compounded
monthly. Find the accumulated amount if the annual rate is (a) 5 % (b) 7,5 % or (c) 10 %.
Solution PV  10000 . n  40 .
0.05
(a) i 
. FV  10000(1  i) 40  10000(1.004167) 40  11809.51.
12
0.075
. FV  10000(1  i) 40  10000(1.00625) 40  12830.27 .
(b) i 
12
0 .1
. FV  10000(1  i) 40  10000(1.0083333) 40  13936.86
(c) i 
12
Exercise 5 What present value will amount to € 20000.00 at 8 %annual interest
compounded daily in (a) 1 year (b) 2 years (c) 3 years ? Assume an exact interest year.
Solution FV  20000 . r  0.08 i 
20000
0.08 365
(1 
)
365
FV
20000
20000

 17043.18 .
(b) n  730 . PV 
=
n
0.08 730 (1.000219178) 730
(1  i)
(1 
)
365
(a) PV 
FV

(1  i) n
0.08
. n  365 .
365
20000
=
 18462.49 .
(1.000219178) 365
(c) n  1095 . PV 
FV

(1  i) n
20000
20000
 15732.97 .
=
0.08 1095 (1.000219178)1095
(1 
)
365
17
18