STARRY GOLD ACADEMY
INSTITUTE OF CHARTERED ACCOUNTANTS OF
NIGERIA
MAY 2015 MOCK EXAMINATION
QUANTITATIVE TECHNIQUE
TOPICS
SIMPLE INTEREST
COMPOUND INTEREST
DISCOUNTING
ANNUITIES
PRESENT AND FUTURE VALUES OF ANNUITIES
SINKING FUNDS AND AMORTIZATION
QUESTIONS
1. Find the simple interest on the loan of N142,500 in 6 years at 4% per annum.
2. Find the principal which produces simple interest of N37,800 in 6 years at
7% per annum.
3. Find the rate per annum at which N80,000 generates simple interest of
48,000 in 2 years.
4. Calculate the time in which the sum of N65, 000 generates simple interest of
N9,100 at 3.5% per annum.
COMPOUND INTEREST
1. Find the compound interest on N25, 000 in 6 years at 2% per annum.
2. Find to the nearest naira, the compound interest on N400, 000 in 6
months at 10% per annum.
3. Find the compound interest on N 90, 000 in 3 years at 4% per annum.
DISCOUNTING
1. A buyer paid the sum of N800 for an article over which he enjoyed 5% cash
discount. How much was the marked price?
2. An article was marked at N25.40 per unit. The seller allows for a 2%
discount on purchase of 20 – 30 units at a time, 3% for 31 – 40 units and 4%
for 41 units and above. Find out how much a buyer of the following
quantities will pay;
(i)
28 units (ii) 35 units and (iii) 60 units.
ANNUITIES
1. Tade Bakare has just won a bonanza promising N100, 000 for 10 years.
The first payment being due is a year’s time. If in the alternative, he is to
obtain the value of his winnings now, how much do you advise he
request for?
Note: Compounding rate is 10%.
PRESENT AND FUTURE VALUES OF ANNUITIES
1. What is the future value of N100, 000 receivable at the end of each year for
10 years if the rate of interest is 10%.
2. How much must you deposit in a savings account now if you have to fulfil a
debt obligation of N10,000 at the beginning of each of the next 5 years.
Interest rate is 10% payable annually.
3. Calculate the future value of N10,000 deposited at the beginning of each of
the next 5 years.
SINKING FUNDS AND AMORTIZATION.
In 1975, when the country was relatively buoyant, Mr Pat took a car loan of
N9000 with 60 months payment at 9% per annum.
a. What is his monthly payment?
b. If he intends to pay off the balance after making 12 payments, how much
cash will he need?
c. If he pays the loan off after 12 payments, how much interest will he have
paid?
d. Suppose he pays the balance after having made 24 payments, how much
cash will he need?
e. In part (d), how much interest will he have paid so far?
SUGGESTED SOLUTIONS
SIMPLE INTEREST
Find the simple interest on the loan of N142,500 in 6 years at 4% per
annum.
SOLUTION
By formular,
SI = P x R x T / 100
From the Question,
SI = ?
P = 142, 500
T = 6years
R = 4%
SI = 142, 500 X 6 X 4 / 100
SI = 3,420,000 / 100
SI = N34,200.
Find the principal which produces simple interest of N37,800 in 6 years at
7% per annum.
By formular,
SI = P x R x T / 100
From the Question,
SI = N37,800
P=?
T = 6years
R = 7%
37,800 = P X 7 X 6 / 100
37,800 = 42 X P / 100
By cross multiplication,
37,800 X 100 = 42 X P
3,780,000 = 42 X P
P = 3,780,000 / 42
P = N90,000.
Find the rate per annum at which N80,000 generates simple interest of
48,000 in 2 years.
By formular,
SI = P x R x T / 100
From the Question,
SI = N48,000
P = N80,000
T = 2years
R = ?%
SI = P X R X T / 100
48,000 = 80,000 X R X 2 / 100
48,000 = 160,000 X R / 100
By cross multiplication,
48,000 X 100 = 160,000 X R
4,800,000 = 160,000 X R
R = 4,800,000 / 160,000
R = 30%
Calculate the time in which the sum of N65, 000 generates simple interest
of N9,100 at 3.5% per annum.
By formular,
SI = P x R x T / 100
From the Question,
SI = N9,100
P = N65,000
T = 2years
R = 3.5%
SI = P X R X T / 100
9,100 = 65,000 X 3.5 X T / 100
9,100 = 227,500 X T / 100
By cross multiplication,
9,100 X 100 = 227,500 X T
910,000 = 227,500 X T
T = 910,000 / 227,500
T = 4years.
KEYS: SI = Simple Interest; P = Principal; T = Time and R = Rate.
COMPOUND INTEREST
Find the compound interest on N25, 000 in 6 years at 2% per annum.
By formular,
CI = P[ 1 + R/1OO]n – P
From the Question,
CI = ?
P = N25,000
R = 2%
T = 6years
CI = P[ 1 + R/1OO]t – P
CI = 25,000 [ 1 + 2/100]6 – 25,000
CI = 25,000 [ 1 + 0.02]6 – 25,000
CI = 25,000 [ 1.02]6 – 25,000
CI = 25,000 [ 1.126] – 25,000
CI = 28,154 – 25,000
CI = N3,154.
Find to the nearest naira, the compound interest on N400, 000 in 6
months at 10% per annum.
By formular,
CI = P[ 1 + R/1OO]n – P
From the Question,
CI = ?
P = N400,000
R = 10%
T = 6months = 0.5year
CI = P[ 1 + R/1OO]t – P
CI = 400,000 [ 1 + 10/100]0.5 – 400,000
CI = 400,000 [ 1 + 0.1]0.5 – 400,000
CI = 400,000 [ 1.1]0.5 – 400,000
CI = 400,000 [ 1.048] – 400,000
CI = 419,524 – 400,000
CI = N19,524.
Find the compound interest on N 90, 000 in 3 years at 4% per annum.
By formular,
CI = P[ 1 + R/1OO]n – P
From the Question,
CI = ?
P = N90,000
R = 4%
T = 3years.
CI = P[ 1 + R/1OO]t – P
CI = 90,000 [ 1 + 4/100]3 – 90,000
CI = 90,000 [ 1 + 0.04]3 – 90,000
CI = 90,000 [ 1.04]3 – 90,000
CI = 90,000 [ 1.125] – 90,000
CI = 101,250 – 90,000
CI = N11,250.
DISCOUNTING
A buyer paid the sum of N800 for an article over which he enjoyed 5%
cash discount. How much was the marked price?
SOLUTION
Amount paid = N800
Cash Discount = 5%
Cash Discount in naira = 5% of 800
= 5 /100 X 800
= N40
Cost price = Amount paid + Discount given
= 800 + 40
= N840.
Therefore, the marked price is N840.
An article was marked at N25.40 per unit. The seller allows for a 2% discount
on purchase of 20 – 30 units at a time, 3% for 31 – 40 units and 4% for 41 units
and above. Find out how much a buyer of the following quantities will pay;
(ii)
28 units (ii) 35 units and (iii) 60 units.
SOLUTION
Marked price = N25.40
For 28 units, the applicable Discount is 2%
Discount
= 2% of 25.40
= 2/100 X 25.40
= 0.02 X 25.40
= 0.508
Amount paid = Marked price – Discount
= 25.40 – 0.508
= N24.892
For 35 units, the applicable Discount is 3%
Discount
= 3% of 25.40
= 3/100 X 25.40
= 0.03 X 25.40
= 0.762
Amount paid = Marked price – Discount
= 25.40 – 0.762
= N24.638
For 60 units, the applicable Discount is 4%
Discount
= 4% of 25.40
= 4/100 X 25.40
= 0.04 X 25.40
= 1.016
Amount paid = Marked price – Discount
= 25.40 – 1.016
= N24.384
ANNUITIES
Tade Bakare has just won a bonanza promising N100, 000 for 10 years.
The first payment being due is a year’s time. If in the alternative, he is
to obtain the value of his winnings now, how much do you advise he
request for?
Note: Compounding rate is 10%.
SOLUTION
The problem requires us to derive the Present Value of N100,000 for 10 years at
the end of each year.
By formular,
P = A[ 1 – ( 1 + r)-n / r]
From the Question,
P=?
A = N100, 000
r = 10% = 0.1
n = 10
P = 100,000 [ 1 – ( 1 + 0.1)-10 / 0.1]
P = 100,000 [ 1 – (1.1)-10 / 0.1]
P = 100,000 [ 1 – 0.3855 / 0.1]
P = 100,000 [ 6.1446]
P = N614,460.
Thus, Tade should request he is paid (not less than) N614,460.
PRESENT AND FUTURE VALUES OF ANNUITIES
What is the future value of N100, 000 receivable at the end of each year for
10 years if the rate of interest is 10%.
SOLUTION
By formular,
FV = A[ (1 + r)n – 1 / r ]
From the Question,
FV = ?
A = N100,000
r = 10% = 0.1
n = 10
FV = 100,000 [ (1 + 0.1)10 – 1 / 0.1]
FV = 100,000 [ (1.1)10 – 1 / 0.1]
FV = 100,000 [ (2.594 – 1 / 0.1]
FV = 100,000 [ 1.594 / 0.1]
FV = 100,000 [ 15.94]
FV = N1,593,742.
How much must you deposit in a savings account now if you have to fulfil a
debt obligation of N10,000 at the beginning of each of the next 5 years.
Interest rate is 10% payable annually.
SOLUTION
By formular,
P = A[( 1 + r) – (1 + r)-n / r ]
From the Question,
P=?
A = N10,000
r = 10% = 0.1
n = 5.
P = 10,000[( 1 + 0.1) – (1+ 0.1)-5 / 0.1]
P = 10,000 [ ( 1.1) – ( 1.1)-5 / 0.1]
P = 10,000 [ (1.1) – 0.6830 / 0.1]
P = 10,000 [ 1.1 – 0.6830 / 0.1]
P = 10,000 [ 0.417 / 0.1]
P = 10,000 [ 4.17]
P = N41,700.
Calculate the future value of N10,000 deposited at the beginning of each of
the next 5 years.
By formular,
FV = A[ ( 1 + r)n – (1 + r) / r ]
From the Question,
FV = ?
A = 10,000
r = 10% = 0.1
n=5
FV = 10,000 [ (1 + 0.1)5 – (1 + 0.1) / 0.1]
FV = 10,000 [ (1.1)5 – (1.1) / 0.1]
FV = 10,000 [ 1.611 – 1.1 / 0.1]
FV = 10,000 [ 0.511 / 0.1]
FV = 10,000 [ 5.11]
FV = N51,100
SINKING FUNDS AND AMORTIZATION.
In 1975, when the country was relatively buoyant, Mr Pat took a car loan
of N9000 with 60 months payment at 9% per annum.
a. What is his monthly payment?
b. If he intends to pay off the balance after making 12 payments, how
much cash will he need?
c. If he pays the loan off after 12 payments, how much interest will he
have paid?
d. Suppose he pays the balance after having made 24 payments, how
much cash will he need?
e. In part (d), how much interest will he have paid so far?
SOLUTION
a. The rate per period of a month is
r = 9% = 9/100 = 0.09 = 0.09 / 12 = 0.0075,
Payment period = 60,
P = N9,000
n = Number of payment period.
Using the formular,
R=P[
r / 1 – ( 1 + r)-n ]
R = 9,000[
0.0075 / 1 – ( 1 + 0.0075)-60 ]
R = 9,000[
0.0075 / 1 – (1.0075)-60 ]
R = 9,000[
0.0075 / 1 – 0.6837 ]
R = 9,000[
0.0075 / 0.3613 ]
R = 9,000[
0.02076 ]
R = 9,000[
0.02076 ]
R = 9,000 X 0.02076
R = N186,825.
b. Mr. Pat wants to pay off the balance after making 12 payments implies that
he has
60 – 12 = 48.
Therefore, the monthly payments to use here n = 48
Using n = 48, the formular to use is;
P = R [ 1 – ( 1 + r)-n / r]
Recall, R = 186.83; n = 48; r = 0.0075
Substituting these values into the above formular ,
P = N7,507.72.
That is, he needs N7, 507.72 cash to settle the remaining debt.
c. Amount paid after 12 payments = 12 X 186.825
= N2,241.96
Cash needed to offset the loan balance as obtained in (b) above is N7, 507.72
Therefore,
The total amount paid = 2,241.96 + 7,507.72 = N9, 749.68
And
The interest paid = amount paid – initial worth of car
= 9,749.68 – 9,000
= N749.68
d. As explained in (b) above n = 60 – 24 = 36 and the cash needed is;
P = 186.825 [ 1 – (1 + 0.0075)-36 / 0.0075 ]
P = N5,875.21
e. Amount paid to date (after 24 payments) =
= 24 X 186.825
= N4,483.92
Cash needed to offset balance of loan = 5,875.21
Total amount paid = 5,875.21 + 4,483.92 = 10,359.13
Interest paid = 10,359.13 – 9,000
= N1,359.13
TOPICS
USES OF CALCULUS
DIFFERENTIATION
INTEGRATION
IDENTIFICATION OF STATIONARY POINTS
OPTIMIZATION TECHNIQUES
REVIEW QUESTIONS
1. What is calculus? Explain briefly its major types.
2. Itemize the major uses of calculus
3. What is differential calculus? Itemize the major rules that govern the
differential calculus.
4. Differentiate the following functions
a. y=f(x)= 2x2 – x
b. y=f(x) = 3x2 + 4x + 8
c. y=f(x) = 3x2 - 4x - 8
d. y=f(x) = 3x2 - 4x - 8
e. y = (x2 + 2x)(4x3 + x2)
f.
f(x) = x2 – 4
X3 + 2
g. y= f(x) = sin x
h. y= f(x) = cos x
i. y=f(x) = tan x
j. y=f(x) = cot x
k. y=f(x) = ex
l. y=f(x) = ekx
m. y=log x
n. y= loge(3x2 – 5)
5. What do you understand by Integration?
6. Differentiate clearly between Definite and Indefinite integral.
7. Integrate the following functions;
a. (6x2 + 8x + 3)dx
b. 6x(3x2 + 4)5dx
c. X2(4x3 + 2)8dx
d. (8x + 5)10dx
8. Find the minimum and maximum points of the function
f(x) = x3 – 21x2 + 30x + 15
2
9. A manufacturer total cost is C(q) = 0.1q3 – 0.5q2 + 500q + 200 naira,
where q is the number of units produced.
a. Estimate the cost of manufacturing the 4th unit
b. Compute the actual cost of manufacturing the 4th unit
c. What is the average cost at this level of production i.e. q =4?
10.Given demand function, P = 200 – x/1000 and cost function, TC = 12,000
+ 40x.
a. Derive the profit function
b. Find the level of output that maximizes profit
c. What is the amount of the profit maximized?
SUGGESTED SOLUTIONS
1. What is calculus? Explain briefly its major types.
The term Calculus simply means a branch of mathematics that is majorly
concerned with the rate of change of a quantity to another.
It has two major types such as;
i.
ii.
Differential calculus (Differentiation)
Integral calculus (Integration)
Itemize the major uses of calculus
The major uses of calculus are;
i.
ii.
iii.
iv.
v.
vi.
It is used to determine the rate of change within a given function.
It also helps to determine the degree of dependence between a variable
and another.
It is purely used in marginal analysis
It helps to determine the relationship between limit and continuity.
It helps to distinguish clearly between average and instantaneous rate of
change of a function with respect to the dependent variable at a specific
point.
It can be applied to solve various business problems of the world etc.
What is differential calculus? Itemize the major rules that govern the
differential calculus.
The term- Differentiation simply means the rate of change. That is, it means the
rate at which a variable changes as a result of corresponding change in other
variable. Variables here are regarded as dependent and independent variables. They
can also be called endogeneous and exogeneous variables. It is a division of
calculus that is more concerned with rate of change. It can otherwise be called
SLOPE, DERIVATIVE, RATE OF CHANGE etc.
In summary, the term differentiation simply denotes the rate of change and it is
commonly called dy/dx.
The major rules of differentiation are;
As defined, differentiation simply means the rate of change i.e dy/dx.
Thus, there are basic rules that are commonly employed to solve various
differential calculus problems. Some of these basic rules are;
i.
ii.
iii.
iv.
v.
vi.
vii.
Constant rule
Sum and Difference rule
Product rule
Quotient rule
Trigonometric rule e.g. sine and cosine rules
Logarithmic rule
Chain rule or function of function rule etc.
However, the basic principle that is commonly employed to solve differentiation
problems is called FIRST PRINCIPLE.
Differentiate the following functions
y=f(x)= 2x2 – x
y = f(x) = 2x2 – x
Using the formular, y + ∆y = x + ∆x
y + ∆y = 2(x + ∆x)2 - x + ∆x
y + ∆y = 2(x + ∆x)( x + ∆x) - x + ∆x
y + ∆y = 2(x2 + x∆x + x∆x + ∆x2) - x + ∆x
y + ∆y = 2x2 + 4x∆x + 2(∆x)2 - x + ∆x
∆y = 2x2 + 4x∆x + 2(∆x)2 - x + ∆x – y
∆y = 2x2 + 4x∆x + 2(∆x)2 - x - ∆x - 2x2 + x
∆y = 4x∆x + 2(∆x)2 - ∆x
∆y = ∆x(4x + 2∆x – 1)
∆y/∆x = 4x + 2∆x – 1
Take limit of both sides such as ∆y/∆x = dy/dx and ∆x = 0
Therefore, dy/dx = 4x + 2(0) – 1
Dy/dx = 4x – 1.
y=f(x) = 3x2 + 4x + 8
Using General principle; arn-1
Therefore, dy/dx = 2.3x2-1 + 1.4x1-1 + 0
The derivative of a constant value is 0.
dy/dx = 6x + 4
y=f(x) = 3x2 - 4x - 8
Therefore, dy/dx = 2.3x2-1 - 1.4x1-1 - 0
Dy/dx = 6x – 4
y=f(x) = 3x2 - 4x – 8
Therefore, dy/dx = 2.3x2-1 - 1.4x1-1 - 0
Dy/dx = 6x – 4
y = (x2 + 2x)(4x3 + x2)
This problem is presented in product form. Hence, we have to apply Product rule
to solve it.
dy/dx = Udv/dx + V du/dx.
Given that, y = (x2 + 2x)(4x3 + x2)
Let U = x2 + 2x, du/dx = 2x + 2
Also, V = 4x3 + x2, dv/dx = 12x2 + 2x
 Applyng the Product rule,
 dy/dx= Udv/dx + Vdu/dx
dy/dx= (x2 + 2x)(12x2 + 2x) + (4x3 + x2)( 2x + 2)
YOU CAN EXPAND AND COLLECT LIKE TERMS
f(x) = x2 – 4
X3 + 2
However, this problem is given in division form. Therefore, the right
formula to use is Quotient rule.
Thus, if y = u/v
dy/dx = Vdu/dx – Udv/dx
v2
f(x) = x2 – 4
X3 + 2
Put U = x2 – 4 and V= x3 + 2
du/dx = 2x and dv/dx = 3x2
hence using quotient rule,
dy/dx = (x3 + 2)(2x) – (x2 – 4)(3x2)
(x3 + 2)2
YOU CAN EXPAND AND COLLECT LIKE TERMS
y= f(x) = sin x
dy/dx = cos x
y= f(x) = cos x
dy/dx = -sinx
y=f(x) = tan x
dy/dx = sec2x
y=f(x) = cot x
dy/dx = -cosec2x
y=f(x) = ex
dy/dx = ex
y=f(x) = ekx
dy/dx = k ekx.
y=log x
that is y= In f(x)
Therefore, dy/dx = f’(x)/f(x).
Thus, dy/dx = 1/x
y= loge(3x2 – 5)
dy/dx = f’(x)/f(x) = 6x/3x2 – 5.
What do you understand by Integration?
The word- Integration simply means anti-differentiation. It is a part of calculus that
does not apply or obey all the techniques involved in differentiation. Rather, it is
an opposite direction of differentiation, hence anti-differentiation. The symbol
commonly used in its analysis is ʃ .
Differentiate clearly between Definite and Indefinite integral.
Basically, there are two types of integral calculus namely;
i.
ii.
definite integral
indefinite integral
Thus, the major distinction between definite integral and indefinite integral is
that; while definite integral is more concerned with a function whose
integrand is definite in nature such as ʃf(x)dx; indefinite integral on the other
hand is closely related to those functions whose integrands are not definite
i.e. they range between two extreme values such as ʃabf(x)dx.
Integrate the following functions;
ʃ (6x2 + 8x + 3)dx
Using the integration formula; anxn+1/ n+1 + C
ʃdy = 6x2+1/2+1 + 8x1+1/1+1 + 3x0+1/0+1
ʃdy = 6x3 /3 + 8x2/2 + 3x/1
ʃdy = 2x3 + 4x2 + 3x + C
ʃ6x(3x2 + 4)5dx
Let U = 3x2+4
du/dx = 6x
dx = du/6x
ʃ6x U5 . du/6x
ʃ U5 du
= U5+1 / 5+1
= U6 /6
Recall, U = 3x2+4
Therefore, = (3x2 +4)6 /6 + C
ʃ x2(4x3 + 2)8dx
Let U = 4x3 + 2
du/dx = 12x2
dx = du/12x2
ʃx2(U)8. du/12x2. Note x2 cancels x2
ʃ U8.du/12
= 1/12 ʃ U8du
= 1/12 . U8+1/8+1
= 1/12 . U9/9
Recall that U = 4x3 + 2
= 1/12 .( 4x3 + 2)9 /9
= (4x3 + 2)9/108 + C
ʃ (8x + 5)10dx
Let U = 8x + 5
ʃ (U)10dx
U10+1/10+1
U11/11
Recall that U = 8x + 5
= (8x + 5)11/11 + C
Find the minimum and maximum points of the function
f(x) = x3 – 21x2 + 30x + 15
2
Using the first derivative test, stationary points occur at f’(x) or dy/dx = 0,
such that
f(x) = x3 – 21x2 + 30x + 15
2
’
f (x) = dy/dx = 3x2 – 21x + 30
This equation is equivalent to zero, that is
dy/dx = 0,
Therefore, 3x2 – 21x + 30 = 0
On factorizing, we have
(x – 2)(x – 5) = 0
x-2 = 0 or x-5 = 0
Therefore, x = 2 or 5.
Now, when x = 2,
f(2) = 23 – 21(22) + 30(2) + 15 = 41
2
When x = 5,
f(5) = 53 – 21(52) + 30(5) + 15 = 27.5
2
Hence (5, 27.5) are the stationary points. To determine which of them is a
minimum point, we have to differentiate f’(x), that is we find the second
order derivative such that;
d (f’(x)) = f’’(x) = 6x – 21
dx
But
f’’(2) = 6(2) – 21 = 12 – 21 = -9 ˂ 0.
Hence (2, 41) is a maximum point.
However,
f’’(5) = 6(5) – 21 = 30 – 21 = 9 > 0 and (5, 27.5) is a minimum point.
Therefore the minimum value of the function f(x) is f(5) = 27.5 while the
maximum value of f(x) is f(2) = 41.
A manufacturer total cost is C(q) = 0.1q3 – 0.5q2 + 500q + 200 naira,
where q is the number of units produced.
Estimate the cost of manufacturing the 4th unit
This Question is centered on the application of differential calculus to solve
various problems in business.
Given TC = 0.1q3 – 0.5q2 + 500q + 200
a) The marginal cost
MC = dTC/dq = 0.3q2 – q + 500
Please note that the cost of manufacturing the fourth unit is the marginal cost
when 3 units are being produced and that Marginal Cost is the derivative of
Total Cost. That is, the cost of manufacturing the 4th unit is;
MC(3) = 0.3(3)2 – 3 + 500 = 2.7 -3 + 500 = 499.7 naira
Compute the actual cost of manufacturing the 4th unit
b). The actual cost of manufacturing the 4th unit =
TC(4) – TC(3)
TC = 0.1q3 – 0.5q2 + 500q + 200
TC(4) = 0.1(4)3 – 0.5(4)2 + 500(4) + 200
TC(4) = 2198.4
TC(3) = 0.1(3)3 – 0.5(3)2 + 500(3) + 200
TC(3) = 1698.2
Therefore, the actual cost of producing 4th unit is;
TC(4) – TC(3) = 2198.4 – 1698.2 = N500.2
What is the average cost at this level of production i.e. q =4?
a) Average cost,
AC = Total cost/Units produced
AC = 0.1q3/q – 0.5q2/q + 500q/q + 200/q
AC = 0.1q2 – 0.5q + 500 + 200/q
When q = 4
AC = 0.1(42) – 0.5(4) + 500 + 50 = N549.60
However, MC(4) = 0.3(42) – 4 + 500 = N496 which shows that MC is less than the
AC. Hence the manufacturer should produce more so that the AC is minimal.
Given demand function, P = 200 – x/1000 and cost function, TC = 12,000 +
40x.
Derive the profit function
Profit has been defined as the excess of revenue over cost. That is, Profit =
TR – TC.
Given demand function, P = 200 – x/1000 and cost function, TC = 12,000 +
40x.
Note that, TR = P. x
Therefore, TR = (200 – x/1000) x
TR = 200x – x2/1000
Also, TC = 12,000 + 40x
Therefore, Profit function is
Π = Profit = TR – TC
= 200x – x2/1000 – (12,000 + 40x)
= 200x - x2/1000 – 12,000 – 40x
This is what we call profit function.
Recall, we can still go ahead to determine the level of output that maximizes
the profit. This can be gotten by differentiating the profit function and solve
mathematically and then substitute the value obtained into the profit function
to get the maximum profit.
TOPICS
1.
2.
3.
4.
5.
6.
7.
DATA COLLECTION
METHODS OF DATA COLLECTION
PRESENTATION OF DATA
FREQUENCY DISTRIBUTIONS
HISTOGRAMS
FREQUENCY POLYGONS
OGIVES
OBJECTIVE
1. Which of the following is NOT a graphical representation of data?
A. Histogram
B. Cumulative frequency curve
C. Frequency polygon
D. Frequency distribution
E. Z- chart
2. Which of the following is NOT a merit of personal interview method?
A. Instant answers can be received or collected
B. Non- response rate is very low
C. Personal bias is likely to enter the results because of interviewer’s presence
D. It yields high percentage responses
E. If properly used, it is convenient for some intensive surveys
3. Primary data obtained directly from the respondents are referred to as
……………………… data.
4. The number of times a particular value occurs in a given data set is known as
its…………………………..
5. A frequency polygon is a
A. Rectangle whose sides correspond to frequency
B. Line chart
C. Histogram
D. Bar chart
E. Series of straight lines joining the midpoints in histogram
6. A procedure in the form of report involving a combination of text and figures is
known as ………………
7. Data presentation in form of graphs and charts is known as ………………
8. Which of the following is/are qualities of a good questionnaire?
I. Each question in the questionnaire must be precise and unambiguous.
II. Avoidance of leading questions in the questionnaire
III. A questionnaire must be lengthy in order to accommodate many questions.
A. I only
B. I and II
C. I and III
D. II and III
E. I, II and III
9. Data can be classified into any of the following EXCEPT
I. Qualitative and Quantitative
II. Primary and Secondary
III. Continuous and Discrete
IV. Ordinal and Non-numeric
A. I and II
B. IV only
C. I and III
D. II and III
E. I, II and III
10. In the construction of an ogive, the vertical axis represents the................
11.The statistical tools that can be further used for further statistical analysis is?
A. Mean and median
B. Mean and mode
C. Mean and standard deviation
D. Range and standard deviation
12. The statistical tools that consider the use of all data during computation are?
A. Mean and median
B. Mean and mode
C. Mean and standard deviation
D. Range and standard deviation
13. The measure of normality or symmetry of a data set is called the ............
14. Which ONE of the following pairs of measures is used for estimation on a
histogram?
A. (Mean, Mode)
B. (Mean, Median)
C. (Mode, Median)
D. (Mean, First quartile)
E. (Median, Harmonic mean)
15. Chronological classification of data deals with classification with respect
to...............
16. The following are the factors which determine the choice between primary and
secondary data EXCEPT
A. Direct observation
B. The purpose of the enquiry
C. Time required for the exercise
D. Funds at the disposal of investigation
E. Nature of statistical investigation
17. Which of the following is NOT an advantage of primary data?
A. It produces the desired information in detail
B. The degree of bias can be reduced with quality control procedures
C. The magnitude of sampling error and the nature of the non-sampling error are
known and can be described
D. Primary data are more correct and reliable than secondary data
E. The use of primary data will reduce cost (in terms of money and resources)
18. Decision-making is important in order to
A. Make profit for a business
B. Solve an identified problem
C. Please the customer
D. Perform a task
E. Please the management
19. A firm hires a set of 50 casual workers paying each N800 per day and another
set of 30
labourers paying each N600 per day. Find the daily mean wage (in N) of all the 80
workers.
A. 725
B. 700
C. 650
D. 750
E. 675
THEORY
1. What is data? List its basic features or characteristics.
2. List and briefly explain the methods commonly employed in collecting or
gathering data.
3. How can data collected be presented?
4. The record below shows the scores obtained by some students in a
Mathematics test:
12
15
20
25
30
20
30
15
12
12
15
12
25
12
20
12
20
12
15
25
12
15
12
15
12
15
12
20
20
30
30
25
15
30
20
15
25
30
15
12
12
15
12
15
12
30
20
20
12
15
a. Represent the data on pie chart, bar chart.
b. Using the data above to prepare simple frequency distribution table.
c. Using the data above to prepare cumulative frequency distribution table.
d. Using the data above to construct Histogram.
5. Consider the weights of a number of students given below measured in
Kilograms:
18
34
20
39
36
32
29
19
25
38
25
22
25
28
37
40
27
39
23
19
30
19
21
25
40
27
32
18
19
40
26
18
39
31
19
29
30
40
21
26
40
20
37
33
26
35
29
35
25
18
Using a class interval of 18-20, 21-23, 24-26, ….
a. Represent the data on pie chart, bar chart.
b. Using the data above to prepare simple frequency distribution table.
c. Using the data above to prepare cumulative frequency distribution table.
d. Using the data above to construct Histogram.
e. Using the data above to construct frequency polygon.
SUGGESTED SOLUTIONS
1. D
2. C
3. REFER TO VIDEO LECTURE
4. FREQUENCY
5. E
6. APPENDIX
7. GRAPHICAL PRESENTATION OF DATA
8. C
9. B
10. REFER TO VIDEO LECTURE
11. C
12. C
13. NORMAL DISTRIBUTION
14. C
15. REFER TO VIDEO LECTURE
16. A
17. C
18. B
19. A
THEORY
What is data? List its basic features or characteristics.
Data can simply be defined as those facts and statistics collected together for
reference or analysis purposes.
It could also be defined as a collection of facts, such as values or measurements
which can be numbers, words or observations.
However, the major characteristics of a good data are;
i.
ii.
iii.
iv.
v.
vi.
vii.
Accuracy
Validity
Reliability
Timeliness
Relevance
Completeness
Consistency
List and briefly explain the methods commonly employed in collecting or
gathering data.
There are several methods that are usually employed in data collection. Some of
these methods are;
QUESTIONNAIRE
This is the commonest method of data collection. It involves the printing and
distribution of questionnaire to targeted audience on post or self-administered.
ADVANTAGES
(1) It is cheaper to conduct even with geographically diverse respondents
(2) Interviewer bias is easily eliminated
(3) No prior arrangement with respondent is required
(4) It gives the respondents the chance of having more time to think and give
accurate responses.
DISADVANTAGES
(1) It may be filled by wrong persons especially in mailed questionnaire
(2) The respondent may not have the opportunity to clarify ambiguous questions
(3) Low response rate may be recorded bios not all respondents will return the
questionnaire fully filled.
(4) Questionnaire can get lost in transit.
INTERVIEW
Interview is an instrument used to elicit information from the respondent through
verbal, face to face interaction between the interviewee(s) or respondents and the
interviewer (the researcher).
Interview is done with the sole purpose of asking relevant questions in order to
answer research question or test hypothesis.
ADVANTAGES
(1) The method allows for accurate information to be obtained as opportunity is
provided for explanation and clarification of question.
(2) It can be used to collect information even from illustrates
(3) There is high response rate arising from persuasion co-operation and
motivation.
(4) It allows for face-to-face contact and opportunity to record non-verbal
communication which may improve the quality of response.
(5) Interview can be done on the phone.
DISADVANTAGES
(1) The method is time consuming
(2) It is relatively expensive due to distance
(3) It is liable to interviewer’s bias.
E-INTERVIEW
Due to the development in technology and distance, interview some times are done
via electronic without facial contact between different locations.
Example: Attending an interview section of a company via internet or through the
organization’s website.
How can data collected be presented?
When appropriate data have been collected, such data collected can be presented or
represented in various forms of diagrams or charts as;
i.
ii.
iii.
Pie chart
Bar chart
Histogram etc.
The record below shows the scores obtained by some students in a
Mathematics test:
12
15
12
30
12
15
12
15
25
15
20
25
12
15
12
25
12
15
30
15
30
20
12
20
12
20
12
15
15
30
30
20
12
25
20
15
12
20
30
20
12
15
20
15
12
12
25
30
12
15
SCORES
TALLY
FREQUENCY
12
IIIII IIIII IIIII I
16
15
IIIII IIIII III
13
20
IIIII IIII
9
25
IIIII
5
30
IIIII II
7
The data above is ungrouped data because each entry is recorded on its own.
The separate values are identified as unique entities without being combined
with each other.
Representation of data on Pie chart.
Computation of sectoral angles for each scores.
Scores
Frequency
12
16
15
13
20
9
25
5
30
7
50
Angle for each score is done below;
Formula = Frequency of a given score/ Total frequencies X 360
Angle for score 12 = 16/50 X 360 = 115.20
Angle for score 15 = 13/50 X 360 = 93.60
Angle for score 20 = 9/50 X 360 = 64.80
Angle for score 25 = 5/50 X 360 = 360
Angle for score 30 = 7/50 X 360 = 50.40
Next is to draw a circle and then divide it into sectors accordingly.
PLS REFER TO VIDEO LECTURE FOR DIAGRAMATIC
ILLUSTRATION OF PIE CHART.
REPRESENTATION OF DATA ON BAR CHART
Bar chart is one of the most popular types of diagram used to show statistical
information.
It consists of rectangular bars of equal width whose heights or lengths are
proportional to the quantities or sizes of the items that are being represented.
Each bar is however separated by equal gaps.
In the question above, the scores is plotted on the Horizontal axis while the
frequency is plotted on the Vertical axis.
PLS REFER TO VIDEO LECTURE FOR DIAGRAMATIC
ILLUSTRATION OF BAR CHART.
REPRESENTATION OF DATA ON HISTOGRAM
A histogram is a graphical representation of a frequency distribution.
They are rectangular bars placed side by side.
The vertical axis represents the frequency while the horizontal axis
represents the variable (scores) being represented.
The area of each rectangular bar of a histogram is proportional to the
corresponding frequency.
The histogram differs from the bar chart in the following ways;
i.
No gaps between the bars
ii.
The histogram can be used for grouped data.
PLS REFER TO VIDEO LECTURE FOR DIAGRAMATIC
ILLUSTRATION OF HISTOGRAM.
SIMPLE FREQUENCY DISTRIBUTION
Scores
Frequency
12
16
15
13
20
9
25
5
30
7
50
CUMULATIVE FREQUENCY DISTRIBUTION
Scores
Frequency
Cum. Freque
12
16
16 + 0 = 16
15
13
16 + 13 = 29
20
9
29 + 9 = 38
25
5
38 + 5 = 43
30
7
43 + 7 = 50.
50
Consider the weights of a number of students given below measured in
Kilograms:
18
34
20
39
36
32
29
19
25
22
25
28
37
40
27
39
30
19
21
25
40
27
32
18
26
18
39
31
19
29
30
40
40
20
37
33
26
35
29
35
Using a class interval of 18-20, 21-23, 24-26, ….
25
23
19
21
25
Class Interval
18 – 20
21 – 23
24 – 26
27 – 29
30 – 32
33 – 35
36 – 38
39 – 41
Frequency
Tally
IIIII IIIII I
IIII
IIIII III
IIIII II
IIIII
IIII
IIII
IIIII II
38
19
40
26
18
11
4
8
7
5
4
4
7
Angle for each class is done below;
Formula = Frequency of a given class/ Total frequencies X 360
Angle for 18 – 20 = 11/50 X 360 = 79.20
Angle for 21 – 23 = 4/50 X 360 = 28.80
Angle for 24 – 26 = 8/50 X 360 = 57.60
Angle for 27 – 29 = 7/50 X 360 = 50.40
Angle for 30 – 32 = 5/50 X 360 = 360
Angle for 33 – 35 = 4/50 X 360 = 28.80
Angle for 36 – 38 = 4/50 X 360 = 28.80
Angle for 39 – 41 = 7/50 X 360 = 50.40
Next is to draw a circle and then divide it into sectors accordingly.
PLS REFER TO VIDEO LECTURE FOR DIAGRAMATIC
ILLUSTRATION OF PIE CHART.
REPRESENTATION OF DATA ON BAR CHART
Bar chart is one of the most popular types of diagram used to show statistical
information.
It consists of rectangular bars of equal width whose heights or lengths are
proportional to the quantities or sizes of the items that are being represented.
Each bar is however separated by equal gaps.
In the question above, the class interval is plotted on the Horizontal axis
while the frequency is plotted on the Vertical axis.
PLS REFER TO VIDEO LECTURE FOR DIAGRAMATIC
ILLUSTRATION OF BAR CHART.
REPRESENTATION OF DATA ON HISTOGRAM
A histogram is a graphical representation of a frequency distribution.
They are rectangular bars placed side by side.
The vertical axis represents the frequency while the horizontal axis
represents the variable (class interval) being represented.
The area of each rectangular bar of a histogram is proportional to the
corresponding frequency.
The histogram differs from the bar chart in the following ways;
iii. No gaps between the bars
iv.
The histogram can be used for grouped data.
PLS REFER TO VIDEO LECTURE FOR DIAGRAMATIC
ILLUSTRATION OF HISTOGRAM.
SIMPLE FREQUENCY DISTRIBUTION
Class interval
Frequency
18 - 20
11
21 - 23
4
24 - 26
8
27 - 29
7
30 – 32
5
33 – 35
4
36 - 38
39 – 41
4
7
50
CUMULATIVE FREQUENCY DISTRIBUTION
Class interval
Frequency
Cum. Freq
18 - 20
11
11 + 0 = 11
21 - 23
4
11 + 4 = 15
24 - 26
8
15 + 8 = 23
27 - 29
7
23 + 7 = 30
30 – 32
5
30 + 5 = 35
33 – 35
4
35 + 4 = 39
36 - 38
4
39 + 4 = 43
39 – 41
7
43 + 7 = 50
50
HOW TO CONSTRUCT A FREQUENCY POLYGON
A frequency polygon is a line graph of a frequency distribution.
The frequency distribution is plotted against the class marks of the
respective class intervals.
The graph is obtained by joining the mid points (class mark) of the tops of
the histogram by line segments.
In drawing the frequency polygon, one class is added to each end of the
frequency distribution.
The added classes each has zero frequency; as a result, they will touch the
horizontal axis.
PLS REFER TO VIDEO LECTURE FOR DIAGRAMATIC
ILLUSTRATION OF FREQUENCY POLYGON.
TOPICS
ARITHMETIC MEAN
MEDIAN
MODE
REVIEW QUESTIONS
1. Find the arithmetic mean, median and mode of the numbers 42, 56, 38, 41,
86 and 56.
2. The record below shows the scores obtained by some students in a
Mathematics test:
12
15
20
25
30
20
30
15
12
12
15
12
25
12
20
12
20
12
15
25
12
15
12
15
12
15
12
20
20
30
30
25
15
30
20
15
25
30
15
12
12
15
12
15
12
30
20
20
12
15
Compute the arithmetic mean, median and mode.
3. Consider the weights of a number of students given below measured in
Kilograms:
18
34
20
39
36
32
29
19
25
38
25
22
25
28
37
40
27
39
23
19
30
19
21
25
40
27
32
18
19
40
26
18
39
31
19
29
30
40
21
26
40
20
37
33
26
35
29
35
25
18
Using a class interval of 18-20, 21-23, 24-26, ….
Compute the arithmetic mean, median and mode.
4. Find the arithmetic mean, median and mode of the following numbers; 16,
13, 10, 23, 36, 9, 8, 48 and 24.
SUGGESTED SOLUTIONS
Find the arithmetic mean, median and mode of the numbers 42, 56, 38, 41,
86 and 56.
X: 42, 56, 38, 41, 86 and 56.
Arithmetic mean (X bar) = ΣX / n
= 42 + 56 + 38 + 41 + 86 + 56 / 6
= 319 / 6
= 53.2
Median = The number that falls into the middle position after the data have
been arranged accordingly.
Re-arrangement = 38, 41, 42, 56, 56, 86.
Therefore, Median = 42 + 56 / 2
= 98 / 2
= 49.
Mode= The number that occurs most frequently.
If we consider the data given above, it is obvious that every of the data appears
once.
Therefore, it can be concluded that there is no mode.
The record below shows the scores obtained by some students in a
Mathematics test:
12
15
20
25
30
20
30
15
12
15
12
25
12
20
12
20
12
15
12
15
12
15
12
15
12
20
20
30
25
15
30
20
15
25
30
15
12
15
12
15
12
30
20
20
12
Compute the arithmetic mean, median and mode.
12
25
30
12
15
At first, we have to prepare a frequency distribution table as done below;
SCORES
TALLY
FREQUENCY
12
15
20
25
30
Computation of Mean
X
F
12
16
15
13
20
9
25
5
30
7
50
Mean = ΣFX / n
= 902 / 50
= 18.04
IIIII IIIII IIIII I
IIIII IIIII III
IIIII IIII
IIIII
IIIII II
16
13
9
5
7
FX
192
195
180
125
210
902
Median = n + 1 / 2
= 50 + 1 / 2
= 51 / 2
= 25.5th
From the table, 25.5th falls into Scores of 15 marks.
Mode = Class with the highest frequency.
Therefore, from the table, mode = Scores of 12 marks.
Find the arithmetic mean, median and mode of the following numbers; 16,
13, 10, 23, 36, 9, 8, 48 and 24.
X = 16, 13, 10, 23, 36, 9, 8, 48 and 24.
Mean = ΣX / n
= 16+ 13+ 10+ 23+ 36+ 9+ 8+ 48 + 24 / 9
= 187 / 9
= 20.8
Median = The number that falls into the middle position after the data have
been arranged accordingly.
Re-arrangement = 8, 9, 10, 13, 16, 23, 24, 36, 48
From the table, the median = 16.
Mode = This is the number that occurs most frequently.
If we consider the data given above, it is obvious that every of the data appears
once.
Therefore, it can be concluded that there is no mode.
Consider the weights of a number of students given below measured in
Kilograms:
18
34
20
39
36
32
29
19
25
38
25
22
25
28
37
40
27
39
23
19
30
19
21
25
40
27
32
18
19
40
26
18
39
31
19
29
30
40
21
26
40
20
37
33
26
35
29
35
25
18
Using a class interval of 18-20, 21-23, 24-26, ….
Compute the arithmetic mean, median and mode.
At first, we have to prepare a frequency distribution table as done below;
Class Interval
Tally
Frequency
18 – 20
IIIII IIIII I
11
21 – 23
IIII
4
24 – 26
IIIII III
8
27 – 29
IIIII II
7
30 – 32
IIIII
5
33 – 35
IIII
4
36 – 38
39 – 41
IIII
IIIII II
4
7
Computation of Mean
Class Interval
F
X
FX
CB
18 – 20
11
19
209
17.5 –20.5
21 – 23
4
22
88
20.5 – 23.5
24 – 26
8
25
200
23.5 – 26.5
27 – 29
7
28
196
26.5 – 29.5
30 – 32
5
31
155
29.5 – 32.5
33 – 35
4
34
136
32.5 – 35.5
36 – 38
4
37
148
35.5 – 38.5
39 – 41
7
40
280
38.5 – 41.5
50
Mean = ΣFX / n
= 1412 / 50
= 28.24
Median = Li + [n/2 – Σfb] x C
fm
where;
Li = Lower Class Boundary of the median class
N = total sum of frequencies
Σfb = sum of frequencies before median class
Fm = Frequency of the median class
C = class size.
Median class = n + 1 / 2
= 50 + 1 / 2
= 51 / 2
1412
= 25.5th
From the table, median class = 27 – 29.
Therefore,
Median = 26.5 + [ 50/2 – 23] x 3
7
= 26.5 + [ 25 – 23] x 3
7
= 26.5 + 2/7 x 3
= 26.5 + 0.86
= 27.4
Mode = Li + [ fi/ fi + f2] x C
Modal class = The class with the highest frequency.
From the table, the modal class = 18 – 20
Therefore,
Mode = 17.5 + [0/ 0 + 4] x 3
= 17.5 + 0
= 17.5